Thermodynamic solutions

Borgnakke and Sonntag
CONTENT
SUBSECTION PROB NO.
Concept Problems 1-18
Properties and Units 19-22
Force and Energy 23-34
Specific Volume 35-40
Pressure 41-56
Manometers and Barometers 57-77
Temperature 78-83
Review problems 84-89
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In-Text Concept Questions

2.a
Make a control volume around the turbine in the steam power plant in Fig. 1.1 and
list the flows of mass and energy that are there.
Solution:
We see hot high pressure steam flowing in
at state 1 from the steam drum through a
flow control (not shown). The steam leaves
at a lower pressure to the condenser (heat
exchanger) at state 2. A rotating shaft gives
a rate of energy (power) to the electric
generator set.
W
T
1
2
2.b
Take a control volume around your kitchen refrigerator and indicate where the
components shown in Figure 1.6 are located and show all flows of energy transfer.
Solution:
The valve and the
cold line, the
evaporator, is
inside close to the
inside wall and
usually a small
blower distributes
cold air from the
freezer box to the
refrigerator room.
cb
W
.
Q
.
Q leak The black grille in
the back or at the
bottom is the
condenser that
gives heat to the
room air.
The compressor
sits at the bottom.

2.c
Why do people float high in the water when swimming in the Dead Sea as compared
with swimming in a fresh water lake?
As the dead sea is very salty its density is
higher than fresh water density. The
buoyancy effect gives a force up that equals
the weight of the displaced water. Since
density is higher the displaced volume is
smaller for the same force.

2.d
Density of liquid water is ρ = 1008 – T/2 [kg/m3] with T in oC. If the temperature
increases, what happens to the density and specific volume?
Solution:
The density is seen to decrease as the temperature increases.
∆ρ = – ∆T/2
Since the specific volume is the inverse of the density v = 1/ρ it will increase.
2.e
A car tire gauge indicates 195 kPa; what is the air pressure inside?
The pressure you read on the gauge is a gauge pressure, ∆P, so the absolute
pressure is found as
` P = Po + ∆P = 101 + 195 = 296 kPa

2.f
Can I always neglect ∆P in the fluid above location A in figure 2.12? What does that
depend on?
If the fluid density above A is low relative to the manometer fluid then you
neglect the pressure variation above position A, say the fluid is a gas like air and the
manometer fluid is like liquid water. However, if the fluid above A has a density of
the same order of magnitude as the manometer fluid then the pressure variation with
elevation is as large as in the manometer fluid and it must be accounted for.
2.g
A U tube manometer has the left branch connected to a box with a pressure of 110
kPa and the right branch open. Which side has a higher column of fluid?
Solution:
Since the left branch fluid surface feels 110 kPa and the
right branch surface is at 100 kPa you must go further
down to match the 110 kPa. The right branch has a higher
column of fluid.
cb
Po
Box
H

Concept Problems

2.1
Make a control volume around the whole power plant in Fig. 1.2 and with the help
of Fig. 1.1 list what flows of mass and energy are in or out and any storage of
energy. Make sure you know what is inside and what is outside your chosen C.V.
Solution:
Smoke
stack
Boiler
building
Coal conveyor system
Dock
Turbine house
Storage
gypsum
Coal
storage
flue
gas
cb
Underground
power cable
Welectrical
Hot water
District heating
m
Coal
m
m
Flue gas
Storage for later
Gypsum, fly ash, slag
transport out:
Cold return m
m
Combustion air

2.2
Take a control volume around the rocket engine in Fig. 1.12. Identify the mass flows
and where you have significant kinetic energy and where storage changes.
We have storage in both
tanks are reduced, mass
flows out with modest
velocities.
Energy conversion in the
combustion process.
gas at high pressure
expands towards lower
pressure outside and thus
accelerates to high
velocity with significant
kinetic energy flowing
out.
Control and mixing
Combustion
Oxydizer Fuel
Nozzle
High speed flow out

2.3
Make a control volume that includes the steam flow around in the main turbine loop
in the nuclear propulsion system in Fig.1.3. Identify mass flows (hot or cold) and
energy transfers that enter or leave the C.V.
Solution:
The electrical power
also leaves the C.V.
to be used for lights,
instruments and to
charge the batteries.
1 1Hot steam from generator
c Electric
power gen. WT
cb
Welectrical
3
2
5 4Condensate
to steam gen.
cold 76
Cooling by seawater

2.4
Separate the list P, F, V, v, ρ, T, a, m, L, t, and V into intensive, extensive, and non-
properties.
Solution:
Intensive properties are independent upon mass: P, v, ρ, T
Extensive properties scales with mass: V, m
Non-properties: F, a, L, t, V
Comment: You could claim that acceleration a and velocity V are physical
properties for the dynamic motion of the mass, but not thermal properties.

2.5
An electric dip heater is put into a cup of water and heats it from 20oC to 80oC.
Show the energy flow(s) and storage and explain what changes.
Solution:
Electric power is converted in the heater
element (an electric resistor) so it becomes
hot and gives energy by heat transfer to
the water. The water heats up and thus
stores energy and as it is warmer than the
cup material it heats the cup which also
stores some energy. The cup being
warmer than the air gives a smaller
amount of energy (a rate) to the air as a
heat loss.
Welectric
Q loss
C B

2.6
Water in nature exists in different phases such as solid, liquid and vapor (gas).
Indicate the relative magnitude of density and specific volume for the three phases.
Solution:
Values are indicated in Figure 2.7 as density for common substances. More
accurate values are found in Tables A.3, A.4 and A.5
Water as solid (ice) has density of around 900 kg/m3
Water as liquid has density of around 1000 kg/m3
Water as vapor has density of around 1 kg/m3 (sensitive to P and T)
Ice cube Liquid drop Cloud*
* Steam (water vapor) can not be seen what you see are tiny drops suspended in air from
which we infer that there was some water vapor before it condensed.

2.7
Is density a unique measure of mass distribution in a volume? Does it vary? If so, on
what kind of scale (distance)?
Solution:
Density is an average of mass per unit volume and we sense if it is not evenly
distributed by holding a mass that is more heavy in one side than the other.
Through the volume of the same substance (say air in a room) density varies only
little from one location to another on scales of meter, cm or mm. If the volume
you look at has different substances (air and the furniture in the room) then it can
change abruptly as you look at a small volume of air next to a volume of
hardwood.
Finally if we look at very small scales on the order of the size of atoms the density
can vary infinitely, since the mass (electrons, neutrons and positrons) occupy very
little volume relative to all the empty space between them.

2.8
Density of fibers, rock wool insulation, foams and cotton is fairly low. Why is that?
Solution:
All these materials consist of some solid substance and mainly air or other gas.
The volume of fibers (clothes) and rockwool that is solid substance is low relative
to the total volume that includes air. The overall density is
ρ =
m
V =
msolid + mair
Vsolid + Vair
where most of the mass is the solid and most of the volume is air. If you talk
about the density of the solid only, it is high.

2.9
How much mass is there approximately in 1 L of engine oil? Atmospheric air?
Solution:
A volume of 1 L equals 0.001 m3, see Table A.1. From Table A.4 the density is
885 kg/m3 so we get
m = ρV = 885 kg/m3 × 0.001 m3 = 0.885 kg
For the air we see in Figure 2.7 that density is about 1 kg/m3 so we get
m = ρV = 1 kg/m3 × 0.001 m3 = 0.001 kg
A more accurate value from Table A.5 is ρ = 1.17 kg/m3 at 100 kPa, 25oC.

2.10
Can you carry 1 m3 of liquid water?
Solution:
The density of liquid water is about 1000 kg/m3 from Figure 2.7, see also Table
A.3. Therefore the mass in one cubic meter is
m = ρV = 1000 kg/m3 × 1 m3 = 1000 kg
and we can not carry that in the standard gravitational field.

2.11
A heavy cabinet has four adjustable feet on it. What feature of the feet will ensure
that they do not make dents in the floor?
Answer:
The area that is in contact with the floor supports the total mass in the
gravitational field.
F = PA = mg
so for a given mass the smaller the area is the larger the pressure becomes.

2.12
The pressure at the bottom of a swimming pool is evenly distributed. Suppose we
look at a cast iron plate of 7272 kg lying on the ground with an area of 100 m2. What
is the average pressure below that? Is it just as evenly distributed as the pressure at
the bottom of the pool?
Solution:
The pressure is force per unit area from page 25:
P = F/A = mg/A = 7272 kg × (9.81 m/s2) / 100 m2 = 713.4 Pa
The iron plate being cast can be reasonable plane and flat, but it is stiff and rigid.
However, the ground is usually uneven so the contact between the plate and the
ground is made over an area much smaller than the 100 m2. Thus the local
pressure at the contact locations is much larger than the quoted value above.
The pressure at the bottom of the swimming pool is very even due to the ability of
the fluid (water) to have full contact with the bottom by deforming itself. This is
the main difference between a fluid behavior and a solid behavior.
Iron plate
Ground

2.13
Two divers swim at 20 m depth. One of them swims right in under a supertanker; the
other stays away from the tanker. Who feels a greater pressure?
Solution:
Each one feels the local pressure which is the static pressure only a function of
depth.
Pocean= P0 + ∆P = P0 + ρgH
So they feel exactly the same pressure.

2.14
A manometer with water shows a ∆P of Po/10; what is the column height difference?
Solution:
∆P = Po/10 = ρHg
H = Po/(10 ρ g) =
101.3 × 1000 Pa
10 × 997 kg/m3 × 9.80665 m/s2
= 1.036 m

2.15
A water skier does not sink too far down in the water if the speed is high enough.
What makes that situation different from our static pressure calculations?
The water pressure right under the ski is not a static pressure but a static plus
dynamic pressure that pushes the water away from the ski. The faster you go, the
smaller amount of water is displaced but at a higher velocity.

2.16
What is the smallest temperature in degrees Celsuis you can have? Kelvin?
Solution:
The lowest temperature is absolute zero which is
at zero degrees Kelvin at which point the
temperature in Celsius is negative
TK = 0 K = −273.15 oC

2.17
Convert the formula for water density in In-text Concept Question “e” to be for T in
degrees Kelvin.
Solution:
ρ = 1008 – TC/2 [kg/m3]
We need to express degrees Celsius in degrees Kelvin
TC = TK – 273.15
and substitute into formula
ρ = 1008 – TC/2 = 1008 – (TK – 273.15)/2 = 1144.6 – TK/2

2.18
A thermometer that indicates the temperature with a liquid column has a bulb with a
larger volume of liquid, why is that?
The expansion of the liquid volume with temperature is rather small so by having
a larger volume expand with all the volume increase showing in the very small
diameter column of fluid greatly increases the signal that can be read.

Properties and units

2.19
An apple “weighs” 60 g and has a volume of 75 cm3 in a refrigerator at 8oC.
What is the apple density? List three intensive and two extensive properties of the
apple.
Solution:
ρ =
m
V =
0.06
0.000 075
kg
m3 = 800
kg
m3
Intensive
ρ = 800
kg
m3 ; v =
1
ρ
= 0.001 25
m3
kg
T = 8°C; P = 101 kPa
Extensive
m = 60 g = 0.06 kg
V = 75 cm3 = 0.075 L = 0.000 075 m3

2.20
A steel cylinder of mass 2 kg contains 4 L of liquid water at 25oC at 200 kPa.
Find the total mass and volume of the system. List two extensive and three
intensive properties of the water
Solution:
Density of steel in Table A.3: ρ = 7820 kg/m3
Volume of steel: V = m/ρ =
2 kg
7820 kg/m3 = 0.000 256 m3
Density of water in Table A.4: ρ = 997 kg/m3
Mass of water: m = ρV = 997 kg/m3 ×0.004 m3 = 3.988 kg
Total mass: m = msteel + mwater = 2 + 3.988 = 5.988 kg
Total volume: V = Vsteel + Vwater = 0.000 256 + 0.004
= 0.004 256 m3 = 4.26 L
Extensive properties: m, V
Intensive properties: ρ (or v = 1/ρ), T, P

2.21
A storage tank of stainless steel contains 7 kg of oxygen gas and 5 kg of nitrogen
gas. How many kmoles are in the tank?
Table A.2: MO2 = 31.999 ; MN2 = 28.013
nO2 = mO2 / MO2 =
7
31.999 = 0.21876 kmol
nO2 = mN2 / MN2 =
5
28.013 = 0.17848 kmol
ntot = nO2 + nN2 = 0.21876 + 0.17848 = 0.3972 kmol

2.22
One kilopond (1 kp) is the weight of 1 kg in the standard gravitational field. How
many Newtons (N) is that?
F = ma = mg
1 kp = 1 kg × 9.807 m/s2 = 9.807 N

Force and Energy

2.23
The “standard” acceleration (at sea level and 45° latitude) due to gravity is
9.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in this
gravitational field? How much mass can a force of 1 N support?
Solution:
ma = 0 = ∑ F = F - mg
F = mg = 2 kg × 9.80665 m/s2 = 19.613 N
F = mg =>
m =
F
g =
1 N
9.80665 m/s2 = 0.102 kg
m
F
g

2.24
A steel piston of 2.5 kg is in the standard gravitational field where a force of 25 N
is applied vertically up. Find the acceleration of the piston.
Solution:
Fup = ma = F – mg
a =
F – mg
m =
F
m – g
=
25 N
2.5 kg – 9.807 m/s2
= 0.193 ms-2
g
F

2.25
When you move up from the surface of the earth the gravitation is reduced as g =
9.807 − 3.32 × 10-6 z, with z as the elevation in meters. How many percent is the
weight of an airplane reduced when it cruises at 11 000 m?
Solution:
go= 9.807 ms-2
gH = 9.807 – 3.32 × 10-6 × 11 000 = 9.7705 ms-2
Wo = m go ; WH = m gH
WH/Wo = gH/go =
9.7705
9.807 = 0.9963
Reduction = 1 – 0.9963 = 0.0037 or 0.37%
i.e. we can neglect that for most applications.

2.26
A model car rolls down an incline with a slope so the gravitational “pull” in the
direction of motion is one third of the standard gravitational force (see Problem
2.23). If the car has a mass of 0.06 kg find the acceleration.
Solution:
ma = ∑ F = mg / 3
a = mg / 3m = g/3
= 9.80665 (m/s2) / 3
= 3.27 m/s2
g
This acceleration does not depend on the mass of the model car.

2.27
A van drives at 60 km/h and is brought to a full stop with constant deceleration in
5 seconds. If the total car and driver mass is 2075 kg find the necessary force.
Solution:
Acceleration is the time rate of change of velocity.
a =
dV
dt =
60 × 1000
3600 × 5
= 3.333 m/s2
ma = ∑ F ;
Fnet = ma = 2075 kg × 3.333 m/s2 = 6916 N

2.28
An escalator brings four people of total 300 kg, 25 m up in a building. Explain what
happens with respect to energy transfer and stored energy.
Solution:
The four people (300
kg) have their potential
energy raised, which is
how the energy is
stored. The energy is
supplied as electrical
power to the motor that
pulls the escalator with a
cable.

2.29
A car of mass 1775 kg travels with a velocity of 100 km/h. Find the kinetic
energy. How high should it be lifted in the standard gravitational field to have a
potential energy that equals the kinetic energy?
Solution:
Standard kinetic energy of the mass is
KIN = ½ m V2 = ½ × 1775 kg ×





100 × 1000
3600
2
m2/s2
= ½ × 1775 × 27.778 Nm = 684 800 J
= 684.8 kJ
Standard potential energy is
POT = mgh
h = ½ m V2 / mg =
684 800 Nm
1775 kg × 9.807 m/s2 = 39.3 m

2.30
A 1500-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to
a speed of 75 km/h. What are the force and total time required?
Solution:
a =
dV
dt =
∆V
∆t
=> ∆t =
∆V
a =
(75 − 20) 1000
3600 × 5
= 3.82 sec
F = ma = 1500 kg × 4 m/s2 = 6000 N

2.31
On the moon the gravitational acceleration is approximately one-sixth that on the
surface of the earth. A 5-kg mass is “weighed” with a beam balance on the
surface on the moon. What is the expected reading? If this mass is weighed with a
spring scale that reads correctly for standard gravity on earth (see Problem 2.23),
what is the reading?
Solution:
Moon gravitation is: g = gearth/6
mm
m
Beam Balance Reading is 5 kg Spring Balance Reading is in kg units
This is mass comparison Force comparison length ∝ F ∝ g
Reading will be
5
6
kg

2.32
The escalator cage in Problem 2.28 has a mass of 500 kg in addition to the people.
How much force should the cable pull up with to have an acceleration of 1 m/s2 in
the upwards direction?
Solution:
The total mass moves upwards with an
acceleration plus the gravitations acts
with a force pointing down.
ma = ∑ F = F – mg
F = ma + mg = m(a + g)
= (500 + 300) kg × (1 + 9.81) m/s2
= 8648 N
F
g

2.33
A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration
of 2 m/s2 relative to the ground at a location where the local gravitational
acceleration is 9.5 m/s2. Find the required force.
Solution:
F = ma = Fup − mg
Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N g
Fup

2.34
A bottle of 12 kg steel has 1.75 kmole of liquid propane. It accelerates horizontal
with 3 m/s2, what is the needed force?
Solution:
The molecular weight for propane is M = 44.094 from Table A.2. The force
must accelerate both the container mass and the propane mass.
m = msteel + mpropane = 12 + (1.75 × 44.094) = 90.645 kg
ma = ∑ F ⇒
F = ma = 90.645 kg × 3 m/s2 = 271.9 N

Specific Volume

2.35
A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800
kg/m3. If the system is decelerated with 2g what is the needed force?
Solution:
m = mtank + mgasoline
= 15 kg + 0.3 m3 × 800 kg/m3
= 255 kg
F = ma = 255 kg × 2 × 9.81 m/s2
= 5003 N
cb

2.36
A power plant that separates carbon-dioxide from the exhaust gases compresses it
to a density of 110 kg/m3 and stores it in an un-minable coal seam with a porous
volume of 100 000 m3. Find the mass they can store.
Solution:
m = ρ V = 110 kg/m3 × 100 000 m3 = 11 × 10 6 kg
Just to put this in perspective a power plant that generates 2000 MW by burning
coal would make about 20 million tons of carbon-dioxide a year. That is 2000
times the above mass so it is nearly impossible to store all the carbon-dioxide
being produced.

2.37
A 1 m3 container is filled with 400 kg of granite stone, 200 kg dry sand and 0.2
m3 of liquid 25°C water. Use properties from tables A.3 and A.4. Find the
average specific volume and density of the masses when you exclude air mass and
volume.
Solution:
Specific volume and density are ratios of total mass and total volume.
mliq = Vliq/vliq = Vliq ρliq = 0.2 m3 × 997 kg/m3 = 199.4 kg
mTOT = mstone + msand + mliq = 400 + 200 + 199.4 = 799.4 kg
Vstone = mv = m/ρ = 400 kg/ 2750 kg/m3 = 0.1455 m3
Vsand = mv = m/ρ = 200/ 1500 = 0.1333 m3
VTOT = Vstone + Vsand + Vliq
= 0.1455 + 0.1333 + 0.2 = 0.4788 m3
v = VTOT / mTOT = 0.4788/799.4 = 0.000599 m3/kg
ρ = 1/v = mTOT/VTOT = 799.4/0.4788 = 1669.6 kg/m3

2.38
One kilogram of diatomic oxygen (O2 molecular weight 32) is contained in a 500-
L tank. Find the specific volume on both a mass and mole basis (v and v ).
Solution:
From the definition of the specific volume
v =
V
m =
0.5
1 = 0.5 m3/kg
v =
V
n =
V
m/M = M v = 32 × 0.5 = 16 m3/kmol

2.39
A tank has two rooms separated by a membrane. Room A has 1 kg air and volume
0.5 m3, room B has 0.75 m3 air with density 0.8 kg/m3. The membrane is broken
and the air comes to a uniform state. Find the final density of the air.
Solution:
Density is mass per unit volume
m = mA + mB = mA + ρBVB = 1 + 0.8 × 0.75 = 1.6 kg
V = VA + VB = 0.5 + 0.75 = 1.25 m3
ρ =
m
V =
1.6
1.25 = 1.28 kg/m3
A B
cb

2.40
A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the
rest of the volume is air with density 1.15 kg/m3. Find the mass of air and the
overall (average) specific volume.
Solution:
mair = ρ V = ρair ( Vtot −
mgranite
ρ
)
= 1.15 [ 5 -
900
2400 ] = 1.15 × 4.625 = 5.32 kg
v =
V
m =
5
900 + 5.32 = 0.005 52 m3/kg
Comment: Because the air and the granite are not mixed or evenly distributed in
the container the overall specific volume or density does not have much meaning.

Pressure

2.41
The hydraulic lift in an auto-repair shop has a cylinder diameter of 0.2 m. To what
pressure should the hydraulic fluid be pumped to lift 40 kg of piston/arms and 700
kg of a car?
Solution:
Force acting on the mass by the gravitational field
F↓ = ma = mg = 740 × 9.80665 = 7256.9 N = 7.257 kN
Force balance: F↑ = ( P - P0 ) A = F↓ => P = P0 + F↓ / A
A = π D2 (1 / 4) = 0.031416 m2
P = 101 kPa +
7.257 kN
0.031416 m2 = 332 kPa

2.42
A valve in a cylinder has a cross sectional area of 11 cm2 with a pressure of 735
kPa inside the cylinder and 99 kPa outside. How large a force is needed to open
the valve?
Fnet = PinA – PoutA
= (735 – 99) kPa × 11 cm2
= 6996 kPa cm2
= 6996 ×
kN
m2 × 10-4 m2
= 700 N
cb
Pcyl

2.43
A hydraulic lift has a maximum fluid pressure of 500 kPa. What should the
piston-cylinder diameter be so it can lift a mass of 850 kg?
Solution:
With the piston at rest the static force balance is
F↑ = P A = F↓ = mg
A = π r2 = π D2/4
PA = P π D2/4 = mg ⇒ D2 =
4mg
P π
D = 2
mg
Pπ
= 2
850 kg × 9.807 m/s2
500 kPa × π × 1000 (Pa/kPa)
= 0.146 m

2.44
A laboratory room keeps a vacuum of 0.1 kPa. What net force does that put on the
door of size 2 m by 1 m?
Solution:
The net force on the door is the difference between the forces on the two sides as
the pressure times the area
F = Poutside A – Pinside A = ∆P A = 0.1 kPa × 2 m × 1 m = 200 N
Remember that kPa is kN/m2.
Pabs = Po - ∆P
∆P = 0.1 kPa

2.45
A vertical hydraulic cylinder has a 125-mm diameter piston with hydraulic fluid
inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity,
find the piston mass that will create a pressure inside of 1500 kPa.
Solution:
Force balance:
F↑ = PA = F↓ = P0A + mpg;
P0 = 1 bar = 100 kPa
A = (π/4) D2 = (π/4) × 0.1252 = 0.01227 m2
cb
g
Po
mp = (P − P0)
A
g = ( 1500 − 100 ) × 1000 ×
0.01227
9.80665 = 1752 kg

2.46
A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg
resting on the stops, as shown in Fig. P2.46. With an outside atmospheric pressure
of 100 kPa, what should the water pressure be to lift the piston?
Solution:
The force acting down on the piston comes from gravitation and the
outside atmospheric pressure acting over the top surface.
Force balance: F↑ = F↓ = PA = mpg + P0A
Now solve for P (divide by 1000 to convert to kPa for 2nd
term)
P = P0 +
mpg
A = 100 kPa +
100 × 9.80665
0.01 × 1000
kPa
= 100 kPa + 98.07 kPa = 198 kPa
Water
cb

2.47
A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the
gun-powder is burned a pressure of 7 MPa is created in the gas behind the ball.
What is the acceleration of the ball if the cylinder (cannon) is pointing
horizontally?
Solution:
The cannon ball has 101 kPa on the side facing the atmosphere.
ma = F = P1 × A − P0 × A = (P1 − P0 ) × A
= (7000 – 101) kPa × π ( 0.152
/4 ) m2
= 121.9 kN
a =
F
m =
121.9 kN
5 kg = 24 380 m/s2

2.48
Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up
relative to the horizontal direction.
Solution:
ma = F = ( P1 - P0 ) A - mg sin 400
ma = (7000 - 101 ) kPa × π × ( 0.152 / 4 ) m2
- 5 × 9.807 × 0.6428 N
= 121.9 kN - 31.52 N = 121.87 kN
a =
F
m =
121.87 kN
5 kg = 24 374 m/s2

2.49
A large exhaust fan in a laboratory room keeps the pressure inside at 10 cm water
relative vacuum to the hallway. What is the net force on the door measuring 1.9 m
by 1.1 m?
Solution:
The net force on the door is the difference between the forces on the two sides as
F = Poutside A – Pinside A = ∆P × A
= 10 cm H2O × 1.9 m × 1.1 m
= 0.10 × 9.80638 kPa × 2.09 m2
= 2049 N
Table A.1: 1 m H2O is 9.80638 kPa and kPa is kN/m2.

2.50
A tornado rips off a 100 m2 roof with a mass of 1000 kg. What is the minimum
vacuum pressure needed to do that if we neglect the anchoring forces?
Solution:
The net force on the roof is the difference between the forces on the two sides as
F = Pinside A – PoutsideA = ∆P A
That force must overcome the gravitation mg, so the balance is
∆P A = mg
∆P = mg/A = (1000 kg × 9.807 m/s2 )/100 m2 = 98 Pa = 0.098 kPa
Remember that kPa is kN/m2.

2.51
A 2.5 m tall steel cylinder has a cross sectional area of 1.5 m2. At the bottom with
a height of 0.5 m is liquid water on top of which is a 1 m high layer of gasoline.
This is shown in Fig. P2.51. The gasoline surface is exposed to atmospheric air at
101 kPa. What is the highest pressure in the water?
Solution:
The pressure in the fluid goes up with the depth as
P = Ptop + ∆P = Ptop + ρgh
and since we have two fluid layers we get
P = Ptop + [(ρh)gasoline + (ρh)water] g
Air
Water
1 m
0.5 m
Gasoline
The densities from Table A.4 are:
ρgasoline = 750 kg/m3; ρwater = 997 kg/m3
P = 101 + [750 × 1 + 997 × 0.5]
9.807
1000 = 113.2 kPa

2.52
What is the pressure at the bottom of a 5 m tall column of fluid with atmospheric
pressure 101 kPa on the top surface if the fluid is
a) water at 20°C b) glycerine 25°C or c) gasoline 25°C
Solution:
Table A.4: ρH2O = 997 kg/m3; ρGlyc = 1260 kg/m3; ρgasoline = 750 kg/m3
∆P = ρgh P = Ptop + ∆P
a) ∆P = ρgh = 997× 9.807× 5 = 48 888 Pa
P = 101 + 48.99 = 149.9 kPa
b) ∆P = ρgh = 1260× 9.807× 5 = 61 784 Pa
P = 101 + 61.8 = 162.8 kPa
c) ∆P = ρgh = 750× 9.807× 5 = 36 776 Pa
P = 101 + 36.8 = 137.8 kPa

2.53
At the beach, atmospheric pressure is 1025 mbar. You dive 15 m down in the
ocean and you later climb a hill up to 250 m elevation. Assume the density of
water is about 1000 kg/m3 and the density of air is 1.18 kg/m3. What pressure do
you feel at each place?
Solution:
∆P = ρgh,
Units from A.1: 1 mbar = 100 Pa (1 bar = 100 kPa).
Pocean= P0 + ∆P = 1025 × 100 + 1000 × 9.81 × 15
= 2.4965 × 105 Pa = 250 kPa
Phill = P0 - ∆P = 1025 × 100 - 1.18 × 9.81 × 250
= 0.99606 × 105 Pa = 99.61 kPa

2.54
A steel tank of cross sectional area 3 m2 and 16 m tall weighs 10 000 kg and it is
open at the top. We want to float it in the ocean so it sticks 10 m straight down by
pouring concrete into the bottom of it. How much concrete should I put in?
Solution:
The force up on the tank is from the water
pressure at the bottom times its area. The
force down is the gravitation times mass and
the atmospheric pressure.
F↑ = PA = (ρoceangh + P0)A
F↓ = (mtank + mconcrete)g + P0A
The force balance becomes
Air
Ocean
Concrete
10 m
F↑ = F↓ = (ρoceangh + P0)A = (mtank + mconcrete)g + P0A
Solve for the mass of concrete
mconcrete = (ρoceanhA - mtank) = 997 × 10 × 3 – 10 000 = 19 910 kg
Notice: The first term is the mass of the displaced ocean water. The net force
up is the weight (mg) of this mass called bouyancy, P0 cancel.

2.55
A piston, mp= 5 kg, is fitted in a cylinder, A = 15 cm2, that contains a gas. The
setup is in a centrifuge that creates an acceleration of 25 m/s2 in the direction of
piston motion towards the gas. Assuming standard atmospheric pressure outside
the cylinder, find the gas pressure.
Solution:
Force balance: F↑ = F↓ = P0A + mpg = PA
P = P0 +
mpg
A
= 101.325 +
5 × 25
1000 × 0.0015
kPa kg m/s2
Pa m2
= 184.7 kPa
gasg
Po

2.56
Liquid water with density ρ is filled on top of a thin piston in a cylinder with
cross-sectional area A and total height H, as shown in Fig. P2.56. Air is let in
under the piston so it pushes up, spilling the water over the edge. Derive the
formula for the air pressure as a function of piston elevation from the bottom, h.
Solution:
Force balance
H
h
P0 Piston: F↑ = F↓
PA = P0A + mH2Og
P = P0 + mH2Og/A
P = P0 + (H − h)ρg h, V air
P
P0

Manometers and Barometers

2.57
You dive 5 m down in the ocean. What is the absolute pressure there?
Solution:
The pressure difference for a column is from Eq.2.2 and the density of water is
from Table A.4.
∆P = ρgH
= 997 kg/m3 × 9.81 m/s2 × 5 m
= 48 903 Pa = 48.903 kPa
Pocean= P0 + ∆P
= 101.325 + 48.903
= 150 kPa

2.58
A barometer to measure absolute pressure shows a mercury column height of 725
mm. The temperature is such that the density of the mercury is 13 550 kg/m3.
Find the ambient pressure.
Solution:
Hg : L = 725 mm = 0.725 m; ρ = 13 550 kg/m3
The external pressure P balances the column of height L so from Fig. 2.10
P = ρ L g = 13 550 kg/m3 × 9.80665 m/s2 × 0.725 m × 10-3 kPa/Pa
= 96.34 kPa
This is a more common type that
does not involve mercury as the
wall mounted unit to the left.

2.59
The density of atmospheric air is about 1.15 kg/m3, which we assume is constant.
How large an absolute pressure will a pilot see when flying 2000 m above ground
level where the pressure is 101 kPa.
Solution:
Assume g and ρ are constant then the pressure difference to carry a
column of height 1500 m is from Fig.2.10
∆P = ρgh = 1.15 kg/m3 × 9.807 ms-2 × 2000 m
= 22 556 Pa = 22.6 kPa
The pressure on top of the column of air is then
P = P0 – ∆P = 101 – 22.6 = 78.4 kPa

2.60
A differential pressure gauge mounted on a vessel shows 1.25 MPa and a local
barometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure
inside the vessel.
Solution:
Convert all pressures to units of kPa.
Pgauge = 1.25 MPa = 1250 kPa;
P0 = 0.96 bar = 96 kPa
P = Pgauge + P0 = 1250 + 96 = 1346 kPa

2.61
A manometer shows a pressure difference of 1 m of liquid mercury. Find ∆P in kPa.
Solution:
Hg : L = 1 m; ρ = 13 580 kg/m3 from Table A.4 (or read Fig 2.8)
The pressure difference ∆P balances the column of height L so from Eq.2.2
∆P = ρ g L = 13 580 kg/m3 × 9.80665 m/s2 × 1.0 m × 10-3 kPa/Pa
= 133.2 kPa

2.62
Blue manometer fluid of density 925 kg/m3 shows a column height difference of
3 cm vacuum with one end attached to a pipe and the other open to P0 = 101 kPa.
What is the absolute pressure in the pipe?
Solution:
Since the manometer shows a vacuum we have
PPIPE = P0 - ∆P
∆P = ρgh = 925 × 9.807 × 0.03
= 272.1 Pa = 0.272 kPa
PPIPE = 101 – 0.272 = 100.73 kPa
cb
Po
Pipe

2.63
What pressure difference does a 10 m column of atmospheric air show?
Solution:
The pressure difference for a column is from Eq.2.2
∆P = ρgH
So we need density of air from Fig.2.7, ρ = 1.2 kg/m3
∆P = 1.2 kg/m3 × 9.81 ms-2 × 10 m = 117.7 Pa = 0.12 kPa

2.64
The absolute pressure in a tank is 85 kPa and the local ambient absolute pressure
is 97 kPa. If a U-tube with mercury, density 13550 kg/m3, is attached to the tank
to measure the vacuum, what column height difference would it show?
Solution:
∆P = P0 - Ptank = ρg H
H = ( P0 - Ptank ) / ρg = [(97 - 85 ) × 1000 ] / (13550 × 9.80665)
= 0.090 m = 90 mm
H

2.65
The pressure gauge on an air tank shows 75 kPa when the diver is 10 m down in
the ocean. At what depth will the gauge pressure be zero? What does that mean?
Ocean H20 pressure at 10 m depth is
Pwater = P0 + ρLg = 101.3 +
997 × 10 × 9.80665
1000 = 199 kPa
Air Pressure (absolute) in tank
Ptank = 199 + 75 = 274 kPa
Tank Pressure (gauge) reads zero at H20 local pressure
274 = 101.3 +
997 × 9.80665
1000 L
L = 17.66 m
At this depth you will have to suck the
air in, it can no longer push itself
through a valve.

2.66
An exploration submarine should be able to go 4000 m down in the ocean. If the
ocean density is 1020 kg/m3 what is the maximum pressure on the submarine
hull?
Solution:
Assume we have atmospheric pressure inside the submarine then the pressure
difference to the outside water is
∆P = ρLg = (1020 kg/m3 × 4000 m × 9.807 m/s2) / 1000
= 40 012 kPa ≈ 40 MPa

2.67
A submarine maintains 101 kPa inside it and it dives 240 m down in the ocean
having an average density of 1030 kg/m3. What is the pressure difference
between the inside and the outside of the submarine hull?
Solution:
Assume the atmosphere over the ocean is at 101 kPa, then ∆P is from the 240 m
column water.
∆P = ρLg
= (1030 kg/m3 × 240 m × 9.807 m/s2) / 1000 = 2424 kPa

2.68
Assume we use a pressure gauge to measure the air pressure at street level and at
the roof of a tall building. If the pressure difference can be determined with an
accuracy of 1 mbar (0.001 bar) what uncertainty in the height estimate does that
corresponds to?
Solution:
ρair = 1.169 kg/m3 from Table A.5
∆P = 0.001 bar = 100 Pa
L =
∆P
ρg
=
100
1.169 × 9.807
= 8.72 m

2.69
A barometer measures 760 mmHg at street level and 735 mmHg on top of a
building. How tall is the building if we assume air density of 1.15 kg/m3?
Solution:
∆P = ρgH
H = ∆P/ρg =
760 – 735
1.15 × 9.807
mmHg
kg/m2s2
133.32 Pa
mmHg = 295 m

2.70
An absolute pressure gauge attached to a steel cylinder shows 135 kPa. We want
to attach a manometer using liquid water a day that Patm = 101 kPa. How high a
fluid level difference must we plan for?
Solution:
Since the manometer shows a pressure difference we have
∆P = PCYL - Patm = ρ L g
L = ∆P / ρg =
(135 – 101) kPa
997 kg m-3 × 10 × 9.807 m/s2
1000 Pa
kPa
= 3.467 m
H

2.71
A U-tube manometer filled with water, density 1000 kg/m3, shows a height
difference of 25 cm. What is the gauge pressure? If the right branch is tilted to
make an angle of 30° with the horizontal, as shown in Fig. P2.71, what should the
length of the column in the tilted tube be relative to the U-tube?
Solution:
Same height in the two sides in the direction of g.
∆P = F/A = mg/A = Vρg/A = hρg
= 0.25 m × 1000 kg/m3 × 9.807m/s2
= 2452.5 Pa
= 2.45 kPa
h = H × sin 30°
⇒ H = h/sin 30° = 2h = 50 cm
30o
H
h

2.72
A pipe flowing light oil has a manometer attached as shown in Fig. P2.72. What
is the absolute pressure in the pipe flow?
Solution:
Table A.3: ρoil = 910 kg/m3; ρwater = 997 kg/m3
PBOT = P0 + ρwater g Htot = P0 + 997 kg/m3 × 9.807 m/s2 × 0.8 m
= Po + 7822 Pa
PPIPE = PBOT – ρwater g H1 – ρoil g H2
= PBOT – 997 × 9.807 × 0.1 – 910 × 9.807 × 0.2
= PBOT – 977.7 Pa – 1784.9 Pa
PPIPE = Po + (7822 – 977.7 – 1784.9) Pa
= Po + 5059.4 Pa = 101.325 + 5.06 = 106.4 kPa

2.73
The difference in height between the columns of a manometer is 200 mm with a
fluid of density 900 kg/m3. What is the pressure difference? What is the height
difference if the same pressure difference is measured using mercury, density
13600 kg/ m3, as manometer fluid?
Solution:
∆P = ρ1gh1 = 900 kg/m3 × 9.807 m/s2 × 0.2 m = 1765.26 Pa = 1.77 kPa
hHg = ∆P/ (ρhg g) = (ρ1 gh1) / (ρhg g) =
900
13600 × 0.2
= 0.0132 m = 13.2 mm

2.74
Two cylinders are filled with liquid water, ρ = 1000 kg/m3, and connected by a
line with a closed valve. A has 100 kg and B has 500 kg of water, their cross-
sectional areas are AA = 0.1 m2 and AB = 0.25 m2 and the height h is 1 m. Find
the pressure on each side of the valve. The valve is opened and water flows to an
equilibrium. Find the final pressure at the valve location.
Solution:
VA = vH2OmA = mA/ρ = 0.1 = AAhA => hA = 1 m
VB = vH2OmB = mB/ρ = 0.5 = ABhB => hB = 2 m
PVB = P0 + ρg(hB+H) = 101325 + 1000 × 9.81 × 3 = 130 755 Pa
PVA = P0 + ρghA = 101325 + 1000 × 9.81 × 1 = 111 135 Pa
Equilibrium: same height over valve in both
Vtot = VA + VB = h2AA + (h2 - H)AB ⇒ h2 =
hAAA + (hB+H)AB
AA + AB
= 2.43 m
PV2 = P0 + ρgh2 = 101.325 + (1000 × 9.81 × 2.43)/1000 = 125.2 kPa

2.75
Two piston/cylinder arrangements, A and B, have their gas chambers connected
by a pipe. Cross-sectional areas are AA = 75 cm2 and AB = 25 cm2 with the piston
mass in A being mA = 25 kg. Outside pressure is 100 kPa and standard
gravitation. Find the mass mB so that none of the pistons have to rest on the
bottom.
Solution:
P
Po
o
cb
Force balance for both pistons: F↑ = F↓
A: mPAg + P0AA = PAA
B: mPBg + P0AB = PAB
Same P in A and B gives no flow between them.
mPAg
AA
+ P0 =
mPBg
AB
+ P0
=> mPB = mPA AA/ AB = 25 × 25/75 = 8.33 kg

2.76
Two hydraulic piston/cylinders are of same size and setup as in Problem 2.75, but
with negligible piston masses. A single point force of 250 N presses down on
piston A. Find the needed extra force on piston B so that none of the pistons have
to move.
Solution:
AA = 75 cm2 ;
AB = 25 cm2
No motion in connecting pipe: PA = PB
Forces on pistons balance
Po
Po
cb
A B
FB
FA
PA = P0 + FA / AA = PB = P0 + FB / AB
FB = FA ×
AB
AA
= 250 ×
25
75 = 83.33 N

2.77
A piece of experimental apparatus is located where g = 9.5 m/s2 and the
temperature is 5°C. An air flow inside the apparatus is determined by measuring
the pressure drop across an orifice with a mercury manometer (see Problem 2.79
for density) showing a height difference of 200 mm. What is the pressure drop in
kPa?
Solution:
∆P = ρgh ; ρHg = 13600 kg/m3
∆P = 13 600 kg/m3 × 9.5 m/s2 × 0.2 m = 25840 Pa = 25.84 kPa
g
Air

Temperature

2.78
What is a temperature of –5oC in degrees Kelvin?
Solution:
The offset from Celsius to Kelvin is 273.15 K,
so we get
TK = TC + 273.15 = -5 + 273.15
= 268.15 K

2.79
The density of mercury changes approximately linearly with temperature as
ρHg = 13595 − 2.5 T kg/ m3 T in Celsius
so the same pressure difference will result in a manometer reading that is
influenced by temperature. If a pressure difference of 100 kPa is measured in the
summer at 35°C and in the winter at −15°C, what is the difference in column
height between the two measurements?
Solution:
The manometer reading h relates to the pressure difference as
∆P = ρ L g ⇒ L =
∆P
ρg
The manometer fluid density from the given formula gives
ρsu = 13595 − 2.5 × 35 = 13507.5 kg/m3
ρw = 13595 − 2.5 × (−15) = 13632.5 kg/m3
The two different heights that we will measure become
Lsu =
100 × 103
13507.5 × 9.807
kPa (Pa/kPa)
(kg/m3) m/s2 = 0.7549 m
Lw =
100 × 103
13632.5 × 9.807
kPa (Pa/kPa)
(kg/m3) m/s2 = 0.7480 m
∆L = Lsu - Lw = 0.0069 m = 6.9 mm

2.80
A mercury thermometer measures temperature by measuring the volume
expansion of a fixed mass of liquid Hg due to a change in the density, see
problem 2.35. Find the relative change (%) in volume for a change in
temperature from 10°C to 20°C.
Solution:
From 10°C to 20°C
At 10°C : ρHg = 13595 – 2.5 × 10 = 13570 kg/m3
At 20°C : ρHg = 13595 – 2.5 × 20 = 13545 kg/m3
The volume from the mass and density is: V = m/ρ
Relative Change =
V20– V10
V10
=
(m/ρ20) - (m/ρ10)
m/ρ10
=
ρ10
ρ20
– 1 =
13570
13545 – 1 = 0.0018 (0.18%)

2.81
Density of liquid water is ρ = 1008 – T/2 [kg/m3] with T in oC. If the temperature
increases 10oC how much deeper does a 1 m layer of water become?
Solution:
The density change for a change in temperature of 10oC becomes
∆ρ = – ∆T/2 = –5 kg/m3
from an ambient density of
ρ = 1008 – T/2 = 1008 – 25/2 = 995.5 kg/m3
Assume the area is the same and the mass is the same m = ρV = ρAH, then we
have
∆m = 0 = V∆ρ + ρ∆V ⇒ ∆V = - V∆ρ/ρ
and the change in the height is
∆H =
∆V
A =
H∆V
V =
-H∆ρ
ρ
=
-1 × (-5)
995.5 = 0.005 m
barely measurable.

2.82
Using the freezing and boiling point temperatures for water in both Celsius and
Fahrenheit scales, develop a conversion formula between the scales. Find the
conversion formula between Kelvin and Rankine temperature scales.
Solution:
TFreezing = 0 oC = 32 F; TBoiling = 100 oC = 212 F
∆T = 100 oC = 180 F ⇒ ToC = (TF - 32)/1.8 or TF = 1.8 ToC + 32
For the absolute K & R scales both are zero at absolute zero.
TR = 1.8 × TK

2.83
The atmosphere becomes colder at higher elevation. As an average the standard
atmospheric absolute temperature can be expressed as Tatm = 288 - 6.5 × 10−3 z,
where z is the elevation in meters. How cold is it outside an airplane cruising at
12 000 m expressed in Kelvin and in Celsius?
Solution:
For an elevation of z = 12 000 m we get
Tatm = 288 - 6.5 × 10−3 z = 210 K
To express that in degrees Celsius we get
TC = T – 273.15 = −63.15oC

2.84
Repeat problem 2.77 if the flow inside the apparatus is liquid water, ρ ≅ 1000
kg/m3, instead of air. Find the pressure difference between the two holes flush
with the bottom of the channel. You cannot neglect the two unequal water
columns.
Solution: Balance forces in the manometer:
P P1. 2
·
h
h1
2
H
(H - h2) - (H - h1) = ∆hHg = h1 - h2
P1A + ρH2Oh1gA + ρHg(H - h1)gA
= P2A + ρH2Oh2gA + ρHg(H - h2)gA
⇒ P1 - P2 = ρH2O(h2 - h1)g + ρHg(h1 - h2)g
P1 - P2 = ρHg∆hHgg - ρH2O∆hHgg = 13 600 × 0.2 × 9.5 - 1000 × 0.2 × 9.5
= 25 840 - 1900 = 23940 Pa = 23.94 kPa

2.85
A dam retains a lake 6 m deep. To construct a gate in the dam we need to know
the net horizontal force on a 5 m wide and 6 m tall port section that then replaces
a 5 m section of the dam. Find the net horizontal force from the water on one side
and air on the other side of the port.
Solution:
Pbot = P0 + ∆P
∆P = ρgh = 997× 9.807× 6 = 58 665 Pa = 58.66 kPa
Neglect ∆P in air
Fnet = Fright – Fleft = Pavg A - P0A
Pavg = P0 + 0.5 ∆P Since a linear pressure variation with depth.
Fnet = (P0 + 0.5 ∆P)A - P0A = 0.5 ∆P A = 0.5 × 58.66 × 5 × 6 = 880 kN
F
Fleft
right

2.86
In the city water tower, water is pumped up to a level 25 m above ground in a
pressurized tank with air at 125 kPa over the water surface. This is illustrated in
Fig. P2.86. Assuming the water density is 1000 kg/m3 and standard gravity, find
the pressure required to pump more water in at ground level.
Solution:
∆P = ρ L g
= 1000 kg/m3 × 25 m × 9.807 m/s2
= 245 175 Pa = 245.2 kPa
Pbottom = Ptop + ∆P
= 125 + 245.2
= 370 kPa
cb

2.87
The main waterline into a tall building has a pressure of 600 kPa at 5 m elevation
below ground level. How much extra pressure does a pump need to add to ensure
a water line pressure of 200 kPa at the top floor 150 m above ground?
Solution:
The pump exit pressure must balance the top pressure plus the column
∆P. The pump inlet pressure provides part of the absolute pressure.
Pafter pump = Ptop + ∆P
∆P = ρgh = 997 kg/m3 × 9.807 m/s2 × (150 + 5) m
= 1 515 525 Pa = 1516 kPa
Pafter pump = 200 + 1516 = 1716 kPa
∆Ppump = 1716 – 600 = 1116 kPa

2.88
Two cylinders are connected by a piston as shown in Fig. P2.88. Cylinder A is
used as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg and
there is standard gravity. What is the gas pressure in cylinder B?
Solution:
Force balance for the piston: PBAB + mpg + P0(AA - AB) = PAAA
AA = (π/4)0.12 = 0.00785 m2; AB = (π/4)0.0252 = 0.000 491 m2
PBAB = PAAA - mpg - P0(AA - AB) = 500× 0.00785 - (25 × 9.807/1000)
- 100 (0.00785 - 0.000 491) = 2.944 kN
PB = 2.944/0.000 491 = 5996 kPa = 6.0 MPa
P
B
GAS
A Oil
Po
cb

2.89
A 5-kg piston in a cylinder with diameter of 100 mm is loaded with a linear
spring and the outside atmospheric pressure of 100 kPa as shown in Fig. P2.89.
The spring exerts no force on the piston when it is at the bottom of the cylinder
and for the state shown, the pressure is 400 kPa with volume 0.4 L. The valve is
opened to let some air in, causing the piston to rise 2 cm. Find the new pressure.
Solution:
A linear spring has a force linear proportional to displacement. F = k x, so
the equilibrium pressure then varies linearly with volume: P = a + bV, with an
intersect a and a slope b = dP/dV. Look at the balancing pressure at zero volume
(V -> 0) when there is no spring force F = PA = PoA + mpg and the initial state.
These two points determine the straight line shown in the P-V diagram.
Piston area = AP = (π/4) × 0.12 = 0.00785 m2
400
106.2
2
1
0 0.4
P
V
0.557
2P
a = P0 +
mpg
Ap
= 100 kPa +
5 × 9.80665
0.00785 Pa
= 106.2 kPa intersect for zero volume.
V2 = 0.4 + 0.00785 × 20 = 0.557 L
P2 = P1 +
dP
dV ∆V
= 400 +
(400-106.2)
0.4 - 0 (0.557 - 0.4)
= 515.3 kPa

2 Borgnakke and Sonntag
CONTENT CHAPTER 3
SUBSECTION PROB NO.
Concept problems 1-15
Phase diagrams, triple and critical points 16-24
General tables 25-64
Ideal gas 65-84
Compressibility factor 85-97
Equations of state 98-106
Linear interpolation 126-130
Computer tables 131-135

3.a
If the pressure is smaller than Psat at a given T, what is the phase?
Refer to the phase diagrams in Figures 3.6
and 3.7. For a lower P you are below the
vaporization curve and that is the superheated
vapor region. You have the gas phase.
ln P
T
Vapor
L
S
Critical Point
S
3.b
An external water tap has the valve activated by a long spindle so the closing
mechanism is located well inside the wall. Why is that?
Solution:
By having the spindle inside the wall the coldest location with water when the valve
is closed is kept at a temperature above the freezing point. If the valve spindle was
outside there would be some amount of water that could freeze while it is trapped
inside the pipe section potentially rupturing the pipe.

3.c
What is the lowest temperature (approximately) at which water can be liquid?
Look at the phase diagram in Fig. 3.7. At the
border between ice I, ice III and the liquid region
is a triple point which is the lowest T where you
can have liquid. From the figure it is estimated to
be about 255 K i.e. at -18oC.
T ≈ 255 K ≈ - 18°C
ln P
T
V
L
S
CR.P.
lowest T liquid

3.d
Some tools should be cleaned in water at a least 150oC. How high a P is needed?
Solution:
If I need liquid water at 150oC I must have a pressure that is at least the saturation
pressure for this temperature.
Table B.1.1: 150oC, Psat = 475.9 kPa.

3.e
Water at 200 kPa has a quality of 50%. Is the volume fraction Vg/Vtot < 50% or
> 50%?
This is a two-phase state at a given pressure without looking in the table we know
that vf is much smaller than vg.
From the definition of quality we get the masses from total mass, m, as
mf = (1 – x) m, mg = x m
The volumes are
Vf = mf vf = (1 – x) m vf, Vg = mg vg = x m vg
So when half the mass is liquid and the other half is vapor the liquid volume is
much smaller that the vapor volume.

3.f
Why are most of the compressed liquid or solid regions not included in the printed
tables?
For the compressed liquid and the solid phases the specific volume and thus
density is nearly constant. These surfaces are very steep nearly constant v and there
is then no reason to fill up a table with the same value of v for different P and T.
3.g
Why is it not typical to find tables for Ar, He, Ne or air like an Appendix B table?
The temperature at which these substances are close to the two-phase
region is very low. For technical applications with temperatures around
atmospheric or higher they are ideal gases. Look in Table A.2 and we can see the
critical temperatures as
Ar : 150.8 K He: 5.19 K Ne: 44.4 K
It requires a special refrigerator in a laboratory to bring a substance down
to these cryogenic temperatures.

3.h
What is the percent change in volume as liquid water freezes? Mention some effects
in nature and for our households the volume change can have.
The density of water in the different phases can be found in Tables A.3 and A.4
and in Table B.1.
From Table B.1.1 vf = 0.00100 m3/kg
From Table B.1.5 vi = 0.0010908 m3/kg
Percent change: 100
vi – vf
vf
= 100 ×
0.0010908 – 0.001
0.001 = 9.1 % increase
Liquid water that seeps into cracks or other confined spaces and then freezes
will expand and widen the cracks. This is what destroys any porous material exposed
to the weather on buildings, roads and mountains. It can burst water pipes and crack
engine blocks (that is why you put anti-freeze in it).

3.i
How accurate is it to assume that methane is an ideal gas at room conditions?
From Table A.2: Tc = 190.4 K, Pc = 4.60 MPa
So at room conditions we have much higher T > Tc and P << Pc so this is the
ideal gas region. To confirm look in Table B.7.2
100 kPa, 300 K, v = 1.55215 m3/kg
Find the compressibility factor (R from Table A.5) as
Z = Pv/RT =
100 kPa × 1.55215 m3/kg
0.5183 kJ/kg-K × 300 K
= 0.99823
so Z is 1 with an accuracy of 0.2% better than most measurements can be done.

3.j
I want to determine a state of some substance, and I know that P = 200 kPa; is it
helpful to write PV = mRT to find the second property?
NO. You need a second property.
Notice that two properties are needed to determine a state. The EOS can give you a
third property if you know two, like (P,T) gives v just as you would get by entering a
table with a set (P,T). This EOS substitutes for a table when it is applicable.

3.k
A bottle at 298 K should have liquid propane; how high a pressure is needed? (use
Fig. D.1)
To have a liquid the pressure must be higher than or equal to the saturation pressure.
There is no printed propane table so we use the compressibility chart and Table A.2
Propane Table A.2: Tc = 369.8 K, Pc = 4.25 MPa
The reduced temperature is:
Tr =
T
Tc
=
298
369.8 = 0.806,
for which we find in Fig. D.1: Pr sat = 0.25
P = Pr sat Pc = 0.25 × 4.25 MPa = 1.06 MPa

3.l
A bottle at 298 K should have liquid propane; how high a pressure is needed? (use
the software)
To have a liquid the pressure must be higher than or equal to the saturation pressure.
There is no printed propane table but the software has propane included
Start CATT3, select cryogenic substances, propane
select calculator, select T, x = (25C, 0) P = 0.9518 MPa

Concept Problems

3.1
Are the pressures in the tables absolute or gauge pressures?
Solution:
The behavior of a pure substance depends on the absolute pressure, so P in the tables
is absolute.

3.2
What is the minimum pressure for which I can have liquid carbon dioxide?
Look at the phase diagram in Fig. 3.6. The minimum P in the liquid phase is at the
triple point. From Table 3.2 this is at 520 kPa (a similar around 4-500 kPa is seen in
Fig. 3.6).

3.3
When you skate on ice, a thin liquid film forms under the skate. How can that be?
The ice is at some temperature below the freezing temperature for the
atmospheric pressure of 100 kPa = 0.1 MPa and thus to the left of the fusion line in
the solid ice I region of Fig. 3.7. As the skate comes over the ice the pressure is
increased dramatically right under the blade so it brings the state straight up in the
diagram crossing the fusion line and brings it into a liquid state at same temperature.
The very thin liquid film under the skate changes the friction to be viscous rather
than a solid to solid contact friction. Friction is thus significantly reduced.
Comment: The latest research has shown that the pressure may not be enough to
generate the liquid, but that such a liquid layer always exist on an ice surface, maybe
only a few molecules thick (dependent upon temperature).

3.4
At a higher elevation like in mountains the pressure is lower, what effect does that
have for cooking food?
A lower pressure means that water will boil at a lower temperature, see the
vaporization line in Fig. 3.7, or in Table B.1.2 showing the saturated temperature as
a function of the pressure. You therefore must increase the cooking time a little.

3.5
Water at room temperature and room pressure has v ≈ 1 × 10n m3/kg what is n?
See Table B.1.1 or B.1.2 to determine it is in the liquid phase
(you should know this already).
Table A.4 or from B1.1 at 20oC: n = –3 (v = 0.00100 m3/kg)
3.6
In Example 3.1 b is there any mass at the indicated specific volume? Explain.
This state is a two-phase mixture of liquid and vapor. There is no mass at the
indicated state, the v value is an average for all the mass, so there is some mass at
the saturated vapor state (fraction is the quality x) and the remainder of the mass is
saturated liquid (fraction 1-x).

3.7
Sketch two constant-pressure curves (500 kPa and 30 000 kPa) in a T-v diagram and
indicate on the curves where in the water tables you see the properties.
P
0.5
MPa
30
C.P.
v
T
v
500 kPa
30 MPa
B
1
4
B.1.3
B
1
4 B.1.3
B.1.3
B.1.2
B.1.1
B.1.5
B.1.5
B.1.3
The 30 MPa line in Table B.1.4 starts at 0oC and table ends at 380oC, the line is
continued in Table B.1.3 starting at 375oC and table ends at 1300oC.
The 500 kPa line in Table B.1.4 starts at 0.01oC and table ends at the saturated
liquid state (151.86oC). The line is continued in Table B.1.3 starting at the saturated
vapor state (151.86oC) continuing up to 1300oC.

3.8
If I have 1 L ammonia at room pressure and temperature (100 kPa, 20oC) how much
mass is that?
Ammonia Tables B.2:
B.2.1 Psat = 857.5 kPa at 20oC so superheated vapor.
B.2.2 v = 1.4153 m3/kg under subheading 100 kPa
m =
V
v =
0.001 m3
1.4153 m3/kg
= 0.000 706 kg = 0.706 g

3.9
Locate the state of ammonia at 200 kPa, -10oC. Indicate in both the P-v and the T-v
diagrams the location of the nearest states listed in the printed Table B.2
T
C.P.
v
200 kPa
P
C.P.
v
T
200
290.9
-18.9
-10
0
-18.9 C
-10 C 150
kPa

3.10
Why are most of the compressed liquid or solid regions not included in the printed
tables?
For the compressed liquid and the solid phases the specific volume and thus
density is nearly constant. These surfaces are very steep nearly constant v and there
is then no reason to fill up a table with the same value of v for different P and T.

3.11
How does a constant v-process look like for an ideal gas in a P-T diagram?
For an ideal gas: Pv = RT so then P = (R/v) T
Constant v is a straight line with slope (R/v) in the P-T diagram
P
T
VaporLiquid
Critical point
S
P = RT/v

3.12
If v = RT/P for an ideal gas what is the similar equation for a liquid?
The equation for a liquid is: v = Constant = vo
If you include that v increases a little with T then: v = vo + C (T − To)
where C is a small constant.

3.13
How accurate (find Z) is it to assume propane is an ideal gas a room conditions?
The propane table is not printed in appendix B so to get this information we either
must use the computer software or the generalized charts.
From Table A.2 Pc = 4250 kPa, Tc = 370 K
The reduced properties: Pr =
101
4250 = 0.024, Tr =
293
370 = 0.792
From Fig. D.1: Z = 0.98
So it is an ideal gas within 2% accuracy.
A modest pressure natural
gas container.

3.14
With Tr = 0.80 what is the ratio of vg/vf using Fig. D.1 or Table D.4
Since the two specific volumes are at the same P and T we have
Pvf = Zf RT and Pvg = Zg RT
so the ratio is
vg
vf
=
Zg
Zf
=
0.807
0.042 = 19.2

3.15
To solve for v given (P, T) in Eq 3.9 what mathematical problem do you have?
From Eq. 3.9 you notice it is non-linear in specific volume. Multiplying through
with the common denominator we get an equation cubic in v. We thus have to
find the roots for a cubic equation (zero points in a polynomial of order three). If
you have ever seen the exact solution to that you realize it will be easier to solve
by trial and error and select the proper root(s) (it cannot be negative).

Phase Diagrams, Triple and Critical Points

3.16
Carbon dioxide at 280 K can be in different phases. Indicate the pressure range
you have for each of the three phases (vapor, liquid and solid).
Look at the P-T phase diagram in Fig. 3.6 at 280 K:
P < 4000 kPa vapor
4000 kPa < P < 400 MPa liquid
400 MPa < P solid
ln P
T
V
L
S
280 K

3.17
Modern extraction techniques can be based on dissolving material in supercritical
fluids such as carbon dioxide. How high are pressure and density of carbon
dioxide when the pressure and temperature are around the critical point? Repeat
for ethyl alcohol.
Solution:
CO2 :
Table A.2: Pc = 7.38 MPa, Tc = 304 K, vc = 0.00212 m3/kg
ρc = 1/vc = 1/0.00212 = 472 kg/m3
C2H5OH:
Table A.2: Pc = 6.14 MPa, Tc = 514 K, vc = 0.00363 m3/kg
ρc = 1/vc = 1/0.00363 = 275 kg/m3

3.18
The ice cap on the North Pole could be 1000 m thick with a density of 920 kg/m3.
Find the pressure at the bottom and the corresponding melting temperature.
Solution:
ρICE = 920 kg/m3
∆P = ρgH = 920 kg/m3 × 9.80665 m/s2 × 1000 = 9 022 118 Pa
P = Po + ∆P = 101.325 + 9022 = 9123 kPa
See figure 3.7 liquid solid interphase => TLS = −1°C

3.19
Find the lowest temperature at which it is possible to have water in the liquid
phase. At what pressure must the liquid exist?
Solution:
There is no liquid at lower temperatures
than on the fusion line, see Fig. 3.7,
saturated ice III to liquid phase boundary is
at
T ≈ 263K ≈ - 10°C and
P ≈ 210 MPa
ln P
T
V
L
S
CR.P.
lowest T liquid

3.20
Water at 27°C can exist in different phases dependent upon the pressure. Give the
approximate pressure range in kPa for water being in each one of the three phases
vapor, liquid or solid.
Solution:
The phases can be seen in Fig. 3.7, a sketch
of which is shown to the right.
T = 27 °C = 300 Κ
From Fig. 3.6:
PVL ≈ 4 × 10−3 MPa = 4 kPa,
PLS = 103 MPa
ln P
T
V
L
S
CR.P.
S
0 < P < 4 kPa VAPOR
0.004 MPa < P < 1000 MPa LIQUID
P > 1000 MPa SOLID (ICE)

3.21
Dry ice is the name of solid carbon dioxide. How cold must it be at atmospheric
(100 kPa) pressure? If it is heated at 100 kPa what eventually happens?
Solution:
The phase boundaries are shown in Figure 3.6
At 100 kPa the carbon dioxide is solid if T < 190 K
It goes directly to a vapor state without becoming a liquid hence its name.
The 100 kPa is below
the triple point.
ln P
T
V
L
S
100
kPa

3.22
What is the lowest temperature in Kelvin for which you can see metal as a liquid
if the metal is a. silver b. copper
Solution:
Assume the two substances have a phase diagram similar to Fig. 3.6, then
we can see the triple point data in Table 3.2
Ta = 961oC = 1234 K
Tb = 1083oC = 1356 K

3.23
A substance is at 2 MPa, 17°C in a rigid tank. Using only the critical properties
can the phase of the mass be determined if the substance is nitrogen, water or
propane?
Solution:
Find state relative to critical point properties which are from Table A.2:
a) Nitrogen N2 : 3.39 MPa 126.2 K
b) Water H2O : 22.12 MPa 647.3 K
c) Propane C3H8 : 4.25 MPa 369.8 K
State is at 17 °C = 290 K and 2 MPa < Pc
for all cases:
N2 : T >> Tc Superheated vapor P < Pc
H2O : T << Tc ; P << Pc
you cannot say.
C3H8 : T < Tc ; P < Pc you cannot say
ln P
T
Vapor
Liquid Cr.P.
a
c
b

3.24
Give the phase for the following states.
Solution:
a. CO2 T = 40°C P = 0.5 MPa Table A.2
T > Tc => also P << Pc
superheated vapor assume ideal gas Table A.5
b. Air T = 20°C P = 200 kPa Table A.2
superheated vapor assume ideal gas Table A.5
c. NH3 T = 170°C P = 600 kPa Table B.2.2 or A.2
T > Tc => superheated vapor
P C.P.
v
T
v
T
a,b,c
a, b, c
P = const.

General Tables

3.25
Give the phase for the following states.
Solution:
a. H2O T = 260°C P = 5 MPa Table B.1.1 or B.1.2
B.1.1 For given T read: Psat = 4.689 MPa
P > Psat => compressed liquid
B.1.2 For given P read: Tsat = 264°C
T < Tsat => compressed liquid
b. H2O T = −2°C P = 100 kPa Table B.1.1 T < Ttriple point
Table B.1.5 at −2°C read: Psat = 0.518 kPa
since P > Psat => compressed solid
Note state b in P-v, see the
3-D figure, is up on the
solid face.
P C.P.
v
T
C.P.
v
T
a
b
a
b
P = const.
P
T
v
V
L
S
C.P.
a
b

3.26
Determine the phase of the substance at the given state using Appendix B tables
a) Water 100°C, 500 kPa
b) Ammonia -10°C, 150 kPa
c) R-410a 0°C, 350 kPa
Solution:
a) From Table B.1.1 Psat(100°C) = 101.3 kPa
500 kPa > Psat then it is compressed liquid
OR from Table B.1.2 Tsat(500 kPa) = 152°C
100°C < Tsat then it is subcooled liquid = compressed liquid
b) Ammonia NH3 :
Table B.2.1: P < Psat(-10 °C) = 291 kPa
Superheated vapor
c) R-410a
Table B.4.1: P < Psat(0 °C) = 799 kPa
Superheated vapor.
The S-L fusion line goes slightly to the
left for water. It tilts slightly to the right
for most other substances.
ln P
T
Vapor
L Cr.P.
a
b,cS

3.27
Determine whether water at each of the following states is a compressed liquid, a
superheated vapor, or a mixture of saturated liquid and vapor.
a. P = 10 MPa, v = 0.003 m3/kg b. 1 MPa, 190°C
c. 200°C, 0.1 m3/kg d. 10 kPa, 10°C
Solution:
For all states start search in table B.1.1 (if T given) or B.1.2 (if P given)
a. P = 10 MPa, v = 0.003 m3/kg so look in B.1.2 at 10 MPa
vf = 0.001452; vg = 0.01803 m3/kg,
=> vf < v < vg => so mixture of liquid and vapor.
b. 1 MPa, 190°C : Only one of the two look-ups is needed
B.1.1: P < Psat = 1254.4 kPa so it is superheated vapor
B.1.2: T > Tsat = 179.91°C so it is superheated vapor
c. 200°C, 0.1 m3/kg: look in B.1.1
vf = 0.001156 m3/kg ; vg = 0.12736 m3/kg,
=> vf < v < vg => so mixture of liquid and vapor.
d. 10 kPa, 10°C : Only one of the two look-ups is needed
From B.1.1: P > Pg = 1.2276 kPa so compressed liquid
From B.1.2: T < Tsat = 45.8°C so compressed liquid
States shown are
placed relative to the
two-phase region, not
to each other.
P C.P.
v
T
C.P.
v
Ta
d
c
b
d
a c
b
P = const.

3.28
For water at 100 kPa with a quality of 10%, find the volume fraction of vapor.
This is a two-phase state at a given pressure:
Table B.1.2: vf = 0.001 043 m3/kg, vg = 1.6940 m3/kg
From the definition of quality we get the masses from total mass, m, as
mf = (1 – x) m, mg = x m
The volumes are
Vf = mf vf = (1 – x) m vf, Vg = mg vg = x m vg
So the volume fraction of vapor is
Fraction =
Vg
V =
Vg
Vg + Vf
=
x m vg
x m vg + (1 – x)m vf
=
0.1 × 1.694
0.1 × 1.694 + 0.9 × 0.001043
=
0.1694
0.17034 = 0.9945
Notice that the liquid volume is only about 0.5% of the total. We could also have
found the overall v = vf + xvfg and then V = m v.

3.29
Determine whether refrigerant R-410a in each of the following states is a
compressed liquid, a superheated vapor, or a mixture of saturated liquid and
vapor.
Solution:
All cases are seen in Table B.4.1
a. 50°C, 0.05 m3/kg From table B.4.1 at 50°C vg = 0.00707 m3/kg
since v > vg we have superheated vapor
b. 1.0 MPa, 20°C From table B.4.1 at 20°C Pg = 909.9 kPa
since P < Pg we have superheated vapor
c. 0.1 MPa, 0.1 m3/kg From table B.4.1 at 0.1 MPa (use 101 kPa)
vf = 0.00074 and vg = 0.2395 m3/kg
as vf < v < vg we have a mixture of liquid &
vapor
d −20°C, 200 kPa superheated vapor, P < Pg = 400 kPa at -20°C
3.30
Place the states in Problem 3.29 in a sketch of the P-v diagram.
States shown are
to each other.
P C.P.
v
T
C.P.
v
T
a, b
dc d
a, b
c
P = const.

3.31
Refer to Fig. 3.18. How much is the change in liquid specific volume for water at
20oC as you move up from state i towards state j reaching 15 000 kPa?
State “i”, here “a”, is saturated liquid and up is then compressed liquid states
a Table B.1.1: vf = 0.001 002 m3/kg at 2.34 kPa
b Table B.1.4: vf = 0.001 002 m3/kg at 500 kPa
c Table B.1.4: vf = 0.001 001 m3/kg at 2000 kPa
d Table B.1.4: vf = 0.001 000 m3/kg at 5000 kPa
e Table B.1.4: vf = 0.000 995 m3/kg at 15 000 kPa
f Table B.1.4: vf = 0.000 980 m3/kg at 50 000 kPa
Notice how small the changes in v are for very large changes in P.
v
T
v
P
a
b
c
T = 20 Co
d
e
f
f-a
P
T
v
V
L
S
C.P.
a
f

3.32
Fill out the following table for substance ammonia:
Solution:
P [kPa] T [ oC] v [m3/kg] x
a) 1200 50 0.1185 Undefined
b) 2033 50 0.0326 0.5
a) B.2.1 v > vg => superheated vapor Look in B.2.2
b) B.2.1 P = Psat = 2033 kPa
v = vf + x vfg = 0.001777 + 0.5 × 0.06159 = 0.0326 m3/kg
3.33
Place the two states a-b listed in Problem 3.32 as labeled dots in a sketch of the P-
v and T-v diagrams.
Solution:
T
C.P.
v
ab
P = const.
P C.P.
v
T
ab
2033
50
1200

3.34
Give the phase and the missing property of P, T, v and x.
a. R-134a T = -20oC, P = 150 kPa
b. R-134a P = 300 kPa, v = 0.072 m3/kg
Solution:
a) B.5.1 P > Psat = 133.7 kPa ⇒ compressed liquid
v ~ vf = 0.000738 m3/kg
x = undefined
b) B.5.2 v > vg at 300 kPa ⇒ superheated vapor
T = 10 + (20 - 10) ( 0.072 - 0.07111
0.07441 - 0.07111) = 12.7°C
x = undefined
T
C.P.
v
a
b P = const.
P
v
a
T
b

3.35
Fill out the following table for substance water:
Solution:
P [kPa] T [ oC] v [m3/kg] x
a) 500 20 0.001002 Undefined
b) 500 151.86 0.20 0.532
c) 1400 200 0.14302 Undefined
d) 8581 300 0.01762 0.8
a) Table B.1.1 P > Psat so it is compressed liquid => Table B.1.4
b) Table B.1.2 vf < v < vg so two phase L + V
x =
v - vf
vfg
= (0.2 – 0.001093) / 0.3738 = 0.532
T = Tsat = 151.86oC
c) Only one of the two look-up is needed
Table B.1.1 200oC P < Psat = => superheated vapor
Table B.1.2 1400 kPa T > Tsat = 195oC
Table B.1.3 sub-table for 1400 kPa gives the state properties
d) Table B.1.1 since quality is given it is two-phase
v = vf + x × vfg = 0.001404 + 0.8 × 0.02027 = 0.01762 m3/kg
3.36
Place the four states a-d listed in Problem 3.35 as labeled dots in a sketch of the
P-v and T-v diagrams.
Solution:
T
C.P.
v
d
a
c
b
P = const.
P C.P.
v
Ta
d
c
b
500
1400
8581
20
152
200
300

3.37
Determine the specific volume for R-410a at these states:
a. –15oC, 500 kPa
b. 20oC, 1000 kPa
c. 20oC, quality 25%
a) Table B.4.1: P > Psat = 480.4 kPa, so compressed liquid.
v ≈ vf = 0.000815 m3/kg
b) Table B.4.1: P < Psat = 1444 kPa, so superheated vapor
Table B.4.2: v = 0.02838 m3/kg
c) Table B.4.1: vf = 0.000815 m3/kg, vfg = 0.01666 m3/kg so
v = vf + x vfg = 0.000815 + 0.25 × 0.01666 = 0.00498 m3/kg

3.38
Give the missing property of P, T, v and x for CH4 at:
a. T = 155 K, v = 0.04 m3/kg
b. T = 350 K, v = 0.25 m3/kg
a) B.7.1 v < vg = 0.04892 m3/kg ⇒ 2-phase
x =
v - vf
vfg
=
0.04 - 0.002877
0.04605 = 0.806
P = Psat = 1296 kPa
b) B.7.1 T > Tc and v >> vc ⇒ superheated vapor B.7.2
located between 600 & 800 kPa
P = 600 + 200
0.25 - 0.30067
0.2251 - 0.30067 = 734 kPa
T
C.P.
v
P = const.
P
v
T
a b a
b

3.39
Give the specific volume of carbon-dioxide at -20oC for 2000 kPa and repeat for
1400 kPa.
Table B.3.1: -20oC Psat = 1969 kPa,
at 2000 kPa state is compressed liquid: v = vf = 0.000969 m3/kg
at 1400 kPa state is superheated vapor: v = 0.0296 m3/kg
The 2000 kPa is above and
the 1400 kPa is below
the vaporization line.
ln P
T
V
L
S
b
a

3.40
Calculate the following specific volumes
a. CO2 10°C, 80% quality
b. Water 4 MPa, 90% quality
c. Nitrogen 120 K, 60% quality
Solution:
All states are two-phase with quality given. The overall specific
volume is given by Eq.3.1 or 3.2
v = vf + x vfg = (1-x)vf + x vg
a. CO2 10°C, 80% quality in Table B.3.1
v = 0.001161 + x × 0.00624 = 0.006153 m3/kg
b. Water 4 MPa, 90% quality in Table B.1.2
v = 0.001252(1-x) + x × 0.04978 = 0.04493 m3/kg
c. Nitrogen 120 K, 60% quality in Table B.6.1
v = 0.001915 + x × 0.00608 = 0.005563 m3/kg

3.41
Give the phase and P, x for nitrogen at:
a) T = 120 K, v = 0.006 m3/kg
b) T = 140 K, v = 0.002 m3/kg
Solution:
a) Table B.6.1 v < vg = 0.00799 m3/kg so state is two-phase L + V
P = Psat = 2513 kPa
x =
v - vf
vfg
= (0.006 – 0.001915)/0.00608 = 0.6719
b) Table B.6.1 T > Tc = 126.2 K so go to B.6.2 superheated vapor.
Look at any entry for 140 K then you see a too large v meaning P is
higher. Last one
P = 10 000 kPa, x = undefined
T
c.p.
v
P = const.
P
v
T
a
b
a
b

3.42
You want a pot of water to boil at 105oC. How heavy a lid should you put on the
15 cm diameter pot when Patm = 101 kPa?
Solution:
Table B.1.1 at 105oC : Psat = 120.8 kPa
A =
π
4 D2 =
π
4 0.152 = 0.01767 m2
Fnet = (Psat –Patm) A = (120.8 - 101) kPa × 0.01767 m2
= 0.3498 kN = 350 N
Fnet = mlid g
mlid = Fnet/g =
350
9.807 = 35.7 kg
Some lids are
clamped on, the
problem deals with
one that stays on due
to its weight.

3.43
Water at 120oC with a quality of 25% has its temperature raised 20oC in a constant
volume process. What is the new quality and pressure?
Solution:
State 1 from Table B.1.1 at 120oC
v = vf + x vfg = 0.001060 + 0.25 × 0.8908 = 0.22376 m3/kg
State 2 has same v at 140oC also from Table B.1.1
x =
v - vf
vfg
=
0.22376 - 0.00108
0.50777 = 0.4385
P = Psat = 361.3 kPa
T
C.P.
v
P C.P.
v
T
198.5
361.3
120
140
120 C
140 C

3.44
A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100°C.
The vessel is now heated. If a safety pressure valve is installed, at what pressure
should the valve be set to have a maximum temperature of 200°C?
Solution:
Process: v = V/m = constant
State 1: v1 = 1/2 = 0.5 m3/kg
from Table B.1.1
it is 2-phase
State 2: 200°C, 0.5 m3/kg
Table B.1.3 between 400
and 500 kPa so interpolate
C.P.T
v
100 C
500 kPa
400 kPa
P ≅ 400 +
0.5-0.53422
0.42492-0.53422 × (500 - 400) = 431.3 kPa

3.45
Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this
state the piston is 0.1 m from the cylinder bottom. How much is this distance and
the temperature if the water is cooled to occupy half the original volume?
Solution:
State 1: B 1.2 v1 = vg (200 kPa) = 0.8857 m3/kg, T1 = 120.2°C
Process: P = constant = 200 kPa
State 2: P, v2 = v1/2 = 0.44285 m3/kg
Table B.1.2 v2 < vg so two phase T2 = Tsat = 120.2°C
Height is proportional to volume
h2 = h1 × v2/v1 = 0.1 × 0.5 = 0.05m
T
C.P.
v
12
P = 200 kPa
P C.P.
v
T
2
200
120
1

3.46
Saturated liquid water at 60°C is put under pressure to decrease the volume by 1%
keeping the temperature constant. To what pressure should it be compressed?
Solution:
State 1: T = 60°C , x = 0.0; Table B.1.1: v = 0.001017 m3/kg
Process: T = constant = 60°C
State 2: T, v = 0.99 × vf (60°C) = 0.99×0.001017 = 0.0010068 m3/kg
Between 20 & 30 MPa in Table B.1.4, P ≅ 23.8 MPa
P
C.P.
v
T
v
20 MPa
1
2
1
2
30 MPa

3.47
Water at 200 kPa with a quality of 25% has its temperature raised 20oC in a constant
pressure process. What is the new quality and volume?
Solution:
State 1 from Table B.1.2 at 200 kPa
v = vf + x vfg = 0.001061 + 0.25 × 0.88467 = 0.22223 m3/kg
State 2 has same P from Table B.1.2 at 200 kPa
T2 = Tsat + 20 = 120.23 + 20 = 140.23oC
so state is superheated vapor
x = undefined
v2 = 0.88573 + (0.95964 – 0.88573)
20
150 - 120.23 = 0.9354 m3/kg
Interpolate between the saturated vapor state (the first entry in table B.1.3 for 200
kPa) and the 150oC superheated vapor state also at 200 kPa.
T
C.P.
v
200 kPa
P C.P.
v
T
200
120
140
120.2 C
140 C

3.48
In your refrigerator the working substance evaporates from liquid to vapor at -20
oC inside a pipe around the cold section. Outside (on the back or below) is a black
grille inside which the working substance condenses from vapor to liquid at +40
oC. For each location find the pressure and the change in specific volume (v) if
the substance is ammonia.
Solution:
The properties come from the saturated tables where each phase change takes
place at constant pressure and constant temperature.
Substance TABLE T Psat , kPa ∆v = vfg, m3/kg
Ammonia B.2.1 40 oC 1555 0.0814
Ammonia B.2.1 -20 oC 190 0.622
T
C.P.
v
1 2
P C.P.
v
T
2
-20
1
40
40C 34
34

3.49
Repeat the previous problem with the substances
a) R-134a b) R-410a
In your refrigerator, the working substance evaporates from liquid to vapor at -20
oC inside a pipe around the cold section. Outside (on the back or below) is a black
grille inside which the working substance condenses from vapor to liquid at +40
oC. For each location find the pressure and the change in specific volume (v).
Solution:
Substance TABLE T Psat , kPa ∆v = vfg, m3/kg
R-134a B.5.1 40 oC 1017 0.019
R-134a B.5.1 -20 oC 134 0.146
R-410a B.4.1 40 oC 2421 0.00865
R-410a B.4.1 -20 oC 400 0.064
T
C.P.
v
1 2
P C.P.
v
T
2
-20
1
40
40C 34
34

3.50
Repeat Problem 3.48 with CO2, condenser at + 20oC and evaporator at -30oC.
Solution:
Substance TABLE T Psat , kPa ∆v = vfg
CO2 B.3.1 20 oC 5729 0.00386
CO2 B.3.1 -30 oC 1428 0.026
T
C.P.
v
1 2
P C.P.
v
T
2
-20
1
40
40C 34
34

3.51
A glass jar is filled with saturated water at 500 kPa, quality 25%, and a tight lid is
put on. Now it is cooled to −10°C. What is the mass fraction of solid at this
temperature?
Solution:
Constant volume and mass ⇒ v1 = v2 = V/m
From Table B.1.2: v1 = 0.001093 + 0.25 × 0.3738 = 0.094543
From Table B.1.5: v2 = 0.0010891 + x2 × 446.756 = v1 = 0.094543
⇒ x2 = 0.0002 mass fraction vapor
xsolid =1 - x2 = 0.9998 or 99.98 %
P C.P.
v
T
C.P.
v
T
1
2
1
2
P
T
v
V
L
S
C.P.
1
2

3.52
Two tanks are connected as shown in Fig. P3.52, both containing water. Tank A is at
200 kPa, v = 0.5 m3/kg, V
A = 1 m3 and tank B contains 3.5 kg at 0.5 MPa, 400°C.
The valve is now opened and the two come to a uniform state. Find the final specific
volume.
Solution:
Control volume: both tanks. Constant total volume and mass process.
A B
sup. vapor
State A1: (P, v) mA = VA/vA = 1/0.5 = 2 kg
State B1: (P, T) Table B.1.3 vB = 0.6173 m3/kg
⇒ VB = mBvB = 3.5 × 0.6173 = 2.1606 m3
Final state: mtot = mA + mB = 5.5 kg
Vtot = VA + VB = 3.1606 m3
v2 = Vtot/mtot = 0.5746 m3/kg

3.53
Saturated vapor R-134a at 50oC changes volume at constant temperature. Find the
new pressure, and quality if saturated, if the volume doubles. Repeat the question
for the case the volume is reduced to half the original volume.
Solution:
1: (T, x) B.4.1: v1 = vg = 0.01512 m3/kg, P1 = Psat = 1318 kPa
2: v2 = 2v1 = 0.03024 m3/kg superheated vapor
Interpolate between 600 kPa and 800 kPa
P2 = 600 + 200 ×
0.03024 – 0.03974
0.02861 – 0.03974 = 771 kPa
3: v3 = v1/2 = 0.00756 m3/kg < vg : two phase
x3 =
v3 - vf
vfg
=
0.00756 – 0.000908
0.01422 = 0.4678
P3 = Psat = 1318 kPa
T
C.P.
v
1 2
P = 1318 kPa
P C.P.
v
T
2
1318
50
13
3

3.54
A steel tank contains 6 kg of propane (liquid + vapor) at 20°C with a volume of
0.015 m3. The tank is now slowly heated. Will the liquid level inside eventually rise
to the top or drop to the bottom of the tank? What if the initial mass is 1 kg instead
of 6 kg?
Solution:
Constant volume and mass v2 = v1 =
V
m =
0.015 m3
6 kg = 0.0025 m3/kg
v
T
20°C
C.P.
vc
a b
VaporLiq.
A.2: vc = 0.00454 m3/kg > v1
eventually reaches sat. liquid.
⇒ level rises to top
If m = 1 kg ⇒ v1 = 0.015 m3/kg > vc
then it will reach saturated vapor.
⇒ level falls

3.55
Saturated water vapor at 60°C has its pressure decreased to increase the volume
by 10% keeping the temperature constant. To what pressure should it be
expanded?
Solution:
Initial state: v = 7.6707 m3/kg from table B.1.1
Final state: v = 1.10 × vg = 1.1 × 7.6707 = 8.4378 m3/kg
Interpolate at 60°C between saturated (P = 19.94 kPa) and superheated vapor
P = 10 kPa in Tables B.1.1 and B.1.3
P ≅ 19.941 + (10 − 19.941)
8.4378 − 7.6707
15.3345 − 7.6707
= 18.9 kPa
P C.P.
v
T
C.P.
v
T
P = 10 kPa
60 Co
10 kPa
Comment: T,v ⇒ P = 18 kPa (software) v is not linear in P, more like 1/P,
so the linear interpolation in P is not very accurate.

3.56
Ammonia at 20oC with a quality of 50% and total mass 2 kg is in a rigid tank
with an outlet valve at the bottom. How much liquid (mass) can you take out
through the valve assuming the temperature stays constant?
Solution:
The bottom has liquid until the state inside becomes saturated vapor.
V = m1v1 = 2 × 0.5 (0.001638 + 0.14922) = 0.15086 m3
m2 = V/v2 = 0.15086 m3/ 0.14922 m3/kg = 1.0198 kg
m = m1 – m2 = 2 – 1.0198 = 0.989 kg
P TC.P. C.P.
P = 857 kPa
20 Co
T
v v

3.57
A sealed rigid vessel of 2 m3 contains a saturated mixture of liquid and vapor R-
134a at 10°C. If it is heated to 50°C, the liquid phase disappears. Find the
pressure at 50°C and the initial mass of the liquid.
Solution:
Process: constant volume and constant mass.
P
v
2
1
State 2 is saturated vapor, from table B.5.1
P2 = Psat(50°C) = 1.318 MPa
State 1: same specific volume as state 2
v1 = v2 = 0.015124 m3/kg
v1 = 0.000794 + x1 × 0.048658
⇒ x1 = 0.2945
m = V/v1 = 2/0.015124 = 132.24 kg; mliq = (1 - x1)m = 93.295 kg

3.58
A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up
by 5°C per hour due to a failure in the refrigeration system. How long time will it
take before the methane becomes single phase and what is the pressure then?
Solution: Use Table B.7.1
Assume rigid tank v = constant = v1
v1 = 0.002439 + 0.25 × 0.30367 = 0.078366 m3/kg
We then also see that v1 > vc = 0.00615 m3/kg
All single phase when v = vg => T ≅ 145 K
∆t =
∆T
(5°C/h) ≅
145 – 120
5 = 5 hours P = Psat= 824 kPa

3.59
Ammonia at 10oC and mass 0.1 kg is in a piston cylinder with an initial volume of 1
m3. The piston initially resting on the stops has a mass such that a pressure of 900
kPa will float it. Now the ammonia is slowly heated to 50oC. Find the final pressure
and volume.
Solution:
C.V. Ammonia, constant mass.
Process: V = constant unless P = Pfloat
State 1: T = 10 oC, v1 =
V
m =
1
10 = 0.1 m3/kg
From Table B.2.1 vf < v < vg
x1 =
v - vf
vfg
=
0.1 - 0.0016
0.20381 = 0.4828 V
P
2
1
1a
P
P
1
2
State 1a: P = 900 kPa, v = v1 = 0.1 m3/kg < vg at 900 kPa
This state is two-phase T1a = 21.52oC
Since T2 > T1a then v2 > v1a
State 2: 50oC and on line(s) means
P2 = 900 kPa which is superheated vapor.
Table B.2.2 : v2 = 0.16263 m3/kg
V2 = mv2 = 1.6263 m3

3.60
A 400-m3 storage tank is being constructed to hold LNG, liquified natural gas,
which may be assumed to be essentially pure methane. If the tank is to contain
90% liquid and 10% vapor, by volume, at 100 kPa, what mass of LNG (kg) will
the tank hold? What is the quality in the tank?
Solution:
CH4 is in the section B tables.
From Table B.7.1: vf ≅ 0.002366 m3/kg, (interpolated)
From Table B.7.2: vg ≅ 0.55665 m3/kg (first entry 100 kPa)
mliq =
Vliq
vf
=
0.9 × 400
0.002366 = 152 155.5 kg; mvap =
Vvap
vg
=
0.1 × 400
0.55665 = 71.86 kg
mtot = 152 227 kg, x = mvap / mtot = 4.72 × 10-4

3.61
A boiler feed pump delivers 0.05 m3/s of water at 240°C, 20 MPa. What is the mass
flowrate (kg/s)? What would be the percent error if the properties of saturated liquid
at 240°C were used in the calculation? What if the properties of saturated liquid at
20 MPa were used?
Solution:
State 1: (T, P) compressed liquid seen in B.1.4: v = 0.001205 m3/kg
m
.
= V
.
/v = 0.05/0.001205 = 41.5 kg/s
vf (240°C) = 0.001229 m3/kg ⇒ m
.
= 40.68 kg/s error 2%
vf (20 MPa) = 0.002036 m3/kg ⇒ m
.
= 24.56 kg/s error 41%
vf (20 MPa) = 0.002036 m3/kg ⇒ m
.
= 24.56 kg/s error 41%
P C.P.
v
T
C.P.
v
P = 20 MPa
240 Co
20 MPa
240
The constant T line is nearly vertical for the liquid phase in the P-v diagram.
The state is at so high P, T that the saturated liquid line is not extremely steep.

3.62
A piston/cylinder arrangement is loaded with a linear spring and the outside
atmosphere. It contains water at 5 MPa, 400°C with the volume being 0.1 m3. If the
piston is at the bottom, the spring exerts a force such that Plift = 200 kPa. The system
now cools until the pressure reaches 1200 kPa. Find the mass of water, the final state
(T2, v2) and plot the P–v diagram for the process.
Solution:
P
v
5000
1200
200
1
2
a
? 0.057810
1: Table B.1.3 ⇒ v1= 0.05781 m3/kg
m = V/v1 = 0.1/0.05781 = 1.73 kg
Straight line: P = Pa + C × v
v2 = v1
P2 - Pa
P1 - Pa
= 0.01204 m3/kg
v2 < vg(1200 kPa) so two-phase T2 = 188°C
⇒ x2 = (v2 - 0.001139)/0.1622 = 0.0672

3.63
A pressure cooker (closed tank) contains water at 100°C with the liquid volume
being 1/10 of the vapor volume. It is heated until the pressure reaches 2.0 MPa.
Find the final temperature. Has the final state more or less vapor than the initial
state?
Solution:
State 1: Vf = mf vf = Vg/10 = mgvg/10 ;
Table B.1.1: vf = 0.001044 m3/kg, vg = 1.6729 m3/kg
x1 =
mg
mg + mf
=
10 mfvf / vg
mf + 10 mfvf / vg
=
10 vf
10 vf + vg
=
0.01044
0.01044 + 1.6729 = 0.0062
v1 = 0.001044 + 0.0062×1.67185 = 0.01141 m3/kg
State 2: v2 = v1 = 0.01141 m3/kg < vg(2MPa) from B.1.2 so two-phase
P
v
2
1
At state 2: v2 = vf + x2 vfg
0.01141 = 0.001177 + x2 × 0.09845
=> x2 = 0.104
More vapor at final state
T2 = Tsat(2MPa) = 212.4°C

3.64
A pressure cooker has the lid screwed on tight. A small opening with A = 5 mm2
is covered with a petcock that can be lifted to let steam escape. How much mass
should the petcock have to allow boiling at 120oC with an outside atmosphere at
101.3 kPa?
Table B.1.1.: Psat = 198.5 kPa
F = mg = ∆P × A
m = ∆P × A/g
=
(198.5-101.3)×1000×5×10-6
9.807
= 0.0496 kg = 50 g

Ideal Gas Law

3.65
What is the relative (%) change in P if we double the absolute temperature of an
ideal gas keeping mass and volume constant? What will it be if we double V
having m, T constant.
Ideal gas law: PV = mRT
State 2: P2V = mRT2 = mR2T1 = 2P1V ⇒ P2 = 2P1
Relative change = ∆P/P1 = P1/P1 = 1 = 100%
State 3: P3V3 = mRT1 = P1V1 ⇒ P3 = P1V1/V3 = P1/2
Relative change = ∆P/P1 = -P1/2P1 = -0.5 = -50%
TP
V
T
3
2
1
T1
2
V
2
1 3

3.66
A 1-m3 tank is filled with a gas at room temperature 20°C and pressure 100 kPa.
How much mass is there if the gas is a) air, b) neon or c) propane ?
Solution:
Use Table A.2 to compare T and P to the critical T and P with
T = 20°C = 293.15 K ; P = 100 kPa << Pc for all
Air : T >> TC,N2; TC,O2 = 154.6 K so ideal gas; R= 0.287 kJ/kg K
Neon: T >> Tc = 44.4 K so ideal gas; R = 0.41195 kJ/kg K
Propane: T < Tc = 370 K, but P << Pc = 4.25 MPa
so gas R = 0.18855 kJ/kg K
All states are ideal gas states so the ideal gas law applies
PV = mRT
a) m =
PV
RT =
100 × 1
0.287 × 293.15
= 1.189 kg
b) m =
100 × 1
0.41195 × 293.15
= 0.828 kg
c) m =
100 × 1
0.18855 × 293.15
= 1.809 kg

3.67
Calculate the ideal gas constant for argon and hydrogen based on table A.2 and
verify the value with Table A.5
argon: R = R
_
/ M = 8.3145 / 39.948 = 0.2081 same as Table A.5
hydrogen: R = R
_
/ M = 8.3145 / 2.016 = 4.124256 same as Table A.5

3.68
A pneumatic cylinder (a piston cylinder with air) must close a door with a force of
500 N. The cylinder cross-sectional area is 5 cm2 and its volume is 50 cm3. What
is the air pressure and its mass?
F = PA – PoA ⇒
P = Po + F/A = 100 kPa +
500 N
0.0005 m2 × 1000 N/kN
= 1100 kPa
m =
P1V1
RT1
=
1100 kPa × 0.00005 m3
0.287 kJ/kgK × 298 K
= 0.00064 kg = 0.64 g
Comment: Dependent upon your understanding of the problem you could also have
neglected the atmospheric pressure to get 1000 kPa and 0.58 g for answers.

3.69
Is it reasonable to assume that at the given states the substance behaves as an
ideal gas?
Solution:
a) Oxygen, O2 at 30°C, 3 MPa Ideal Gas ( T » Tc = 155 K from A.2)
b) Methane, CH4 at 30°C, 3 MPa Ideal Gas ( T » Tc = 190 K from A.2)
c) Water, H2O at 30°C, 3 MPa NO compressed liquid P > Psat (B.1.1)
d) R-134a at 30°C, 3 MPa NO compressed liquid P > Psat (B.5.1)
e) R-134a at 30°C, 100 kPa Ideal Gas P is low < Psat (B.5.1)
ln P
T
Vapor
Liq.
Cr.P.
a, b
c, d
e

3.70
Helium in a steel tank is at 250 kPa, 300 K with a volume of 0.1 m3. It is used to fill
a balloon. When the pressure drops to 150 kPa, the flow of helium stops by itself. If
all the helium still is at 300 K, how big a balloon did I get?
Solution:
State 1: m = V/v assume ideal gas so
m =
P1V1
RT1
=
250 × 0.1
2.0771 × 300
= 0.0401 kg
State 2: Same mass so then (T2 = T1)
V2 =
mRT2
P2
=
P1V1 RT2
RT1 P2
= V1
P1
P2
= 0.1
250
150 = 0.16667 m3
c
i
r
c
u
s
t
h
e
r
m
o
cb
The balloon volume is
Vballoon = V2 – V1 = 0.16667 – 0.1 = 0.06667 m3

3.71
A hollow metal sphere of 150-mm inside diameter is weighed on a precision
beam balance when evacuated and again after being filled to 875 kPa with an
unknown gas. The difference in mass is 0.0025 kg, and the temperature is 25°C.
What is the gas, assuming it is a pure substance listed in Table A.5 ?
Solution:
Assume an ideal gas with total volume: V =
π
6(0.15)3
= 0.001767 m3
M =
mR
_
T
PV =
0.0025 × 8.3145 × 298.2
875 × 0.001767
= 4.009 ≈ MHe
=> Helium Gas

3.72
A spherical helium balloon of 10 m in diameter is at ambient T and P, 15oC and
100 kPa. How much helium does it contain? It can lift a total mass that equals the
mass of displaced atmospheric air. How much mass of the balloon fabric and cage
can then be lifted?
We need to find the masses and the balloon volume
V =
π
6 D3 =
π
6 103 = 523.6 m3
mHe = ρV =
V
v =
PV
RT =
100 kPa × 523.6 m3
2.0771 kJ/kgK × 288 K
= 87.5 kg
mair =
PV
RT =
100 kPa × 523.6 m3
0.287 kJ/kgK × 288 K
= 633 kg
mlift = mair – mHe = 633-87.5 = 545.5 kg

3.73
A glass is cleaned in 45oC hot water and placed on the table bottom up. The room
air at 20oC that was trapped in the glass gets heated up to 40oC and some of it
leaks out so the net resulting pressure inside is 2 kPa above ambient pressure of
101 kPa. Now the glass and the air inside cools down to room temperature. What
is the pressure inside the glass?
Solution:
1 air: 40oC, 103 kPa
2 air: 20oC, ?
Constant Volume: V1 = V2,
AIR
Slight amount
of liquid water
seals to table top
Constant Mass m1 = m2
Ideal Gas P1V1 = m1RT1 and P2V2 = m1RT2
Take Ratio
P2 = P1
T1
T2
= 103 ×
20 + 273
40 + 273 = 96.4 kPa
This is a vacuum relative to atm pressure so the glass is pressed against table.

3.74
Air in an internal combustion engine has 227oC, 1000 kPa with a volume of 0.1 m3.
Now combustion heats it to 1500 K in a constant volume process. What is the mass
of air and how high does the pressure become?
The mass comes from knowledge of state 1 and ideal gas law
m =
P1V1
RT1
=
1000 kPa × 0.1 m3
0.287 kJ/kgK × (227 + 273) K
= 0.697 kg
The final pressure is found from the ideal gas law written for state 1 and state 2 and
then eliminate the mass, gas constant and volume (V2 = V1) between the equations
P1 V1 = m RT1 and P2 V2 = m RT2
P2 = P1 × T2/T1 = 1000 ×
1500
500 = 3000 kPa

3.75
Air in an automobile tire is initially at −10°C and 190 kPa. After the automobile is
driven awhile, the temperature gets up to 10°C. Find the new pressure. You must
make one assumption on your own.
Solution:
Assume constant volume and that air is an ideal
gas
P2 = P1 × T2/T1
= 190 ×
283.15
263.15 = 204.4 kPa

3.76
A rigid tank of 1 m3 contains nitrogen gas at 600 kPa, 400 K. By mistake
someone lets 0.5 kg flow out. If the final temperature is 375 K what is then the
final pressure?
Solution:
m =
PV
RT =
600 × 1
0.2968 × 400
= 5.054 kg
m2 = m - 0.5 = 4.554 kg
P2 =
m2RT2
V =
4.554 × 0.2968 × 375
1 = 506.9 kPa

3.77
Assume we have 3 states of saturated vapor R-134a at +40 oC, 0 oC and -40 oC.
Calculate the specific volume at the set of temperatures and corresponding saturated
pressure assuming ideal gas behavior. Find the percent relative error = 100(v - vg)/vg
with vg from the saturated R-134a table.
Solution:
R-134a. Table values from Table B.5.1 Psat, vg(T)
Ideal gas constant from Table A.5: RR-134a = 0.08149 kJ/kg K
T Psat , kPa vg vID.G. = RT / Psat error %
-40 o
C 51.8 0.35696 0.36678 2.75
0 o
C 294 0.06919 0.07571 9.4
40 o
C 1017 0.02002 0.02509 25.3
v
P
v
T
1
2 1
2
33

3.78
Do Problem 3.77, but for the substance R-410a.
Solution:
R-410a. Table values from Table B.4.1 Psat, vg(T)
Ideal gas constant from Table A.5: RR-410a = 0.1146 kJ/kg K
-40 o
C 175.0 0.14291 0.1527 6.8
0 o
C 798.7 0.03267 0.03919 20
40 o
C 2420.7 0.00967 0.01483 53.3
v
P
v
T
1
2 1
2
33

3.79
Do Problem 3.77, but for the substance ammonia.
Solution:
NH3. Table values from Table B.2.1 Psat, vg(T)
Ideal gas constant from Table A.5: Rammonia = 0.4882 kJ/kg K
-40 o
C 71.7 1.5526 1.5875 2.25
0 o
C 429.6 0.28929 0.3104 7.3
40 o
C 1555 0.08313 0.09832 18.3
v
P
v
T
1
2 1
2
33

3.80
A 1 m3 rigid tank has propane at 100 kPa, 300 K and connected by a valve to
another tank of 0.5 m3 with propane at 250 kPa, 400 K. The valve is opened and
the two tanks come to a uniform state at 325 K. What is the final pressure?
Solution:
Propane is an ideal gas (P << Pc) with R = 0.1886 kJ/kgK from Tbl. A.5
mA =
PAVA
RTA
=
100 × 1
0.1886 × 300
= 1.7674 kg
mB =
PBVB
RTB
=
250 × 0.5
0.1886 × 400
= 1.6564 kg
V2 = VA + VB = 1.5 m3
m2 = mA + mB = 3.4243 kg
P2 =
m2RT2
V2
=
3.4243 × 0.1886 × 325
1.5 = 139.9 kPa

3.81
A vacuum pump is used to evacuate a chamber where some specimens are dried at
50°C. The pump rate of volume displacement is 0.5 m3/s with an inlet pressure of
0.1 kPa and temperature 50°C. How much water vapor has been removed over a 30-
min period?
Solution:
Use ideal gas since P << lowest P in steam tables.
From table A.5 we get R = 0.46152 kJ/kg K
m = m
.
∆t with mass flow rate as: m
.
= V
.
/v = PV
.
/RT (ideal gas)
⇒ m = PV
.
∆t/RT =
0.1 × 0.5 × 30×60
(0.46152 × 323.15)
= 0.603 kg

3.82
A 1-m3 rigid tank with air at 1 MPa, 400 K is connected to an air line as shown in
Fig. P3.82. The valve is opened and air flows into the tank until the pressure
reaches 5 MPa, at which point the valve is closed and the temperature inside is
450K.
a. What is the mass of air in the tank before and after the process?
b. The tank eventually cools to room temperature, 300 K. What is the pressure
inside the tank then?
Solution:
P, T known at both states and assume the air behaves as an ideal gas.
mair1
=
P1
V
RT1
=
1000 × 1
0.287 × 400
= 8.711 kg
mair2
=
P2
V
RT2
=
5000 × 1
0.287 × 450
= 38.715 kg
Process 2 → 3 is constant V, constant mass cooling to T3
P3
= P2
× (T3
/T2
) = 5000 × (300/450) = 3.33 MPa

3.83
A cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then
filled with carbon dioxide gas at 20°C. To what pressure should it be charged if there
should be 1.2 kg of carbon dioxide?
Solution:
Assume CO2 is an ideal gas, table A.5: R = 0.1889 kJ/kg K
Vcyl = A × L =
π
4(0.2)2 × 1 = 0.031416 m3
P V = mRT => P =
mRT
V
⇒ P =
1.2 kg × 0.1889 kJ/kg Κ × (273.15 + 20) K
0.031416 m3 = 2115 kPa

3.84
Ammonia in a piston/cylinder arrangement is at 700 kPa, 80°C. It is now cooled at
constant pressure to saturated vapor (state 2) at which point the piston is locked with
a pin. The cooling continues to −10°C (state 3). Show the processes 1 to 2 and 2 to 3
on both a P–v and T–v diagram.
Solution:
State 1: T, P from table B.2.2 this is superheated vapor.
State 2: T, x from table B.2.1
State 3: T, v two-phase
700
290
P
v
2
3
1
v
T
2
3
180
14
-10

Compressibility Factor

3.85
How close to ideal gas behavior (find Z) is ammonia at saturated vapor, 100 kPa?
How about saturated vapor at 2000 kPa?
Table B.2.2: v1 = 1.1381 m3/kg, T1 = -33.6oC, P1 = 100 kPa
v2 = 0.06444 m3/kg, T2 = 49.37oC, P2 = 2000 kPa
Table A.5: R = 0.4882 kJ/kg K
Extended gas law: Pv = ZRT so we can calculate Z from this
Z1 =
P1v1
RT1
=
100 × 1.1381
0.4882 × (273.15 - 33.6)
= 0.973
Z2 =
P2v2
RT2
=
2000 × 0.06444
0.4882 × (273.15 + 49.37)
= 0.8185
So state 1 is close to ideal gas and state 2 is not so close.
ln Pr
Z
T = 2.0r
T = 0.7r
T = 0.7r
0.1 1
T = 1.2r
1 2

3.86
Carbon dioxide at 60oC is pumped at a very high pressure 10 MPa into an oil well
to reduce the oil viscosity for better oil flow. What is its compressibility?
Table B.3.2: v = 0.00345 m3/kg
Z = Pv/RT =
10 000 kPa × 0.00345 m3/kg
0.1889 kJ/kg-K × 333 K
= 0.55

3.87
Find the compressibility for carbon dioxide at 60oC and 10 MPa using Fig. D.1
Solution:
Table A.2 CO2: Tc = 304.1 K Pc = 7.38 MPa
Tr = T/Tc = 333/304.1 = 1.095
Pr = P/Pc = 10/7.38 = 1.355
From Figure D.1: Z ≈ 0.45
Compare with table B.3.2: v = 0.00345 m3/kg
Z = Pv/RT =
10 000 kPa × 0.00345 m3/kg
0.1889 kJ/kg-K × 333 K
= 0.55
ln Pr
Z
T = 2.0r
T = 0.7r
T = 0.7r
0.1 1
T = 1.1r

3.88
What is the percent error in specific volume if the ideal gas model is used to
represent the behavior of superheated ammonia at 40°C, 500 kPa? What if the
generalized compressibility chart, Fig. D.1, is used instead?
Solution:
NH3 T = 40°C = 313.15 K, Tc = 405.5 K, Pc = 11.35 MPa from Table A.1
Table B.2.2: v = 0.2923 m3/kg
Ideal gas: v =
RT
P =
0.48819 × 313
500 = 0.3056 m3/kg ⇒ 4.5% error
Figure D.1: Tr =
313.15
405.5 = 0.772, Pr =
0.5
11.35 = 0.044 ⇒ Z = 0.97
v =
ZRT
P = 0.2964 m3/kg ⇒ 1.4% error

3.89
A cylinder fitted with a frictionless piston contains butane at 25°C, 500 kPa. Can
the butane reasonably be assumed to behave as an ideal gas at this state ?
Solution
Butane 25°C, 500 kPa, Table A.2: Tc = 425 K; Pc = 3.8 MPa
Tr =
25 + 273
425 = 0.701; Pr =
0.5
3.8 = 0.13
Look at generalized chart in Figure D.1
Actual Pr > Pr, sat = 0.1 => liquid!! not a gas
The pressure should be less than 380 kPa to have a gas at that T.

3.90
Estimate the saturation pressure of chlorine at 300 K.
Solution:
We do not have a table in the B section for Chlorine so we must use the generalized
chart.
Table A.2: Pc = 7.98 MPa, Tc = 416.9 K
Tr = T/Tc = 300 / 416.9 = 0.7196
Figure D.1: Pr sat = 0.13 (same estimation from Table D.4)
P = Pc Pr sat = 7.98 × 0.13 = 1.04 MPa
If you use the CATT3 program then you will find Pr sat = 0.122 and P = 973 kPa

3.91
A bottle with a volume of 0.1 m3 contains butane with a quality of 75% and a
temperature of 300 K. Estimate the total butane mass in the bottle using the
generalized compressibility chart.
Solution:
We need to find the property v the mass is: m = V/v
so find v given T1 and x as : v = vf + x vfg
Table A.2: Butane Tc = 425.2 K Pc = 3.8 MPa = 3800 kPa
Tr = 300/425.2 = 0.705 =>
From Fig. D.1 or table D.4: Zf ≈ 0.02; Zg ≈ 0.9; Pr sat = 0.1
ln Pr
Z
T = 2.0r
f
T = 0.7r
T = 0.7r
0.1 1
g
P = Psat = Pr sat × Pc = 0.1× 3.80 ×1000 = 380 kPa
vf = ZfRT/P = 0.02 × 0.14304 × 300/380 = 0.00226 m3/kg
vg = ZgRT/P = 0.9 × 0.14304 × 300/380 = 0.1016 m3/kg
v = 0.00226 + 0.75 × (0.1016 – 0.00226) = 0.076765 m3/kg
m =
V
v =
0.1
0.076765 = 1.303 kg

3.92
Find the volume of 2 kg of ethylene at 270 K, 2500 kPa using Z from Fig. D.1
Ethylene Table A.2: Tc = 282.4 K, Pc = 5.04 MPa
Table A.5: R = 0.2964 kJ/kg K
The reduced temperature and pressure are:
Tr =
T
Tc
=
270
282.4 = 0.956, Pr =
P
Pc
=
2.5
5.04 = 0.496
Enter the chart with these coordinates and read: Z = 0.76
V =
mZRT
P =
2 kg × 0.76 × 0.2964 kJ/kg-K × 270 K
2500 kPa = 0.0487 m3
ln Pr
Z
T = 2.0r
T = 0.7r
T = 0.7r
0.1 1
T = 1.2r
T = 0.96r
0.5

3.93
With Tr = 0.85 and a quality of 0.6, find the compressibility factor using Fig. D.1
For the saturated states we will use Table D.4 instead of the figure. There
we can see at Tr = 0.85
Zf = 0.062, Zg = 0.747
Z = (1 – x) Zf + xZg = (1 – 0.6) 0.062 + 0.6 × 0.747 = 0.473
ln Pr
Z
T = 2.0r
T = 0.7r
T = 0.7r
0.1 1
T = 1.2r
T = 0.85r
0.5

3.94
Argon is kept in a rigid 5 m3 tank at −30°C, 3 MPa. Determine the mass using the
compressibility factor. What is the error (%) if the ideal gas model is used?
Solution:
No Argon table, so we use generalized chart Fig. D.1
Tr = 243.15/150.8 = 1.612, Pr = 3000/4870 = 0.616 => Z ≅ 0.96
m =
PV
ZRT =
3000 × 5
0.96 × 0.2081 × 243.2
= 308.75 kg
Ideal gas Z = 1
m =
PV
RT = 296.4 kg 4% error

3.95
Refrigerant R-32 is at -10 oC with a quality of 15%. Find the pressure and specific
volume.
Solution:
For R-32 there is no section B table printed. We will use compressibility chart.
From Table A.2: Tc = 351.3 K ; Pc = 5.78 MPa ;
From Table A.5: R = 0.1598 kJ/kg K
Tr = T/Tc = 263/351.3 = 0.749
From Table D.4 or Figure D.1, Zf ≈ 0.029 ; Zg ≈ 0.86 ; Pr sat ≈ 0.16
P = Pr sat Pc = 0.16 × 5780 = 925 kPa
v = vf + x vfg = (Zf + x × Zfg) RT/P
= [0.029 + 0.15 × (0.86 – 0.029)] × 0.1598 kJ/kg-K × 263 K/ 925 kPa
= 0.007 m3/kg
ln Pr
Z
T = 2.0r
T = 0.7r
T = 0.7r
0.1 1

3.96
To plan a commercial refrigeration system using R-123 we would like to know how
much more volume saturated vapor R-123 occupies per kg at -30 oC compared to the
saturated liquid state.
Solution:
For R-123 there is no section B table printed. We will use compressibility chart.
From Table A.2 Tc = 456.9 K ; Pc = 3.66 MPa ; M = 152.93
Tr = T/Tc = 243/456.9 = 0.53
R = R
_
/M = 8.31451 / 152.93 = 0.0544
The value of Tr is below the range in Fig. D.1 so use the table D.4
Table D.4, Zg = 0.979 Zf = 0.00222
Zfg = 0.979 − 0.0022 = 0.9768; Pr = Pr sat = 0.0116
P = Pr × Pc = 42.5 kPa
vfg = Zfg RT/P = 0.9768 × 0.0544 × 243 / 42.5 = 0.304 m3/kg
Comment: If you check with the software the solution is off by a factor of 6. The
linear interpolation is poor and so is the approximation for Pr sat so the real
saturation pressure should be 6.75 kPa. Also the very small value of Zf is
inaccurate by itself, minute changes in the curve gives large relative variations.

3.97
A new refrigerant R-125 is stored as a liquid at -20 oC with a small amount of vapor.
For a total of 1.5 kg R-125 find the pressure and the volume.
Solution:
As there is no section B table use compressibility chart.
Table A.2: R-125 Tc = 339.2 K Pc = 3.62 MPa
Tr = T / Tc = 253.15 / 339.2 = 0.746
We can read from Figure D.1 or a little more accurately interpolate from table
D.4 entries:
Pr sat = 0.16 ; Zg = 0.86 ; Zf = 0.029
P = Pr sat Pc = 0.16 × 3620 kPa = 579 kPa
Vliq = Zf mliq RT/P = 0.029 ×1.5 kg × 0.06927 kJ/kgK × 253.15 K/ 579 kPa
= 0.0013 m3
ln Pr
Z
T = 2.0r
T = 0.7r
T = 0.7r
0.1 1
sat vapor
sat liq.

Equations of State
For these problems see appendix D for the equation of state (EOS) and chapter 14.

3.98
Determine the pressure of nitrogen at 160 K, v = 0.00291 m3/kg using ideal gas,
van der Waal Equation of State and the nitrogen table.
Nitrogen from table A.2: Tc = 126.2 K, Pc = 3390 kPa,
Ideal gas: P =
RT
v =
0.2968 × 160
0.00291
kJ/kg-K × K
m3/kg
= 16 319 kPa
For van der Waal equation of state from Table D.1 we have
b =
1
8
RTc
Pc
= 0.125 ×
0.2968 × 126.2
3390 = 0.001 381 m3/kg,
a = 27 b2 Pc = 27 × (0.001 381)2 × 3390 = 0.174 562 kPa (m3/kg)2
The EOS is: P =
RT
v − b
−
a
v2 =
0.2968 × 160
0.00291 - 0.001 381 −
0.174 562
0.002912 = 10 444 kPa
Table B.6.2: P = 10 000 kPa.

3.99
Determine the pressure of nitrogen at 160 K, v = 0.00291 m3/kg using Redlich-
Kwong Equation of State and the nitrogen table.
Tr = T/Tc = 160/126.2 = 1.26783
For Redlich-Kwong EOS we have the parameters from Table D.1
b = 0.08664
RTc
Pc
= 0.08664 ×
0.2968 × 126.2
3390 = 0.000 957 3 m3/kg,
a = 0.42748 T
-1/2
r
R2T
2
c
Pc
= 0.42748 ×
0.29682 × 126.22
1.26781/2 × 3390
= 0.157 122 kPa (m3/kg)2
The equation is:
P =
RT
v − b
−
a
v2 + bv
=
0.2968 × 160
0.00291 - 0.000 9573 −
0.157122
0.002912 + 0.000 9573 × 0.00291
= 10 357 kPa
Table B.6.2: P = 10 000 kPa.

3.100
Determine the pressure of nitrogen at 160 K, v = 0.00291 m3/kg using Soave
Equation of State and the nitrogen table.
Tr = T/Tc = 160/126.2 = 1.26783
For Soave EOS see Appendix D (very close to Redlich-Kwong)
P =
RT
v − b
−
a
v2 + bv
where the parameters are from Table D.1 and D.4
ω = 0.039
f = 0.48 + 1.574ω – 0.176ω2 = 0.54112
ao = 0.42748 [1 + f (1 - T
1/2
r ) ] 2 = 0.371184
b = 0.08664
RTc
Pc
= 0.08664 ×
0.2968 × 126.2
3390 = 0.000 957 3 m3/kg,
a = 0.371184
R2T
2
c
Pc
= 0.371184 ×
0.29682 × 126.22
3390 = 0.153 616 kPa (m3/kg)2
P =
RT
v − b
−
a
v2 + bv
=
0.2968 × 160
0.00291 - 0.000 9573 −
0.153 616
0.002912 + 0.000 9573 × 0.00291
= 10 669 kPa
Nitrogen Table B.6.2: P = 10 000 kPa.

3.101
Carbon dioxide at 60oC is pumped at a very high pressure 10 MPa into an oil well
to reduce the oil viscosity for better oil flow. We want to find its specific volume
from the CO2 table, ideal gas and van der Waals equation of state by iteration.
Table B.3.2: v = 0.00345 m3/kg.
Ideal gas: v =
RT
P =
0.1889 × (60 + 273.15)
10 000
kJ/kg-K × K
kPa = 0.006 293 m3/kg
Carbon dioxide from table A.2: Tc = 304.1 K, Pc = 7380 kPa,
b =
1
8
RTc
Pc
= 0.125 ×
0.1889 × 304.1
7380 = 0.000 972 98 m3/kg,
a = 27 b2 Pc = 27 × (0.000 972 98)2 × 7380 = 0.188 6375 kPa (m3/kg)2
The EOS is: P =
RT
v − b
−
a
v2
Since it is nonlinear in v we use trial and error starting with the Table entry.
v = 0.00345 : P =
0.1889 × 333.15
0.00345 - 0.000 972 98 −
0.188 6375
0.003452 = 9557.8 kPa low
v = 0.003 P =
0.1889 × 333.15
0.003 - 0.000 972 98 −
0.188 6375
0.0032 = 10 086.8 kPa high
v = 0.00307 : P =
0.1889 × 333.15
0.00307 - 0.000 972 98 −
0.188 6375
0.003072 = 9995.4 OK.
v = 0.00307 m3/kg

3.102
Solve the previous problem using the Redlich-Kwong equation of state. Notice
this becomes trial and error.
Tr = T/Tc = 333.15/304.1 = 1.09553
For Redlich-Kwong EOS we have the parameters from Table D.1
b = 0.08664
RTc
Pc
= 0.08664 ×
0.1889 × 304.1
7380 = 0.000 6744 m3/kg,
a = 0.42748 T
-1/2
r
R2T
2
c
Pc
= 0.42748 ×
0.18892 × 304.12
1.095531/2 × 7380
= 0.18262 kPa (m3/kg)2
The equation is:
P =
RT
v − b
−
a
v2 + bv
10 000 =
0.1889 × 333.15
v - 0.000 6744 −
0.18262
v2 + 0.000 6744 × v
Trial and error on v (start guided by Table B.3.2):
v = 0.0035 m3/kg P = 9772.8 kPa so v smaller
v = 0.0033 m3/kg P = 10044.7 kPa so v larger
v = 0.0034 m3/kg P = 9906.5 kPa
linear interpolation gives
v = 0.00333 m3/kg P = 10002.7 kPa OK.

3.103
Solve Problem 3.101 using the Soave equation of state, ω = 0.239. Notice this
becomes trial and error.
Tr = T/Tc = 333.15/304.1 = 1.09553
For Soave EOS see Appendix D (very close to Redlich-Kwong)
P =
RT
v − b
−
a
v2 + bv
where the parameters are from Table D.1 and D.4
ω = 0.239
f = 0.48 + 1.574ω – 0.176ω2 = 0.84613
ao = 0.42748 [1 + f (1 - T
1/2
r ) ] 2 = 0.394381
b = 0.08664
RTc
Pc
= 0.08664 ×
0.1889 × 304.1
7380 = 0.000 6744 m3/kg,
a = 0.394381
R2T
2
c
Pc
= 0.394381 ×
0.18892 × 304.12
7380 = 0.176 342 kPa (m3/kg)2
The equation is:
P =
RT
v − b
−
a
v2 + bv
10 000 =
0.1889 × 333.15
v - 0.000 6744 −
0.176 342
v2 + 0.000 6744 × v
Trial and error on v (start guided by Table B.3.2):
v = 0.0035 m3/kg P = 10 202 kPa so v larger
v = 0.0036 m3/kg P = 10 051 kPa so v close
v = 0.00363 m3/kg P = 10 006 kPa OK

3.104
A tank contains 8.35 kg methane in 0.1 m3 at 250 K. Find the pressure using ideal
gas, van der Waal EOS and the methane table.
The state is given by (T = 250 K, v = V/m = 0.1/8.35 = 0.011976 m3/kg)
Table A.2 or B.7.2:
Tc = 190.4 K,
Pc = 4600 kPa
Tr =
T
Tc
=
250
190.4 = 1.313
P
T
Vapor
Liquid
Crit.P. CH4
state
S
Ideal gas model: P =
RT
v =
0.5183 × 250
0.011976 = 10 820 kPa
b =
1
8
RTc
Pc
= 0.125 ×
0.5183 × 190.4
4600 = 0.002 681 64 m3/kg,
a = 27 b2 Pc = 27 × (0.002 681 64)2 × 4600 = 0.893 15 kPa (m3/kg)2
The EOS is: P =
RT
v − b
−
a
v2 =
0.5183 × 250
0.011976 - 0.002 681 64 −
0.89315
0.0119762 = 7714 kPa
Locating the state in Table B.7.2: P = 8000 kPa, very close

3.105
Do the previous problem using the Redlich-Kwong equation of state.
Table A.2 or B.7.2:
Tc = 190.4 K,
Pc = 4600 kPa
Tr =
T
Tc
=
250
190.4 = 1.313
P
T
Vapor
Liquid
Crit.P. CH4
state
S
RT
v =
0.5183 × 250
0.011976 = 10 820 kPa
For Redlich-Kwong equation of state we have the parameters from Table D.1
b = 0.08664
RTc
Pc
= 0.08664 ×
0.5183 × 190.4
4600 = 0.001 858 7 m3/kg,
a = 0.42748 T
-1/2
r
R2T
2
c
Pc
= 0.42748 ×
0.51832 × 190.42
1.3131/2 × 4600
= 0.789809 kPa (m3/kg)2
The equation is:
P =
RT
v − b
−
a
v2 + bv
=
0.5183 × 250
0.011976 - 0.0018587 −
0.789809
0.0119762 + 0.0018587 × 0.011976
= 8040 kPa
Locating the state in Table B.7.2: P = 8000 kPa, very close

3.106
Do Problem 3.104 using the Soave EOS.
A tank contains 8.35 kg methane in 0.1 m3 at 250 K. Find the pressure using ideal
gas, van der Waal EOS and the methane table.
Table A.2 or B.7.2:
Tc = 190.4 K,
Pc = 4600 kPa
Tr =
T
Tc
=
250
190.4 = 1.313
P
T
Vapor
Liquid
Crit.P. CH4
state
S
RT
v =
0.5183 × 250
0.011976 = 10 820 kPa
For Soave EOS we have the parameters from Table D.1 and D.4
ω = 0.011
f = 0.48 + 1.574ω – 0.176ω2 = 0.49729
ao = 0.42748 [1 + f (1 - T
1/2
r ) ] 2 = 0.367714
b = 0.08664
RTc
Pc
= 0.08664 ×
0.5183 × 190.4
4600 = 0.001 8587 m3/kg,
a = 0.367714
R2T
2
c
Pc
= 0.367714 ×
0.51832 × 190.42
4600 = 0.778 482 kPa (m3/kg)2
The equation is:
P =
RT
v − b
−
a
v2 + bv
=
0.5183 × 250
0.011976 - 0.001 8587 −
0.778 482
0.0119762 + 0.0018587 × 0.011976
= 8108.7 kPa
Locating the state in Table B.7.2: P = 8000 kPa

Review Problems

3.107
Determine the quality (if saturated) or temperature (if superheated) of the
following substances at the given two states:
Solution:
a) Water, H2O, use Table B.1.1 or B.1.2
1) 120°C, 1 m3/kg v > vg superheated vapor, T = 120 °C
2) 10 MPa, 0.01 m3/kg => two-phase v < vg
x = ( 0.01 – 0.001452 ) / 0.01657 = 0.516
b) Nitrogen, N2, table B.6
1) 1 MPa, 0.03 m3/kg => superheated vapor since v > vg
Interpolate between sat. vapor and superheated vapor B.6.2:
T ≅ 103.73 + (120-103.73) ×
0.03−0.02416
0.03117−0.02416
= 117 K
2) 100 K, 0.03 m3/kg => sat. liquid + vapor as two-phase v < vg
v = 0.03 = 0.001452 + x × 0.029764 ⇒ x = 0.959
States shown are
to each other.
P C.P.
v
T
C.P.
v
T
a1, b1
b2 a1, b1
b2
P = const.
a2
a2

3.108
Give the phase and the missing properties of P, T, v and x.
Solution:
a. R-410a T = 10°C v = 0.01 m3/kg
Table B.4.1 v < vg = 0.02383 m3/kg
sat. liquid + vapor. P = Psat = 1085.7 kPa,
x = (v - vf)/vfg = (0.01 - 0.000886)/0.02295 = 0.2713
b. H2O T = 350°C v = 0.2 m3/kg
Table B.1.1 at given T: v > vg = 0.00881 m3/kg
sup. vapor P ≅ 1.40 MPa, x = undefined
c. R-410a T = - 5 °C P = 600 kPa
sup. vapor (P < Pg = 678,9 kPa at -5°C)
Table B.4.2:
v = 0.04351 m3/kg at –8.67°C
v = 0.04595 m3/kg at 0°C
=> v = 0.04454 m3/kg at -5°C
d. R-134a P = 294 kPa, v = 0.05 m3/kg
Table B.5.1: v < vg = 0.06919 m3/kg
two-phase T = Tsat = 0°C
x = (v - vf)/vfg = (0.05 - 0.000773)/0.06842 = 0.7195
States shown are
to each other.
P C.P.
v
T
C.P.
v
T
b,c
a, d b,c
P = const.
a, d

3.109
Find the phase, quality x if applicable and the missing property P or T.
Solution:
a. H2O T = 120°C v = 0.5 m3/kg
Table B.1.1 at given T: v < vg = 0.89186
sat. liq. + vap. P = Psat = 198.5 kPa,
x = (v - vf)/vfg = (0.5 - 0.00106)/0.8908 = 0.56
b. H2O P = 100 kPa v = 1.8 m3/kg
Table B.1.2 at given P: v > vg = 1.694
sup. vap., interpolate in Table B.1.3
T =
1.8 − 1.694
1.93636 − 1.694
(150 – 99.62) + 99.62 = 121.65 °C
c. H2O T = 263 K v = 0.2 m3/kg
Table B.1.5 at given T = -10 °C: v < vg = 466.757
sat. solid + vap., P = Psat = 0.26 kPa,
x = (v - vi)/vig = (200 - 0.001)/466.756 = 0.4285
States shown are
to each other.
P C.P.
v
T
C.P.
v
T
b
a b
P = const.
a
c c

3.110
Find the phase, quality x if applicable and the missing property P or T.
Solution:
a. NH3 P = 800 kPa v = 0.2 m3/kg;
Superheated Vapor (v > vg at 800 kPa)
Table B 2.2 interpolate between 70°C and 80°C
T = 71.4°C
b. NH3 T = 20°C v = 0.1 m3/kg
Table B.2.1 at given T: v < vg = 0.14922
sat. liq. + vap. , P = Psat = 857.5 kPa,
x = (v - vf)/vfg = (0.1 - 0.00164)/0.14758 = 0.666
P C.P.
v
T
C.P.
v
T b
P = const.
a
a
b

3.111
Give the phase and the missing properties of P, T, v and x. These may be a little
more difficult if the appendix tables are used instead of the software.
a) R-410a at T = 10°C, v = 0.02 m3/kg: Table B.4.1 v > vg at 10°C
=> sup. vap. Table B.4.2 interpolate between sat. and sup. both at 10°C
P = 680.7 + (600 - 680.7)
0.036-0.03471
0.04018-0.03471 = 661.7 kPa
b) H2O v = 0.2 m3/kg , x = 0.5: Table B.1.1
sat. liq. + vap. v = (1-x) vf + x vg => vf + vg = 0.4 m3/kg
since vf is so small we find it approximately where vg = 0.4 m3/kg.
vf + vg = 0.39387 at 150°C, vf + vg = 0.4474 at 145°C.
An interpolation gives T ≅ 149.4°C, P ≅ 468.2 kPa
c) H2O T = 60°C, v = 0.001016 m3/kg: Table B.1.1 v < vf = 0.001017
=> compr. liq. see Table B.1.4
v = 0.001015 at 5 MPa so P ≅ 0.5(5000 + 19.9) = 2.51 MPa
d) NH3 T = 30°C, P = 60 kPa : Table B.2.1 P < Psat
=> sup. vapor interpolate in Table B.2.2
v = 2.94578 + (1.95906 - 2.94578)
60 - 50
75 - 50 = 2.551 m3/kg
v is not linearly proportional to P (more like 1/P) so the computer table
gives a more accurate value of 2.45 m3/kg
e) R-134a v = 0.005 m3/kg , x = 0.5: sat. liq. + vap. Table B.5.1
v = (1-x) vf + x vg => vf + vg = 0.01 m3/kg
vf + vg = 0.010946 at 65°C, vf + vg = 0.009665 at 70°C.
An interpolation gives: T ≅ 68.7°C, P = 2.06 MPa
States shown are
to each other.
P C.P.
v
T
C.P.
v
T
a
b, e
a
P = const.
d
b, e
c
c
d

3.112
Refrigerant-410a in a piston/cylinder arrangement is initially at 50°C, x = 1. It is
then expanded in a process so that P = Cv
−1 to a pressure of 100 kPa. Find the final
temperature and specific volume.
Solution:
State 1: 50°C, x = 1 Table B.4.1: P1 = 3065.2 kPa, v1 = 0.00707 m3/kg
Process: Pv = C = P1v1; => P2 = C/v2= P1v1/v2
State 2: 100 kPa and v2 = v1P1/P2 = 0.2167 m3/kg
v2 < vg at 100 kPa, T2 ≅ -51.65°C from Table B.3.2
Notice T is not constant
v
P
v
T
1
2
1
2
The first part of the process may be in the sup. vapor region.

3.113
Consider two tanks, A and B, connected by a valve, as shown in Fig. P3.113.
Each has a volume of 200 L and tank A has R-410a at 25°C, 10% liquid and 90%
vapor by volume, while tank B is evacuated. The valve is now opened and
saturated vapor flows from A to B until the pressure in B has reached that in A, at
which point the valve is closed. This process occurs slowly such that all
temperatures stay at 25°C throughout the process. How much has the quality
changed in tank A during the process?
Solution:
A B
vacuum
State A1: Table B.4.1 vf = 0.000944 m3/kg, vg = 0.01514 m3/kg
mA1 =
Vliq1
vf 25°C
+
Vvap1
vg 25°C
=
0.1 × 0.2
0.000944 +
0.9 × 0.2
0.01514
= 21.186 + 11.889 = 33.075 kg
xA1 =
11.889
33.075 = 0.3594 ;
State B2: Assume A still two-phase so saturated P for given T
mB2 =
VB
vg 25°C
=
0.2
0.01514 = 13.210 kg
State A2: mass left is mA2 = 33.075 - 13.210 = 19.865 kg
vA2 =
0.2
19.865 = 0.010068 = 0.000944 + xA2 × 0.01420
xA2 = 0.6425 ∆x = 0.283

3.114
Water in a piston/cylinder is at 90°C, 100 kPa, and the piston loading is such that
pressure is proportional to volume, P = CV. Heat is now added until the temperature
reaches 200°C. Find the final pressure and also the quality if in the two-phase
region.
Solution:
Final state: 200°C , on process line P = CV
1
2
v
P
State 1: Table B.1.1: v1 = 0.001036 m3/kg
P2 = P1v2/v1 from process equation
Check state 2 in Table B.1.1
vg(T2) = 0.12736; Pg(T2) = 1.5538 MPa
If v2 = vg(T2) ⇒ P2 = 12.3 MPa > Pg not OK
If sat. P2 = Pg(T2) = 1553.8 kPa ⇒ v2 = 0.0161 m3kg < vg sat. OK,
P2 = 1553.8 kPa, x2 = (0.0161 - 0.001156) / 0.1262 = 0.118

3.115
A tank contains 2 kg of nitrogen at 100 K with a quality of 50%. Through a volume
flowmeter and valve, 0.5 kg is now removed while the temperature remains constant.
Find the final state inside the tank and the volume of nitrogen removed if the
valve/meter is located at
a. The top of the tank
b. The bottom of the tank
Solution
Table B.6.1:
v1 = 0.001452 + x1 × 0.029764 = 0.016334 m3/kg
Vtank = m1v1 = 0.0327 m3
m2 = m1 - 0.5 = 1.5 kg
v2 = Vtank/m2 = 0.0218 < vg(T)
x2 =
0.0218-0.001452
0.031216-0.001452 = 0.6836
Top: flow out is sat. vap. vg = 0.031216 m3/kg, Vout = moutvg = 0.0156 m3
Bottom: flow out is sat. liq. vf = 0.001452 Vout = moutvf = 0.000726 m3

3.116
A spring-loaded piston/cylinder contains water at 500°C, 3 MPa. The setup is
such that pressure is proportional to volume, P = CV. It is now cooled until the
water becomes saturated vapor. Sketch the P-v diagram and find the final
pressure.
Solution:
Process: m is constant and P = C0V = C0m v = C v
P = Cv ⇒ C = P1/v1 = 3000/0.11619 = 25820 kPa kg/m3
State 2: x2 = 1 & P2 = Cv2 (on process line)
P
v
1
2
Trial & error on T2sat or P2sat:
Here from B.1.2:
at 2 MPa vg = 0.09963 ⇒ C = 20074 (low)
2.5 MPa vg = 0.07998 ⇒ C = 31258 (high)
2.25 MPa vg = 0.08875 ⇒ C = 25352 (low)
Interpolate to get the right C ⇒ P2 = 2270 kPa

3.117
A container with liquid nitrogen at 100 K has a cross sectional area of 0.5 m2.
Due to heat transfer, some of the liquid evaporates and in one hour the liquid level
drops 30 mm. The vapor leaving the container passes through a valve and a heater
and exits at 500 kPa, 260 K. Calculate the volume rate of flow of nitrogen gas
exiting the heater.
Solution:
Properties from table B.6.1 for volume change, exit flow from table B.6.2:
∆V = A × ∆h = 0.5 × 0.03 = 0.015 m3
∆mliq = -∆V/vf = -0.015/0.001452 = -10.3306 kg
∆mvap = ∆V/vg = 0.015/0.0312 = 0.4808 kg
mout = 10.3306 - 0.4808 = 9.85 kg
vexit = 0.15385 m3
/kg
V
.
= m
.
vexit = (9.85 / 1 h)× 0.15385 m3
/kg
= 1.5015 m3
/h = 0.02526 m
3
/min

3.118
For a certain experiment, R-410a vapor is contained in a sealed glass tube at 20°C. It
is desired to know the pressure at this condition, but there is no means of measuring
it, since the tube is sealed. However, if the tube is cooled to −20°C small droplets of
liquid are observed on the glass walls. What is the initial pressure?
Solution:
Control volume: R-410a fixed volume (V) & mass (m) at 20°C
Process: cool to -20°C at constant v, so we assume saturated vapor
State 2: v2 = vg at -20°C = 0.06480 m3/kg
State 1: 20°C, v1 = v2 = 0.06480 m3/kg
interpolate between 400 and 500 kPa in Table B.4.2
=> P1 = 485 kPa
v
P
v
T
1
300 kPa
250 kPa
P
-20 C
20 C
-20 C
20 C
300 kPa
250 kPa
1
2
2

3.119
A cylinder/piston arrangement contains water at 105°C, 85% quality with a
volume of 1 L. The system is heated, causing the piston to rise and encounter a
linear spring as shown in Fig. P3.119. At this point the volume is 1.5 L, piston
diameter is 150 mm, and the spring constant is 100 N/mm. The heating continues,
so the piston compresses the spring. What is the cylinder temperature when the
pressure reaches 200 kPa?
Solution:
P1 = 120.8 kPa, v1 = vf + x vfg = 0.001047 + 0.85*1.41831 = 1.20661
m = V1/ v1 =
0.001
1.20661 = 8.288×10-4 kg
v2 = v1 (V2 / V1) = 1.20661× 1.5 = 1.8099
& P = P1 = 120.8 kPa ( T2 = 203.5°C )
P3 = P2 + (ks/Ap
2) m(v3-v2) linear spring
P
v
1 2
3200
1 1.5 liters
Ap = (π/4) × 0.152 = 0.01767 m2 ; ks = 100 kN/m (matches P in kPa)
200 = 120.8 + (100/0.01767 2 ) × 8.288×10-4(v3-1.8099)
200 = 120.8 + 265.446 (v3 – 1.8099)=> v3 = 2.1083 m3
/kg
T3 ≅ 600 + 100 × (2.1083 – 2.01297)/(2.2443-2.01297) ≅ 641°C

3.120
Determine the mass of methane gas stored in a 2 m3 tank at −30°C, 2 MPa. Estimate
the percent error in the mass determination if the ideal gas model is used.
Solution:
Table B.7
Methane Table B.7.1 at −30°C = 243.15 K > Tc = 190.6 K, so superheated
vapor in Table B.7.2. Linear interpolation between 225 and 250 K at 2 MPa.
⇒ v ≅ 0.05289 +
243.15-225
250-225 ×(0.06059 - 0.05289) = 0.05848 m3/kg
m = V/v = 2/0.05848 = 34.2 kg
Ideal gas assumption
v =
RT
P =
0.51835 × 243.15
2000 = 0.06302 m3/kg
m =
V
v =
2
0.06302 = 31.74 kg
Error:
∆m = 2.46 kg ; 7.2% too large
Comment: The compressibility of the methane Z = 0.93.

3.121
A cylinder containing ammonia is fitted with a piston restrained by an external force
that is proportional to cylinder volume squared. Initial conditions are 10°C, 90%
quality and a volume of 5 L. A valve on the cylinder is opened and additional
ammonia flows into the cylinder until the mass inside has doubled. If at this point
the pressure is 1.2 MPa, what is the final temperature?
Solution:
State 1 Table B.2.1: v1 = 0.0016 + 0.9(0.205525 - 0.0016) = 0.18513 m3/kg
P1 = 615 kPa; V1 = 5 L = 0.005 m3
m1 = V/v = 0.005/0.18513 = 0.027 kg
State 2: P2 = 1.2 MPa, Flow in so: m2 = 2 m1 = 0.054 kg
Process: Piston Fext = KV2 = PA => P = CV2 => P2 = P1 (V2/V1)2
From the process equation we then get:
V2 = V1 (P2/P1)
1/2
= 0.005 (
1200
615 )
1/2
= 0.006984 m3
v2 = V/m =
0.006984
0.054 = 0.12934 m3/kg
At P2, v2: T2 = 70.9°C

3.122
A cylinder has a thick piston initially held by a pin as shown in Fig. P3.122. The
cylinder contains carbon dioxide at 200 kPa and ambient temperature of 290 K.
The metal piston has a density of 8000 kg/m3 and the atmospheric pressure is 101
kPa. The pin is now removed, allowing the piston to move and after a while the
gas returns to ambient temperature. Is the piston against the stops?
Solution:
Force balance on piston determines equilibrium float pressure.
Piston mp = Ap × l × ρ ρpiston = 8000 kg/m3
Pext on CO2
= P0 +
mpg
Ap
= 101 +
Ap × 0.1 × 9.807 × 8000
Ap × 1000
= 108.8 kPa
Pin released, as P1 > Pext piston moves up, T2 = To & if piston at stops,
then V2 = V1 × Η2/Η1 = V1 × 150 / 100
Ideal gas with T2 = T1 then gives
⇒ P2 = P1 × V1 / V2 = 200 ×
100
150 = 133 kPa > Pext
⇒ piston is at stops, and P2 = 133 kPa

3.123
What is the percent error in pressure if the ideal gas model is used to represent the
behavior of superheated vapor R-410a at 60°C, 0.03470 m3
/kg? What if the
generalized compressibility chart, Fig. D.1, is used instead (iterations needed)?
Solution:
Real gas behavior: P = 1000 kPa from Table B.4.2
Ideal gas constant: R = R
_
/M = 8.31451/72.585 = 0.1146 kJ/kg K
P = RT/v = 0.1146 × (273.15 + 60) / 0.0347
= 1100 kPa which is 10% too high
Generalized chart Fig D.1 and critical properties from A.2:
Tr = 333.2/(273.15 + 71.3) = 0.967; Pc = 4901 kPa
Assume P = 1000 kPa => Pr = 0.204 => Z ≅ 0.92
v = ZRT/P = 0.92 × 0.1146 × 333.15 /1000 = 0.03512 too high
Assume P = 1050 kPa => Pr = 0.214 => Z ≅ 0.915
v = ZRT/P = 0.915 × 0.1146 × 333.15 / 1050 = 0.03327 too low
P ≅ 1000 + ( 1050 − 1000 ) ×
0.03470 − 0.03512
0.03327 − 0.03512
= 1011 kPa 1.1 % high

3.124
An initially deflated and flat balloon is connected by a valve to a 12 m3 storage
tank containing helium gas at 2 MPa and ambient temperature, 20°C. The valve is
opened and the balloon is inflated at constant pressure, Po = 100 kPa, equal to
ambient pressure, until it becomes spherical at D
1 = 1 m. If the balloon is larger
than this, the balloon material is stretched giving a pressure inside as
P = P0 + C






1 −
D1
D
D1
D
The balloon is inflated to a final diameter of 4 m, at which point the pressure
inside is 400 kPa. The temperature remains constant at 20°C. What is the
maximum pressure inside the balloon at any time during this inflation process?
What is the pressure inside the helium storage tank at this time?
Solution:
At the end of the process we have D = 4 m so we can get the constant C as
P = 400 = P0 + C ( 1 –
1
4 )
1
4 = 100 + C × 3/16 => C = 1600 kPa
The pressure is: P = 100 + 1600 ( 1 – X –1
) X –1
; X = D / D1
Differentiate to find max:
dP
dD = C ( - X –2
+ 2 X –3
) / D1 = 0
=> - X –2
+ 2 X –3
= 0 => X = 2
at max P => D = 2D1 = 2 m; V =
π
6 D3
= 4.18 m3
Pmax = 100 + 1600 ( 1 -
1
2 )
1
2 = 500 kPa
Helium is ideal gas A.5: m =
PV
RT =
500 × 4.189
2.0771 × 293.15
= 3.44 kg
mTANK, 1 =
PV
RT =
2000 × 12
2.0771 × 293.15
= 39.416 kg
mTANK, 2 = 39.416 – 3.44 = 35.976 kg
PT2 = mTANK, 2 RT/V = ( mTANK, 1 / mTANK, 2 ) × P1 = 1825.5 kPa

3.125
A piston/cylinder arrangement, shown in Fig. P3.125, contains air at 250 kPa,
300°C. The 50-kg piston has a diameter of 0.1 m and initially pushes against the
stops. The atmosphere is at 100 kPa and 20°C. The cylinder now cools as heat is
transferred to the ambient.
a. At what temperature does the piston begin to move down?
b. How far has the piston dropped when the temperature reaches ambient?
Solution:
Piston Ap =
π
4 × 0.12 = 0.00785 m2
Balance forces when piston floats:
Pfloat = Po +
mpg
Ap
= 100 +
50 × 9.807
0.00785 × 1000
= 162.5 kPa = P2 = P3
To find temperature at 2 assume ideal gas:
2
1
P
V
P2
Vstop
3
T2 = T1 ×
P2
P1
= 573.15 ×
162.5
250 = 372.5 K
b) Process 2 -> 3 is constant pressure as piston floats to T3 = To = 293.15 K
V2 = V1 = Ap × H = 0.00785 × 0.25 = 0.00196 m3 = 1.96 L
Ideal gas and P2 = P3 => V3 = V2 ×
T3
T2
= 1.96 ×
293.15
372.5 = 1.54 L
∆H = (V2 -V3)/A = (1.96-1.54) × 0.001/0.00785 = 0.053 m = 5.3 cm

Linear Interpolation

3.126
Find the pressure and temperature for saturated vapor R-410a with v = 0.1 m3/kg
Solution:
Table B.4.1 Look at the saturated vapor column vg and it is found between −35°
C and −30°C. We must then do a linear interpolation between these values.
T = −35 + [ –30 – (–35)]
0.1 − 0.11582
0.09470 − 0.11582
= −35 + 5 × 0.749 = −31.3°C
P = 218.4 + (269.6 – 218.4) × 0.749 = 256.7 kPa
v
T
2
1-35
-30
0.09470 0.11582
0.1
v
P
2
1
0.09470 0.11582
0.1
218.4
269.6
To understand the interpolation equation look at the smaller and larger
triangles formed in the figure. The ratio of the side of the small triangle in v as
(0.11582 - 0.1) to the side of the large triangle (0.11582 − 0.09470) is equal to
0.749. This fraction of the total ∆P = 269.6 − 218.4 or ∆T = −30 − (−35) is
added to the lower value to get the desired interpolated result.

3.127
Use a linear interpolation to estimate properties of ammonia to fill out the table
below
P [kPa] T [ °C] v [m3/kg] x
a) 550 0.75
b) 80 20
c) 10 0.4
Solution:
a) Find the pressures in Table B.2.1 that brackets the given pressure.
T = 5 + (10 – 5)
550 − 515.9
615.2 − 515.9
= 5 + 5 × 0.341 = 6.7 °C
vf = 0.001583 + (0.0016 – 0.001583) 0.341 = 0.001589 m3/kg
vg = 0.24299 + (0.20541 – 0.24299) 0.341 = 0.230175 m3/kg
v = vf + xvfg = 0.001589 + 0.75(0.230175 – 0.001589)
= 0.1729 m3/kg
b) Interpolate between 50 and 100 kPa to get properties at 80 kPa
v = 2.8466 + (1.4153 – 2.8466)
80 − 50
100 − 50
= 2.8466 + ( − 1.4313) × 0.6 = 1.9878 m3/kg
x: Undefined
c) Table B.2.1: v > vg so the state is superheated vapor.
Table B.2.2 locate state between 300 and 400 kPa.
P = 300 + (400 – 300)
0.4 - 0.44251
0.32701 − 0.44251
= 300 + 100 × 0.368 = 336.8 kPa
x: Undefined

3.128
Use a linear interpolation to estimate Tsat at 900 kPa for nitrogen. Sketch by hand
the curve Psat(T) by using a few table entries around 900 kPa from table B.6.1. Is
your linear interpolation over or below the actual curve?
Solution:
The 900 kPa in Table B.6.1 is located between 100 and 105 K.
T = 100 + (105 – 100)
900 − 779.2
1084.6 − 779.2
= 100 + 5 × 0.3955 = 102 K
The actual curve has a positive second derivative (it curves up) so T is
slightly underestimated by use of the chord between the 100 K and the 105 K
points, as the chord is above the curve.
P
T
100 105 110
779.2
1084.6
1467.6
900

3.129
Use a double linear interpolation to find the pressure for superheated R-134a at
13°C with v = 0.3 m3/kg.
Solution:
Table B.5.2: Superheated vapor
At 10°C, 0.3 m3/kg
P = 50 + (100 – 50) ×
0.3 - 0.45608
0.22527 - 0.45608 = 83.8 kPa
At 20°C, 0.3 m3/kg
P = 50 + (100 – 50) ×
0.3 - 0.47287
0.23392 - 0.47287 = 86.2 kPa
Interpolating at 13°C,
P = 83.8 + (3/10) × (86.2 − 83.8) = 84.5 kPa
This could also be interpolated as following:
At 13°C, 50 kPa, v = 0.45608 + (3/10) × 0.0168 = 0.4611 m3/kg
At 13°C, 100 kPa, v = 0.22527 + (3/10) × 0.0087 = 0.2279 m3/kg
Interpolating at 0.3 m3/kg.
P = 50 + (100 – 50) ×
0.3 - 0.4611
0.2279 - 0.4611 = 84.5 kPa

3.130
Find the specific volume for CO2 at 0oC and 625 kPa.
Solution:
The state is superheated vapor in Table B.3.2 between 400 and 800 kPa.
v = 0.12552 + (0.06094 – 0.12552)
625 − 400
800 − 400
= 0.12552 + ( − 0.06458) × 0.5625 = 0.0892 m3/kg
v
0.12552
0.06094
P
400 625 800

Computer Tables

3.131
Use the computer software to find the properties for water at the 4 states in
Problem 3.35
Start the software, click the tab for water as the substance, and click the
small calculator icon. Select the proper CASE for the given properties.
CASE RESULT
a) 1 (T, P) Compressed liquid, x = undefined, v = 0.001002 m3/kg
b) 5 (P, v) Two-phase, T = 151.9°C, x = 0.5321
c) 1 (T, P) Sup. vapor, x = undefined, v = 0.143 m3/kg
d) 4 (T, x) P = Psat = 8581 kPa, v = 0.01762 m3/kg

3.132
Use the computer software to find the properties for ammonia at the 2 states listed
in Problem 3.32
Start the software, click the tab for cryogenic substances, and click the tab
for the substance ammonia. Then click the small calculator icon and select the
proper CASE for the given properties.
CASE RESULT
a) 2 (T, v) Sup. vapor, x = undefined, P = 1200 kPa
b) 4 (T, x) Two-phase, P = 2033 kPa, v = 0.03257 m3/kg

3.133
Use the computer software to find the properties for ammonia at the 3 states listed
in Problem 3.127
Start the software, click the tab for cryogenic substances, select ammonia
and click the small calculator icon. Select the proper CASE for the given
properties.
CASE RESULT
a) 8 (P, x) T = 6.795°C, v = 0.1719 m3/kg
b) 1 (T, P) Sup. vapor, x = undefined, v = 1.773 m3/kg
c) 2 (T, v) Sup. vapor, x = undefined, P = 330.4 kPa

3.134
Find the value of the saturated temperature for nitrogen by linear interpolation in
table B.6.1 for a pressure of 900 kPa. Compare this to the value given by the
computer software.
The 900 kPa in Table B.6.1 is located between 100 and 105 K.
T = 100 + (105 – 100)
900 − 779.2
1084.6 − 779.2
= 100 + 5 × 0.3955 = 101.98 K
The actual curve has a positive second derivative (it curves up) so T is
slightly underestimated by use of the chord between the 100 K and the 105 K
points, as the chord is above the curve.
From the computer software:
CASE: 8 (P,x) T = -171°C = 102.15 K
So we notice that the curvature has only a minor effect.
P
T
100 105 110
779.2
1084.6
1467.6
900

3.135
Use the computer software to sketch the variation of pressure with temperature in
Problem 3.44. Extend the curve a little into the single-phase region.
P was found for a number of temperatures. A small table of (P, T) values were
entered into a spreadsheet and a graph made as shown below. The
superheated vapor region is reached at about 140°C and the graph shows a
small kink at that point.
T
P
80
130
180
230
280
330
380
430
100 110 120 130 140 150 160

CONTENT
SUBSECTION PROB NO.
Force displacement work 18-27
Boundary work: simple one-step process 28-46
Polytropic process 47-60
Boundary work: multistep process 61-74
Other types of work and general concepts 75-86
Rates of work 87-101
Heat transfer rates 102-115

4.a
The electric company charges the customers per kW-hour. What is that in SI
units?
Solution:
The unit kW-hour is a rate
multiplied with time. For the
standard SI units the rate of
energy is in W and the time is
in seconds. The integration in
Eq.4.21 becomes
1 kW- hour = 1000 W × 60
min
hour hour × 60
s
min = 3 600 000 Ws
= 3 600 000 J = 3.6 MJ
4.b
Torque and energy and work have the same units (N m). Explain the difference.
Solution:
Work = force × displacement, so units are N × m. Energy in transfer
Energy is stored, could be from work input 1 J = 1 N m
Torque = force × arm static, no displacement needed

4.c
What is roughly the relative magnitude of the work in the process 1-2c versus the
process 1-2a shown in figure 4.8?
By visual inspection the area below the curve 1-2c is roughly 50% of the
rectangular area below the curve 1-2a. To see this better draw a straight line from
state 1 to point f on the axis. This curve has exactly 50% of the area below it.
4.d
Helium gas expands from 125 kPa, 350 K and 0.25 m3 to 100 kPa in a polytropic
process with n = 1.667. Is the work positive, negative or zero?
The boundary work is: W = ⌡⌠P dV
P drops but does V go up or down?
The process equation is: PVn = C
so we can solve for P to show it in a P-V diagram
P = CV-n
as n = 1.667 the curve drops as V goes up we see
V2 > V1 giving dV > 0
and the work is then positive.
P
V
W
1
2

4.e
An ideal gas goes through an expansion process where the volume doubles.
Which process will lead to the larger work output: an isothermal process or a
polytropic process with n = 1.25?
The process equation is: PVn = C
The polytropic process with n = 1.25 drops the pressure faster than the isothermal
process with n = 1 and the area below the curve is then smaller.
P
V
W
1
2
n = 1

4.1
A car engine is rated at 160 hp. What is the power in SI units?
Solution:
The horsepower is an older unit for power
usually used for car engines. The
conversion to standard SI units is given in
Table A.1
1 hp = 0.7355 kW = 735.5 W
1 hp = 0.7457 kW for the UK horsepower
160 hp = 160 × 745.7 W = 119 312 W = 119.3 kW

4.2
Two engines provide the same amount of work to lift a hoist. One engine can
provide 3 F in a cable and the other 1 F, What can you say about the motion of the
point where the force F acts in the two engines?
Since the two work terms are the same we get
W = ∫ F dx = 3 F x1 = 1 F x2
x2 = 3 x1
so the lower force has a larger displacement.

4.3
Two hydraulic piston/cylinders are connected through a hydraulic line so they
have roughly the same pressure. If they have diameters of D1 and D2 = 2D1
respectively, what can you say about the piston forces F1 and F2?
For each cylinder we have the total force as: F = PAcyl = P π D2/4
F1 = PAcyl 1 = P π D
2
1/4
F2 = PAcyl 2 = P π D
2
2/4 = P π 4 D
2
1/4 = 4 F1
cb
12
F
2
F 1
The forces are the total force
acting up due to the cylinder
pressure. There must be other
forces on each piston to have a
force balance so the pistons do
not move.

4.4
Normally pistons have a flat head, but in diesel engines pistons can have bowls in
them and protruding ridges. Does this geometry influence the work term?
The shape of the surface does not influence the displacement
dV = An dx
where An is the area projected to the plane normal to the direction of motion.
An = Acyl = π D2/4
Work is
dW = F dx = P dV = P An dx = P Acyl dx
and thus unaffected by the surface shape.
Semi-spherical
head is made to
make room for
larger valves.
Piston
Bowl
Ridge
x
normal plane

4.5
CV A is the mass inside a piston-cylinder, CV B is that
plus the piston, outside which is the standard atmosphere.
Write the process equation and the work term for the two
CVs.
Po
m gp
Am
Solution:
C.V. A: Process: P = Po + mpg/Acyl = C,
1W2 = ∫ P dV = P1 ∫ m dv = P1m(v2 – v1)
C.V. B: Process: P = Po = C,
1W2 = ∫ Po dV = Po ∫ m dv = Pom(v2 – v1)

4.6
Assume a physical set-up as in Fig. P4.5. We now heat the
cylinder. What happens to P, T and v (up, down or c
What transf
onstant)?
ers do we have for Q and W (pos., neg., or zero)?
Po
mp
Solution:
Process: P = Po + mpg/Acyl = C
Heat in so T increases, v increases and Q is positive.
As the volume increases the work is positive. 1W2 = ∫ P dV

4.7
For a buffer storage of natural gas (CH4) a large bell in a container can move up and
down keeping a pressure of 105 kPa inside. The sun then heats the container and the
gas from 280 K to 300 K during 4 hours. What happens to the volume and what is
the sign of the work term?
Solution
The process has constant pressure
Ideal gas: PV = mRT as T increases then V increases
1W2 = ∫ P dV > 0 so positive.

4.8
A drag force on an object moving through a medium (like a car through air or a
submarine through water) is Fd = 0.225 A ρV2. Verify the unit becomes Newton.
Solution:
Fd = 0.225 A ρV2
Units = m2 × ( kg/m3 ) × ( m2/ s2 ) = kg m / s2 = N

4.9
The sketch shows a physical situation, show the possible process in a P-v diagram.
a) b) c)
Po
mp
R-410a
V
P
1
P
1
Vstop
V
P
1
P
1
V1
V
P
1
P
1
VstopV1

4.10
For the indicated physical set-up in a-b and c above write a process equation and the
expression for work.
a) P = P1 and V ≥ Vstop or V = Vstop and P ≤ P1
1W2 = P1(V2 – V1) [ P1 = Pfloat ]
b) P = A + BV; 1W2 =
1
2 (P1 + P2)(V2 – V1)
c) P = P1 and V ≤ Vstop or V = Vstop and P ≥ P1
1W2 = P1(V2 – V1) [ P1 = Pfloat ]
a) b) c)
Po
mp
R-410a
V
P
1
P
1
Vstop
V
P
1
P
1
V1
V
P
1
P
1
VstopV1

4.11
Assume the physical situation as in Fig. P4.9b; what is the work term a, b, c or d?
a: 1w2 = P1(v2 – v1) b: 1w2 = v1(P2 – P1)
c: 1w2 =
1
2 (P1 + P2)(v2 – v1) d: 1w2 =
1
2 (P1 – P2)(v2 + v1)
Solution: work term is formula c the area under the process curve in a
P-v diagram.

4.12
The sketch in Fig. P4.12 shows a physical situation; show the possible process in a
P-v diagram.
a) b) c)
Solution:
v
P
1P
1
vstop
v
P
1P
1
v 1
v
P
1P
1
v stop
Po
mp
R-410a

4.13
What can you say about the beginning state of the R-410a in Fig. P4.9 versus the
case in Fig. P4.12 for the same piston-cylinder?
For the case where the piston floats as in Fig. P4.9 the pressure of the R-410a
must equal the equilibrium pressure that floats (balance forces on) the piston.
The situation in Fig. P4.12 is possible if the R-410a pressure equals or exceeds
the float pressure.

4.14
Show how the polytropic exponent n can be evaluated if you know the end state
properties, (P1, V1) and (P2, V2).
Polytropic process: PVn = C
Both states must be on the process line: P2V
n
2 = C = P1V
n
1
Take the ratio to get:
P1
P2
=





V2
V1
n
and then take ln of the ratio
ln





P1
P2
= ln





V2
V1
n
= n ln





V2
V1
now solve for the exponent n
n = ln





P1
P2
/ln





V2
V1

4.15
A piece of steel has a conductivity of k = 15 W/mK and a brick has k = 1 W/mK.
How thick a steel wall will provide the same insulation as a 10 cm thick brick?
The heat transfer due to conduction is from Eq. 4.18
.
Q = − kA
dT
dx ≈ kA
∆T
∆x
For the same area and temperature difference the heat transfers become
the same for equal values of (k / ∆x) so
(
k
∆x
)brick = (
k
∆x
)steel =>
∆xsteel = ∆xbrick
ksteel
kbrick
= 0.1 m ×
15
1 = 1.5 m

4.16
A thermopane window, see Fig. 4.27, traps some gas between the two glass
panes. Why is this beneficial?
The gas has a very low conductivity relative to a liquid or solid so the heat
transfer for a given thickness becomes smaller. The gap is furthermore made so
small that possible natural convection motion is reduced to a minimum. It
becomes a trade off to minimize the overall heat transfer due to conduction and
convection. Typically these windows can be manufactured with an E-glaze to
reduce radiation loss (winter) or gain (summer).

4.17
On a chilly 10oC fall day a house, 20oC inside, loses 6 kW by heat transfer. What
transfer happens on a 30oC warm summer day assuming everything else is the
same?
The heat transfer is
.
Q = CA ∆T where the details of the heat transfer is in the
factor C. Assuming those details are the same then it is the temperature difference
that changes the heat transfer so
.
Q = CA ∆T = 6 kW = CA (20 – 10) => CA = 0.6
kW
K
Then
.
Q = CA ∆T = 0.6
kW
K × (20 – 30) K = – 6 kW (it goes in)

Force displacement work

4.18
A piston of mass 2 kg is lowered 0.5 m in the standard gravitational field. Find
the required force and work involved in the process.
Solution:
F = ma = 2 kg × 9.80665 m/s2 = 19.61 N
W = ∫ F dx = F ∫ dx = F ∆x = 19.61 N × 0.5 m = 9.805 J

4.19
A hydraulic cylinder of area 0.01 m2 must push a 1000 kg arm and shovel 0.5 m
straight up. What pressure is needed and how much work is done?
F = mg = 1000 kg × 9.81 m/s2
= 9810 N = PA
P = F/A = 9810 N/ 0.01 m2
= 981 000 Pa = 981 kPa
W = ⌡⌠F dx = F ∆x = 9810 N × 0.5 m = 4905 J

4.20
An escalator raises a 100 kg bucket of sand 10 m in 1 minute. Determine the total
amount of work done during the process.
Solution:
The work is a force with a displacement and force is constant: F = mg
W = ∫ F dx = F ∫ dx = F ∆x = 100 kg × 9.80665 m/s2 × 10 m = 9807 J

4.21
A bulldozer pushes 500 kg of dirt 100 m with a force of 1500 N. It then lifts the
dirt 3 m up to put it in a dump truck. How much work did it do in each situation?
Solution:
W = ∫ F dx = F ∆x
= 1500 N × 100 m
= 150 000 J = 150 kJ
W = ∫ F dz = ∫ mg dz = mg ∆Z
= 500 kg × 9.807 m/s2 × 3 m
= 14 710 J = 14.7 kJ

4.22
A hydraulic cylinder has a piston of cross sectional area 15 cm2 and a fluid
pressure of 2 MPa. If the piston is moved 0.25 m how much work is done?
Solution:
The work is a force with a displacement and force is constant: F = PA
W = ∫ F dx = ∫ PA dx = PA ∆x
= 2000 kPa × 15 × 10-4 m2 × 0.25 m = 0.75 kJ
Units: kPa m2 m = kN m-2 m2 m = kN m = kJ

4.23
A linear spring, F = ks(x − x0), with spring constant ks = 500 N/m, is stretched
until it is 100 mm longer. Find the required force and work input.
Solution:
F = ks(x - x0) = 500 × 0.1 = 50 N
W = ∫ F dx =
⌡⌠ ks(x - x0)d(x - x0) = ks(x - x0)2/2
= 500
N
m × (0.12/2) m2 = 2.5 J

4.24
Two hydraulic cylinders maintain a pressure of 1200 kPa. One has a cross
sectional area of 0.01 m2 the other 0.03 m2. To deliver a work of 1 kJ to the
piston how large a displacement (V) and piston motion H is needed for each
cylinder? Neglect Patm.
Solution:
W = ∫ F dx = ∫ P dV = ∫ PA dx = PA* H = P∆V
∆V =
W
P =
1 kJ
1200 kPa = 0.000 833 m3
Both cases the height is H = ∆V/A
H1 =
0.000833
0.01 = 0.0833 m
H2 =
0.000833
0.03 = 0.0278 m
cb
12
F
2
F 1

4.25
Two hydraulic piston/cylinders are connected with a line. The master cylinder has
an area of 5 cm2 creating a pressure of 1000 kPa. The slave cylinder has an area
of 3 cm2. If 25 J is the work input to the master cylinder what is the force and
displacement of each piston and the work out put of the slave cylinder piston?
Solution:
W = ∫ Fx dx = ∫ P dv = ∫ P A dx = P A ∆x
∆xmaster =
W
PA =
25
1000×5×10-4 = 0.05 m
A∆x = ∆V = 5 ×10-4× 0.05 = 2.5 ×10-5 m = ∆Vslave = A ∆x
∆xslave = ∆V/A = 2.5 × 10-5 / 3 ×10-4 = 0.0083 33 m
Fmaster = P A = 1000× 5 ×10-4 ×103 = 500 N
Fslave = P A = 1000 ×103× 3 ×10-4 = 300 N
Wslave = F ∆x = 300 × 0.08333 = 25 J
Master Slave

4.26
The rolling resistance of a car depends on its weight as: F = 0.006 m g. How long
will a car of 1400 kg drive for a work input of 25 kJ?
Solution:
Work is force times distance so assuming a constant force we get
W = ⌡⌠ F dx = F x = 0.006 mgx
Solve for x
x =
W
0.006 mg =
25 kJ
0.006 × 1400 kg × 9.807 m/s2 = 303.5 m

4.27
The air drag force on a car is 0.225 A ρV2. Assume air at 290 K, 100 kPa and a
car frontal area of 4 m2 driving at 90 km/h. How much energy is used to
overcome the air drag driving for 30 minutes?
The formula involves density and velocity and work involves distance so:
ρ =
1
v =
P
RT =
100
0.287 ×290
= 1.2015
kg
m3
V = 90
km
h = 90 ×
1000
3600
m
s = 25 m/s
∆x = V ∆t = 25 m/s × 30 min × 60 s/min = 45 000 m
Now
F = 0.225 A ρ V2 = 0.225 × 4 × 1.2015 × 252
= 675.8 m2 kg
m3 ×
m2
s2 = 676 N
W = F ∆x = 676 N × 45 000 m = 30 420 000 J = 30.42 MJ

Boundary work simple 1 step process

4.28
The R-410a in Problem 4.12 c is at 1000 kPa, 50oC with mass 0.1 kg. It is cooled
so the volume is reduced to half the initial volume . The piston mass and
gravitation is such that a pressure of 400 kPa will float the piston. Find the work
in the process.
If the volume is reduced the piston must drop and thus float with P = 400 kPa.
The process therefore follows a process curve shown in the P-V diagram.
Table B.4.2: v1 = 0.03320 m3/kg
1W2 = ⌡⌠PdV = area
= Pfloat (V2 – V1) = –Pfloat V1/2
= –400 kPa × 0.1 kg × 0.0332 m3/kg /2
= – 0.664 kJ
v
P
1P
1
v stop
400
2

4.29
A steam radiator in a room at 25°C has saturated water vapor at 110 kPa flowing
through it, when the inlet and exit valves are closed. What is the pressure and the
quality of the water, when it has cooled to 25oC? How much work is done?
Solution: Control volume radiator.
After the valve is closed no more flow, constant volume and mass.
1: x1 = 1, P1 = 110 kPa ⇒ v1 = vg = 1.566 m3/kg from Table B.1.2
2: T2 = 25oC, ?
Process: v2 = v1 = 1.566 m3/kg = [0.001003 + x2 × 43.359] m3/kg
x2 =
1.566 – 0.001003
43.359 = 0.0361
State 2 : T2 , x2 From Table B.1.1 P2 = Psat = 3.169 kPa
1W2 = ⌡⌠PdV = 0

4.30
A constant pressure piston cylinder contains 0.2 kg water as saturated vapor at
400 kPa. It is now cooled so the water occupies half the original volume. Find the
work in the process.
Solution:
Table B.1.2 v1= 0.4625 m3/kg V1 = mv1 = 0.0925 m3
v2 = v1/ 2 = 0.23125 m3/kg V2 = V1 / 2 = 0.04625 m3
Process: P = C so the work term integral is
W = ∫ PdV = P(V2-V1) = 400 kPa × (0.04625 – 0.0925) m3 = -18.5 kJ
T
C.P.
v
12
P = 400 kPa
P C.P.
v
T
400
144
2 1
cb

4.31
Find the specific work in Problem 3.47.
Solution:
v = vf + x vfg = 0.001061 + 0.25 × 0.88467 = 0.22223 m3/kg
T2 = Tsat + 20 = 120.23 + 20 = 140.23oC
so state is superheated vapor
v2 = 0.88573 + (0.95964 – 0.88573)
20
150 - 120.23 = 0.9354 m3/kg
Process P = C: 1w2 = ∫ P dv = P1(v2 – v1)
= 200 (0.9354 – 0.22223)
= 142.6 kJ/kg
T
C.P.
v
200 kPa
P C.P.
v
T
200
120
140
120.2 C
140 C

4.32
A 400-L tank A, see figure P4.32, contains argon gas at 250 kPa, 30oC. Cylinder B,
having a frictionless piston of such mass that a pressure of 150 kPa will float it, is
initially empty. The valve is opened and argon flows into B and eventually reaches a
uniform state of 150 kPa, 30oC throughout. What is the work done by the argon?
Solution:
Take C.V. as all the argon in both A and B. Boundary movement work done in
cylinder B against constant external pressure of 150 kPa. Argon is an ideal gas, so
write out that the mass and temperature at state 1 and 2 are the same
PA1VA = mARTA1 = mART2 = P2( VA + VB2)
=> VB2 =
250 × 0.4
150 - 0.4 = 0.2667 m3
1W2 = ⌡⌠
1
2
PextdV = Pext(VB2 - VB1) = 150 kPa (0.2667 - 0) m3 = 40 kJ
V
P
1 2
B
B
B
A
Argon
Po
g
Notice there is a pressure loss in the valve so the pressure in B is always 150 kPa
while the piston floats.

4.33
A piston cylinder contains 1.5 kg water at 200 kPa, 150oC. It is now heated in a
process where pressure is linearly related to volume to a state of 600 kPa, 350oC.
Find the final volume and the work in the process.
Take as CV the 1.5 kg of water.
m2 = m1 = m ;
Process Eq.: P = A + BV (linearly in V)
State 1: (P, T) => v1 = 0.95964 m3/kg,
P C.P.
2
600
1
T200
v
State 2: (P, T) => v2 = 0.47424 m3/kg, V2 = mv2 = 0.7114 m3
From process eq.:
1W2 = ∫ P dV = area =
m
2 (P1 + P2)(v2 − v1)
=
1.5
2 kg (200 + 600) kPa (0.47424 – 0.95964) m3/kg = −291.24 kJ
Notice volume is reduced so work is negative.

4.34
A cylinder fitted with a frictionless piston contains 5 kg of superheated refrigerant
R-134a vapor at 1000 kPa, 140°C. The setup is cooled at constant pressure until
the R-134a reaches a quality of 25%. Calculate the work done in the process.
Solution:
Constant pressure process boundary work. State properties from Table B.5.2
State 1: v = 0.03150 m3/kg ,
State 2: v = 0.000871 + 0.25 × 0.01956 = 0.00576 m3/kg
Interpolated to be at 1000 kPa, numbers at 1017 kPa could have
been used in which case: v = 0.00566 m3/kg
1W2 = ∫ P dV = P (V2-V1) = mP (v2-v1)
= 5 × 1000 (0.00576 - 0.03150) = -128.7 kJ
T
C.P.
v
1
2
P = 1000 kPa
P C.P.
v
T
1000
39
1402 1
cb

4.35
A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01 m3. A
constant pressure process gives 54 kJ of work out. Find the final volume and
temperature of the air.
Solution:
W = ∫ P dV = P ∆V
∆V = W/P =
54
600 = 0.09 m3
V2 = V1 + ∆V = 0.01 + 0.09 = 0.1 m3
Assuming ideal gas, PV = mRT, then we have
T2 =
P2 V2
mR =
P2 V2
P1 V1
T1=
V2
V1
T1 =
0.1
0.01 × 290 = 2900 K

4.36
A piston/cylinder has 5 m of liquid 20oC water on top of the piston (m = 0) with
cross-sectional area of 0.1 m2, see Fig. P2.56. Air is let in under the piston that rises
and pushes the water out over the top edge. Find the necessary work to push all the
water out and plot the process in a P-V diagram.
Solution:
P1 = Po + ρgH
= 101.32 + 997 × 9.807 × 5 / 1000 = 150.2 kPa
∆V = H × A = 5 × 0.1 = 0.5 m3
1W2 = AREA = ∫ P dV = ½ (P1 + Po )(Vmax -V1)
= ½ (150.2 + 101.32) kPa × 0.5 m3
= 62.88 kJ
H2O
Po
Air
cb V
P
2
1
P
P
1
0
V V1 max

4.37
Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this
state the piston is 0.1 m from the cylinder bottom and cylinder area is 0.25 m2.
The temperature is then changed to 200oC. Find the work in the process.
Solution:
State 1 from B.1.2 (P, x): v1 = vg = 0.8857 m3/kg (also in B.1.3)
State 2 from B.1.3 (P, T): v2 = 1.0803 m3/kg
Since the mass and the cross sectional area is the same we get
h2 =
v2
v1
× h1 =
1.0803
0.8857 × 0.1 = 0.122 m
Process: P = C so the work integral is
W = ∫ PdV = P(V2 - V1) = PA (h2 - h1)
W = 200 kPa × 0.25 m2 × (0.122 − 0.1) m = 1.1 kJ
T
C.P.
v
1
2
P = 200 kPa
P C.P.
v
T
200
120
20021
cb

4.38
A piston cylinder contains 1 kg of liquid water at 20oC and 300 kPa, as shown in
Fig. P4.38. There is a linear spring mounted on the piston such that when the
water is heated the pressure reaches 3 MPa with a volume of 0.1 m3.
a) Find the final temperature
b) Plot the process in a P-v diagram.
c) Find the work in the process.
Solution:
Take CV as the water. This is a constant mass:
m2 = m1 = m ;
State 1: Compressed liquid, take saturated liquid at same temperature.
B.1.1: v1 = vf(20) = 0.001002 m3/kg,
State 2: v2 = V2/m = 0.1/1 = 0.1 m3/kg and P = 3000 kPa from B.1.3
=> Superheated vapor close to T = 400oC
Interpolate: T2 = 404oC
Work is done while piston moves at linearly varying pressure, so we get:
1W2 = ∫ P dV = area = Pavg (V2 − V1) =
1
2 (P1 + P2)(V2 - V1)
= 0.5 (300 + 3000)(0.1 − 0.001) = 163.35 kJ
T
C.P.
v
2
1
300 kPa
P C.P.
v
T
300
20
2
1

4.39
Find the specific work in Problem 3.53 for the case the volume is reduced.
Saturated vapor R-134a at 50oC changes volume at constant temperature. Find the
new pressure, and quality if saturated, if the volume doubles. Repeat the question
for the case the volume is reduced to half the original volume.
Solution:
R-134a 50oC
Table B.4.1: v1 = vg = 0.01512 m3/kg, v2 = v1 / 2 = 0.00756 m3/kg
1W2 = ∫ PdV = 1318.1 kPa (0.00756 – 0.01512) m3/kg = -9.96 kJ/kg
T
C.P.
v
12
P = 1318 kPa
P C.P.
v
T
2
1318
50
1
cb

4.40
A piston/cylinder contains 1 kg water at 20oC with volume 0.1 m3. By mistake
someone locks the piston preventing it from moving while we heat the water to
saturated vapor. Find the final temperature, volume and the process work.
Solution
1: v1 = V/m = 0.1 m3/1 kg = 0.1 m3/kg (two-phase state)
2: Constant volume: v2 = vg = v1
V2 = V1 = 0.1 m3
1W2 = ∫ P dV = 0
State 2: (v2, x2 = 1)
T2 = Tsat = 210 + 5
0.1 - 0.10324
0.09361 - 0.10324 = 211.7°C

4.41
Ammonia (0.5 kg) is in a piston cylinder at 200 kPa, -10oC is heated in a process
where the pressure varies linear with the volume to a state of 120oC, 300 kPa.
Find the work the ammonia gives out in the process.
Solution:
Take CV as the Ammonia, constant mass.
Continuity Eq.: m2 = m1 = m ;
Process: P = A + BV (linear in V)
State 1: Superheated vapor v1 = 0.6193 m3/kg
State 2: Superheated vapor v2 = 0.63276 m3/kg
Work is done while piston moves at increasing pressure, so we get
1W2 = ∫ P dV = area = Pavg (V2 − V1) =
1
2 (P1 + P2)m(v2 − v1)
= ½(200 + 300) × 0.5 (0.63276 − 0.6193) = 1.683 kJ
P
2P2
1
P
1
v

4.42
Air in a spring loaded piston/cylinder has a pressure that is linear with volume, P
= A + BV. With an initial state of P = 150 kPa, V = 1 L and a final state of 800
kPa and volume 1.5 L it is similar to the setup in Problem 4.38. Find the work
done by the air.
Solution:
Knowing the process equation: P = A + BV giving a linear variation of
pressure versus volume the straight line in the P-V diagram is fixed by the two
points as state 1 and state 2. The work as the integral of PdV equals the area
under the process curve in the P-V diagram.
State 1: P1 = 150 kPa V1 = 1 L = 0.001 m3
State 2: P2 = 800 kPa V2 = 1.5 L = 0.0015 m3
Process: P = A + BV linear in V
⇒ 1W2 = ⌡⌠
1
2
PdV = (P1 + P2
2 )(V2 - V1)
P
VW1
2
=
1
2 (150 + 800) kPa (1.5 - 1) × 0.001 m3 = 0.2375 kJ

4.43
Air, 3 kg, is in a piston cylinder similar to Fig. P.4.5 at 27oC, 300 kPa. It is now
heated to 500 K. Plot the process path in a P-v diagram, and find the work in the
process.
Solution:
Ideal gas PV = mRT
State 1: T1, P1 ideal gas so P1V1 = mRT1
V1 = mR T1 / P1 = 3 × 0.287 × 293.15/300 = 0.8413 m3
State 2: T2, P2 = P1 and ideal gas so P2V2 = mRT2
V2 = mR T2 / P2 = 3 × 0.287 × 600/300 = 1.722 m3
1
W2
= ⌡⌠ PdV = P (V2 - V1) = 300 (1.722 – 0.8413) = 264.2 kJ
T
v
2
1
300 kPa
P
v
T
300
293
2
1
T
1
2
600

4.44
Find the work for Problem 3.62.
A piston/cylinder arrangement is loaded with a linear spring and the outside
atmosphere. It contains water at 5 MPa, 400°C with the volume being 0.1 m3. If the
piston is at the bottom, the spring exerts a force such that Plift = 200 kPa. The system
now cools until the pressure reaches 1200 kPa. Find the mass of water, the final state
(T2, v2) and plot the P–v diagram for the process.
Solution :
P
v
5000
1200
200
1
2
a
? 0.057810
1: 5 MPa, 400°C ⇒ v1= 0.05781 m3/kg
m = V/v1 = 0.1/0.05781 = 1.73 kg
Straight line: P = Pa + Cv
v2 = v1
P2 - Pa
P1 - Pa
= 0.01204 m3/kg
v2 < vg(1200 kPa) so two-phase T2 = 188°C
⇒ x2 =
v2 - 0.001139
0.1622 = 0.0672
The P-V coordinates for the two states are then:
(P1 = 5 MPa, V1 = 0.1 m3), (P2 = 1200 kPa, V2 = mv2 = 0.02083 m3)
P vs. V is linear so 1W2 = ⌡⌠PdV =
1
2 (P1 + P2)(V2 - V1)
=
1
2 (5000 + 1200)(0.02083 - 0.1) kPa-m3
= -245.4 kJ

4.45
Heat transfer to a block of 1.5 kg ice at -10oC melts it to liquid at 10oC in a
kitchen. How much work does the water gives out?
Work is done against the atmosphere due to volume change in the process.
State 1: Compressed solid, B.1.5, v1 = 0.0010891 m3/kg
State 2: Compressed liquid B.1.1 v2 = 0.001000 m3/kg
1
W2
= ⌡⌠ PdV = Po (V2 − V1) = Po m (v2 − v1)
= 101.325 kPa × 1.5 kg × (0.001 – 0.0010891) m3/kg
= – 0.0135 kJ
Notice the work is negative, the volume is reduced!

4.46
A piston cylinder contains 0.5 kg air at 500 kPa, 500 K. The air expands in a
process so P is linearly decreasing with volume to a final state of 100 kPa, 300 K.
Find the work in the process.
Solution:
Process: P = A + BV (linear in V, decreasing means B is negative)
From the process: 1
W2
= ⌡⌠ PdV = AREA =
1
2 (P1 + P2)(V2 - V1)
V1 = mR T1/ P1 = 0.5 × 0.287 × (500/500) = 0.1435 m3
V2 = mR T2/ P2 = 0.5 × 0.287 × (300/100) = 0.4305 m3
1
W2
=
1
2 × (500 + 100) kPa × (0.4305 - 0.1435) m3 = 86.1 kJ
T
v
2
1
P
v
T
100 300
2
1
T 12
500500
cb

Polytropic process

4.47
A nitrogen gas goes through a polytropic process with n = 1.3 in a piston/cylinder. It
starts out at 600 K, 600 kPa and ends at 800 K. Is the work positive, negative or zero?
The work is a boundary work so it is
W = ⌡⌠ PdV = ⌡⌠ Pm dv = AREA
so the sign depends on the sign for dV (or dv). The process looks like the following
The actual
process is on a
steeper curve
than n = 1.
-1.3P P = C v T
-0.3T = C v2
T 22
1 1T
1
v v
As the temperature increases we notice the volume decreases so
dv < 0 W < 0
Work is negative and goes into the nitrogen gas.

4.48
Consider a mass going through a polytropic process where pressure is directly
proportional to volume (n = − 1). The process start with P = 0, V = 0 and ends with P
= 600 kPa, V = 0.01 m3. Find the boundary work done by the mass.
Solution:
The setup has a pressure that varies linear with volume going through the
initial and the final state points. The work is the area below the process curve.
0.01
600
P
V
0
0
W
W = ⌡⌠ PdV = AREA
=
1
2 (P1 + P2)(V2 - V1)
=
1
2 (P2 + 0)( V2 - 0)
=
1
2 P2 V2 =
1
2 × 600 × 0.01 = 3 kJ

4.49
process with n = 1.667. How much work does it give out?
Solution:
Process equation: PV
n
= constant = P1V
n
1 = P2V
n
2
Solve for the volume at state 2
V2 = V1 (P1/P2)
1/n
= 0.25 ×





125
100
0.6
= 0.2852 m3
Work from Eq.4.4
1W2 =
P2V2- P1 V1
1-n =
100× 0.2852 - 125× 0.25
1 - 1.667 kPa m3 = 4.09 kJ
The actual process is
on a steeper curve
than n = 1.
P 1
2 n = 1
W
V

4.50
Air at 1500 K, 1000 kPa expands in a polytropic process, n = 1.5, to a pressure of
200 kPa. How cold does the air become and what is the specific work out?
n
= constant = P1V
n
1 = P2V
n
2
Solve for the temperature at state 2 by using ideal gas (PV = mRT)
T2
T1
=
P2V2
P1V1
=
P2
P1
×





P1
P2
1/n
=





P2
P1
(n-1)/n
T2 = T1 (P2/P1)
(n-1)/n
= 1500 ×





200
1000
(1.5-1)/1.5
= 877.2 K
Work from Eq.4.5
1w2 =
P2 v2 - P1 v1
1-n =
R(P2 - T1)
1-n
=
0.287(877.2 - 1500)
1 - 1.5 = 357.5 kJ/kg
on a steeper curve
than n = 1.
P 1
2 n = 1
W
V

4.51
The piston/cylinder shown in Fig. P4.51 contains carbon dioxide at 300 kPa,
100°C with a volume of 0.2 m3. Mass is added at such a rate that the gas
compresses according to the relation PV1.2 = constant to a final temperature of
200°C. Determine the work done during the process.
Solution:
From Eq. 4.4 for the polytopic process PVn = const ( n =/ 1 )
1W2 = ⌡⌠
1
2
PdV =
P2V2 - P1V1
1 - n
Assuming ideal gas, PV = mRT
1W2 =
mR(T2 - T1)
1 - n ,
But mR =
P1V1
T1
=
300 × 0.2
373.15
kPa m3
K = 0.1608 kJ/K
1W2 =
0.1608(473.2 - 373.2)
1 - 1.2
kJ K
K = -80.4 kJ

4.52
Air goes through a polytropic process from 125 kPa, 325 K to 300 kPa and 500 K.
Find the polytropic exponent n and the specific work in the process.
Solution:
Process: Pvn = Const = P1v
n
1 = P2 v
n
2
Ideal gas Pv = RT so
v1 =
RT
P =
0.287 × 325
125 = 0.7462 m3/kg
v2 =
RT
P =
0.287 × 500
300 = 0.47833 m3/kg
From the process equation
(P2/ P1) = (v1/ v2)
n
=> ln(P2/ P1) = n ln(v1/ v2)
n = ln(P2/ P1) / ln(v1/ v2) =
ln 2.4
ln 1.56 = 1.969
The work is now from Eq.4.4 per unit mass
1w2 =
P2v2-P1v1
1-n =
R(T2 - T1)
1-n =
0.287(500 - 325)
1-1.969 = −51.8 kJ/kg

4.53
A gas initially at 1 MPa, 500°C is contained in a piston and cylinder arrangement
with an initial volume of 0.1 m3. The gas is then slowly expanded according to the
relation PV = constant until a final pressure of 100 kPa is reached. Determine the
work for this process.
Solution:
By knowing the process and the states 1 and 2 we can find the relation
between the pressure and the volume so the work integral can be performed.
Process: PV = C ⇒ V2 = P1V1/P2 = 1000 × 0.1/100 = 1 m3
For this process work is integrated to Eq.4.5
1W2 = ∫ P dV = ⌡⌠ CV-1dV = C ln(V2/V1)
1W2 = P1V1 ln
V2
V1
= 1000 kPa × 0.1 m3 × ln (1/0.1)
= 230.3 kJ
P
VW
1
2

4.54
A balloon behaves so the pressure is P = C2 V1/3, C2 = 100 kPa/m. The balloon
is blown up with air from a starting volume of 1 m3 to a volume of 3 m3. Find the
final mass of air assuming it is at 25oC and the work done by the air.
Solution:
The process is polytropic with exponent n = -1/3.
P1 = C2 V1/3 = 100 × 11/3 = 100 kPa
P2 = C2 V1/3 = 100 × 31/3 = 144.22 kPa
P
VW
1
2
1W2 = ∫ P dV =
P2V2 - P1V1
1 - n (Equation 4.4)
=
144.22 × 3 - 100 × 1
1 - (-1/3) kPa-m3 = 249.5 kJ
m2 =
P2V2
RT2
=
144.22 × 3
0.287 × 298
kPa-m3
kJ/kg = 5.056 kg

4.55
A piston cylinder contains 0.1 kg nitrogen at 100 kPa, 27oC and it is now
compressed in a polytropic process with n = 1.25 to a pressure of 250 kPa. What
is the work involved?
Take CV as the nitrogen. m2 = m1 = m ;
Process Eq.: Pvn = Constant (polytropic)
From the ideal gas law and the process equation we can get:
State 2: T2 = T1 (P2 / P1)
n-1
n
= 300.15 ( 250
100 )
0.25
1.25
= 360.5 K
From process eq.:
1W2 = ∫ P dV = area =
m
1−n
(P2v2 − P1v1) =
mR
1−n
(T2 − T1)
=
0.1 × 0.2968
1 - 1.25 (360.5 – 300.15) = −7.165 kJ
v
P
2
1
T
P = C T5 -1.25P P = C v T
-0.25T = C v2
T 22
1 1T
1
v

4.56
A piston cylinder contains 0.1 kg air at 100 kPa, 400 K which goes through a
polytropic compression process with n = 1.3 to a pressure of 300 kPa. How much
work has the air done in the process?
Solution:
Process: Pvn
= Const.
T2 = T1 ( P2 V2 / P1V1) = T1 ( P2 / P1)(P1 / P2 )
1/n
= 400 × (300/100)
(1 - 1/1.3)
= 515.4 K
Work term is already integrated giving Eq.4.4
1W2 =
1
1 − n
(P2 V2 − P1V1) =
mR
1 − n
( T2 − T1) Since Ideal gas,
=
0.1 × 0.287
1 − 1.3
× (515.4 − 400) = −11 kJ
P
VW
1
2
n = 1

4.57
A balloon behaves such that the pressure inside is proportional to the diameter
squared. It contains 2 kg of ammonia at 0°C, 60% quality. The balloon and
ammonia are now heated so that a final pressure of 600 kPa is reached.
Considering the ammonia as a control mass, find the amount of work done in the
process.
Solution:
Process : P ∝ D2, with V ∝ D3 this implies P ∝ D2 ∝ V2/3 so
PV -2/3 = constant, which is a polytropic process, n = −2/3
From table B.2.1: V1 = mv1 = 2(0.001566 + 0.6 × 0.28783) = 0.3485 m3
V2 = V1





P2
P1
3/2
= 0.3485





600
429.3
3/2
= 0.5758 m3
1W2 = ∫ P dV =
P2V2 - P1V1
=
600 × 0.5758 - 429.3 × 0.3485
1 - (-2/3) kPa-m3 = 117.5 kJ
P
VW
1
2

4.58
Consider a piston cylinder with 0.5 kg of R-134a as saturated vapor at -10°C. It is
now compressed to a pressure of 500 kPa in a polytropic process with n = 1.5. Find
the final volume and temperature, and determine the work done during the process.
Solution:
Take CV as the R-134a which is a control mass. m2 = m1 = m
Process: Pv
1.5
= constant until P = 500 kPa
1: (T, x) v1 = 0.09921 m3/kg, P = Psat = 201.7 kPa from Table B.5.1
2: (P, process) v2 = v1 (P1/P2)
(1/1.5)
= 0.09921× (201.7/500)
2/3
= 0.05416 m3/kg
Given (P, v) at state 2 from B.5.2 it is superheated vapor at T2 = 79°C
Process gives P = C v
-1.5
, which is integrated for the work term, Eq.(4.4)
1W2 = ∫ P dV =
m
1 - 1.5 (P2v2 - P1v1)
=
2
- 0.5 kg × (500 × 0.05416 - 201.7 × 0.09921) kPa-m3/kg
= -7.07 kJ

4.59
A piston/cylinder contains water at 500°C, 3 MPa. It is cooled in a polytropic
process to 200°C, 1 MPa. Find the polytropic exponent and the specific work in
the process.
Solution:
Polytropic process: Pvn = C
Both states must be on the process line: P2v
n
2 = C = P1v
n
1
Take the ratio to get:
P1
P2
=





v2
v1
n
and then take ln of the ratio: ln





P1
P2
= ln





v2
v1
n
= n ln





v2
v1
now solve for the exponent n
n = ln





P1
P2
/ln





v2
v1
=
1.0986
0.57246 = 1.919
Table B.1.3 v1 = 0.11619 m3/kg, v2 = 0.20596 m3/kg
1w2 = ∫ P dv =
P2v2 - P1v1
=
1000 × 0.20596 - 3000 × 0.11619
1 - 1.919 kPa-m3/kg
= 155.2 kJ/kg

4.60
A spring loaded piston/cylinder assembly contains 1 kg water at 500oC, 3 MPa.
The setup is such that the pressure is proportional to volume, P = CV. It is now
cooled until the water becomes saturated vapor. Sketch the P-v diagram and find
the final state by iterations and the work in the process.
Solution :
polytropic process with n = −1
P = Cv ⇒ C = P1/v1 = 3000/0.11619 = 25820 kPa kg/m3
2
1
P
v
C
Here from B.1.2:
at 2 MPa vg = 0.09963 ⇒ C = P/vg = 20074 (low)
2.5 MPa vg = 0.07998 ⇒ C = P/vg = 31258 (high)
2.25 MPa vg = 0.08875 ⇒ C = P/vg = 25352 (low)
Now interpolate to match the right slope C:
P2 = 2250 + 250
25 820 – 25 352
31 258 – 25 352 = 2270 kPa,
v2 = P2/C = 2270/25820 = 0.0879 m3/kg
P is linear in V so the work becomes (area in P-v diagram)
1W2 = ∫ P dv = m
1
2(P1 + P2)(v2 - v1)
= 1 kg ×
1
2 (3000 + 2270)(0.0879 - 0.11619) kPa-m3
= − 74.5 kJ

4.61
Consider a two-part process with an expansion from 0.1 to 0.2 m3 at a constant
pressure of 150 kPa followed by an expansion from 0.2 to 0.4 m3 with a linearly
rising pressure from 150 kPa ending at 300 kPa. Show the process in a P-V
diagram and find the boundary work.
Solution:
By knowing the pressure versus volume variation the work is found. If we
plot the pressure versus the volume we see the work as the area below the
process curve.
150
300
0.1 0.2 0.4
P
V
1 2
3
1W3 = 1W2 + 2W3 = ⌡⌠
1
2
PdV + ⌡⌠
2
3
PdV
= P1 (V2 – V1) +
1
2 (P2 + P3)(V3-V2)
= 150 kPa (0.2 – 1.0) m3 +
1
2 (150 + 300) kPa (0.4 – 0.2) m3
= (15 + 45) kJ = 60 kJ

4.62
A helium gas is heated at constant volume from a state of 100 kPa, 300 K to 500
K. A following process expands the gas at constant pressure to three times the
initial volume. What is the specific work in the combined process?
The two processes are:
1 -> 2: Constant volume V2 = V1
2 -> 3: Constant pressure P3 = P2
V
P
3
1
2
P
P
1
2
Use ideal gas approximation for helium.
State 1: T, P => v1 = RT1/P1
State 2: V2 = V1 => P2 = P1 (T2/T1)
State 3: P3 = P2 => V3 = 3V2; T3 = T2 v3/v2 = 500 × 3 = 1500 K
We find the work by summing along the process path.
1w3 = 1w2 + 2w3 = 2w3 = P3(v3 - v2) = R(T3 - T2)
= 2.0771 kJ/kg-K × (1500 - 500) K = 2077 kJ/kg

4.63
Find the work in Problem 3.59.
Ammonia at 10oC with a mass of 10 kg is in a piston cylinder arrangement with
an initial volume of 1 m3. The piston initially resting on the stops has a mass such
that a pressure of 900 kPa will float it. The ammonia is now slowly heated to
50oC. Find the work in the process.
State 1: T = 10oC, v1 =
V
m =
1
10 = 0.1 m3/kg
x1 = (v - vf)/vfg = (0.1 - 0.0016)/0.20381
= 0.4828 V
P
2
1
1a
P
P
1
2
cb
State 1a: P = 900 kPa, v = v1 = 0.1 < vg at 900 kPa
State 2: 50oC and on line(s) means 900 kPa which is superheated vapor.
From Table B.2.2 linear interpolation between 800 and 1000 kPa:
v2 = 0.1648 m3/kg, V2 = mv2 = 1.648 m3
1W2 = ∫ P dV = Pfloat (V2 - V1) = 900 kPa (1.648 - 1.0) m3
= 583.2 kJ

4.64
A piston/cylinder arrangement shown in Fig. P4.64 initially contains air at 150 kPa,
400°C. The setup is allowed to cool to the ambient temperature of 20°C.
a. Is the piston resting on the stops in the final state? What is the final
pressure in the cylinder?
b. What is the specific work done by the air during this process?
Solution:
State 1: P1 = 150 kPa, T1 = 400°C = 673.2 K
State 2: T2 = T0 = 20°C = 293.2 K
For all states air behave as an ideal gas.
a) If piston at stops at 2, V2 = V1/2 and pressure less than Plift = P1
⇒ P2 = P1 ×
V1
V2
×
T2
T1
= 150 kPa × 2 ×
293.2
673.2 = 130.7 kPa < P1
⇒ Piston is resting on stops at state 2.
b) Work done while piston is moving at constant Pext = P1.
1W2 = ∫ Pext dV = P1 (V2 − V1) ; V2 =
1
2 V1 =
1
2 m RT1/P1
1w2 = 1W2/m = RT1 (
1
2− 1 ) = −
1
2 × 0.287 kJ/kg-K × 673.2 K
= −96.6 kJ/kg
V
P
1
2
1a
P
P
2
1
V
T
1
2
1a
T
T
2
1a
T1

4.65
A cylinder containing 1 kg of ammonia has an externally loaded piston. Initially the
ammonia is at 2 MPa, 180°C and is now cooled to saturated vapor at 40°C, and then
further cooled to 20°C, at which point the quality is 50%. Find the total work for the
process, assuming a piecewise linear variation of P versus V.
Solution:
o
C
C
C2
1
P
3
180
40
20
857
1555
2000
v
o
o
cb
State 1: (T, P) Table B.2.2
v1 = 0.10571 m3/kg
State 2: (T, x) Table B.2.1 sat. vap.
P2 = 1555 kPa,
v2 = 0.08313 m3/kg
State 3: (T, x) P3 = 857 kPa, v3 = (0.001638 + 0.14922)/2 = 0.07543 m3/kg
Sum the work as two integrals each evaluated by the area in the P-v diagram.
1W3 = ⌡⌠
1
3
PdV ≈ (
P1 + P2
2 ) m(v2 - v1) + (
P2 + P3
2 ) m(v3 - v2)
=
2000 + 1555
2 kPa × 1 kg × (0.08313 - 0.10571) m3/kg
+
1555 + 857
2 kPa × 1 kg × (0.07543 - 0.08313) m3/kg
= -49.4 kJ

4.66
A piston cylinder has 1.5 kg of air at 300 K and 150 kPa. It is now heated up in a
two step process. First constant volume to 1000 K (state 2) then followed by a
constant pressure process to 1500 K, state 3. Find the final volume and the work
in the process.
Solution:
V
P
3
1
2
P
P
1
2
Use ideal gas approximation for air.
State 1: T, P => V1 = mRT1/P1 = 1.5×0.287×300/150 = 0.861 m3
State 2: V2 = V1 => P2 = P1 (T2/T1) = 150×1000/300 = 500 kPa
State 3: P3 = P2 => V3 = V2 (T3/T2) = 0.861×1500/1000 = 1.2915 m3
1W3 = 1W2 + 2W3 = 2W3 = P3(V3 - V2)
= 500 kPa × (1.2915 - 0.861) m3 = 215.3 kJ

4.67
The refrigerant R-410a is contained in a piston/cylinder as shown in Fig. P4.67,
where the volume is 11 L when the piston hits the stops. The initial state is −30°C,
150 kPa with a volume of 10 L. This system is brought indoors and warms up to
15°C.
a. Is the piston at the stops in the final state?
b. Find the work done by the R-410a during this process.
Solution:
Initially piston floats, V < Vstop so the piston moves at constant Pext = P1 until it
reaches the stops or 15°C, whichever is first.
a) From Table B.4.2: v1 = 0.17732 m3/kg,
m = V/v =
0.010
0.17732 = 0.056395 kg
R-410a
Po
mp
1a
2
1
P
V
P1
Vstop
Check the temperature at state 1a: P1a = 150 kPa, v = Vstop/m.
v1a = V/m =
0.011
0.056395 = 0.19505 m3/kg => T1a = -9.4°C & T2 = 15°C
Since T2 > T1a then it follows that P2 > P1 and the piston is against stop.
b) Work done at constant Pext = P1.
1W2 = ∫ Pext dV = Pext(V2 - V1) = 150(0.011 - 0.010) = 0.15 kJ

4.68
A piston/cylinder contains 1 kg of liquid water at 20°C and 300 kPa. Initially the
piston floats, similar to the setup in Problem 4.67, with a maximum enclosed
volume of 0.002 m3 if the piston touches the stops. Now heat is added so a final
pressure of 600 kPa is reached. Find the final volume and the work in the process.
Solution:
Take CV as the water which is a control mass: m2 = m1 = m ;
Table B.1.1: 20°C => Psat = 2.34 kPa
State 1: Compressed liquid v = vf(20) = 0.001002 m3/kg
State 1a: vstop = 0.002 m3/kg , 300 kPa
State 2: Since P2 = 600 kPa > Plift then piston is pressed against the stops
v2 = vstop = 0.002 m3/kg and V = 0.002 m3
For the given P : vf < v < vg so 2-phase T = Tsat = 158.85 °C
Work is done while piston moves at Plift = constant = 300 kPa so we get
1W2 = ∫ P dV = m Plift(v2 -v1) = 1 kg × 300 kPa × (0.002 − 0.001002) m3
= 0.30 kJ
V
P
1
2
H O
Po
2
cb

4.69
A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m3.
Stops in the cylinder restricts the enclosed volume to 0.5 m3, similar to the setup
in Problem 4.67. The water is now heated to 200°C. Find the final pressure,
volume and the work done by the water.
Solution:
Initially the piston floats so the equilibrium
lift pressure is 200 kPa
1: 200 kPa, v1= 0.1/50 = 0.002 m3/kg,
2: 200°C, on line
Check state 1a:
vstop = 0.5/50 = 0.01 m3/kg
=>
Table B.1.2: 200 kPa , vf < vstop < vg
1a
2
1
P
V
P1
Vstop
State 1a is two phase at 200 kPa and Tstop ≈ 120.2 °C so as T2 > Tstop the
state is higher up in the P-V diagram with
v2 = vstop < vg = 0.127 m3/kg (at 200°C)
State 2 two phase => P2 = Psat(T2) = 1.554 MPa, V2 = Vstop = 0.5 m3
1W2 = 1Wstop = 200 kPa × (0.5 – 0.1) m3 = 80 kJ

4.70
A piston/cylinder assembly (Fig. P4.70) has 1 kg of R-134a at state 1 with 110°C,
600 kPa, and is then brought to saturated vapor, state 2, by cooling while the
piston is locked with a pin. Now the piston is balanced with an additional constant
force and the pin is removed. The cooling continues to a state 3 where the R-134a
is saturated liquid. Show the processes in a P-V diagram and find the work in
each of the two steps, 1 to 2 and 2 to 3.
Solution :
CV R-134a This is a control mass.
Properties from table B.5.1 and 5.2
State 1: (T,P) B.5.2 => v = 0.04943 m3/kg
State 2: given by fixed volume v2 = v1 and x2 = 1.0 so from B.5.1
v2 = v1 = vg = 0.04943 m3/kg => T = 10°C
State 3 reached at constant P (F = constant) v3 = vf = 0.000794 m3/kg
P
V
1
2
3
cb
Since no volume change from 1 to 2 => 1W2 = 0
2W3 = ∫P dV = P(V3 -V2) = mP(v3 -v2) Constant pressure
= 415.8 kPa × (0.000794 - 0.04943) m3/kg × 1 kg
= -20.22 kJ

4.71
Find the work in Problem 3.84.
Ammonia at 10oC with a mass of 10 kg is in a piston cylinder arrangement with
an initial volume of 1 m3. The piston initially resting on the stops has a mass such
that a pressure of 900 kPa will float it. The ammonia is now slowly heated to
50oC. Find the work in the process.
State 1: T = 10oC, v1 =
V
m =
1
10 = 0.1 m3/kg
x1 = (v - vf)/vfg = (0.1 - 0.0016)/0.20381
= 0.4828 V
P
2
1
1a
P
P
1
2
cb
State 1a: P = 900 kPa, v = v1 = 0.1 < vg at 900 kPa
State 2: 50oC and on line(s) means 900 kPa which is superheated vapor.
From Table B.2.2 linear interpolation between 800 and 1000 kPa:
v2 = 0.1648 m3/kg, V2 = mv2 = 1.648 m3
1W2 = ∫ P dV = Pfloat (V2 - V1) = 900 kPa × (1.648 - 1.0) m3
= 583.2 kJ

4.72
10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at
100 kPa, with a quality of 50%. It is now heated so the volume triples. The mass
of the piston is such that a cylinder pressure of 200 kPa will float it, see Fig.
P4.72.
a) Find the final temperature and volume of the water.
b) Find the work given out by the water.
Solution:
Take CV as the water m2 = m1 = m;
Process: v = constant until P = Plift then P is constant.
State 1: v1 = vf + x vfg = 0.001043 + 0.5 × 1.69296 = 0.8475 m3/kg
State 2: v2, P2 ≤ Plift => v2 = 3 × 0.8475 = 2.5425 m3/kg;
T2 = 829°C ; V2 = m v2 = 25.425 m3
1W2 = ∫ P dV = Plift × (V2 - V1)
= 200 kPa × 10 kg × (2.5425 – 0.8475) m3/kg = 3390 kJ
H2O
Po
cb
V
P
2
1P
P
1
2

4.73
A piston cylinder setup similar to Problem 4.72 contains 0.1 kg saturated liquid and
vapor water at 100 kPa with quality 25%. The mass of the piston is such that a
pressure of 500 kPa will float it. The water is heated to 300°C. Find the final
pressure, volume and the work, 1W2.
Solution:
Take CV as the water: m2 = m1 = m
Process: v = constant until P = Plift
To locate state 1: Table B.1.2
v1 = 0.001043 + 0.25×1.69296 = 0.42428 m3/kg
1a: v1a = v1 = 0.42428 m3/kg > vg at 500 kPa
so state 1a is superheated vapor T1a = 200°C
P1
Plift
V
P
1
21a
cb
State 2 is 300°C so heating continues after state 1a to 2 at constant P =>
2: T2, P2 = Plift = 500 kPa => Table B.1.3 v2 = 0.52256 m3/kg ;
V2 = mv2 = 0.05226 m3
1W2 = Plift (V2 - V1) = 500 kPa × (0.05226 - 0.04243) m3 = 4.91 kJ

4.74
A piston cylinder contains air at 1000 kPa, 800 K with a volume of 0.05 m3. The
piston is pressed against the upper stops and it will float at a pressure of 750 kPa.
Then the air is cooled to 400 K. What is the process work?
We need to find state 2. Let us see if we proceed past state 1a during the cooling.
T1a = T1 Pfloat / P1 = 800 × 750 / 100 = 600 K
so we do cool below T1a. That means the piston is floating. Write the ideal gas
law for state 1 and 2 to get
V2 =
mRT2
P2
=
P1V1T2
P2T1
=
1000 × 0.05 × 400
750 × 800
= 0.0333 m3
1W2 = 1aW2 = ∫ P dV = P2 (V2 - V1)
= 750 kPa × (0.0333 – 0.05) m3 = –12.5 kJ
Air
Po
mp
1a
2
1
P
V
P
Vstop
2

Other types of work and general concepts

4.75
Electric power is volts times ampere (P = V i). When a car battery at 12 V is
charged with 6 amp for 3 hours how much energy is delivered?
Solution:
W = ⌡⌠ W
.
dt = W
.
∆t = V i ∆t
= 12 V × 6 Amp × 3 × 3600 s
= 777 600 J = 777.6 kJ
Remark: Volt times ampere is also watts, 1 W = 1 V × 1 Amp. = 1 J/s

4.76
A copper wire of diameter 2 mm is 10 m long and stretched out between two
posts. The normal stress (pressure) σ = E(L – Lo)/Lo , depends on the length L
versus the un-stretched length Lo and Young’s modulus E = 1.1 × 106 kPa. The
force is F = Aσ and measured to be 110 N. How much longer is the wire and how
much work was put in?
Solution:
F = As = AE ∆L/ Lo and ∆L = FLo /AE
A =
π
4D2 =
π
4 × 0.0022 = 3.142 ×10-6 m2
∆L =
110 ×10
3.142×10-6 ×1.1 ×106 ×103 = 0.318 m
1W2 = ∫ F dx = ∫ A s dx = ∫ AE
x
Lo
dx
=
AE
Lo
½ x2 where x = L - Lo
=
3.142×10 -6 ×1.1 ×106 ×103
10 × ½ × 0.3182 = 17.47 J

4.77
A 0.5-m-long steel rod with a 1-cm diameter is stretched in a tensile test. What is
the required work to obtain a relative strain of 0.1%? The modulus of elasticity of
steel is 2 × 108 kPa.
Solution :
−1W2 =
AEL0
2 (e)2, A =
π
4 (0.01)2 = 78.54 × 10-6 m2
−1W2 =
78.54×10-6 × 2×108 × 0.5
2 m2-kPa-m (10-3)2
= 3.93 J

4.78
A soap bubble has a surface tension of S = 3 × 10-4 N/cm as it sits flat on a rigid
ring of diameter 5 cm. You now blow on the film to create a half sphere surface of
diameter 5 cm. How much work was done?
1W2 = ∫ F dx = ∫ S dA = S ∆A
= 2 × S × (
π
2 D2 -
π
4 D 2)
= 2 × 3 × 10-4 N/cm × 100 cm/m ×
π
2 0.052 m2 ( 1- 0.5 )
= 1.18 × 10-4 J
Notice the bubble has 2 surfaces.
A1 =
π
4 D 2 ,
A2 = ½ π D2

4.79
A film of ethanol at 20°C has a surface tension of 22.3 mN/m and is maintained
on a wire frame as shown in Fig. P4.79. Consider the film with two surfaces as a
control mass and find the work done when the wire is moved 10 mm to make the
film 20 × 40 mm.
Solution :
Assume a free surface on both sides of the frame, i.e., there are two surfaces
20 × 30 mm
W = −⌡⌠ S dA = −22.3×10-3 N/m × 2 (800 − 600) ×10-6 m2
= −8.92×10-6 J = −8.92 µJ

4.80
Assume we fill a spherical balloon from a bottle of helium gas. The helium gas
provides work ∫ PdV that stretches the balloon material ∫ S dA and pushes back
the atmosphere ∫ Po dV. Write the incremental balance for dWhelium = dWstretch +
dWatm to establish the connection between the helium pressure, the surface
tension S and Po as a function of radius.
WHe = ∫ P dV = ∫ S dA + ∫ Po dV
dWHe = P dV = S dA + Po dV
dV = d (
π
6 D3 ) =
π
6 × 3D2 dD
dA = d ( 2 × π × D2) = 2π (2D) dD
P
π
2 D2 dD = S (4π)D dD + Po
π
2 D2 dD
Divide by
π
2 D2 to recognize
PHe = Po + 8
S
D

4.81
Assume a balloon material with a constant surface tension of S = 2 N/m. What is
the work required to stretch a spherical balloon up to a radius of r = 0.5 m?
Neglect any effect from atmospheric pressure.
Assume the initial area is small, and that we have 2 surfaces inside and out
W = −∫ S dA = −S (A2 − A1)
= − S(A2) = −S( 2× π D
2
2 )
= −2 N/m × 2 × π × 1 m2 = -12.57 J
Win = −W = 12.57 J

4.82
A sheet of rubber is stretched out over a ring of radius 0.25 m. I pour liquid water
at 20oC on it, as in Fig. P4.82, so the rubber forms a half sphere (cup). Neglect the
rubber mass and find the surface tension near the ring?
Solution:
F ↑ = F ↓ ; F ↑ = SL
The length is the perimeter, 2πr, and there is two surfaces
S × 2 × 2πr = mH2o g = ρH2o Vg = ρH2o×
1
12 π (2r) 3g = ρH2o× π
2
3 r 3
S = ρH2o
1
6 r2 g
= 997 kg/m3 ×
1
6 × 0.252 m2 × 9.81 m/s2
= 101.9 N/m

4.83
Consider a window-mounted air conditioning unit used in the summer to cool
incoming air. Examine the system boundaries for rates of work and heat transfer,
including signs.
Solution :
Air-conditioner unit, steady operation with no change of temperature of AC unit.
Cool side Hot side
Inside Outside
C 30°C25°C
37°C15°C
- electrical work (power) input operates unit,
+Q rate of heat transfer from the room which actually is room air being
cooled,
a larger -Q rate of heat transfer (sum of the other two energy rates) out to
the outside air.

4.84
Consider a light bulb that is on. Explain where we have rates of work and heat
transfer (include modes) that moves energy.
Solution:
Electrical power comes in, that is rate of work.
In the wire filament this electrical work is converted to internal energy so the wire
becomes hot and radiates energy out. There is also some heat transfer by
conduction-convection to the gas inside the bulb so that become warm.
The gas in turn heats the glass by conduction/convection and some of the
radiation may be absorbed by the glass (often the glass is coated white)
All the energy leaves the bulb as a combination of radiation over a range of
wavelengths some of which you cannot see and conduction convection heat
transfer to the air around the bulb.

4.85
Consider a household refrigerator that has just been filled up with room-
temperature food. Define a control volume (mass) and examine its boundaries for
rates of work and heat transfer, including sign.
a. Immediately after the food is placed in the refrigerator
b. After a long period of time has elapsed and the food is cold
Solution:
I. C.V. Food.
a) short term.: -Q from warm food to cold refrigerator air. Food cools.
b) Long term: -Q goes to zero after food has reached refrigerator T.
II. C.V. refrigerator space, not food, not refrigerator system
a) short term: +Q from the warm food, +Q from heat leak from room into
cold space. -Q (sum of both) to refrigeration system. If not equal the
refrigerator space initially warms slightly and then cools down to preset T.
b) long term: small -Q heat leak balanced by -Q to refrigeration system.
Note: For refrigeration system CV any Q in from refrigerator space plus
electrical W input to operate system, sum of which is Q rejected to the room.

4.86
A room is heated with an electric space heater on a winter day. Examine the
following control volumes, regarding heat transfer and work , including sign.
a) The space heater.
b) Room
c) The space heater and the room together
Solution:
a) The space heater.
Electrical work (power) input, and equal (after system warm up) Q out to the
room.
b) Room
Q input from the heater balances Q loss to the outside, for steady (no
temperature change) operation.
c) The space heater and the room together
Electrical work input balances Q loss to the outside, for steady operation.

Rates of work

4.87
A 100 hp car engine has a drive shaft rotating at 2000 RPM. How much torque is
on the shaft for 25% of full power?
Solution:
Power = 0.25 × 100 hp = 0.25 × 73.5 kW (if SI hp) = 18.375 kW = Tω
ω = angular velocity (rad/s) = RPM 2 π /60
T = Power/ω =
power × 60
RPM × 2π =
18.375 kW × 60 s/min
2000 × 2π rad/min = 87.73 Nm
We could also have used UK hp to get 0.25 × 74.6 kW. then T = 89 Nm.

4.88
A car uses 25 hp to drive at a horizontal level at constant 100 km/h. What is the
traction force between the tires and the road?
Solution:
We need to relate the rate of work to the force and velocity
dW = F dx =>
dW
dt = W
.
= F
dx
dt = FV
F = W
.
/ V
W
.
= 25 hp = 25 × 0.7355 kW = 18.39 kW
V = 100 ×
1000
3600 = 27.78 m/s
F = W
.
/ V =
18.39 kW
27.78 m/s = 0.66 kN
Units: kW / (ms−1) = kW s m−1 = kJ s−1s m−1 = kN m m−1 = kN

4.89
An escalator raises a 100 kg bucket of sand 10 m in 1 minute. Determine the rate
of work done during the process.
Solution:
The work is a force with a displacement and force is constant: F = mg
W = ∫ F dx = F ∫ dx = F ∆x = 100 kg × 9.80665 m/s2 × 10 m = 9807 J
The rate of work is work per unit time
.
W =
W
∆t
=
9807 J
60 s = 163 W

4.90
A crane lifts a bucket of cement with a total mass of 450 kg vertically up with a
constant velocity of 2 m/s. Find the rate of work needed to do that.
Solution:
Rate of work is force times rate of displacement. The force is due to gravity (a
= 0) alone.
.
W = FV = mg × V = 450 kg × 9.807 ms−2 × 2 ms−1 = 8826 J/s
.
W = 8.83 kW

4.91
A force of 1.2 kN moves a truck with 60 km/h up a hill. What is the power?
Solution:
W
.
= F V = 1.2 kN × 60 (km/h)
= 1.2 × 103 × 60 ×
103
3600
Nm
s
= 20 000 W = 20 kW

4.92
A piston/cylinder of cross sectional area 0.01 m2 maintains constant pressure. It
contains 1 kg water with a quality of 5% at 150oC. If we heat so 1 g/s liquid turns
into vapor what is the rate of work out?
Vvapor = mvapor vg , Vliq = mliq vf
mtot = constant = mvapor mliq
Vtot = Vvapor + Vliq
m
.
tot = 0 = m
.
vapor + m
.
liq ⇒ m
.
liq = -m
.
vapor
V
.
tot = V
.
vapor + V
.
q = m
.
vaporvg + m
.
iqvfli l
= m
.
vapor (vg- vf ) = m
.
vapor vfg
W
.
= PV
.
= P m
.
vapor vfg
= 475.9 kPa × 0.001 kg/s × 0.39169 m3/kg = 0.1864 kW
= 186 W

4.93
Consider the car with the rolling resistance as in problem 4.26. How fast can it
drive using 30 hp?
F = 0.006 mg
Power = F × V = 30 hp = W
.
V = W
.
/ F =
W
.
0.006 mg =
30 ×0.7457 ×1000
0.006 ×1200 ×9.81
= 271.5 m/s
Comment: This is a very high velocity, the rolling resistance is low relative to the
air resistance.

4.94
Consider the car with the air drag force as in problem 4.27. How fast can it drive
using 30 hp?
ρ =
1
v =
P
RT =
100
0.287 ×290
= 1.2015
kg
m3 and A = 4 m2
Drag force: Fdrag = 0.225 A ρ V2
Power for drag force: W
.
drag = 30 hp × 0.7457 = 22.371 kW
W
.
drag = Fdrag V = 0.225 × 4 × 1.2015 × V3
V3 = W
.
drag /(0.225 × 4 × 1.2015) = 20 688
V = 27.452 m/s = 27.452 ×
3600
1000 = 98.8 km/h

4.95
Consider a 1400 kg car having the rolling resistance as in problem 4.26 and air
resistance as in problem 4.27. How fast can it drive using 30 hp?
Ftot = Frolling + Fair = 0.006 mg + 0.225 AρV2
m = 1400 kg , A = 4 m2
ρ = P/RT = 1.2015 kg/m3
W
.
= FV = 0.006 mgV + 0.225 ρAV3
Nonlinear in V so solve by trial and error.
W
.
= 30 hp = 30 × 0.7355 kW = 22.06 kW
= 0.0006 × 1400 × 9.807 V + 0.225 × 1.2015 × 4 V3
= 82.379V + 1.08135 V3
V = 25 m/s ⇒ W
.
= 18 956 W
V = 26 m/s W
.
= 21 148 W
V = 27 m/s W
.
= 23 508 W
Linear interpolation
V = 26.4 m/s = 95 km/h

4.96
A current of 10 amp runs through a resistor with a resistance of 15 ohms. Find the
rate of work that heats the resistor up.
Solution:
.
W = power = E i = R i2 = 15 × 10 × 10 = 1500 W
R

4.97
A battery is well insulated while being charged by 12.3 V at a current of 6 A.
Take the battery as a control mass and find the instantaneous rate of work and the
total work done over 4 hours.
Solution :
Battery thermally insulated ⇒ Q = 0
For constant voltage E and current i,
Power = E i = 12.3 × 6 = 73.8 W [Units V × A = W]
W = ∫ power dt = power ∆t
= 73.8 × 4 × 60 × 60 = 1 062 720 J = 1062.7 kJ

4.98
A torque of 650 Nm rotates a shaft of diameter 0.25 m with ω = 50 rad/s. What
are the shaft surface speed and the transmitted power?
Solution:
V = ωr = ω D/2 = 50 × 0.25 / 2 = 6.25 m/s
Power = Tω = 650 × 50 Nm/s = 32 500 W = 32.5 kW
Recall also Power = FV = (T/r) V = T V/r = Tω

4.99
Air at a constant pressure in a piston cylinder is at 300 kPa, 300 K and a volume
of 0.1 m3. It is heated to 600 K over 30 seconds in a process with constant piston
velocity. Find the power delivered to the piston.
Solution:
Process: P = constant :
Boundary work: dW = P dV =>
.
W = P
.
V
V2 = V1× (T2/T1) = 0.1 × (600/300) = 0.2 m3
.
W = P
∆V
∆t
= 300 ×
0.2 - 0.1
30 kPa
m3
s = 1 kW
Q
P
V
Remark: Since we do not know the area we do not know the velocity

4.100
A pressure of 650 kPa pushes a piston of diameter 0.25 m with V = 5 m/s. What is
the volume displacement rate, the force and the transmitted power?
A =
π
4 D2 = 0.049087 m2
.
V = AV = 0049087 m2 × 5 m/s = 0.2454 m3/s
F = P A = 650 kPa × 0.049087 m2 = 31.9 kN
.
W = power = F V = P
.
V = 650 kPa × 0.2454 m3/s = 159.5 kW
P
V

4.101
Assume the process in Problem 4.61 takes place with a constant rate of change in
volume over 2 minutes. Show the power (rate of work) as a function of time.
Solution:
W = ∫ P dV since 2 min = 120 sec
.
W = P (∆V / ∆t)
(∆V / ∆t) = 0.3 m3/ 120 s = 0.0025 m3/s
150
300
0.1 0.2 0.4
P
V
1 2
3
0.1 0.2 0.4
W
V
1 2
3
kW
0.75
0.375

Heat Transfer rates

4.102
Find the rate of conduction heat transfer through a 1.5 cm thick hardwood board,
k = 0.16 W/m K, with a temperature difference between the two sides of 20oC.
One dimensional heat transfer by conduction, we do not know the area so
we can find the flux (heat transfer per unit area W/m2).
.
q =
.
Q/A = k
∆T
∆x
= 0.16
W
m K ×
20
0.015
K
m = 213 W/m2

4.103
A pot of steel, conductivity 50 W/m K, with a 5 mm thick bottom is filled with
15°C liquid water. The pot has a diameter of 20 cm and is now placed on an
electric stove that delivers 250 W as heat transfer. Find the temperature on the
outer pot bottom surface assuming the inner surface is at 15°C.
Solution :
Steady conduction through the bottom of the steel pot. Assume the inside
surface is at the liquid water temperature.
.
Q = k A
∆T
∆x
⇒ ∆Τ =
.
Q ∆x / kΑ
∆T = 250 W × 0.005 m /(50 W/m-K ×
π
4 × 0.22 m2) = 0.796 K
T = 15 + 0.796 ≅ 15.8°C
cb

4104
The sun shines on a 150 m2 road surface so it is at 45°C. Below the 5 cm thick
asphalt, average conductivity of 0.06 W/m K, is a layer of compacted rubbles at a
temperature of 15°C. Find the rate of heat transfer to the rubbles.
Solution :
This is steady one dimensional conduction through the asphalt layer.
.
Q = k A
∆T
∆x
= 0.06
W
m-K ×150 m2 ×
45-15
0.05
K
m
= 5400 W

4.105
A water-heater is covered up with insulation boards over a total surface area of 3
m2. The inside board surface is at 75°C and the outside surface is at 18°C and the
board material has a conductivity of 0.08 W/m K. How thick a board should it be
to limit the heat transfer loss to 200 W ?
Solution :
Steady state conduction through a single layer
board.
.
Q cond = k A
∆T
∆x
⇒ ∆x = k Α ∆Τ/
.
Q
∆x = 0.08
W
m K × 3m2 ×
75 − 18
200
K
W
= 0.068 m

4.106
A large condenser (heat exchanger) in a power plant must transfer a total of 100
MW from steam running in a pipe to sea water being pumped through the heat
exchanger. Assume the wall separating the steam and seawater is 4 mm of steel,
conductivity 15 W/m K and that a maximum of 5°C difference between the two
fluids is allowed in the design. Find the required minimum area for the heat
transfer neglecting any convective heat transfer in the flows.
Solution :
Steady conduction through the 4 mm steel wall.
.
Q = k A
∆T
∆x
⇒ Α =
.
Q ∆x / k∆Τ
A = 100 × 106
W × 0.004 m / (15 W/mK × 5 K)
= 480 m2
Condensing
water
Sea
water
cb

4.107
A 2 m2 window has a surface temperature of 15oC and the outside wind is
blowing air at 2oC across it with a convection heat transfer coefficient of h = 125
W/m2K. What is the total heat transfer loss?
Solution:
.
Q = h A ∆T = 125 W/m2K × 2 m2 × (15 – 2) K = 3250 W
as a rate of heat transfer out.
2 Co
15 Co

4.108
You drive a car on a winter day with the atmospheric air at −15°C and you keep
the outside front windshield surface temperature at +2°C by blowing hot air on
the inside surface. If the windshield is 0.5 m2 and the outside convection
coefficient is 250 W/m2K find the rate of energy loos through the front
windshield. For that heat transfer rate and a 5 mm thick glass with k = 1.25 W/m
K what is then the inside windshield surface temperature?
Solution :
The heat transfer from the inside must match the loss on the outer surface
to give a steady state (frost free) outside surface temperature.
.
Q conv = h A ∆Τ = 250 × 0.5 × [2 − ( −15)]
= 250 × 0.5 × 17 = 2125 W
This is a substantial amount of power.
.
Q cond = k A
∆T
∆x
⇒ ∆Τ =
.
Q
kA ∆x
∆Τ =
2125 W
1.25 W/mK × 0.5 m2 × 0.005 m = 17 K
Tin = Tout + ∆T = 2 + 17 = 19°C
-15 C
o
Windshield
2 C
o
T = ?
Warm air

4.109
The brake shoe and steel drum on a car continuously absorbs 25 W as the car
slows down. Assume a total outside surface area of 0.1 m2 with a convective heat
transfer coefficient of 10 W/m2 K to the air at 20°C. How hot does the outside
brake and drum surface become when steady conditions are reached?
Solution :
.
Q = hA∆Τ ⇒ ∆Τ =
.
Q
hA
∆T = ( ΤBRAKE − 20 ) =
25
10 × 0.1
= 25 °C
TBRAKE = 20 + 25 = 45°C

4.110
Due to a faulty door contact the small light bulb (25 W) inside a refrigerator is
kept on and limited insulation lets 50 W of energy from the outside seep into the
refrigerated space. How much of a temperature difference to the ambient at 20°C
must the refrigerator have in its heat exchanger with an area of 1 m2 and an
average heat transfer coefficient of 15 W/m2 K to reject the leaks of energy.
Solution :
.
Q tot = 25 + 50 = 75 W to go out
.
Q = hA ∆T = 15 × 1 × ∆T = 75 W
∆T =
.
Q
hA =
75
15 × 1
= 5 °C
OR T must be at least 25 °C

4.111
The black grille on the back of a refrigerator has a surface temperature of 35°C with a
total surface area of 1 m2. Heat transfer to the room air at 20°C takes place with an
average convective heat transfer coefficient of 15 W/m2 K. How much energy can be
removed during 15 minutes of operation?
Solution :
.
Q = hA ∆T; Q =
.
Q ∆t = hA ∆T ∆t
Q = 15 W/m2 K × 1 m2 × (35-20) Κ × 15 min × 60 s/min
= 202 500 J = 202.5 kJ

4.112
A wall surface on a house is at 30°C with an emissivity of ε = 0.7. The
surrounding ambient to the house is at 15°C, average emissivity of 0.9. Find the
rate of radiation energy from each of those surfaces per unit area.
Solution :
.
Q /A = εσAT4, σ = 5.67 × 10 –8
W/m2K4
a)
.
Q/A = 0.7 × 5.67 × 10-8
× ( 273.15 + 30)4 = 335 W/m
2
b)
.
Q/A = 0.9 × 5.67 × 10-8
× 288.154 = 352 W/m
2

4.113
A radiant heating lamp has a surface temperature of 1000 K with ε = 0.8. How
large a surface area is needed to provide 250 W of radiation heat transfer?
Radiation heat transfer. We do not know the ambient so let us find the area
for an emitted radiation of 250 W from the surface
.
Q = εσ 4AT
A =
.
Q
εσT4 =
250
0.8 × 5.67 × 10-8 × 10004
W
W/m2
= 0.0055 m2

4.114
A log of burning wood in the fireplace has a surface temperature of 450°C.
Assume the emissivity is 1 (perfect black body) and find the radiant emission of
energy per unit surface area.
Solution :
.
Q /A = 1 × σ T4
= 5.67 × 10 –8
× ( 273.15 + 450)4
= 15 505 W/m2
= 15.5 kW/m
2

4.115
A radiant heat lamp is a rod, 0.5 m long and 0.5 cm in diameter, through which
400 W of electric energy is deposited. Assume the surface has an emissivity of
0.9 and neglect incoming radiation. What will the rod surface temperature be ?
Solution :
For constant surface temperature outgoing power equals electric power.
.
Qrad = εσAT4
=
.
Qel ⇒
T4
=
.
Qel / εσA = 400 / (0.9 × 5.67 ×10 –8
× 0.5 × π × 0.005)
= 9.9803 ×1011
K4
⇒ T ≅ 1000 K OR 725 °C

Review Problems

4.116
A nonlinear spring has the force versus displacement relation of F = kns(x − x0)n
.
If the spring end is moved to x
1 from the relaxed state, determine the formula for
the required work.
Solution:
In this case we know F as a function of x and can integrate
W = ⌡⌠Fdx =
⌡
⌠ kns(x - xo)n d(x - xo) =
kns
n + 1 (x1 - xo)n+1

4.117
A vertical cylinder (Fig. P4.117) has a 61.18-kg piston locked with a pin trapping
10 L of R-410a at 10°C, 90% quality inside. Atmospheric pressure is 100 kPa,
and the cylinder cross-sectional area is 0.006 m2. The pin is removed, allowing
the piston to move and come to rest with a final temperature of 10°C for the R-
410a. Find the final pressure, final volume and the work done by the R-410a.
Solution:
State 1: (T, x) from table B.4.1
v1 = 0.0008 + 0.9 × 0.03391 = 0.03132 m3/kg
m = V1/v1 = 0.010/0.03132 = 0.319 kg
Force balance on piston gives the equilibrium pressure
P2 = P0 + mPg/ AP = 100 +
61.18 × 9.807
0.006 × 1000
= 200 kPa
oP
pm cb
R-
410a
g
State 2: (T,P) in Table B.4.2 v2 = 0.13129 m3/kg
V2 = mv2 = 0.319 kg × 0.13129 m3/kg = 0.04188 m3 = 41.88 L
1W2 =
⌡⌠Pequil dV = P2(V2-V1) = 200 kPa (0.04188- 0.010) m3 = 6.38 kJ

4.118
Two kilograms of water is contained in a piston/cylinder (Fig. P4.110) with a
massless piston loaded with a linear spring and the outside atmosphere. Initially
the spring force is zero and P1 = Po = 100 kPa with a volume of 0.2 m3. If the
piston just hits the upper stops the volume is 0.8 m3 and T = 600°C. Heat is now
added until the pressure reaches 1.2 MPa. Find the final temperature, show the P–
V diagram and find the work done during the process.
Solution:
V
P
2
3
1
V1
P1
Vstop
3
,
2
,
State 1: v1 = V/m = 0.2 / 2 = 0.1 m3/kg
Process: 1 → 2 → 3 or 1 → 3’
State at stops: 2 or 2’
v2 = Vstop/m = 0.4 m3/kg & T2 = 600°C
Table B.1.3 ⇒ Pstop = 1 MPa < P3
since Pstop < P3 the process is as 1 → 2 → 3
State 3: P3 = 1.2 MPa, v3 = v2 = 0.4 m3/kg ⇒ T3 ≅ 770°C
W13 = W12 + W23 =
1
2(P1 + P2)(V2 - V1) + 0
=
1
2(100 + 1000) kPa × (0.8 - 0.2) m3
= 330 kJ

4.119
A piston/cylinder contains butane, C4H10, at 300°C, 100 kPa with a volume of
0.02 m3. The gas is now compressed slowly in an isothermal process to 300 kPa.
a. Show that it is reasonable to assume that butane behaves as an ideal gas
during this process.
b. Determine the work done by the butane during the process.
Solution:
a) Tr1 =
T
Tc
=
573.15
425.2 = 1.35; Pr1 =
P
Pc
=
100
3800 = 0.026
From the generalized chart in figure D.1 Z1 = 0.99
Tr2 =
T
Tc
=
573.15
425.2 = 1.35; Pr2 =
P
Pc
=
300
3800 = 0.079
From the generalized chart in figure D.1 Z2 = 0.98
Ideal gas model is adequate for both states.
b) Ideal gas T = constant ⇒ PV = mRT = constant
W = ⌡⌠P dV = P1V1 ln
P1
P2
= 100kPa × 0.02 m3 × ln
100
300
= −2.2 kJ

4.120
Consider the process described in Problem 3.116. With 1 kg water as a control mass,
determine the boundary work during the process.
A spring-loaded piston/cylinder contains water at 500°C, 3 MPa. The setup is such
that pressure is proportional to volume, P = CV. It is now cooled until the water
becomes saturated vapor. Sketch the P-v diagram and find the final pressure.
Solution :
P = Cv ⇒ C = P1/v1 = 3000/0.11619 = 25820 kPa kg/m3
2
1
P
v
C
Here from B.1.2:
at 2 MPa vg = 0.09963 ⇒ C = P/vg = 20074 (low)
2.5 MPa vg = 0.07998 ⇒ C = P/vg = 31258 (high)
2.25 MPa vg = 0.08875 ⇒ C = P/vg = 25352 (low)
Now interpolate to match the right slope C:
P2 = 2270 kPa, v2 = P2/C = 2270/25820 = 0.0879 m3/kg
P is linear in V so the work becomes (area in P-v diagram)
1w2 = ∫ P dv =
1
2(P1 + P2)(v2 - v1)
=
1
2 (3000 + 2270)(0.0879 - 0.11619) = - 74.5 kJ/kg

4.121
A cylinder having an initial volume of 3 m3 contains 0.1 kg of water at 40°C. The
water is then compressed in an isothermal quasi-equilibrium process until it has a
quality of 50%. Calculate the work done in the process splitting it into two steps.
Assume the water vapor is an ideal gas during the first step of the process.
Solution: C.V. Water
State 2: (40°C, x = 1) Tbl B.1.1 => PG = 7.384 kPa, vG = 19.52 m3/kg
State 1: v1 = V1/m = 3 / 0.1 = 30 m3/kg ( > vG )
so H2O ~ ideal gas from 1-2 so since constant T
P1 = PG
vG
v1
= 7.384 ×
19.52
30 = 4.8 kPa
V2 = mv2 = 0.1 × 19.52 = 1.952 m3
T
C.P.
v
12
P
P C.P.
v
T
3
7.38
40
2
1
3
sat
P1
Process T = C: and ideal gas gives work from Eq.4.5
1W2 =⌡⌠
1
2
PdV = P1V1ln
V2
V1
= 4.8 × 3.0 × ln
1.952
3 = −6.19 kJ
v3 = 0.001008 + 0.5 × 19.519 = 9.7605 => V3 = mv3 = 0.976 m3
P = C = Pg: This gives a work term as
2W3 = ⌡⌠
2
3
PdV = Pg (V3−V2) = 7.384 kPa (0.976 - 1.952) m3 = −7.21 kJ
Total work:
1W3 = 1W2 + 2W3 = − 6.19 − 7.21 = -13.4 kJ

4.122
A piston/cylinder (Fig. P4.72) contains 1 kg of water at 20°C with a volume of
0.1 m3. Initially the piston rests on some stops with the top surface open to the
atmosphere, Po and a mass so a water pressure of 400 kPa will lift it. To what
temperature should the water be heated to lift the piston? If it is heated to
saturated vapor find the final temperature, volume and the work, 1W2.
Solution:
(a) State to reach lift pressure of P = 400 kPa, v = V/m = 0.1 m3/kg
Table B.1.2: vf < v < vg = 0.4625 m3/kg
=> T = T sat = 143.63°C
(b) State 2 is saturated vapor at 400 kPa since state 1a is two-phase.
V
P
1
2
H O
Po
2
1a
v2 = vg = 0.4625 m3/kg , V2 = m v2 = 0.4625 m3,
Pressure is constant as volume increase beyond initial volume.
1W2 = ∫ P dV = P (V2 - V1) = Plift (V2 – V1)
= 400 kPa (0.4625 – 0.1) m3 = 145 kJ

4.123
Find the work for Problem 3.112.
Refrigerant-410a in a piston/cylinder arrangement is initially at 50°C, x = 1. It is then
expanded in a process so that P = Cv
−1 to a pressure of 100 kPa. Find the final
temperature and specific volume.
Solution:
Knowing the process (P versus V) and states 1 and 2 allows calculation of W.
State 1: 50°C, x = 1 Table B.4.1: P1 = 3065.2 kPa, v1 = 0.00707 m3/kg
Process: Pv = C = P1v1; => P2 = C/v2= P1v1/v2
State 2: 100 kPa and v2 = v1P1/P2 = 0.2167 m3/kg
v2 < vg at 100 kPa, T2 ≅ -51.65°C from Table B.3.2
Notice T is not constant. It is not an ideal gas in this range
The constant C for the work term is P1v1 so per unit mass we get
1w2 = P1v1 ln
v2
v1
= 3065.2 × 0.0707 × ln
0.2167
0.0707 = 242.7 kJ/kg
v
P
v
T
1
2
1
2

4.124
A cylinder fitted with a piston contains propane gas at 100 kPa, 300 K with a
volume of 0.2 m3. The gas is now slowly compressed according to the relation
PV1.1 = constant to a final temperature of 340 K. Justify the use of the ideal gas
model. Find the final pressure and the work done during the process.
Solution:
The process equation and T determines state 2. Use ideal gas law to say
P2 = P1 (
T2
T1
)
n
n-1 = 100 (
340
300 )
1.1
0.1 = 396 kPa
V2 = V1 (
P1
P2
)
1/n
= 0.2 (
100
396 )
1/1.1
= 0.0572 m3
For propane Table A.2: Tc = 370 K, Pc = 4260 kPa, Figure D.1 gives Z.
Tr1 = 0.81, Pr1 = 0.023 => Z1 = 0.98
Tr2 = 0.92, Pr2 = 0.093 => Z2 = 0.95
Ideal gas model OK for both states, minor corrections could be used. The
work is integrated to give Eq.4.4
1W2 = ∫ P dV =
P2V2-P1V1
1-n =
(396 × 0.0572) - (100 × 0.2)
1 - 1.1 kPa m3
= -26.7 kJ

4.125
Consider the nonequilibrium process described in Problem 3.122. Determine the
work done by the carbon dioxide in the cylinder during the process.
A cylinder has a thick piston initially held by a pin as shown in Fig. P3.122. The
cylinder contains carbon dioxide at 200 kPa and ambient temperature of 290 K. The
metal piston has a density of 8000 kg/m3 and the atmospheric pressure is 101 kPa.
The pin is now removed, allowing the piston to move and after a while the gas returns
to ambient temperature. Is the piston against the stops?
Solution:
Knowing the process (P vs. V) and the states 1 and 2 we can find W.
If piston floats or moves:
P = Plift = Po + ρHg = 101.3 + 8000 × 0.1 × 9.807 / 1000 = 108.8 kPa
Assume the piston is at the stops (since P1 > Plift piston would move)
V2 = V1 × 150 / 100 = (π/4) 0.12 × 0.1× 1.5 = 0.000785× 1.5 = 0.001 1775 m3
For max volume we must have P > Plift so check using ideal gas and constant
T process: P2 = P1 V1/ V2 = 200/1.5 = 133 kPa > Plift and piston is at stops.
1W2 = ∫ Plift dV = Plift (V2 -V1) = 108.8 kPa (0.0011775 - 0.000785) m3
= 0.0427 kJ
Remark: The work is determined by the equilibrium pressure, Plift, and not the
instantaneous pressure that will accelerate the piston (give it kinetic energy). We
need to consider the quasi-equilibrium process to get W.

4.126
The gas space above the water in a closed storage tank contains nitrogen at 25°C,
100 kPa. Total tank volume is 4 m3, and there is 500 kg of water at 25°C. An
additional 500 kg water is now forced into the tank. Assuming constant
temperature throughout, find the final pressure of the nitrogen and the work done
on the nitrogen in this process.
Solution:
The water is compressed liquid and in the process the pressure goes up
so the water stays as liquid. Incompressible so the specific volume does
not change. The nitrogen is an ideal gas and thus highly compressible.
State 1: VH2O 1 = 500 × 0.001003 = 0.5015 m3
VN2 1 = 4.0 - 0.5015 = 3.4985 m3
State 2: VN2 2 = 4.0 - 2 × 0.5015 = 2.997 m3
Process: T = C gives P1V1 = mRT = P2V2
PN2 2 = 100 ×
3.4985
2.997 = 116.7 kPa
Constant temperature gives P = mRT/V i.e. pressure inverse in V for which
the work term is integrated to give Eq.4.5
Wby N2
= ⌡⌠
1
2
PN2
dVN2
= P1V1 ln(V2/V1)
= 100 × 3.4985 × ln
2.997
3.4985 = -54.1 kJ

4.127
Consider the problem of inflating the helium balloon, as described in problem
3.124. For a control volume that consists of the helium inside the balloon
determine the work done during the filling process when the diameter changes
from 1 m to 4 m.
Solution :
Inflation at constant P = P0 = 100 kPa to D1 = 1 m, then
P = P0 + C ( D* -1 - D* -2 ), D* = D / D1,
to D2 = 4 m, P2 = 400 kPa, from which we find the constant C as:
400 = 100 + C[ (1/4) - (1/4)2 ] => C = 1600 kPa
The volumes are: V =
π
6 D3 => V1 = 0.5236 m3; V2 = 33.51 m3
WCV = ⌡⌠
1
2
PdV
= P0(V2 - V1) + ⌡⌠
1
2
C(D* -1 - D* -2)dV
V =
π
6 D3, dV =
π
2 D2 dD =
π
2 D1
3 D* 2 dD*
⇒ WCV = P0(V2 - V1) + 3CV1 ⌡⌠
D1
*=1
D2
*=4
(D*-1)dD*
= P0(V2 - V1) + 3CV1[
D2
* 2 - D1
* 2
2 - (D
*
2
- D
*
1
)]
4
1
= 100 × (33.51 – 0.5236) + 3 × 1600 × 0.5236 [
16-1
2 – (4–1)]
= 14 608 kJ

4.128
Air at 200 kPa, 30°C is contained in a cylinder/piston arrangement with initial
volume 0.1 m3 . The inside pressure balances ambient pressure of 100 kPa plus an
externally imposed force that is proportional to V0.5. Now heat is transferred to
the system to a final pressure of 225 kPa. Find the final temperature and the work
done in the process.
Solution:
C.V. Air. This is a control mass. Use initial state and process to find T2
P1 = P0 + CV1/2; 200 = 100 + C(0.1)1/2, C = 316.23 =>
225 = 100 + CV2
1/2 ⇒ V2 = 0.156 m3
P2V2 = mRT2 =
P1V1
T1
T2 ⇒
T2 = (P2V2 / P1V1) T1 = 225 × 0.156 ×303.15 / (200 ×0.1) = 532 K = 258.9°C
W12 = ∫ P dV = ∫ (P0 + CV1/2) dV
= P0 (V2 - V1) + C ×
2
3 × (V2
3/2 - V1
3/2)
= 100 (0.156 – 0.1) + 316.23 ×
2
3 × (0.1563/2 – 0.13/2)
= 5.6 + 6.32 = 11.9 kJ

4.129
Two springs with same spring constant are installed in a massless piston/cylinder
with the outside air at 100 kPa. If the piston is at the bottom, both springs are
relaxed and the second spring comes in contact with the piston at V = 2 m3. The
cylinder (Fig. P4.129) contains ammonia initially at −2°C, x = 0.13, V = 1 m3,
which is then heated until the pressure finally reaches 1200 kPa. At what pressure
will the piston touch the second spring? Find the final temperature and the total
work done by the ammonia.
Solution :
P
P0 W21
W32
0 2 3
1
2
3
0
V
V
1
cb
State 1: P = 399.7 kPa Table B.2.1
v = 0.00156 + 0.13×0.3106 = 0.0419
At bottom state 0: 0 m3, 100 kPa
State 2: V = 2 m3 and on line 0-1-2
Final state 3: 1200 kPa, on line segment 2.
Slope of line 0-1-2: ∆P/ ∆V = (P1 - P0)/∆V = (399.7-100)/1 = 299.7 kPa/ m3
P2 = P1 + (V2 - V1)∆P/∆V = 399.7 + (2-1)×299.7 = 699.4 kPa
State 3: Last line segment has twice the slope.
P3 = P2 + (V3 - V2)2∆P/∆V ⇒ V3 = V2+ (P3 - P2)/(2∆P/∆V)
V3 = 2 + (1200-699.4)/599.4 = 2.835 m3
v3 = v1V3/V1 = 0.0419×2.835/1 = 0.1188 ⇒ T = 51°C
1W3 = 1W2 + 2W3 =
1
2 (P1 + P2)(V2 - V1) +
1
2 (P3 + P2)(V3 - V2)
= 549.6 + 793.0 = 1342.6 kJ

4.130
A spring-loaded piston/cylinder arrangement contains R-134a at 20°C, 24% quality
with a volume 50 L. The setup is heated and thus expands, moving the piston. It is
noted that when the last drop of liquid disappears the temperature is 40°C. The
heating is stopped when T = 130°C. Verify the final pressure is about 1200 kPa by
iteration and find the work done in the process.
Solution:
C.V. R-134a. This is a control mass.
v
P
P1
P2
P3
1
2
3
State 1: Table B.5.1 =>
v1 = 0.000817 + 0.24*0.03524 = 0.009274
P1 = 572.8 kPa,
m = V/ v1 = 0.050 / 0.009274 = 5.391 kg
Process: Linear Spring
P = A + Bv
State 2: x2 = 1, T2 ⇒ P2 = 1.017 MPa, v2 = 0.02002 m3/kg
Now we have fixed two points on the process line so for final state 3:
P3 = P1 +
P2 - P1
v2 - v1
(v3 - v1) = RHS Relation between P3 and v3
State 3: T3 and on process line ⇒ iterate on P3 given T3
at P3 = 1.2 MPa => v3 = 0.02504 => P3 - RHS = -0.0247
at P3 = 1.4 MPa => v3 = 0.02112 => P3 - RHS = 0.3376
Linear interpolation gives :
P3 ≅ 1200 +
0.0247
0.3376 + 0.0247 (1400-1200) = 1214 kPa
v3 = 0.02504 +
0.0247
0.3376 + 0.0247 (0.02112-0.02504) = 0.02478 m3/kg
W13 = ∫ P dV =
1
2 (P1 + P3)(V3 - V1) =
1
2 (P1 + P3) m (v3 - v1)
=
1
2 5.391 kg (572.8 + 1214) kPa (0.02478 - 0.009274) m3/kg
= 74.7 kJ

CONTENT
SUBSECTION PROB NO.
Concept Problems 1-14
Kinetic and potential energy 15-24
Properties (u,h) from general tables 25-34
Energy equation: simple process 35-67
Energy eqaution: multistep process 68-77
Energy equation: solids and liquids 78-89
Properties (u, h, Cv, Cp), ideal gas 90-101
Energy equation: ideal gas 102-120
Energy equation: polytropic process 121-135
Energy equation in rate form 136-156
Review Problems 157-172
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other
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Copyright Act without the permission of the copyright owner is unlawful.

5.a
In a complete cycle what is the net change in energy and in volume?
For a complete cycle the substance has no change in energy and therefore no storage,
so the net change in energy is zero.
For a complete cycle the substance returns to its beginning state, so it has no change in
specific volume and therefore no change in total volume.
5.b
Explain in words what happens with the energy terms for the stone in Example 5.2. What
would happen if it were a bouncing ball falling to a hard surface?
In the beginning all the energy is potential energy associated with the gravitational
force. As the stone falls the potential energy is turned into kinetic energy and in the
impact the kinetic energy is turned into internal energy of the stone and the water. Finally
the higher temperature of the stone and water causes a heat transfer to the ambient until
ambient temperature is reached.
With a hard ball instead of the stone the impact would be close to elastic transforming
the kinetic energy into potential energy (the material acts as a spring) that is then turned
into kinetic energy again as the ball bounces back up. Then the ball rises up transforming
the kinetic energy into potential energy (mgZ) until zero velocity is reached and it starts
to fall down again. The collision with the floor is not perfectly elastic so the ball does not
rise exactly up to the original height losing a little energy into internal energy (higher
temperature due to internal friction) with every bounce and finally the motion will die
out. All the energy eventually is lost by heat transfer to the ambient or sits in lasting
deformation (internal energy) of the substance.

5.c
Make a list of at least 5 systems that store energy, explaining which form of energy.
A spring that is compressed. Potential energy (1/2)kx2
A battery that is charged. Electrical potential energy. V Amp h
A raised mass (could be water pumped up higher) Potential energy mgH
A cylinder with compressed air. Potential (internal) energy like a spring.
A tank with hot water. Internal energy mu
A fly-wheel. Kinetic energy (rotation) (1/2)Iω2
A mass in motion. Kinetic energy (1/2)mV2
5.d
A constant mass goes through a process where 100 J of heat transfer comes in and 100
J of work leaves. Does the mass change state?
Yes it does.
As work leaves a control mass its volume must go up, v increases
As heat transfer comes in an amount equal to the work out means u is constant
if there are no changes in kinetic or potential energy.

5.e
Water is heated from 100 kPa, 20oC to 1000 kPa, 200oC. In one case pressure is raised
at T = C, then T is raised at P = C. In a second case the opposite order is done. Does
that make a difference for 1Q2 and 1W2?
Yes it does. Both 1Q2 and 1W2 are process dependent. We can illustrate the
work term in a P-v diagram.
P
T
V
L
Cr.P.
S
1000 a
20 200
1
2
100
T
C.P.
v
a
P
v
a
180 C
2 2
20 C 20
200
1100
1000 200 C 100
1553 kPa
1000
1
b
b
In one case the process proceeds from 1 to state “a” along constant T then from
“a” to state 2 along constant P.
The other case proceeds from 1 to state “b” along constant P and then from “b”
to state 2 along constant T.

5.f
A rigid insulated tank A contains water at 400 kPa, 800C. A pipe and valve connect this
to another rigid insulated tank B of equal volume having saturated water vapor at 100
kPa. The valve is opened and stays open while the water in the two tanks comes to a
uniform final state. Which two properties determine the final state?
Continuity eq.: m2 – m1A – m1B = 0 ⇒ m2 = m1A + m1B
Energy eq.: m2u2 – m1Au1A – m1Bu1B = 0 – 0
Process: Insulated: 1Q2 = 0,
Rigid: V2 = C = VA + VB ⇒ 1W2 = 0
From continuity eq. and process: v2 = V2/m2 =
m1A
m2
v1A +
m1B
m2
v1B
From energy eq.: u2 =
m1A
m2
u1A +
m1B
m2
u1B
Final state 2: (v2, u2) both are the mass weighted average of the initial values.

5.g
To determine v or u for some liquid or solid, is it more important that I know P or T?
T is more important, v and u are nearly independent of P.
5.h
To determine v or u for an ideal gas, is it more important that I know P or T?
For v they are equally important ( v = RT/P), but for u only T is important. For an
ideal gas u is a function of T only (independent of P).

5.i
I heat 1 kg of substance at a constant pressure (200 kPa) 1 degree. How much heat is
needed if the substance is water at 10oC, steel at 25oC, air at 325 K, or ice at –10oC.
Heating at constant pressure gives (recall the analysis in Section 5.5, page 141)
1Q2 = H2 – H1 = m(h2 – h1) ≈ m Cp (T2 – T1)
For all cases: 1Q2 = 1 kg × C × 1 K
Water 10oC, 200 kPa (liquid) so A.4: C = 4.18 kJ/kg-K, 1Q2 = 4.18 kJ
Steel 25oC, 200 kPa (solid) so A.3: C = 0.46 kJ/kg-K 1Q2 = 0.46 kJ
Air 325 K, 200 kPa (gas) so A.5: Cp = 1.004 kJ/kg-K 1Q2 = 1.004 kJ
Ice –10oC, 200 kPa (solid) so A.3: C = 2.04 kJ/kg-K 1Q2 = 2.04 kJ
Comment: For liquid water we could have interpolated h2 – h1 from Table B.1.1 and
for ice we could have used Table B.1.5. For air we could have used Table A.7.

Concept Problems

5.1
What is 1 cal in SI units and what is the name given to 1 N-m?
Look in the conversion factor table A.1 under energy:
1 cal (Int.) = 4.1868 J = 4.1868 Nm = 4.1868 kg m2/s2
This was historically defined as the heat transfer needed to bring 1 g of liquid water
from 14.5oC to 15.5oC, notice the value of the heat capacity of water in Table A.4
1 N-m = 1 J or Force times displacement = energy = Joule

5.2
Why do we write ∆E or E2 – E1 whereas we write 1Q2 and 1W2?
∆E or E2 – E1 is the change in the stored energy from state 1 to state 2 and
depends only on states 1 and 2 not upon the process between 1 and 2.
1Q2 and 1W2 are amounts of energy transferred during the process between
1 and 2 and depend on the process path. The quantities are associated with
the process and they are not state properties.
5.3
If a process in a control mass increases energy E2 – E1 > 0 can you say anything about
the sign for 1Q2 and 1W2?
No.
The net balance of the heat transfer and work terms from the energy equation is
E2 – E1 = 1Q2 – 1W2 > 0
but that does not separate the effect of the two terms.

5.4
When you windup a spring in a toy or stretch a rubber band what happens in terms of
work, energy, and heat transfer? Later, when they are released, what happens then?
In both processes work is put into the device and the energy is stored as potential
energy. If the spring or rubber is inelastic some of the work input goes into internal
energy (it becomes warmer or permanently deformed) and not into its potential energy.
Being warmer than the ambient air it cools slowly to ambient temperature.
When the spring or rubber band is released the potential energy is transferred back
into work given to the system connected to the end of the spring or rubber band. If
nothing is connected the energy goes into kinetic energy and the motion is then
dampened as the energy is transformed into internal energy.

5.5
CV A is the mass inside a piston-cylinder, CV B is that
plus the piston outside, which is the standard atmosphere.
Write the energy equation and work term for the two CVs
assuming we have a non-zero Q between state 1 and state
2.
Po
pm g
A
m
CV A: E2 – E1 = mA(e2 – e1) = mA(u2 – u1) = 1Q2 – 1WA2
1WA2 = ∫ P dV = P(V2 – V1)
CV B: E2 – E1 = mA(e2 – e1) + mpist(e2 – e1) = mA(u2 – u1) + mpist(gZ2 – gZ1)
= 1Q2 – 1WB2
1WB2 = ∫ Po dV = Po(V2 – V1)
Notice how the P inside CV A is P = Po + mpistg /Acyl i.e. the first work term is
larger than the second. The difference between the work terms is exactly equal to the potential
energy of the piston sitting on the left hand side in the CV B energy Eq. The two equations
are mathematically identical.
1WA2 = P(V2 – V1) = [Po + mpistg /Acyl ] (V2 – V1) = 1WB2 + mpistg(V2 – V1)/Acyl
= 1WB2 + mpistg(Z2 – Z1)

5.6
Saturated water vapor has a maximum for u and h at around 235oC. Is it similar for other
substances?
Look at the various substances listed in appendix B. Everyone has a maximum u and h
somewhere along the saturated vapor line at different T for each substance. This means
the constant u and h curves are different from the constant T curves and some of them
cross over the saturated vapor line twice, see sketch below.
Constant h lines are
similar to the constant
u line shown.
P C.P.
v
T
C.P.
v
T
P = C
u = C
u = C
Notice the constant u (or h) line becomes parallel to the constant T lines in the
superheated vapor region for low P where it is an ideal gas. In the T-v diagram the
constant u (or h) line becomes horizontal in the ideal gas region.

5.7
Some liquid water is heated so it becomes superheated vapor. Do I use u or h in the
energy equation? Explain.
The energy equation for a control mass is: m(u2 – u1) = 1Q2 – 1W2
The storage of energy is a change in u (when we neglect kinetic and potential energy
changes) and that is always so. To solve for the heat transfer we must know the work
in the process and it is for a certain process (P = C) that the work term combines with
the change in u to give a change in h. To avoid confusion you should always write the
energy equation as shown above and substitute the appropriate expression for the
work term when you know the process equation that allows you to evaluate work.

5.8
Some liquid water is heated so it becomes superheated vapor. Can I use specific heat
to find the heat transfer? Explain.
NO.
The specific heat can not give any information about the energy required to do the
phase change. The specific heat is useful for single phase state changes only.

5.9
Look at the R-410a value for uf at -50oC. Can the energy really be negative? Explain.
The absolute value of u and h are arbitrary. A constant can be added to all u and h
values and the table is still valid. It is customary to select the reference such that u for
saturated liquid water at the triple point is zero. The standard for refrigerants like R-
410a is that h is set to zero as saturated liquid at -40oC, other substances like
cryogenic substances like nitrogen, methane etc. may have different states at which h
is set to zero. The ideal gas tables use a zero point for h as 25oC or at absolute zero, 0
K.

5.10
A rigid tank with pressurized air is used to a) increase the volume of a linear spring
loaded piston cylinder (cylindrical geometry) arrangement and b) to blow up a
spherical balloon. Assume that in both cases P = A + BV with the same A and B.
What is the expression for the work term in each situation?
The expression is exactly the same; the geometry does not matter as long as we have
the same relation between P and V then
1W2 = ∫ P dV = ∫ (A + BV) dV
= A(V2 – V1) + 0.5 B (V
2
2 – V
2
1)
= A(V2 – V1) + 0.5 B (V2 + V1) (V2 – V1)
= 0.5 [ A + B V2 + A + B V1] (V2 – V1)
= 0.5 (P1 + P2) (V2 – V1)
Notice the last expression directly gives the area below the curve in the P-V diagram.

5.11
An ideal gas in a piston-cylinder is heated with 2 kJ during an isothermal process.
How much work is involved?
Energy Eq.: u2 − u1 = 1q2 – 1w2 = 0 since u2 = u1 (isothermal)
Then
1W2 = m 1w2 = 1Q2 = m 1Q2 = 2 kJ
5.12
An ideal gas in a piston-cylinder is heated with 2 kJ during an isobaric process. Is the
work pos., neg., or zero?
As the gas is heated u and T increase and since PV = mRT it follows that the volume
increase and thus work goes out.
w > 0

5.13
You heat a gas 10 K at P = C. Which one in Table A.5 requires most energy? Why?
A constant pressure process in a control mass gives (recall Eq.5.29)
1q2 = u2 − u1 + 1w2 = h2 − h1 ≈ Cp ∆T
The one with the highest specific heat is hydrogen, H2. The hydrogen has the
smallest mass, but the same kinetic energy per mol as other molecules and thus the
most energy per unit mass is needed to increase the temperature.

5.14
A 500 W electric space heater with a small fan inside heats air by blowing it over a
hot electrical wire. For each control volume: a) wire only b) all the room air and c)
total room plus the heater, specify the storage, work and heat transfer terms as +
500W or -500W or 0 W, neglect any Q
.
through the room walls or windows.
Storage Work Heat transfer
Wire 0 W -500 W -500 W
Room air 500 W 0 W 500 W
Tot room 500 W -500 W 0 W

Kinetic and Potential Energy

5.15
A piston motion moves a 25 kg hammerhead vertically down 1 m from rest to a
velocity of 50 m/s in a stamping machine. What is the change in total energy of the
hammerhead?
Solution: C.V. Hammerhead
The hammerhead does not change internal energy (i.e. same P, T), but it does have
a change in kinetic and potential energy.
E2 – E1 = m(u2 – u1) + m[(1/2)V2
2
– 0] + mg (Z2 - 0)
= 0 + 25 kg × (1/2) × (50 m/s)
2
+ 25 kg × 9.80665 m/s2 × (-1) m
= 31250 J – 245.17 J = 31005 J = 31 kJ

5.16
A steel ball weighing 5 kg rolls horizontal with 10 m/s. If it rolls up an incline how
high up will it be when it comes to rest assuming standard gravitation.
C.V. Steel ball.
Energy Eq.: E2 – E1 = 1Q2 – 1W2 = 0 – 0 = 0
E1 = mu1 + mgZ1 + 0.5 mV2
E2 = mu2 + mgZ2 + 0
We assume the steel ball does not change temperature (u2 = u1) so then the energy
equation gives
mu2 + mgZ2 – mu1 – mgZ1 – 0.5 mV2 = 0
mg (Z2 – Z1) = 0.5 mV2
Z2 – Z1 = 0.5 V2/g = 0.5 × 102 (m2/s2) / (9.81 m/s2) = 5.1 m

5.17
A 1200 kg car accelerates from zero to 100 km/h over a distance of 400 m. The road at
the end of the 400 m is at 10 m higher elevation. What is the total increase in the car
kinetic and potential energy?
Solution:
∆KE = ½ m (V
2
2 - V
2
1)
V2 = 100 km/h =
100 × 1000
3600 m/s
= 27.78 m/s
∆KE = ½ ×1200 kg × (27.782 – 02) (m/s)2 = 463 037 J = 463 kJ
∆PE = mg(Z2 – Z1) = 1200 kg × 9.807 m/s2 ( 10 - 0 ) m = 117684 J = 117.7 kJ

5.18
A hydraulic hoist raises a 1750 kg car 1.8 m in an auto repair shop. The hydraulic
pump has a constant pressure of 800 kPa on its piston. What is the increase in
potential energy of the car and how much volume should the pump displace to deliver
that amount of work?
Solution: C.V. Car.
No change in kinetic or internal energy of the car, neglect hoist mass.
E2 – E1 = PE2 - PE1 = mg (Z2 – Z1)
= 1750 × 9.80665 × 1.8 = 30 891 J
The increase in potential energy is work into car
from pump at constant P.
W = E2 – E1 = ∫ P dV = P ∆V ⇒
∆V =
E2 – E1
P =
30891
800 × 1000 = 0.0386 m3

5.19
The rolling resistance of a car depends on its weight as: F = 0.006 mg. How far will a
car of 1200 kg roll if the gear is put in neutral when it drives at 90 km/h on a level
road without air resistance?
Solution:
The car decreases its kinetic energy to zero due to the force (constant) acting over the
distance.
m (1/2V
2
2 –1/2V
2
1) = -1W2 = -∫ F dx = -FL
V2 = 0, V1 = 90
km
h =
90 ×1000
3600 ms-1 = 25 ms-1
-1/2 mV
2
1 = -FL = - 0.006 mgL
L =
0.5 V
2
1
0.0006g =
0.5×252
0.006×9.807
m2/s2
m/s2 = 5311 m
Remark: Over 5 km! The air resistance is much higher than the rolling resistance so
this is not a realistic number by itself.

5.20
A 1200 kg car is accelerated from 30 to 50 km/h in 5 s. How much work is that? If you
continue from 50 to 70 km/h in 5 s; is that the same?
The work input is the increase in kinetic energy.
E2 – E1 = (1/2)m[V
2
2 - V
2
1] = 1W2
= 0.5 × 1200 kg [502 – 302]





km
h
2
= 600 [ 2500 – 900 ] kg





1000 m
3600 s
2
= 74 074 J = 74.1 kJ
The second set of conditions does not become the same
E2 – E1 = (1/2)m[V
2
2 - V
2
1] = 600 [ 702 – 502 ] kg





1000 m
3600 s
2
= 111 kJ

5.21
Airplane takeoff from an aircraft carrier is assisted by a steam driven piston/cylinder
device with an average pressure of 1250 kPa. A 17500 kg airplane should be
accelerated from zero to a speed of 30 m/s with 30% of the energy coming from the
steam piston. Find the needed piston displacement volume.
Solution: C.V. Airplane.
No change in internal or potential energy; only kinetic energy is changed.
E2 – E1 = m (1/2) (V
2
2 - 0) = 17500 kg × (1/2) × 302 (m/s)2
= 7875 000 J = 7875 kJ
The work supplied by the piston is 30% of the energy increase.
W = ∫ P dV = Pavg ∆V = 0.30 (E2 – E1)
= 0.30 × 7875 = 2362.5 kJ
∆V =
W
Pavg
=
2362.5
1250
kJ
kPa = 1.89 m3

5.22
Solve Problem 5.21, but assume the steam pressure in the cylinder starts at 1000 kPa,
dropping linearly with volume to reach 100 kPa at the end of the process.
Solution: C.V. Airplane.
E2 – E1 = m (1/2) (V2
2
- 0)
= 17 500 kg × (1/2) × 30
2
(m/s)2
= 7875 000 J = 7875 kJ
W = 0.30(E2 – E1) = 0.30 × 7875 = 2362.5 kJ
W = ∫ P dV = (1/2)(Pbeg + Pend) ∆V
P
1
2
V
W
1000
100
∆V =
W
Pavg
=
2362.5 kJ
1/2(1000 + 100) kPa = 4.29 m3

5.23
A 25 kg piston is above a gas in a long vertical cylinder. Now the piston is released
from rest and accelerates up in the cylinder reaching the end 5 m higher at a velocity
of 25 m/s. The gas pressure drops during the process so the average is 600 kPa with an
outside atmosphere at 100 kPa. Neglect the change in gas kinetic and potential energy,
and find the needed change in the gas volume.
Solution:
C.V. Piston
(E2 – E1)PIST. = m(u2 – u1) + m[(1/2)V
2
2 – 0] + mg (H2 – 0)
= 0 + 25 × (1/2) × 25
2
+ 25 × 9.80665 × 5
= 7812.5 + 1225.8 = 9038.3 J = 9.038 kJ
Energy equation for the piston is:
E2 – E1 = Wgas - Watm = Pavg ∆Vgas – Po ∆Vgas
(remark ∆Vatm = – ∆Vgas so the two work terms are of opposite sign)
∆Vgas =
9.038
600 – 100
kJ
kPa = 0.018 m3
V
H
Po
g
P
1
2
V
Pavg

5.24
A piston of 2 kg is accelerated to 20 m/s from rest. What constant gas pressure is
required if the area is 10 cm2, the travel 10 cm and the outside pressure is 100 kPa?
C.V. Piston
(E2 – E1)PIST. = m(u2 – u1) + m[(1/2)V
2
2 – 0] + mg (0 – 0)
= (1/2) m V
2
2 = 0.5 × 2 kg × 202 (m/s)2 = 400 J
Energy equation for the piston is:
(E2 – E1) PIST. = Wgas - Watm = Pavg ∆Vgas – Po ∆Vgas
∆Vgas = A L = 10 cm2 × 10 cm = 0.0001 m3
Pavg ∆Vgas = (E2 – E1)PIST. + Po ∆Vgas
Pavg = (E2 – E1)PIST. / ∆Vgas + Po
= 400 J / 0.0001 m3 + 100 kPa
= 4000 kPa + 100 kPa = 4100 kPa

Properties (u, h) from General Tables

5.25
Find the phase and the missing properties of T, P, v, u and x for water at:
a. 500 kPa, 100oC b. 5000 kPa, u = 800 kJ/kg
c. 5000 kPa, v = 0.06 m3/kg d. -6oC, v = 1 m3/kg
Solution:
a) Look in Table B.1.2 at 500 kPa
T < Tsat = 151oC => compressed liquid
Table B.1.4: v = 0.001043 m3/kg, u = 418.8 kJ/kg
b) Look in Table B.1.2 at 5000 kPa
u < uf = 1147.78 kJ/kg => compressed liquid
Table B.1.4: between 180oC and 200oC
T = 180 + (200 - 180)
800 - 759.62
848.08 - 759.62 = 180 + 20×0.4567 = 189.1oC
v = 0.001124 + 0.4567 (0.001153 - 0.001124) = 0.001137 m3/kg
c) Look in Table B.1.2 at 5000 kPa
v > vg = 0.03944 m3/kg => superheated vapor
Table B.1.3: between 400oC and 450oC.
T = 400 + 50×
0.06 - 0.05781
0.0633 - 0.05781 = 400 + 50×0.3989 = 419.95oC
u = 2906.58 + 0.3989 × (2999.64 - 2906.58) = 2943.7 kJ/kg
d) B.1.5: vi < v < vg = 334.14 m3/kg => 2-phase, P = Psat = 887.6 kPa,
x = (v - vi) / vfg = (0.01 - 0.000857)/0.02224 = 0.4111
u = ui + x ufg = 248.34 + 0.4111×148.68 = 309.46 kJ/kg
5.26
States shown are placed
relative to the two-phase
region, not to each other.
P C.P.
v
T
C.P.
v
Tb
c
c
a
b
P = const.
d d
a

5.27
Find the phase and the missing properties of P, T, v, u and x
a. Water at 5000 kPa, u = 3000 kJ/kg
b. Ammonia at 50oC, v = 0.08506 m3/kg
c. Ammonia at 28oC, 1200 kPa
d. R-134a at 20oC, u = 350 kJ/kg
a) Check in Table B.1.2 at 5000 kPa: u > ug = 2597 kJ/kg
Goto B.1.3 it is found very close to 450oC, x = undefined, v = 0.0633 m3/kg
b) Table B.2.1 at 50oC: v > vg = 0.06337 m3/kg, so superheated vapor
Table B.2.2: close to 1600 kPa, u = 1364.9 kJ/kg, x = undefined
c) Table B.2.1 between 25 and 30oC: We see P > Psat = 1167 kPa (30oC)
We conclude compressed liquid without any interpolation.
v = vf = 0.001658 +
28 − 25
5 (0.00168 – 0.001658) = 0.00167 m3/kg
u = uf = 296 +
28 − 25
5 (320.46 – 296.59) = 310.91 kJ/kg
d) Table B.5.1 at 20oC: 227.03 = uf < u < ug = 389.19 kJ/kg so two-phase
x =
u - uf
ufg
=
350 - 227.03
162.16 = 0.7583, P = Psat = 572.8 kPa
v = vf + x vfg = 0.000817 + x × 0.03524 = 0.02754 m3/kg
all states are
relative to the
two-phase
region, not to
each other
P TC.P. C.P.
v
a
P = const.c d a, b
b
dT
c
v

5.28
Find the missing properties and give the phase of the ammonia, NH3.
a. T = 65oC, P = 600 kPa u = ? v = ?
b. T = 20oC, P = 100 kPa u = ? v = ? x = ?
c. T = 50oC, v = 0.1185 m3/kg u = ? P = ? x = ?
Solution:
a) Table B.2.1 P < Psat => superheated vapor Table B.2.2:
v = 0.5 × 0.25981 + 0.5 × 0.26888 = 0.2645 m3/kg
u = 0.5 × 1425.7 + 0.5 × 1444.3 = 1435 kJ/kg
b) Table B.2.1: P < Psat => x = undefined, superheated vapor, from B.2.2:
v = 1.4153 m3/kg ; u = 1374.5 kJ/kg
c) Sup. vap. ( v > vg) Table B.2.2. P = 1200 kPa, x = undefined
u = 1383 kJ/kg
States shown are
to each other.
P C.P.
v
T
C.P.
v
T
a
c
b
a
c
b
1200 kPa
600 kPa

5.29
Find the missing properties of (u, h, and x)
a. H2O T = 120°C, v = 0.5 m3/kg
b. H2O T = 100°C, P = 10 MPa
c. N2 T = 100 K, x = 0.75
d. N2 T = 200 K, P = 200 kPa
e. NH3 T = 100°C, v = 0.1 m3/kg
Solution:
a) Table B.1.1: vf < v < vg => L+V mixture, P = 198.5 kPa,
x = (0.5 - 0.00106)/0.8908 = 0.56,
u = 503.48 + 0.56 × 2025.76 = 1637.9 kJ/kg
b) Table B.1.4: compressed liquid, v = 0.001039 m3/kg, u = 416.1 kJ/kg
c) Table B.6.1: 100 K, x = 0.75
v = 0.001452 + 0.75 × 0.02975 = 0.023765 m3/kg
u = -74.33 + 0.75 ×137.5 = 28.8 kJ/kg
d) Table B.6.2: 200 K, 200 kPa
v = 0.29551 m3/kg ; u = 147.37 kJ/kg
e) Table B.2.1: v > vg => superheated vapor, x = undefined
B.2.2: P = 1600 + 400 ×
0.1 - 0.10539
0.08248-0.10539 = 1694 kPa
States shown are
to each other.
P C.P.
v
T
C.P.
v
T
a d
e
c
b
e
da
c
b
P = const.

5.30
Find the missing properties among (T, P, v, u, h and x if applicable) and indicate the
states in a P-v and a T-v diagram for
a. R-410a P = 500 kPa, h = 300 kJ/kg
b. R-410a T = 10oC, u = 200 kJ/kg
c. R-134a T = 40oC, h = 400 kJ/kg
Solution:
a) Table B.4.1: h > hg = > superheated vapor, look in section 500 kPa and
interpolate
T = 0 + 20 ×
300 − 287.84
306.18 − 287.84
= 20 × 0.66303 = 13.26°C,
v = 0.05651 + 0.66303 × (0.06231-0.05651) = 0.06036 m3/kg,
u = 259.59 + 0.66303 × (275.02 – 259.59) = 269.82 kJ/kg
b) Table B.4.1: u < ug = 255.9 kJ/kg => L+V mixture, P = 1085.7 kPa
x =
u - uf
ufg
=
200 - 72.24
183.66 = 0.6956,
v = 0.000886 + 0.6956 × 0.02295 = 0.01685 m3/kg,
h = 73.21 + 0.6956 × 208.57 = 218.3 kJ/kg
c) Table B.5.1: h < hg => two-phase L + V, look in B.5.1 at 40°C:
x =
h - hf
hfg
=
400 − 256.5
163.3 = 0.87875, P = Psat = 1017 kPa,
v = 0.000 873 + 0.87875 × 0.01915 = 0.0177 m3/kg
u = 255.7 + 0.87875 × 143.8 = 382.1 kJ/kg
States shown are
to each other.
P C.P.
v
T
C.P.
v
Tb, c
a
P = C
b, c
a

5.31
Find the missing properties.
a. H2O T = 250°C, v = 0.02 m3/kg P = ? u = ?
b. N2 T = 120 K, P = 0.8 MPa x = ? h = ?
c. H2O T = −2°C, P = 100 kPa u = ? v = ?
d. R-134a P = 200 kPa, v = 0.12 m3/kg u = ? T = ?
Solution:
a) Table B.1.1 at 250°C: vf < v < vg ⇒ P = Psat = 3973 kPa
x = (v - vf)/ vfg = (0.02 – 0.001251)/0.04887 = 0.38365
u = uf + x ufg = 1080.37 + 0.38365 × 1522.0 = 1664.28 kJ/kg
b) Table B.6.1 P is lower than Psat so it is super heated vapor
=> x = undefined and we find the state in Table B.6.2
Table B.6.2: h = 114.02 kJ/kg
c) Table B.1.1 : T < Ttriple point => B.1.5: P > Psat so compressed solid
u ≅ ui = -337.62 kJ/kg v ≅ vi = 1.09×10-3 m3/kg
approximate compressed solid with saturated solid properties at same T.
d) Table B.5.1 v > vg superheated vapor => Table B.5.2.
T ~ 32.5°C = 30 + (40 – 30) × (0.12 – 0.11889)/(0.12335 - 0.11889)
u = 403.1 + (411.04 – 403.1) × 0.24888 = 405.07 kJ/kg
P
T
v
V
L
S
C.P.
a
b
c d
P C.P.
v
T
C.P.
v
Ta
d
c
b
d
a
c
b
P = C

5.32
Find the missing properties of (P, T, v, u, h and x) and indicate the states in a P-v and
T-v diagram for
a. Water at 5000 kPa, u = 1000 kJ/kg (Table B.1 reference)
b. R-134a at 20oC, u = 300 kJ/kg
c. Nitrogen at 250 K, 200 kPa
Solution:
a) Compressed liquid: B.1.4 interpolate between 220oC and 240oC.
T = 233.3oC, v = 0.001213 m3/kg, x = undefined
b) Table B.5.1: u < ug => two-phase liquid and vapor
x = (u - uf)/ufg = (300 - 227.03)/162.16 = 0.449988 = 0.45
v = 0.000817 + 0.45*0.03524 = 0.01667 m3/kg
c) Table B.6.1: T > Tsat (200 kPa) so superheated vapor in Table B.6.2
x = undefined
v = 0.5(0.35546 + 0.38535) = 0.3704 m3/kg,
u = 0.5(177.23 + 192.14) = 184.7 kJ/kg
States shown are
to each other.
P C.P.
v
T
C.P.
v
T
b c
a
b
c
a
P = const.

5.33
Find the missing properties for CO2 at:
a) 20oC, 2 MPa v = ? and h = ?
b) 10oC, x = 0.5 P = ?, u = ?
c) 1 MPa, v = 0.05 m3/kg, T = ?, h = ?
Solution:
a) Table B.3.1 P < Psat = 5729 kPa so superheated vapor.
Table B.3.2: v = 0.0245 m3/kg, h = 368.42 kJ/kg
b) Table B.3.1 P = Psat = 4502 kPa
u = uf + x ufg = 107.6 + 0.5 × 169.07 = 192.14 kJ/kg
c) Table B.3.1 v > vg ≈ 0.0383 m3/kg so superheated vapor
Table B.3.2: Between 0 and 20oC so interpolate.
T = 0 + 20 ×
0.05 – 0.048
0.0524 – 0.048 = 20 × 0.4545 = 9.09oC
h = 361.14 + (379.63 – 361.14) × 0.4545 = 369.54 kJ/kg

5.34
Saturated liquid water at 20oC is compressed to a higher pressure with constant
temperature. Find the changes in u and h from the initial state when the final pressure
is a) 500 kPa, b) 2000 kPa
Solution:
State 1 is located in Table B.1.1 and the states a-c are from Table B.1.4
State u [kJ/kg] h [kJ/kg] ∆u = u - u1 ∆h = h - h1 ∆(Pv)
1 83.94 83.94
a 83.91 84.41 -0.03 0.47 0.5
b 83.82 85.82 -0.12 1.88 2
For these states u stays nearly constant, dropping slightly as P goes up.
h varies with Pv changes.
v
P
v
T
a
b
1
b,a,1
T = 20 Co
P
T
v
V
L
S
C.P.
1
a
b
cb

Energy Equation: Simple Process

5.35
Saturated vapor R-410a at 0oC in a rigid tank is cooled to -20oC. Find the specific heat
transfer.
Solution:
C.V.: R-410a in tank. m2 = m1 ;
Energy Eq.5.11: (u2 - u1) = 1q2 - 1w2
Process: V = constant, v2 = v1 = V/m => 1w2 = 0/
Table B.4.1: State 1: u1 = 253.0 kJ/kg
State 2: -20oC, v2 = v1 = V/m, look in Table B.4.1 at -20oC
x2 =
v2 - vf2
vfg2
=
0.03267 - 0.000803
0.06400 = 0.4979
u2 = uf2 + x2 ufg2 = 27.92 + x2 ×218.07 = 136.5 kJ/kg
From the energy equation
1q2 = (u2 − u1) = ( 136.5 – 253.0 ) = -116.5 kJ/kg
V
P
1
2

5.36
A 100-L rigid tank contains nitrogen (N2) at 900 K, 3 MPa. The tank is now cooled to
100 K. What are the work and heat transfer for this process?
Solution:
C.V.: Nitrogen in tank. m2 = m1 ;
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: V = constant, v2 = v1 = V/m => 1W2 = 0/
Table B.6.2: State 1: v1 = 0.0900 m3/kg => m = V/v1 = 1.111 kg
u1 = 691.7 kJ/kg
State 2: 100 K, v2 = v1 = V/m, look in Table B.6.2 at 100 K
200 kPa: v = 0.1425 m3/kg; u = 71.7 kJ/kg
400 kPa: v = 0.0681 m3/kg; u = 69.3 kJ/kg
so a linear interpolation gives:
P2 = 200 + 200 (0.09 – 0.1425)/(0.0681 – 0.1425) = 341 kPa
u2 = 71.7 + (69.3 – 71.7)
0.09 – 0.1425
0.0681 – 0.1425 = 70.0 kJ/kg,
1Q2 = m(u2 - u1) = 1.111 kg (70.0 – 691.7) kJ/kg = −690.7 kJ

5.37
Saturated vapor carbon dioxide at 2 MPa in a constant pressure piston cylinder is
heated to 20oC. Find the specific heat transfer.
Solution:
C.V. CO2: m2 = m1 = m;
Energy Eq.5.11 (u2 - u1) = 1q2 - 1w2
Process: P = const. ⇒ 1w2 = ⌡⌠Pdv = P∆v = P(v2 - v1)
State 1: Table B3.2 (or B3.1) h1 = 323.95 kJ/kg
State 2: Table B.3.2 h2 = 368.42 kJ/kg
1q2 = (u2 - u1) + 1w2 = (u2 - u1) + P(v2 - v1) = (h2 - h1)
1q2 = 368.42 – 323.95 = 44.47 kJ/kg
v
P
1 2
v
T
1
2

5.38
Two kg water at 120oC with a quality of 25% has its temperature raised 20oC in a
constant volume process as in Fig. P5.38. What are the heat transfer and work in the
process?
Solution:
C.V. Water. This is a control mass
Energy Eq.: m (u2 − u1 ) = 1Q2 − 1W2
Process : V = constant
1W2 = ∫ P dV = 0
State 1: T, x1 from Table B.1.1
v1 = vf + x1 vfg = 0.00106 + 0.25 × 0.8908 = 0.22376 m3/kg
u1 = uf + x1 ufg = 503.48 + 0.25 × 2025.76 = 1009.92 kJ/kg
State 2: T2, v2 = v1< vg2 = 0.50885 m3/kg so two-phase
x2 =
v2 - vf2
vfg2
=
0.22376 - 0.00108
0.50777 = 0.43855
u2 = uf2 + x2 ufg2 = 588.72 + x2 ×1961.3 = 1448.84 kJ/kg
1Q2 = m(u2 − u1) = 2 ( 1448.84 – 1009.92 ) = 877.8 kJ
T
C.P.
v
P C.P.
v
T
198.5
361.3
120
140
120 C
140 C

5.39
Ammonia at 0°C, quality 60% is contained in a rigid 200-L tank. The tank and
ammonia is now heated to a final pressure of 1 MPa. Determine the heat transfer for
the process.
Solution:
C.V.: NH3
V
P
1
2
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: Constant volume ⇒ v2 = v1 & 1W2 = 0
State 1: Table B.2.1 two-phase state.
v1 = 0.001566 + x1 × 0.28783 = 0.17426 m3/kg
u1 = 179.69 + 0.6 × 1138.3 = 862.67 kJ/kg
m = V/v1 = 0.2/0.17426 = 1.148 kg
State 2: P2 , v2 = v1 superheated vapor Table B.2.2
⇒ T2 ≅ 100°C, u2 ≅ 1490.5 kJ/kg
So solve for heat transfer in the energy equation
1Q2 = m(u2 - u1) = 1.148(1490.5 - 862.67) = 720.75 kJ

5.40
A test cylinder with constant volume of 0.1 L contains water at the critical point. It
now cools down to room temperature of 20°C. Calculate the heat transfer from the
water.
Solution:
C.V.: Water
m2 = m1 = m ;
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: Constant volume ⇒ v2 = v1
Properties from Table B.1.1
State 1: v1 = vc = 0.003155 m3/kg,
u1 = 2029.6 kJ/kg
m = V/v1 = 0.0317 kg
v
P
1
2
State 2: T2, v2 = v1 = 0.001002 + x2 × 57.79
x2 = 3.7×10-5, u2 = 83.95 + x2 × 2319 = 84.04 kJ/kg
Constant volume => 1W2 = 0/
1Q2 = m(u2 - u1) = 0.0317(84.04 - 2029.6) = -61.7 kJ

5.41
A rigid tank holds 0.75 kg ammonia at 70°C as saturated vapor. The tank is now
cooled to 20°C by heat transfer to the ambient. Which two properties determine the
final state. Determine the amount of work and heat transfer during the process.
C.V. The ammonia, this is a control mass.
Process: Rigid tank V = C => v = constant & 1W2 = ⌡⌠
1
2
PdV = 0
Energy Eq.: U2 − U1 = 1Q2 − 1W2 = 1Q2,
State 1: v1 = 0.03787 m3/kg,
u1 = 1338.9 kJ/kg
State 2: T, v => two-phase (straight down in
P-v diagram from state 1)
P
70 C
1
3312
858
2
V
x2 = (v – vf)/vfg = (0.03787 – 0.001638)/0.14758 = 0.2455
u2 = uf + x2 ufg = 272.89 + 0.2455 × 1059.3 = 532.95 kJ/kg
1Q2 = m(u2 − u1) = 0.75 (532.95 − 1338.9) = -604.5 kJ

5.42
A cylinder fitted with a frictionless piston contains 2 kg of superheated refrigerant R-
134a vapor at 350 kPa, 100o
C. The cylinder is now cooled so the R-134a remains at
constant pressure until it reaches a quality of 75%. Calculate the heat transfer in the
process.
Solution:
C.V.: R-134a m2 = m1 = m;
Energy Eq.5.11 m(u2 - u1) = 1Q2 - 1W2
Process: P = const. ⇒ 1W2 = ⌡⌠PdV = P∆V = P(V2 - V1) = Pm(v2 - v1)
V
P
12
V
T
1
2
State 1: Table B.5.2 h1 = (490.48 + 489.52)/2 = 490 kJ/kg
State 2: Table B.5.1 h2 = 206.75 + 0.75 ×194.57 = 352.7 kJ/kg (350.9 kPa)
1Q2 = m(u2 - u1) + 1W2 = m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1)
1Q2 = 2 × (352.7 - 490) = -274.6 kJ

5.43
Water in a 150-L closed, rigid tank is at 100°C, 90% quality. The tank is then cooled
to −10°C. Calculate the heat transfer during the process.
Solution:
C.V.: Water in tank. m2 = m1 ;
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: V = constant, v2 = v1, 1W2 = 0
State 1: Two-phase L + V look in Table B.1.1
v1 = 0.001044 + 0.9 × 1.6719 = 1.5057 m3/kg
u1 = 418.94 + 0.9 × 2087.6 = 2297.8 kJ/kg
State 2: T2, v2 = v1 ⇒ mix of saturated solid + vapor Table B.1.5
v2 = 1.5057 = 0.0010891 + x2 × 466.7 => x2 = 0.003224
u2 = -354.09 + 0.003224 × 2715.5 = -345.34 kJ/kg
m = V/v1 = 0.15/1.5057 = 0.09962 kg
1Q2 = m(u2 - u1) = 0.09962(-345.34 - 2297.8) = -263.3 kJ
P C.P.
v
T
C.P.
v
T
1
1
P = const.
2 2
P
T
v
S + V
L + V
V
L
S
C.P.
1
2

5.44
A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m3. Stops
in the cylinder are placed to restrict the enclosed volume to a maximum of 0.5 m3. The
water is now heated until the piston reaches the stops. Find the necessary heat transfer.
Solution:
C.V. H2O m = constant
Energy Eq.5.11: m(e2 – e1) = m(u2 – u1) = 1Q2 - 1W2
Process : P = constant (forces on piston constant)
⇒ 1W2 = ∫ P dV = P1 (V2 – V1)
Properties from Table B.1.1
State 1: v1 = 0.1/50 = 0.002 m3/kg => 2-phase as v1 < vg
x1 =
v1 – vf
vfg
=
0.002 – 0.001061
0.88467 = 0.001061
h1 = 504.68 + 0.001061 × 2201.96 = 507.02 kJ/kg
State 2: v2= 0.5/50 = 0.01 m3/kg also 2-phase same P
x2 =
v2 – vf
vfg
=
0.01 – 0.001061
0.88467 = 0.01010
h2 = 504.68 + 0.01010 × 2201.96 = 526.92 kJ/kg
Find the heat transfer from the energy equation as
1Q2 = m(u2 – u1) + 1W2 = m(h2 – h1)
1Q2 = 50 kg × (526.92 – 507.02) kJ/kg = 995 kJ
[ Notice that 1W2 = P1 (V2 – V1) = 200 × (0.5 – 0.1) = 80 kJ ]
V
P
1 2
0.1 0.5

5.45
Find the heat transfer for the process in Problem 4.33
Take as CV the 1.5 kg of water. m2 = m1 = m ;
Energy Eq.5.11 m(u2 − u1) = 1Q2 − 1W2
State 1: (P, T) => v1 = 0.95964 m3/kg,
u1 = 2576.87 kJ/kg
P C.P.
2
600
1
T200
v
State 2: (P, T) => v2 = 0.47424 m3/kg, u2 = 2881.12 kJ/kg
From process eq.: 1W2 = ∫ P dV = area =
m
2 (P1 + P2)(v2 − v1)
=
1.5
2 kg (200 + 600) kPa (0.47424 – 0.95964) m3/kg
= −291.24 kJ
From energy eq.: 1Q2 = m(u2 − u1) + 1W2 = 1.5(2881.12 – 2576.87) – 291.24
= 165.14 kJ

5.46
A 10-L rigid tank contains R-410a at −10°C, 80% quality. A 10-A electric current
(from a 6-V battery) is passed through a resistor inside the tank for 10 min, after
which the R-410a temperature is 40°C. What was the heat transfer to or from the tank
during this process?
Solution:
C.V. R-410a in tank. Control mass at constant V.
Energy Eq.: m(u2 - u1) = 1Q2 - 1W2
Process: Constant V ⇒ v2 = v1
=> no boundary work, but electrical work
v
P
1
2
State 1 from table B.4.1
v1 = 0.000827 + 0.8 × 0.04470 = 0.03659 m3/kg
u1 = 42.32 + 0.8 × 207.36 = 208.21 kJ/kg
m = V/v = 0.010/0.03659 = 0.2733 kg
State 2: Table B.4.2 at 40°C and v2 = v1 = 0.03659 m3/kg
=> superheated vapor, so use linear interpolation to get
P2 = 800 + 200 × (0.03659 – 0.04074)/(0.03170 – 0.04074)
= 800 + 200 × 0.45907 = 892 kPa
u2 = 286.83 + 0.45907× (284.35 – 286.83) = 285.69 kJ/kg
1W2 elec = –power × ∆t = –Amp × volts × ∆t = –
10 × 6 × 10 × 60
1000 = –36 kJ
1Q2 = m(u2 – u1) + 1W2 = 0.2733 ( 285.69 – 208.21) – 36 = –14.8 kJ

5.47
A piston/cylinder contains 1 kg water at 20oC with volume 0.1 m3. By mistake
someone locks the piston preventing it from moving while we heat the water to
saturated vapor. Find the final temperature and the amount of heat transfer in the
process.
Solution:
Energy Eq.: m (u2 − u1 ) = 1Q2 − 1W2
Process : V = constant 1W2 = 0
State 1: T, v1 = V1/m = 0.1 m3/kg > vf so two-phase
x1 =
v1 - vf
vfg
=
0.1-0.001002
57.7887 = 0.0017131
u1 = uf + x1 ufg = 83.94 + x1 × 2318.98 = 87.913 kJ/kg
State 2: v2 = v1 = 0.1 & x2 =1
found in Table B.1.1 between 210°C and 215° C
T2 = 210 + 5 ×
0.1-0.10441
0.09479-0.10441 = 210 + 5 × 0.4584 = 212.3°C
u2 = 2599.44 + 0.4584 (2601.06 – 2599.44) = 2600.2 kJ/kg
1Q2 = m(u2 − u1) = 1( 2600.2 – 87.913) = 2512.3 kJ
P T
2
2
1
1
VV

5.48
A piston cylinder contains 1.5 kg water at 600 kPa, 350oC. It is now cooled in a
process where pressure is linearly related to volume to a state of 200 kPa, 150oC. Plot
the P-v diagram for the process and find both the work and the heat transfer in the
process.
Take as CV the 1.5 kg of water.
m2 = m1 = m ;
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
State 1: (P, T) => v1 = 0.47424 m3/kg,
u1 = 2881.12 kJ/kg
P C.P.
1
600
2
200 T
v
State 2: (P, T) => v2 = 0.95964 m3/kg, u2 = 2576.87 kJ/kg
From process eq.: 1W2 = ∫ P dV = area =
m
2 (P1 + P2)(v2 − v1)
=
1.5
2 kg (200 + 600) kPa (0.95964 – 0.47424) m3/kg
= 291.24 kJ
From energy eq.: 1Q2 = m(u2 − u1) + 1W2 = 1.5(2576.87 – 2881.12) + 291.24
= -165.14 kJ

5.49
Two kg water at 200 kPa with a quality of 25% has its temperature raised 20oC in a
constant pressure process. What are the heat transfer and work in the process?
Energy Eq.: m (u2 − u1 ) = 1Q2 − 1W2
Process : P = constant 1W2 = ∫ P dV = mP (v2 − v1)
State 1: Two-phase given P,x so use Table B.1.2
v1 = 0.001061 + 0.25 × 0.88467 = 0.22223 m3/kg
u1 = 504047 + 0.25 × 2025.02 = 1010.725 kJ/kg
T = T + 20 = 120.23 + 20 = 140.23
State 2 is superheated vapor
v2 = 0.88573 +
20
150-120.23 × (0.95964 – 0.88573 ) = 0.9354 m3/kg
u2 = 2529.49 +
20
150-120.23 (2576.87- 2529.49) = 2561.32 kJ/kg
From the process equation we get
1W2 = mP (v2 − v1) = 2 × 200 ( 0.9354 - 0.22223) = 285.3 kJ
1Q2 = m (u2 − u1) + 1W2
= 2 ( 2561.32 – 1010.725 ) + 285.3
= 3101.2 + 285.27 = 3386.5 kJ
V
P
1 2
V
T
1
2

5.50
A water-filled reactor with volume of 1 m3 is at 20 MPa, 360°C and placed inside a
containment room as shown in Fig. P5.50. The room is well insulated and initially
evacuated. Due to a failure, the reactor ruptures and the water fills the containment
room. Find the minimum room volume so the final pressure does not exceed 200 kPa.
Solution:
C.V.: Containment room and reactor.
Mass: m2 = m1 = Vreactor/v1 = 1/0.001823 = 548.5 kg
Energy: m(u2 - u1) = 1Q2 - 1W2 = 0 - 0 = 0
State 1: Table B.1.4 v1 = 0.001823 m3/kg, u1 = 1702.8 kJ/kg
Energy equation then gives u2 = u1 = 1702.8 kJ/kg
State 2: P2 = 200 kPa, u2 < ug => Two-phase Table B.1.2
x2 = (u2 - uf)/ ufg = (1702.8 – 504.47)/2025.02 = 0.59176
v2 = 0.001061 + 0.59176 × 0.88467 = 0.52457 m3/kg
V2 = m2 v2 = 548.5 ×0.52457 = 287.7 m3
P
v
1
T
v
1
2
200 kPa
200
2
u = const
P
T
v
L
C.P.1
2
200 kPa

5.51
A 25 kg mass moves with 25 m/s. Now a brake system brings the mass to a complete
stop with a constant deceleration over a period of 5 seconds. The brake energy is
absorbed by 0.5 kg water initially at 20oC, 100 kPa. Assume the mass is at constant P
and T. Find the energy the brake removes from the mass and the temperature increase
of the water, assuming P = C.
Solution:
C.V. The mass in motion.
E2 - E1= ∆E = 0.5 mV
2
= 0.5 × 25 × 25
2
/1000 = 7.8125 kJ
C.V. The mass of water.
m(u2 - u1) H2O = ∆E = 7.8125 kJ => u2 - u1 = 7.8125 / 0.5 = 15.63 kJ/kg
u2 = u1 + 15.63 = 83.94 + 15.63 = 99.565 kJ/kg
Assume u2 = uf then from Table B.1.1: T2 ≅ 23.7oC, ∆T = 3.7oC
We could have used u2 - u1 = C∆T with C from Table A.4: C = 4.18 kJ/kg K
giving ∆T = 15.63/4.18 = 3.7oC.

5.52
Find the heat transfer for the process in Problem 4.41
Solution:
Energy Eq.: m(u2 – u1) = 1Q2 – 1W2
Process: P = A + BV (linear in V)
State 1: Superheated vapor v1 = 0.6193 m3/kg, u1 = 1316.7 kJ/kg
State 2: Superheated vapor v2 = 0.63276 m3/kg, u2 = 1542.0 kJ/kg
1W2 = ∫ P dV = area = Pavg (V2 − V1) =
1
2 (P1 + P2)m(v2 − v1)
= ½(200 + 300) kPa × 0.5 kg (0.63276 − 0.6193) m3/kg = 1.683 kJ
Heat transfer is found from the energy equation
1Q2 = m(u2 – u1) + 1W2 = 0.5 kg (1542.0 – 1316.7) kJ/kg + 1.683 kJ
= 112.65 + 1.683 = 114.3 kJ
P
2P2
1
P
1
v

5.53
A piston/cylinder arrangement has the piston loaded with outside atmospheric
pressure and the piston mass to a pressure of 150 kPa, shown in Fig. P5.53. It contains
water at −2°C, which is then heated until the water becomes saturated vapor. Find the
final temperature and specific work and heat transfer for the process.
Solution:
C.V. Water in the piston cylinder.
Continuity: m2 = m1,
Energy Eq. per unit mass: u2 - u1 = 1q2 - 1w2
Process: P = constant = P1, => 1w2 = ⌡⌠
1
2
P dv = P1(v2 - v1)
State 1: T1 , P1 => Table B.1.5 compressed solid, take as saturated solid.
v1 = 1.09×10-3 m3/kg, u1 = -337.62 kJ/kg
State 2: x = 1, P2 = P1 = 150 kPa due to process => Table B.1.2
v2 = vg(P2) = 1.1593 m3/kg, T2 = 111.4°C ; u2 = 2519.7 kJ/kg
1w2 = P1(v2 -v1) = 150(1.1593 -1.09×10-3) = 173.7 kJ/kg
1q2 = u2 - u1 + 1w2 = 2519.7 - (-337.62) + 173.7 = 3031 kJ/kg
P
T
v
V
L
S
C.P.
21
S+V
L+V
P C.P.
v
T
C.P.
v
1
2
1
2
P = C

5.54
A constant pressure piston/cylinder assembly contains 0.2 kg water as saturated vapor
at 400 kPa. It is now cooled so the water occupies half the original volume. Find the
heat transfer in the process.
Solution:
C.V. Water. This is a control mass.
Energy Eq.5.11: m(u2 – u1) = 1Q2 – 1W2
Process: P = constant => 1W2 = Pm(v2 – v1)
So solve for the heat transfer:
1Q2 = m(u2 - u1) + 1W2 = m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1)
State 1: Table B.1.2 v1 = 0.46246 m3/kg; h1 = 2738.53 kJ/kg
State 2: v2 = v1 / 2 = 0.23123 = vf + x vfg from Table B.1.2
x2 = (v2 – vf) / vfg = (0.23123 – 0.001084) / 0.46138 = 0.4988
h2 = hf + x2 hfg = 604.73 + 0.4988 × 2133.81 = 1669.07 kJ/kg
Now the heat transfer becomes
1Q2 = 0.2 (1669.07 – 2738.53) = –213.9 kJ
T
C.P.
v
12
P = 400 kPa
P C.P.
v
T
400
144
2 1
cb

5.55
A cylinder having a piston restrained by a linear spring (of spring constant 15 kN/m)
contains 0.5 kg of saturated vapor water at 120°C, as shown in Fig. P5.55. Heat is
transferred to the water, causing the piston to rise. If the piston cross-sectional area is
0.05 m2, and the pressure varies linearly with volume until a final pressure of 500 kPa
is reached. Find the final temperature in the cylinder and the heat transfer for the
process.
Solution:
C.V. Water in cylinder.
Continuity: m2 = m1 = m ;
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
State 1: (T, x) Table B.1.1 => v1 = 0.89186 m3/kg, u1 = 2529.2 kJ/kg
Process: P2 = P1 +
ksm
Ap
2 (v2 - v1) = 198.5 +
15 × 0.5
(0.05)2 (v2 - 0.89186)
State 2: P2 = 500 kPa and on the process curve (see above equation).
=> v2 = 0.89186 + (500 - 198.5) × (0.052/7.5) = 0.9924 m3/kg
(P, v) Table B.1.3 => T2 = 803°C; u2 = 3668 kJ/kg
The process equation allows us to evaluate the work
1W2 = ⌡⌠ PdV =





P1 + P2
2 m(v2 - v1)
=





198.5 + 500
2 × 0.5 × (0.9924 - 0.89186) = 17.56 kJ
Substitute the work into the energy equation and solve for the heat transfer
1Q2 = m(u2 - u1) + 1W2 = 0.5 × (3668 - 2529.2) + 17.56 = 587 kJ
P
v
1
2
T
v
1
2k ms
Ap
2

5.56
A piston cylinder arrangement with a linear spring similar to Fig. P5.55 contains R-
134a at 15oC, x = 0.6 and a volume of 0.02 m3. It is heated to 60oC at which point the
specific volume is 0.03002 m3/kg. Find the final pressure, the work and the heat
transfer in the process.
Take CV as the R-134a.
m2 = m1 = m ; m(u2 − u1) = 1Q2 - 1W2
State 1: T1, x1 => Two phase so Table B.5.1: P1 = Psat = 489.5 kPa
v1 = vf + x1 vfg = 0.000805 + 0.6 × 0.04133 = 0.0256 m3/kg
u1 = uf + x1 ufg = 220.1 + 0.6 × 166.35 = 319.91 kJ/kg
m = V1/v1 = 0.02 m3/ 0.0256 m3/kg = 0.78125 kg
State 2: (T, v) Superheated vapor, Table B.5.2.
P2 = 800 kPa, v2 = 0.03002 m3/kg, u2 = 421.2 kJ/kg
V2 = m v2 = 0.78125 × 0.03002 = 0.02345 m3
Work is done while piston moves at linearly varying pressure, so we get
1W2 = ∫ P dV = area = Pavg (V2 − V1) = 0.5(P2 + P1)( V2 − V1)
= 0.5 × (489.5 + 800) kPa (0.02345 − 0.02) m3 = 2.22 kJ
Heat transfer is found from the energy equation
1Q2 = m(u2 − u1) + 1W2 = 0.78125 × (421.2 – 319.91) + 2.22 = 81.36 kJ
P
2
P2
1
P
1
R-134a vcb

5.57
A closed steel bottle contains CO2 at −20°C, x = 20% and the volume is 0.05 m3. It
has a safety valve that opens at a pressure of 6 MPa. By accident, the bottle is heated
until the safety valve opens. Find the temperature and heat transfer when the valve
first opens.
Solution:
C.V.: CO2 : m2 = m1 = m ;
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: constant volume process ⇒ 1W2 = 0
State 1: (T, x) Table B.3.1
v1 = 0.000969 + 0.2 × 0.01837 = 0.004643 m3/kg
=> m = V/v1 = 0.05/0.004643 = 10.769 kg
u1 = 39.64 + 0.2 × 246.25 = 88.89 kJ/kg
P
2
1
V
State 2: P2 , v2 = v1 => very close to saturated vapor, use 6003 kPa in Table
B.3.1: T ≅ 22°C, x2 = (0.004643 – 0.001332)/0.00341 = 0.971
u2 = 142.03 + 0.971 × 119.89 = 258.44 kJ/kg
1Q2 = m(u2 - u1) = 10.769 (258.44 – 88.89) = 1825.9 kJ

5.58
Superheated refrigerant R-134a at 20°C, 0.5 MPa is cooled in a piston/cylinder
arrangement at constant temperature to a final two-phase state with quality of 50%.
The refrigerant mass is 5 kg, and during this process 500 kJ of heat is removed. Find
the initial and final volumes and the necessary work.
Solution:
C.V. R-134a, this is a control mass.
Continuity: m2 = m1 = m ;
Energy Eq.5.11: m(u2 -u1) = 1Q2 - 1W2 = -500 - 1W2
State 1: T1 , P1 Table B.5.2, v1 = 0.04226 m3/kg ; u1 = 390.52 kJ/kg
=> V1 = mv1 = 0.211 m3
State 2: T2 , x2 ⇒ Table B.5.1
u2 = 227.03 + 0.5 × 162.16 = 308.11 kJ/kg,
v2 = 0.000817 + 0.5 × 0.03524 = 0.018437 m3/kg => V2 = mv2 = 0.0922 m3
1W2 = -500 - m(u2 - u1) = -500 - 5 × (308.11 - 390.52) = -87.9 kJ
v
P
v
T
1
2 12

5.59
A 1-L capsule of water at 700 kPa, 150°C is placed in a larger insulated and otherwise
evacuated vessel. The capsule breaks and its contents fill the entire volume. If the final
pressure should not exceed 125 kPa, what should the vessel volume be?
Solution:
C.V. Larger vessel.
Continuity: m2 = m1 = m = V/v1 = 0.916 kg
Process: expansion with 1Q2 = 0/ , 1W2 = 0/
Energy: m(u2 - u1) = 1Q2 - 1W2 = 0/ ⇒ u2 = u1
State 1: v1 ≅ vf = 0.001091 m3/kg; u1 ≅ uf = 631.66 kJ/kg
State 2: P2 , u2 ⇒ x2 =
631.66 – 444.16
2069.3 = 0.09061
v2 = 0.001048 + 0.09061 × 1.37385 = 0.1255 m3/kg
V2 = mv2 = 0.916 × 0.1255 = 0.115 m3 = 115 L
P
v
1
T
v
1
2
200 kPa
200
2
u = const
P
T
v
L
C.P.1
2
200 kPa

5.60
A piston cylinder contains carbon dioxide at -20oC and quality 75%. It is compressed
in a process where pressure is linear in volume to a state of 3 MPa and 20oC. Find the
specific heat transfer.
CV Carbon dioxide out to the source, both 1Q2 and 1W2
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
Process: P = A + BV => 1W2 = ∫ P dV = ½ m(P1 + P2 ) (v2 - v1)
State 1: Table B.3.1 P = 1969.6 kPa
v1 = 0.000969 + 0.75 × 0.01837 = 0.01475 m3/kg,
u1 = 39.64 + 0.75 × 246.25 = 224.33 kJ/kg,
State 2: Table B.3 v2 = 0.01512 m3/kg, u2 = 310.21 kJ/kg,
1w2 = ½ (P1 + P2 )( v2 - v1) = ½ ×(1969.6 + 3000)( 0.01512 – 0.01475)
= 0.92 kJ/kg
1q2 = u2 − u1 + 1w2 = 310.21 – 224.33 + 0.92 = 86.8 kJ/kg
P
v
1
2

5.61
A rigid tank is divided into two rooms by a membrane, both containing water, shown
in Fig. P5.61. Room A is at 200 kPa, v = 0.5 m3/kg, VA = 1 m3, and room B contains
3.5 kg at 0.5 MPa, 400°C. The membrane now ruptures and heat transfer takes place
so the water comes to a uniform state at 100°C. Find the heat transfer during the
process.
Solution:
C.V.: Both rooms A and B in tank. BA
Continuity Eq.: m2 = mA1 + mB1 ;
Energy Eq.: m2u2 - mA1uA1 - mB1uB1 = 1Q2 - 1W2
State 1A: (P, v) Table B.1.2, mA1 = VA/vA1 = 1/0.5 = 2 kg
xA1 =
v – vf
vfg
=
0.5 - 0.001061
0.88467 = 0.564
uA1 = uf + x ufg = 504.47 + 0.564 × 2025.02 = 1646.6 kJ/kg
State 1B: Table B.1.3, vB1 = 0.6173, uB1 = 2963.2, VB = mB1vB1 = 2.16 m3
Process constant total volume: Vtot = VA + VB = 3.16 m3 and 1W2 = 0/
m2 = mA1 + mB1 = 5.5 kg => v2 = Vtot/m2 = 0.5746 m3/kg
State 2: T2 , v2 ⇒ Table B.1.1 two-phase as v2 < vg
x2 =
v2 – vf
vfg
=
0.5746 – 0.001044
1.67185 = 0.343 ,
u2 = uf + x ufg = 418.91 + 0.343 × 2087.58= 1134.95 kJ/kg
Heat transfer is from the energy equation
1Q2 = m2u2 - mA1uA1 - mB1uB1 = -7421 kJ

5.62
Two kilograms of nitrogen at 100 K, x = 0.5 is heated in a constant pressure process to
300 K in a piston/cylinder arrangement. Find the initial and final volumes and the total
heat transfer required.
Solution:
Take CV as the nitrogen.
Energy Eq.5.11: m(u2 − u1) = 1Q2 - 1W2
Process: P = constant ⇒ 1W2 = ⌡⌠PdV = Pm(v2 - v1)
State 1: Table B.6.1
v1 = 0.001452 + 0.5 × 0.02975 = 0.01633 m3/kg, V1 = 0.0327 m3
h1 = -73.20 + 0.5 × 160.68 = 7.14 kJ/kg
State 2: (P = 779.2 kPa , 300 K) => sup. vapor interpolate in Table B.6.2
v2 = 0.14824 + (0.11115-0.14824)× 179.2/200 = 0.115 m3/kg, V2 = 0.23 m3
h2 = 310.06 + (309.62-310.06) × 179.2/200 = 309.66 kJ/kg
Now solve for the heat transfer from the energy equation
1Q2 = m(u2 - u1) + 1W2 = m(h2 - h1) = 2 × (309.66 - 7.14) = 605 kJ
V
P
1 2
V
T
1
2

5.63
Water in a tank A is at 250 kPa with a quality of 10% and mass 0.5 kg. It is connected
to a piston cylinder holding constant pressure of 200 kPa initially with 0.5 kg water at
400°C. The valve is opened and enough heat transfer takes place to have a final
uniform temperature of 150°C. Find the final P and V, the process work and the
process heat transfer.
C.V. Water in A and B. Control mass goes through process: 1 -> 2
Continuity Eq.: m2 – mA1 – mB1 = 0 => m2 = mA1 + mB1 = 0.5 + 0.5 = 1 kg
Energy Eq.: U2 − U1 = 1Q2 − 1W2
State A1: vA1 = 0.001067 + xA1 × 0.71765 = 0.072832; VA1 = mv = 0.036416 m3
uA1 = 535.08 + 0.1 × 2002.14 = 735.22 kJ/kg
State B1: vB1 = 1.5493 m3/kg; uB1 = 2966.69 kJ/kg
=> VB1 = mB1vB1 = 0.77465 m3
State 2: If V2 > VA1 then P2 = 200 kPa that is the piston floats.
For (T2, P2) = (150°C, 200 kPa) => superheated vapor u2 = 2576.87 kJ/kg
v2 = 0.95964 m3/kg V2 = m2v2 = 0.95964 m3 > VA1 checks OK.
The possible state 2 (P,V)
combinations are shown. State a
is
200 kPa, va =
VA1
m2
= 0.036 m3/kg
and thus two-phase
Ta = 120°C < T2
P
150 C
2467
200 a
V
Process: 1W2 = P2 (V2 – V1) = 200 (0.95964 – 0.77465 – 0.036416) = 29.715 kJ
From the energy Eq.:
1Q2 = m2u2 – mA1uA1 – mB1uB1 + 1W21Q2 = m2u2 – mA1uA1 – mB1uB1 + 1W2
= 1 × 2576.87 – 0.5 × 735.222 – 0.5 × 2966.69 + 29.715= 1 × 2576.87 – 0.5 × 735.222 – 0.5 × 2966.69 + 29.715
= 755.63 kJ= 755.63 kJ

5.64
A 10-m high open cylinder, Acyl = 0.1 m2, contains 20°C water above and 2 kg of
20°C water below a 198.5-kg thin insulated floating piston, shown in Fig. P5.64.
Assume standard g, Po. Now heat is added to the water below the piston so that it
expands, pushing the piston up, causing the water on top to spill over the edge. This
process continues until the piston reaches the top of the cylinder. Find the final state of
the water below the piston (T, P, v) and the heat added during the process.
Solution:
C.V. Water below the piston.
Piston force balance at initial state: F↑ = F↓ = PAA = mpg + mBg + P0A
State 1A,B: Comp. Liq. ⇒ v ≅ vf = 0.001002 m3/kg; u1A = 83.95 kJ/kg
VA1 = mAvA1 = 0.002 m3; mtot = Vtot/v = 1/0.001002 = 998 kg
mass above the piston mB1 = mtot - mA = 996 kg
PA1 = P0 + (mp + mB)g/A = 101.325 +
(198.5 + 996) × 9.807
0.1 × 1000
= 218.5 kPa
State 2A: PA2 = P0 +
mpg
A = 120.8 kPa ; vA2 = Vtot/ mA= 0.5 m3/kg
xA2 = (0.5 - 0.001047)/1.4183 = 0.352 ; T2 = 105°C
uA2 = 440.0 + 0.352 × 2072.34 = 1169.5 kJ/kg
Continuity eq. in A: mA2 = mA1
Energy: mA(u2 - u1) = 1Q2 - 1W2
Process: P linear in V as mB is linear with V
1W2 = ⌡⌠PdV =
1
2(218.5 + 120.82)(1 - 0.002)
= 169.32 kJ
P
1
2
V
W
cb
1Q2 = mA(u2 - u1) + 1W2 = 2170.1 + 169.3 = 2340.4 kJ

5.65
Assume the same setup as in Problem 5.50, but the room has a volume of 100 m3.
Show that the final state is two-phase and find the final pressure by trial and error.
Mass: m2 = m1 = Vreactor/v1 = 1/0.001823 = 548.5 kg
Energy: m(u2 - u1) = 1Q2 - 1W2 = 0 - 0 = 0 ⇒ u2 = u1 = 1702.8 kJ/kg
Total volume and mass => v2 = Vroom/m2 = 0.1823 m3/kg
State 2: u2 , v2 Table B.1.1 see Figure.
Note that in the vicinity of v = 0.1823 m3/kg crossing the saturated vapor line the
internal energy is about 2585 kJ/kg. However, at the actual state 2, u = 1702.8
kJ/kg. Therefore state 2 must be in the two-phase region.
Trial & error v = vf + xvfg ; u = uf + xufg
⇒ u2 = 1702.8 = uf +
v2 - vf
vfg
ufg
Compute RHS for a guessed pressure P2:
v
T
0.184
u=2585
1060 kPa
sat vap
1060 kPa
P2 = 600 kPa: RHS = 669.88 +
0.1823-0.001101
0.31457 × 1897.52 = 1762.9 too large
P2 = 550 kPa: RHS = 655.30 +
0.1823-0.001097
0.34159 × 1909.17 = 1668.1 too small
Linear interpolation to match u = 1702.8 gives P2 ≅ 568.5 kPa

5.66
A piston cylinder has a water volume separated in VA = 0.2 m3 and VB = 0.3 m3 by a
stiff membrane. The initial state in A is 1000 kPa, x = 0.75 and in B it is 1600 kPa and
250°C. Now the membrane ruptures and the water comes to a uniform state at 200°C.
What is the final pressure? Find the work and the heat transfer in the process.
Take the water in A and B as CV.
Continuity: m2 - m1A - m1B = 0
Energy: m2u2 - m1Au1A - m1Bu1B = 1Q2 - 1W2
Process: P2 = Peq = constant = P1A as piston floats and mp, Po do not change
State 1A: Two phase. Table B.1.2
v1A = 0.001127 + 0.75 × 0.19332 = 0.146117 m3/kg,
u1A = 761.67 + 0.75 × 1821.97 = 2128.15 kJ/kg
State 1B: Table B.1.3 v1B = 0.14184 m3/kg, u1B = 2692.26 kJ/kg
=> m1A = V1A/v1A = 1.3688 kg, m1B = V1B/v1B = 2.115 kg
State 2: 1000 kPa, 200oC sup. vapor => v2 = 0.20596 m3/kg, u2 = 2621.9 kJ/kg
m2 = m1A + m1B = 3.4838 kg => V2 = m2v2 = 3.4838 × 0.20596 = 0.7175 m3
So now
1W2 = ∫ P dV = Peq (V2 - V1) = 1000 (0.7175 - 0.5) = 217.5 kJ
1Q2 = m2u2 - m1Au1A - m1Bu1B + 1W2
= 3.4838 × 2621.9 - 1.3688 × 2128.15 - 2.115 × 2692.26 + 217.5 = 744 kJ
Po
pmcb
A:H2O
g
B:H2O

5.67
Two rigid tanks are filled with water. Tank A is 0.2 m3 at 100 kPa, 150oC and tank B
is 0.3 m3 at saturated vapor 300 kPa. The tanks are connected by a pipe with a closed
valve. We open the valve and let all the water come to a single uniform state while we
transfer enough heat to have a final pressure of 300 kPa. Give the two property values
that determine the final state and find the heat transfer.
State A1: u = 2582.75 kJ/kg, v = 1.93636 m3/kg
=> mA1 = V/v = 0.2/1.93636 = 0.1033 kg
State B1: u = 2543.55 kJ/kg, v = 0.60582 m3/kg
=> mB1 = V/v = 0.3 / 0.60582 = 0.4952 kg
The total volume (and mass) is the sum of volumes (mass) for tanks A and B.
m2 = mA1 + mB1 = 0.1033 + 0.4952 = 0.5985 kg,
V2 = VA1 + VB1 = 0.2 + 0.3 = 0.5 m3
=> v2 = V2/m2 = 0.5 /0.5985 = 0.8354 m3/kg
State 2: [P2, v2] = [300 kPa, 0.8354 m3/kg]
=> T2 = 274.76°C and u2 = 2767.32 kJ/kg
The energy equation is (neglecting kinetic and potential energy)
m2 u2 − mAuA1 − mBuB1 = 1Q2 – 1W2 = 1Q2
1Q2 = 0.5985 × 2767.32 – 0.1033 × 2582.75 – 0.4952 × 2543.55
= 129.9 kJ
B A

Energy Equation: Multistep Solution

5.68
A piston cylinder shown in Fig. P5.68 contains 0.5 m3 of R-410a at 2 MPa, 150oC.
The piston mass and atmosphere gives a pressure of 450 kPa that will float the piston.
The whole setup cools in a freezer maintained at -20oC. Find the heat transfer and
show the P-v diagram for the process when T2 = -20oC.
C.V.: R-410a. Control mass.
Continuity: m = constant,
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: F↓ = F↑ = P A = PairA + Fstop
if V < Vstop ⇒ Fstop = 0/
This is illustrated in the P-v diagram shown below. R-410a
Po
State 1: v1 = 0.02247 m3/kg, u1 = 373.49 kJ/kg
⇒ m = V/v = 22.252 kg
State 2: T2 and on line ⇒ compressed liquid, see figure below.
v2 ≅ vf = 0.000 803 m3/kg ⇒ V2 = 0.01787 m3; u2 = uf = 27.92 kJ/kg
1W2 = ⌡⌠PdV = Plift(V2 - V1) = 450 (0.01787 - 0.5) = -217.0 kJ ;
Energy eq. ⇒
1Q2 = 22.252 (27.92 – 373.49) – 217.9 = -7906.6 kJ
2 MPa
450 kPa
P = 2 MPa
P = 450 kPa
v
1
T = -20
2
T
v
150
-16.8
-20
1
2
32.3
P

5.69
A setup as in Fig. P5.68 has the R-410a initially at 1000 kPa, 50oC of mass 0.1 kg.
The balancing equilibrium pressure is 400 kPa and it is now cooled so the volume is
reduced to half the starting volume. Find the work and heat transfer for the process.
Take as CV the 0.1 kg of R-410a.
Energy Eq.5.11 m(u2 − u1) = 1Q2 − 1W2
Process Eq.: P = Pfloat or v = C = v1,
State 1: (P, T) => v1 = 0.0332 m3/kg,
u1 = 292.695 kJ/kg
P C.P.
1
1000
2
400 T
v
State 2: (P, v) => v2 = v1/2 = 0.0166 m3/kg < vg, so it is two-phase.
x2 = (v2 – vf ) /vfg = (0.0166 – 0.000803)/0.064 = 0.2468
u2 = uf + x2 ufg = 27.92 + x2 218.07 = 81.746 kJ/kg
From process eq.: 1W2 = ∫ P dV = area = mP2 (v2 − v1)
= 0.1 × 400 ( 0.0166 – 0.0332) = -0.664 kJ
From energy eq.: 1Q2 = m(u2 − u1) + 1W2 = 0.1× (81.746 – 292.695) - 0.664
= -21.8 kJ

5.70
A vertical cylinder fitted with a piston contains 5 kg of R-410a at 10°C, shown in Fig.
P5.70. Heat is transferred to the system, causing the piston to rise until it reaches a set
of stops at which point the volume has doubled. Additional heat is transferred until the
temperature inside reaches 50°C, at which point the pressure inside the cylinder is 1.4
MPa.
a. What is the quality at the initial state?
b. Calculate the heat transfer for the overall process.
Solution:
C.V. R-410a. Control mass goes through process: 1 -> 2 -> 3
As piston floats pressure is constant (1 -> 2) and the volume is constant for the
second part (2 -> 3). So we have: v3 = v2 = 2 × v1
State 3: Table B.4.2 (P,T) v3 = 0.02249 m3/kg, u3 = 287.91 kJ/kg
v
P
1 2
3
R-410a
Po
cb
So we can then determine state 1 and 2 Table B.4.1:
v1 = 0.011245 = 0.000886 + x1 × 0.02295 => x1 = 0.4514
b) u1 = 72.24 + 0.4514 × 183.66 = 155.14 kJ/kg
State 2: v2 = 0.02249 m3/kg, P2 = P1 = 1086 kPa this is still 2-phase.
We get the work from the process equation (see P-V diagram)
1W3 = 1W2 = ⌡⌠
1
2
PdV = P1(V2 - V1) = 1086 × 5 (0.011245) = 61.1 kJ
The heat transfer from the energy equation becomes
1Q3 = m(u3-u1) + 1W3 = 5(287.91 – 155.14) + 61.1 = 725.0 kJ

5.71
Find the heat transfer in Problem 4.68.
A piston/cylinder contains 1 kg of liquid water at 20°C and 300 kPa. Initially the
piston floats, similar to the setup in Problem 4.64, with a maximum enclosed volume
of 0.002 m3 if the piston touches the stops. Now heat is added so a final pressure of
600 kPa is reached. Find the final volume and the work in the process.
Solution:
Take CV as the water. Properties from table B.1
m2 = m1 = m ; m(u2 - u1) = 1Q2 - 1W2
State 1: Compressed liq. v = vf (20) = 0.001002 m3/kg, u = uf = 83.94 kJ/kg
State 2: Since P > Plift then v = vstop = 0.002 and P = 600 kPa
For the given P : vf < v < vg so 2-phase T = Tsat = 158.85 °C
v = 0.002 = 0.001101 + x × (0.3157-0.001101) => x = 0.002858
u = 669.88 + 0.002858 ×1897.5 = 675.3 kJ/kg
Work is done while piston moves at Plift= constant = 300 kPa so we get
1W2 = ∫ P dV = m Plift (v2 -v1) = 1×300(0.002 - 0.001002) = 0.299 kJ
Heat transfer is found from energy equation
1Q2 = m(u2 - u1) + 1W2 = 1(675.3 - 83.94) + 0.299 = 591.66 kJ
V
P
1
2
H O
Po
2
cb

5.72
10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 100
kPa, with a quality of 50%. It is now heated so the volume triples. The mass of the
piston is such that a cylinder pressure of 200 kPa will float it, as in Fig. 5.72. Find the
final temperature and the heat transfer in the process.
Solution:
Take CV as the water.
Energy Eq.: m(u2 − u1) = 1Q2 − 1W2
Process: v = constant until P = Plift , then P is constant.
State 1: Two-phase so look in Table B.1.2 at 100 kPa
u1 = 417.33 + 0.5 × 2088.72 = 1461.7 kJ/kg,
v1 = 0.001043 + 0.5 × 1.69296 = 0.8475 m3/kg
State 2: v2, P2 ≤ Plift => v2 = 3 × 0.8475 = 2.5425 m3/kg ;
Interpolate: T2 = 829°C, u2 = 3718.76 kJ/kg
=> V2 = mv2 = 25.425 m3
From the process equation (see P-V diagram) we get the work as
1W2 = Plift(V2 −V1) = 200 × 10 (2.5425 − 0.8475) = 3390 kJ
From the energy equation we solve for the heat transfer
1Q2 = m(u2 − u1) + 1W2 = 10×(3718.76 − 1461.7) + 3390 = 25 961 kJ
Po
H2O
cb
V
P
2
1P
P
1
2
cb

5.73
The cylinder volume below the constant loaded piston has two compartments A and B
filled with water. A has 0.5 kg at 200 kPa, 150oC and B has 400 kPa with a quality of
50% and a volume of 0.1 m3. The valve is opened and heat is transferred so the water
comes to a uniform state with a total volume of 1.006 m3.
a) Find the total mass of water and the total initial volume.
b) Find the work in the process
c) Find the process heat transfer.
Solution:
Take the water in A and B as CV.
Continuity: m2 - m1A - m1B = 0
Energy: m2u2 - m1Au1A - m1Bu1B = 1Q2 - 1W2
Process: P = constant = P1A if piston floats
(VA positive) i.e. if V2 > VB = 0.1 m3
State A1: Sup. vap. Table B.1.3 v = 0.95964 m3/kg, u = 2576.9 kJ/kg
=> V = mv = 0.5 × 0.95964 = 0.47982
State B1: Table B.1.2 v = (1-x) × 0.001084 + x × 0.4625 = 0.2318 m3/kg
=> m = V/v = 0.4314 kg
u = 604.29 + 0.5 × 1949.3 = 1578.9 kJ/kg
State 2: 200 kPa, v2 = V2/m = 1.006/0.9314 = 1.0801 m3/kg
Table B.1.3 => close to T2 = 200oC and u2 = 2654.4 kJ/kg
So now
V1 = 0.47982 + 0.1 = 0.5798 m3, m1 = 0.5 + 0.4314 = 0.9314 kg
Since volume at state 2 is larger than initial volume piston goes up and the pressure
then is constant (200 kPa which floats piston).
1W2 = ∫ P dV = Plift (V2 - V1) = 200 (1.006 - 0.57982) = 85.24 kJ
1Q2 = m2u2 - m1Au1A - m1Bu1B + 1W2
= 0.9314 × 2654.4 - 0.5 × 2576.9 - 0.4314 × 1578.9 + 85.24 = 588 kJ

5.74
Calculate the heat transfer for the process described in Problem 4.65.
A cylinder containing 1 kg of ammonia has an externally loaded piston. Initially the
ammonia is at 2 MPa, 180°C and is now cooled to saturated vapor at 40°C, and then
further cooled to 20°C, at which point the quality is 50%. Find the total work for the
process, assuming a piecewise linear variation of P versus V.
Solution:
C.V. Ammonia going through process 1 - 2 - 3. Control mass.
Continuity: m = constant,
Energy Eq.5.11: m(u3 - u1) = 1Q3 - 1W3
Process: P is piecewise linear in V
State 1: (T, P) Table B.2.2: v1 = 0.10571 m3/kg, u1 = 1630.7 kJ/kg
State 2: (T, x) Table B.2.1 sat. vap. P2 = 1555 kPa, v2 = 0.08313 m3/kg
o2
1
P
3
180 C
857
1555
2000
v
40 C
o
o
20 C
State 3: (T, x) P3 = 857 kPa,
v3 = (0.001638+0.14922)/2 = 0.07543 u3 = (272.89 + 1332.2)/2 = 802.7 kJ/kg
Process: piecewise linear P versus V, see diagram. Work is area as:
W13 = ⌡⌠
1
3
PdV ≈ (
P1 + P2
2 ) m(v2 - v1) + (
P2 + P3
2 ) m(v3 - v2)
=
2000 + 1555
2 1(0.08313 - 0.10571) +
1555 + 857
2 1(0.07543 - 0.08313)
= -49.4 kJ
From the energy equation, we get the heat transfer as:
1Q3 = m(u3 - u1) + 1W3 = 1× (802.7 - 1630.7) - 49.4 = -877.4 kJ

5.75
A rigid tank A of volume 0.6 m3 contains 3 kg water at 120oC and the rigid tank B is
0.4 m3 with water at 600 kPa, 200oC. They are connected to a piston cylinder initially
empty with closed valves. The pressure in the cylinder should be 800 kPa to float the
piston. Now the valves are slowly opened and heat is transferred so the water reaches
a uniform state at 250oC with the valves open. Find the final volume and pressure and
the work and heat transfer in the process.
C.V.: A + B + C.
Only work in C, total mass constant.
m2 - m1 = 0 => m2 = mA1 + mB1
U2 - U1 = 1Q2 - 1W2 ;
1W2 = ∫ PdV = Plift (V2 - V1)
A B
C
1A: v = 0.6/3 = 0.2 m3/kg => xA1 = (0.2 - 0.00106)/0.8908 = 0.223327
u = 503.48 + 0.223327 × 2025.76 = 955.89 kJ/kg
1B: v = 0.35202 m3/kg => mB1 = 0.4/0.35202 = 1.1363 kg ; u = 2638.91 kJ/kg
m2 = 3 + 1.1363 = 4.1363 kg and
V2 = VA+ VB + VC = 1 + VC
Locate state 2: Must be on P-V lines shown
State 1a: 800 kPa,
v1a =
VA+VB
m = 0.24176 m3/kg
800 kPa, v1a => T = 173°C too low.
V
P
21a
P2
Assume 800 kPa: 250°C => v = 0.29314 m3/kg > v1a OK
V2 = m2v2 = 4.1363 kg × 0.29314 m3/kg = 1.21 m3
Final state is : 800 kPa; 250°C => u2 = 2715.46 kJ/kg
W = 800(0.29314 - 0.24176) × 4.1363 = 800 × (1.2125 - 1) = 170 kJ
Q = m2u2 - m1u1 + 1W2 = m2u2 - mA1uA1 - mB1uB1 + 1W2
= 4.1363 × 2715.46 - 3 × 955.89 - 1.1363 × 2638.91 + 170
= 11 232 - 2867.7 - 2998.6 + 170 = 5536 kJ

5.76
A piston cylinder setup similar to Problem 4.?? contains 0.1 kg saturated liquid and vapor
water at 100 kPa with quality 25%. The mass of the piston is such that a pressure of 500
kPa will float it. The water is heated to 300°C. Find the final pressure, volume and the
work, 1W2.
Solution:
Take CV as the water: m2 = m1 = m
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: v = constant until P = Plift
To locate state 1: Table B.1.2
v1 = 0.001043 + 0.25×1.69296 = 0.42428 m3/kg
u1 = 417.33 + 0.25×2088.7 = 939.5 kJ/kg
P1
Plift
V
P
1
21a
cb
State 1a: 500 kPa, v1a = v1 = 0.42428 > vg at 500 kPa,
so state 1a is superheated vapor Table B.1.3 T1a = 200°C
State 2 is 300°C so heating continues after state 1a to 2 at constant P = 500 kPa.
2: T2, P2 = Plift => Tbl B.1.3 v2 =0.52256 m3/kg; u2 = 2802.9 kJ/kg
From the process, see also area in P-V diagram
1W2 = Plift m(v2 - v1) = 500 × 0.1 (0.5226 - 0.4243) = 4.91 kJ
1Q2 = m(u2 - u1) + 1W2 = 0.1(2802.9 - 939.5) + 4.91 = 191.25 kJ

5.77
A cylinder/piston arrangement contains 5 kg of water at 100°C with x = 20% and the
piston, mP = 75 kg, resting on some stops, similar to Fig. P5.72. The outside pressure
is 100 kPa, and the cylinder area is A = 24.5 cm2. Heat is now added until the water
reaches a saturated vapor state. Find the initial volume, final pressure, work, and heat
transfer terms and show the P–v diagram.
Solution:
C.V. The 5 kg water.
Continuty: m2 = m1 = m ; Energy: m(u2 - u1) = 1Q2 - 1W2
Process: V = constant if P < Plift otherwise P = Plift see P-v diagram.
P3 = P2 = Plift = P0 + mp g / Ap = 100 +
75 × 9.807
0.00245 × 1000 = 400 kPa
P
v1
2 3
100 C
143 C
o
o
cb
H O
Po
2
cb
State 1: (T,x) Table B.1.1
v1 = 0.001044 + 0.2 × 1.6719, V1 = mv1 = 5 × 0.3354 = 1.677 m3
u1 = 418.91 + 0.2 × 2087.58 = 836.4 kJ/kg
State 3: (P, x = 1) Table B.1.2 => v3 = 0.4625 > v1, u3 = 2553.6 kJ/kg
Work is seen in the P-V diagram (if volume changes then P = Plift)
1W3 = 2W3 = Pextm(v3 - v2) = 400 × 5(0.46246 - 0.3354) = 254.1 kJ
1Q3 = 5 (2553.6 - 836.4) + 254.1 = 8840 kJ

Energy Equation: Solids and Liquids

5.78
I have 2 kg of liquid water at 20oC, 100 kPa. I now add 20 kJ of energy at a constant
pressure. How hot does it get if it is heated? How fast does it move if it is pushed by a
constant horizontal force? How high does it go if it is raised straight up?
a) Heat at 100 kPa.
Energy equation:
E2 – E1 = 1Q2 – 1W2 = 1Q2 – P(V2 – V1) = H2 – H1= m(h2 – h1)
h2 = h1 + 1Q2/m = 83.94 + 20/2 = 94.04 kJ/kg
Back interpolate in Table B.1.1: T2 = 22.5oC
[We could also have used ∆T = 1Q2/mC = 20 / (2*4.18) = 2.4oC]
b) Push at constant P. It gains kinetic energy.
0.5 m V
2
2 = 1W2
V2 = 2 1W2/m = 2 × 20 × 1000 J/2 kg = 141.4 m/s
c) Raised in gravitational field
m g Z2 = 1W2
Z2 = 1W2/m g =
20 000 J
2 kg × 9.807 m/s2 = 1019 m
Comment: Notice how fast (500 km/h) and how high it should by to have the same
energy as raising the temperature just 2 degrees. I.e. in most applications
we can disregard the kinetic and potential energies unless we have very
high V or Z.

5.79
A copper block of volume 1 L is heat treated at 500°C and now cooled in a 200-L oil
bath initially at 20°C, shown in Fig. P5.79. Assuming no heat transfer with the
surroundings, what is the final temperature?
Solution:
C.V. Copper block and the oil bath.
Also assume no change in volume so the work will be zero.
Energy Eq.: U2 - U1 = mmet(u2 - u1)met + moil(u2 - u1)oil = 1Q2 - 1W2 = 0
Properties from Table A.3 and A.4
mmet = Vρ = 0.001 m3 × 8300 kg/m3 = 8.3 kg,
moil = Vρ = 0.2 m3 × 910 kg/m3 = 182 kg
Solid and liquid Eq.5.17: ∆u ≅ Cv ∆T,
Table A.3 and A.4: Cv met = 0.42
kJ
kg K, Cv oil = 1.8
kJ
kg K
The energy equation for the C.V. becomes
mmetCv met(T2 − T1,met) + moilCv oil(T2 − T1,oil) = 0
8.3 × 0.42(T2 − 500) + 182 × 1.8 (T2 − 20) = 0
331.09 T2 – 1743 – 6552 = 0
⇒ T2 = 25 °C

5.80
Because a hot water supply must also heat some pipe mass as it is turned on so it does
not come out hot right away. Assume 80oC liquid water at 100 kPa is cooled to 45oC
as it heats 15 kg of copper pipe from 20 to 45oC. How much mass (kg) of water is
needed?
Solution:
C.V. Water and copper pipe. No external heat transfer, no work.
Energy Eq.5.11: U2 – U1 = ∆Ucu + ∆UH2O = 0 – 0
From Eq.5.18 and Table A.3:
∆Ucu = mC ∆Τ = 15 kg × 0.42
kJ
kg K × (45 – 20) K = 157.5 kJ
mH2O = - ∆Ucu / ∆uH2O
mH2O = ∆Ucu / CH2O(- ∆ΤH2O) =
157.5
4.18 × 35
= 1.076 kg
or using Table B.1.1 for water
mH2O = ∆Ucu / ( u1- u2) =
157.5
334.84 – 188.41 = 1.076 kg
Water
Cu pipe The real problem involves a
flow and is not analyzed by this
simple process.

5.81
In a sink 5 liters of water at 70oC is combined with 1 kg aluminum pots, 1 kg of
flatware (steel) and 1 kg of glass all put in at 20oC. What is the final uniform
temperature neglecting any heat loss and work?
Energy Eq.: U2 - U1 = ∑mi(u2 - u1)i = 1Q2 - 1W2 = 0
For the water: vf = 0.001023 m3/kg, V = 5 L = 0.005 m3; m = V/v = 4.8876 kg
For the liquid and the metal masses we will use the specific heats (Tbl A.3, A.4) so
∑mi(u2 - u1)i = ∑miCv i (T2 - T1)i = T2∑miCv i – ∑miCv iT1 i
noticing that all masses have the same T2 but not same initial T.
∑miCv i = 4.8876 × 4.18 + 1 × 0.9 + 1 × 0.46 + 1 × 0.8 = 22.59 kJ/K
Energy Eq.: 22.59 T2 = 4.8876 × 4.18 ×70 + (1 × 0.9 + 1 × 0.46 + 1 × 0.8) × 20
= 1430.11 + 43.2
T2 = 65.2oC

5.82
A house is being designed to use a thick concrete floor mass as thermal storage
material for solar energy heating. The concrete is 30 cm thick and the area exposed to
the sun during the daytime is 4 m × 6 m. It is expected that this mass will undergo an
average temperature rise of about 3°C during the day. How much energy will be
available for heating during the nighttime hours?
Solution:
C.V.: Control mass concrete.
V = 4 × 6 × 0.3 = 7.2 m3
Concrete is a solid with some properties listed in Table A.3
m = ρV = 2200 kg/m3 × 7.2 m3 = 15 840 kg
Energy Eq.: m(u2 - u1) = 1Q2 - 1W2 = 1Q2
The available heat transfer is the change in U. From Eq.5.18 and C from table A.3
∆U = m C ∆T = 15 840 kg × 0.88
kJ
kg K × 3 K = 41818 kJ = 41.82 MJ

5.83
A closed rigid container is filled with 1.5 kg water at 100 kPa, 55oC, 1 kg of stainless
steel and 0.5 kg of PVC (polyvinyl chloride) both at 20oC and 0.1 kg of air at 400 K,
100 kPa. It is now left alone with no external heat transfer and no water vaporizes.
Find the final temperature and air pressure.
For the liquid and the metal masses we will use the specific heats (Tbl A.3, A.4) so
∑miCv i = 1.5 × 4.18 + 1 × 0.46 + 0.5 × 0.96 + 0.1 × 0.717 = 7.282 kJ/K
Energy Eq.: 7.282 T2 = 1.5 × 4.18 ×55 + (1 × 0.46 + 0.5 × 0.96) × 20
+ 0.1 × 0.717 × (400-273.15) = 372.745 kJ
T2 = 51.2oC
The volume of the air is constant so from PV = mRT it follows that P varies with T
P2 = P1 T2/T1 air = 100 × 324.34 / 400 = 81 kPa

5.84
A car with mass 1275 kg drives at 60 km/h when the brakes are applied quickly to
decrease its speed to 20 km/h. Assume the brake pads are 0.5 kg mass with heat
capacity of 1.1 kJ/kg K and the brake discs/drums are 4.0 kg steel. Further assume
both masses are heated uniformly. Find the temperature increase in the brake
assembly.
Solution:
C.V. Car. Car loses kinetic energy and brake system gains internal u.
No heat transfer (short time) and no work term.
m = constant;
Energy Eq.5.11: E2 - E1 = 0 - 0 = mcar
1
2(V
2
2 − V
2
1) + mbrake(u2 − u1)
The brake system mass is two different kinds so split it, also use Cv from Table
A.3 since we do not have a u table for steel or brake pad material.
msteel Cv ∆T + mpad Cv ∆T = mcar 0.5 (602 − 202)





1000
3600
2
m2/s2
(4 × 0.46 + 0.5 × 1.1)
kJ
K ∆T = 1275 kg × 0.5 × (3200 × 0.077 16) m2/s2
= 157 406 J = 157.4 kJ
=> ∆T = 65.9 °C

5.85
A computer CPU chip consists of 50 g silicon, 20 g copper, 50 g polyvinyl chloride
(plastic). It heats from 15oC to 70oC as the computer is turned on. How much energy
does the heating require?
Energy Eq.: U2 - U1 = ∑mi(u2 - u1)i = 1Q2 - 1W2
For the solid masses we will use the specific heats, Table A.3, and they all have the
same temperature so
∑mi(u2 - u1)i = ∑miCv i (T2 - T1)i = (T2 – T1)∑miCv i
∑miCv i = 0.05 × 0.7 + 0.02 × 0.42 + 0.05 × 0.96 = 0.0914 kJ/K
U2 - U1 = 0.0914 × (70 – 15) = 5.03 kJ

5.86
A 25 kg steel tank initially at –10oC is filled up with 100 kg of milk (assume
properties as water) at 30oC. The milk and the steel come to a uniform temperature of
+5 oC in a storage room. How much heat transfer is needed for this process?
Solution:
C.V. Steel + Milk. This is a control mass.
Energy Eq.5.11: U2 − U1 = 1Q2 − 1W2 = 1Q2
Process: V = constant, so there is no work
1W2 = 0.
Use Eq.5.18 and values from A.3 and A.4 to evaluate changes in u
1Q2 = msteel (u2 - u1)steel + mmilk(u2 - u1)milk
= 25 kg × 0.466
kJ
kg K × [5 − (−10)] Κ + 100 kg ×4.18
kJ
kg K × (5 − 30) Κ
= 172.5 − 10450 = −10277 kJ

5.87
A 1 kg steel pot contains 1 kg liquid water both at 15oC. It is now put on the stove
where it is heated to the boiling point of the water. Neglect any air being heated and
find the total amount of energy needed.
Solution:
Energy Eq.: U2 − U1= 1Q2 − 1W2
: The steel does not change volume
and the change for the liquid is
minimal, so 1W2 ≅ 0.
State 2: T2 = Tsat (1atm) = 100oC
Tbl B.1.1 : u1 = 62.98 kJ/kg, u2 = 418.91 kJ/kg
Tbl A.3 : Cst = 0.46 kJ/kg K
Solve for the heat transfer from the energy equation
1Q2 = U2 − U1 = mst (u2 − u1)st + mH2O (u2 − u1)H2O
= mstCst (T2 – T1) + mH2O (u2 − u1)H2O
1Q2 = 1 kg × 0.46
kJ
kg K ×(100 – 15) K + 1 kg ×(418.91 – 62.98) kJ/kg
= 39.1 + 355.93 = 395 kJ

5.88
A piston cylinder (0.5 kg steel altogether) maintaining a constant pressure has 0.2 kg
R-134a as saturated vapor at 150 kPa. It is heated to 40oC and the steel is at the same
temperature as the R-134a at any time. Find the work and heat transfer for the process.
C.V. The R-134a plus the steel. Constant total mass
m2 = m1 = m ;
U2 – U1 = mR134a(u2 – u1)R134a + msteel(u2 – u1) = 1Q2 – 1W2
State 1: B.5.2 sat. vapor v1 = 0.13139 m3/kg, u1 = 368.06 kJ/kg
State 2: B.5.2 sup. vapor v2 = 0.16592 m3/kg, u2 = 411.59 kJ/kg
V1 = mv1 = 0.2 × 0.13139 = 0.02628 m3
V2 = mv2 = 0.2 × 0.16592 = 0.03318 m3
Steel: A.3, Csteel = 0.46 kJ/kg-K
Process: P = C for the R134a and constant volume for the steel =>
1W2 = ∫P dV = P1(V2 – V1) = 150 kPa (0.03318 – 0.02628) m3
= 1.035 kJ
1Q2 = mR134a (u2 – u1) + msteel (u2 – u1) + 1W2
= mR134a (u2 – u1) + msteelCsteel (T2 – T1) + 1W2
= 0.2 × (411.59 – 368.06) + 0.5 × 0.46 × [40 – (–17.29)] + 1.035
= 8.706 + 13.177 + 1.035 = 22.92 kJ

5.89
An engine consists of a 100 kg cast iron block with a 20 kg aluminum head, 20 kg
steel parts, 5 kg engine oil and 6 kg glycerine (antifreeze). Everything begins at 5oC
and as the engine starts we want to know how hot it becomes if it absorbs a net of
7000 kJ before it reaches a steady uniform temperature.
Energy Eq.: U2 − U1= 1Q2 − 1W2
Process: The steel does not change volume and the change for the liquid is minimal,
so 1W2 ≅ 0.
So sum over the various parts of the left hand side in the energy equation
mFe (u2 − u1) + mAl (u2 − u1)Al + mst (u − u1)st
+ moil (u2 − u1)oil + mgly (u2 − u1)gly = 1Q2
Table A.3 : CFe = 0.42 , CAl = 0.9, Cst = 0.46 all units of kJ/kg K
Table A.4 : Coil = 1.9 , Cgly = 2.42 all units of kJ/kg K
So now we factor out T2 –T1 as u2 − u1 = C(T2 –T1) for each term
[ mFeCFe + mAlCAl + mstCst+ moilCoil + mglyCgly ] (T2 –T1) = 1Q2
T2 –T1 = 1Q2 / Σmi Ci
=
7000
100× 0.42 + 20× 0.9 + 20× 0.46 + 5 ×1.9 + 6 ×2.42
=
7000
93.22 = 75 K
T2 = T1 + 75 = 5 + 75 = 80oC
Exhaust flow
Air intake filter
Coolant flow
Atm.
airShaft
Fan
power
Radiator

Properties (u, h, Cv and Cp), Ideal Gas

5.90
Use the ideal gas air table A.7 to evaluate the heat capacity Cp at 300 K as a slope of
the curve h(T) by ∆h/∆T. How much larger is it at 1000 K and 1500 K.
Solution :
From Eq.5.24:
Cp =
dh
dT =
∆h
∆T
=
h320 - h290
320 - 290 = 1.005 kJ/kg K
1000K Cp =
∆h
∆T
=
h1050 - h950
1050 - 950 =
1103.48 - 989.44
100 = 1.140 kJ/kg K
1500K Cp =
∆h
∆T
=
h1550 - h1450
1550 - 1450 =
1696.45 - 1575.4
100 = 1.21 kJ/kg K
Notice an increase of 14%, 21% respectively.
h
T
300 1000 1500
Cp 300
Cp 1500

5.91
We want to find the change in u for carbon dioxide between 600 K and 1200 K.
a) Find it from a constant Cvo from table A.5
b) Find it from a Cvo evaluated from equation in A.6 at the average T.
c) Find it from the values of u listed in table A.8
Solution :
a) ∆u ≅ Cvo ∆T = 0.653 × (1200 – 600) = 391.8 kJ/kg
b) Tavg =
1
2 (1200 + 600) = 900, θ =
T
1000 =
900
1000 = 0.9
Cpo = 0.45 + 1.67 × 0.9 - 1.27 × 0.92 + 0.39 × 0.93 = 1.2086 kJ/kg K
Cvo = Cpo – R = 1.2086 – 0.1889 = 1.0197 kJ/kg K
∆u = 1.0197 × (1200 – 600) = 611.8 kJ/kg
c) ∆u = 996.64 – 392.72 = 603.92 kJ/kg
u
T
300 600 1200
u
600
u
1200

5.92
We want to find the change in u for carbon dioxide between 50oC and 200oC at a
pressure of 10 MPa. Find it using ideal gas and Table A.5 and repeat using the B
section table.
Solution:
Using the value of Cvo for CO2 from Table A.5,
∆u = Cvo ∆T = 0,653 × (200 – 50) = 97.95 kJ/kg
Using values of u from Table B3.2 at 10 000 kPa, with linear
interpolation between 40oC and 60oC for the 50oC value,
∆u = u200 - u50 = 437.6 – 230.9 = 206.7 kJ/kg
Note: Since the state 50oC, 10 000 kPa is in the dense-fluid supercritical region, a
linear interpolation is quite inaccurate. The proper value for u at this state is
found from the CATT software to be 245.1 instead of 230.9. This results is
∆u = u200 - u50 = 437.6 – 245.1 = 192.5 kJ/kg

5.93
We want to find the change in u for oxygen gas between 600 K and 1200 K.
a) Find it from a constant Cvo from table A.5
b) Find it from a Cvo evaluated from equation in A.6 at the average T.
c) Find it from the values of u listed in table A.8
Solution:
a) ∆u ≅ Cvo ∆T = 0.662 × (1200 − 600) = 397.2 kJ/kg
b) Tavg =
1
2 (1200 + 600) = 900 K, θ =
T
1000 =
900
1000 = 0.9
Cpo = 0.88 − 0.0001 × 0.9 + 0.54 × 0.92 − 0.33 × 0.93 = 1.0767
Cvo = Cpo − R = 1.0767 − 0.2598 = 0.8169 kJ/kg K
∆u = 0.8169 × (1200 − 600) = 490.1 kJ/kg
c) ∆u = 889.72 − 404.46 = 485.3 kJ/kg
u
T
300 600 1200
u
600u
1200

5.94
Estimate the constant specific heats for R-134a from Table B.5.2 at 100 kPa and 125o
C. Compare this to table A.5 and explain the difference.
Solution:
Using values at 100 kPa for h and u at 120oC and 130oC
from Table B5.2, the approximate specific heats at 125oC are
Cp ≈
∆h
∆T
=
521.98 - 511.95
130 - 120 = 1.003 kJ/kg K
compared with 0.852 kJ/kg K for the ideal-gas value at 25oC
from Table A.5.
Cv ≈
∆u
∆T
=
489.36 - 480.16
130 - 120 = 0.920 kJ/kg K
compared with 0.771 kJ/kg K for the ideal-gas value at 25oC
from Table A.5.
There are two reasons for the differences. First, R-134a is not exactly an ideal gas at
the given state, 125oC and 100 kPa. Second and by far the biggest reason for the
differences is that R-134a, chemically CF3CH2 , is a polyatomic molecule with
multiple vibrational mode contributions to the specific heats (see Appendix C), such
that they are strongly dependent on temperature. Note that if we repeat the above
approximation for Cp in Table B.5.2 at 25oC, the resulting value is 0.851 kJ/kg K.

5.95
Water at 150°C, 400 kPa, is brought to 1200°C in a constant pressure process. Find
the change in the specific internal energy, using a) the steam tables, b) the ideal gas
water table A.8, and c) ≈the specific heat from A.5.
Solution:
a)
State 1: Table B.1.3 Superheated vapor u1 = 2564.48 kJ/kg
State 2: Table B.1.3 u2 = 4467.23 kJ/kg
u2 - u1 = 4467.23 - 2564.48 = 1902.75 kJ/kg
b)
Table A.8 at 423.15 K: u1 = 591.41 kJ/kg
Table A.8 at 1473.15 K: u2 = 2474.25 kJ/kg
u2 - u1 = 2474.25 - 591.41 = 1882.8 kJ/kg
c) Table A.5 : Cvo = 1.41 kJ/kgK
u2 - u1 = 1.41 kJ/kgK (1200 – 150) K = 1480.5 kJ/kg
Notice how the
average slope from
150 C to 1200 C is
higher than the one at
25 C ( = Cvo)
u
u
1200 Slope at
25 C
1
u T50
25 1200150

5.96
Nitrogen at 300 K, 3 MPa is heated to 500 K. Find the change in enthalpy using a)
Table B.6, b) Table A.8, and c) Table A.5.
B.6.2 h2 – h1 = 519.29 – 304.94 = 214.35 kJ/kg
A.8 h2 – h1 = 520.75 – 311.67 = 209.08 kJ/kg
A.5 h2 – h1 = Cpo (T2 – T1) = 1.042 kJ/kg-K (500 – 300) K = 208.4 kJ/kg
Comment: The results are listed in order of accuracy (B.6.2 best).

5.97
For a special application we need to evaluate the change in enthalpy for carbon
dioxide from 30oC to 1500oC at 100 kPa. Do this using constant specific heat from
Table A.5 and repeat using Table A.8. Which is the more accurate one?
Solution:
Using constant specific heat:
∆h = Cpo∆T = 0.842 (1500 - 30) = 1237.7 kJ/kg
Using Table A.8:
30°C = 303.15 K => h = 214.38 +
3.15
50 (257.9 – 214.38) = 217.12 kJ/kg
1500°C = 1773.15 K =>
h = 1882.43 +
73.15
100 (2017.67 – 1882.43) = 1981.36 kJ/kg
∆h = 1981.36 – 217.12 = 1764.2 kJ/kg
The result from A.8 is best. For large ∆T or small ∆T at high Tavg, constant specfic
heat is poor approximation.

5.98
Repeat the previous problem but use a constant specific heat at the average
temperature from equation in Table A.6 and also integrate the equation in Table A.6 to
get the change in enthalpy.
Tave =
1
2 (30 + 1500) + 273.15 = 1038.15 K; θ = T/1000 = 1.0382
Table A.6 ⇒ Cpo =1.2513 kJ/kg-K
∆h = Cpo,ave ∆T = 1.2513 × 1470 = 1839 kJ/kg
For the entry to Table A.6:
30°C = 303.15 K => θ1 = 0.30315
1500°C = 1773.15 K => θ2 = 1.77315
∆h = h2- h1 = ∫ Cpo dT
= [0.45 (θ2 - θ1) + 1.67 ×
1
2 (θ2
2 - θ1
2)
–1.27 ×
1
3 (θ2
3 - θ1
3) + 0.39×
1
4 (θ2
4
- θ1
4
)] = 1762.76 kJ/kg

5.99
Reconsider Problem 5.97 and examine if also using Table B.3 would be more accurate
and explain.
Table B.3 does include non-ideal gas effects, however at 100 kPa these effects are
extremely small so the answer from Table A.8 is accurate.
Table B.3. does not cover the 100 kPa superheated vapor states as the saturation
pressure is below the triple point pressure. Secondly Table B.3 does not go up to the
high temperatures covered by Table A.8 and A.9 at which states you do have ideal gas
behavior. Table B.3 covers the region of states where the carbon dioxide is close to the
two-phase region and above the critical point (dense fluid) which are all states where
you cannot assume ideal gas.

5.100
Water at 20°C, 100 kPa, is brought to 200 kPa, 1500°C. Find the change in the
specific internal energy, using the water table and the ideal gas water table in
combination.
Solution:
State 1: Table B.1.1 u1 ≅ uf = 83.95 kJ/kg
State 2: Highest T in Table B.1.3 is 1300°C
Using a ∆u from the ideal gas tables, A.8, we get
u1500 = 3139 kJ/kg u1300 = 2690.72 kJ/kg
u1500 - u1300 = 448.26 kJ/kg
We now add the ideal gas change at low P to the steam tables, B.1.3, ux =
4683.23 kJ/kg as the reference.
u2 - u1 = (u2 - ux)ID.G. + (ux - u1)
= 448.28 + 4683.23 - 83.95 = 5048 kJ/kg

5.101
An ideal gas is heated from 500 to 1500 K. Find the change in enthalpy using constant
specific heat from Table A.5 (room temperature value) and discuss the accuracy of the
result if the gas is
a. Argon b. Oxygen c. Carbon dioxide
Solution:
T1 = 500 K, T2 = 1500 K, ∆h = CP0(T2-T1)
a) Ar : ∆h = 0.520(1500-500) = 520 kJ/kg
Monatomic inert gas very good approximation.
b) O2 : ∆h = 0.922(1500-500) = 922 kJ/kg
Diatomic gas approximation is OK with some error.
c) CO2: ∆h = 0.842(1500-500) = 842 kJ/kg
Polyatomic gas heat capacity changes, see figure 5.11
See also appendix C for more explanation.

Energy Equation: Ideal Gas

5.102
Air is heated from 300 to 350 K at V = C. Find 1q2. What if from 1300 to 1350 K?
Process: V = C 1W2 = Ø
Energy Eq.: u2 − u1 = 1q2 – 0 1q2 = u2 − u1
Read the u-values from Table A.7.1
a) 1q2 = u2 − u1 = 250.32 – 214.36 = 36.0 kJ/kg
b) 1q2 = u2 − u1 = 1067.94 – 1022.75 = 45.2 kJ/kg
case a) Cv ≈ 36/50 = 0.72 kJ/kg K , see A.5
case b) Cv ≈ 45.2/50 = 0.904 kJ/kg K (25 % higher)

5.103
A 250 L rigid tank contains methane at 500 K, 1500 kPa. It is now cooled down to 300
K. Find the mass of methane and the heat transfer using ideal gas.
Solution:
Ideal gas, constant volume
P2 = P1 × (Τ2 / Τ1) = 1500 × 300 / 500 = 900 kPa
m = P1V/RT1 =
1500 × 0.25
0.5183 × 500 = 1.447 kg
Use specific heat from Table A.5
u2 - u1 = Cv (T2 – T1) = 1.736 (300 – 500) = –347.2 kJ/kg
1Q2 = m(u2 - u1) = 1.447(-347.2) = –502.4 kJ

5.104
A rigid tank has 1 kg air at 300 K, 120 kPa and it is heated by an external heater. Use
Table A.7 to find the work and the heat transfer for the process.
CV Air in tank, this is a C.M. at constant volume.
This process is a constant volume process so 1W2 = 0 and 1Q2 comes in.
Energy Eq: m(u2 – u1) = 1Q2 – 1W2 = 1Q2
Process & ideal gas: V2 = V1; P1V1 = mRT1, P2V2 = mRT2
P2 = P1 T2 /T1 = 120 × 1500 / 300 = 600 kPa
Solving using Table A.7 gives:
1Q2 = m(u2 – u1) = 1(1205.25 – 214.36) = 990.89 kJ

5.105
A rigid container has 2 kg of carbon dioxide gas at 100 kPa, 1200 K that is heated to
1400 K. Solve for the heat transfer using a. the heat capacity from Table A.5 and b.
properties from Table A.8
Solution:
C.V. Carbon dioxide, which is a control mass.
Energy Eq.5.11: U2 – U1 = m (u2- u1) = 1Q2 − 1W2
Process: ∆V = 0 ⇒ 1W2 = 0
a) For constant heat capacity we have: u2- u1 = Cvo (T2- T1) so
1Q2 ≅ mCvo (T2- T1) = 2 × 0.653 × (1400 –1200) = 261.2 kJ
b) Taking the u values from Table A.8 we get
1Q2 = m (u2- u1) = 2 × (1218.38 – 996.64) = 443.5 kJ

5.106
Do the previous problem for nitrogen, N2, gas.
A rigid container has 2 kg of carbon dioxide gas at 100 kPa, 1200 K that is heated to
1400 K. Solve for the heat transfer using a. the heat capacity from Table A.5 and b.
properties from Table A.8
Solution:
C.V. Nitrogen gas, which is a control mass.
Energy Eq.5.11: U2 – U1 = m (u2- u1) = 1Q2 − 1W2
Process: ∆V = 0 ⇒ 1W2 = 0
a) For constant heat capacity we have: u2- u1 = Cvo (T2 - T1) so
1Q2 ≅ mCvo (T2- T1) = 2 × 0.745 × (1400 – 1200) = 298 kJ
b) Taking the u values from Table A.8, we get
1Q2 = m (u2- u1) = 2 × (1141.35 – 957) = 368.7 kJ

5.107
A tank has a volume of 1 m3 with oxygen at 15oC, 300 kPa. Another tank contains 4
kg oxygen at 60oC, 500 kPa. The two tanks are connected by a pipe and valve which
is opened allowing the whole system to come to a single equilibrium state with the
ambient at 20oC. Find the final pressure and the heat transfer.
C.V. Both tanks of constant volume.
Continuity Eq.: m2 – m1A – m1B = 0
Energy Eq.: m2u2 – m1Au1A – m1Bu1B = 1
Q2
– 1
W2
Process Eq.: V2 = VA + VB = constant, 1
W2
= 0
State 1A: m1A =
P1AVA
RT1A
=
300 kPa × 1 m3
0.2598 kJ/kg-K × 288.15 K = 4.007 kg
State 1B: VB =
m1BRT1B
P1B
=
4 kg × 0.2598 kJ/kg-K × 333.15 K
500 kPa = 0.6924 m3
State 2: (T2
, v2
= V2
/m2
) V2 = VA + VB = 1 + 0.6924 = 1.6924 m3
m2 = m1A + m1B = 4.007 + 4 = 8.007 kg
P2 =
m2RT2
V2
=
8.007 kg × 0.2598 kJ/kg-K × 293.15 K
1.6924 m3 = 360.3 kPa
Heat transfer from energy equation
1
Q2
= m2u2 – m1Au1A – m1Bu1B = m1A(u2 – u1A) + m1B(u2 – u1B)
= m1ACv (T2 – T1A) + m1B Cv (T2 – T1B)
= 4.007 kg×0.662 kJ/kg-K×(20 –15) K + 4 kg×0.662 kJ/kg-K ×(20 –60) K
= - 92.65 kJ

5.108
A piston cylinder contains 3 kg of air at 20oC and 300 kPa. It is now heated up in a
constant pressure process to 600 K.
Solution:
Ideal gas PV = mRT
State 1: T1, P1
State 2: T2, P2 = P1
P2V2 = mRT2 V2 = mR T2 / P2 = 3×0.287×600 / 300 = 1.722 m3
Process: P = constant,
1
W2
= ⌡⌠ PdV = P (V2 - V1) = 300 (1.722 – 0.8413) = 264.2 kJ
Energy equation becomes
U2 - U1 = 1
Q2
- 1
W2
= m(u2 - u1)
1
Q2
= U2 - U1 + 1
W2
= 3(435.097 – 209.45) + 264.2 = 941 kJ
T
v
2
1
300 kPa
P
v
T
300
293
2
1
T1
2
600

5.109
A 10-m high cylinder, cross-sectional area 0.1 m2, has a massless piston at the bottom
with water at 20°C on top of it, shown in Fig. P5.109. Air at 300 K, volume 0.3 m3,
under the piston is heated so that the piston moves up, spilling the water out over the
side. Find the total heat transfer to the air when all the water has been pushed out.
Solution:
H2O
Po
cb
air
V
P
2
1
P
P1
0
V V1 max
The water on top is compressed liquid and has volume and mass
VH2O = Vtot - Vair = 10 × 0.1 - 0.3 = 0.7 m3
mH2O = VH2O/vf = 0.7 / 0.001002 = 698.6 kg
The initial air pressure is then
P1 = P0 + mH2Og/A = 101.325 +
698.6 × 9.807
0.1 × 1000 = 169.84 kPa
and then mair = PV/RT =
169.84 × 0.3
0.287 × 300 = 0.592 kg
State 2: No liquid over piston: P2 = P0 = 101.325 kPa, V2 = 10×0.1 = 1 m3
State 2: P2, V2 ⇒ T2 =
T1P2V2
P1V1
=
300×101.325×1
169.84×0.3 = 596.59 K
The process line shows the work as an area
1W2 = ⌡⌠PdV =
1
2 (P1 + P2)(V2 - V1) =
1
2 (169.84 + 101.325)(1 - 0.3) = 94.91 kJ
The energy equation solved for the heat transfer becomes
1Q2 = m(u2 - u1) + 1W2 ≅ mCv(T2 - T1) + 1W2
= 0.592 × 0.717 × (596.59 - 300) + 94.91 = 220.7 kJ
Remark: we could have used u values from Table A.7:
u2 - u1 = 432.5 - 214.36 = 218.14 kJ/kg versus 212.5 kJ/kg with Cv.

5.110
A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01 m3. A constant
pressure process gives 18 kJ of work out. Find the final temperature of the air and the
heat transfer input.
Solution:
C.V AIR control mass
Continuity Eq.: m2 – m1 = 0
Energy Eq.: m (u2 − u1) = 1Q2 - 1W2
Process: P = C so 1W2 = ∫ P dV = P(V2 – V1)
1 : P1 , T1,V1 2 : P1 = P2 , ?
m1 = P1V1/RT1 = 600 ×0.01 / 0.287 ×290 = 0.0721 kg
1W2 = P(V2 – V1) = 18 kJ
V2 – V1 = 1W2 / P = 18 kJ / 600 kPa = 0.03 m3
V2 = V1 + 1W2 / P = 0.01 + 0.03 = 0.04 m3
Ideal gas law : P2V2 = mRT2
T2 = P2V2 / mR =
P2V2
P1V1
T1 =
0.04
0.01 × 290 = 1160 K
Energy equation with u’s from table A.7.1
1Q2 = m (u2 − u1 ) + 1W2
= 0.0721 ( 898.04 – 207.19 ) + 18
= 67.81 kJ

5.111
An insulated cylinder is divided into two parts of 1 m3 each by an initially locked
piston, as shown in Fig. P5.111. Side A has air at 200 kPa, 300 K, and side B has air at
1.0 MPa, 1000 K. The piston is now unlocked so it is free to move, and it conducts
heat so the air comes to a uniform temperature TA = TB. Find the mass in both A and
B, and the final T and P.
C.V. A + B Force balance on piston: PAA = PBA
So the final state in A and B is the same.
State 1A: Table A.7 uA1 = 214.364 kJ/kg,
mA = PA1VA1/RTA1 = 200 × 1/(0.287 × 300) = 2.323 kg
State 1B: Table A.7 uB1 = 759.189 kJ/kg,
mB = PB1VB1/RTB1 = 1000 × 1/(0.287 × 1000) = 3.484 kg
For chosen C.V. 1Q2 = 0 , 1W2 = 0 so the energy equation becomes
mA(u2 - u1)A + mB(u2 - u1)B = 0
(mA + mB)u2 = mAuA1 + mBuB1
= 2.323 × 214.364 + 3.484 × 759.189 = 3143 kJ
u2 = 3143/(3.484 + 2.323) = 541.24 kJ/kg
From interpolation in Table A.7: ⇒ T2 = 736 K
P = (mA + mB)RT2/Vtot = 5.807 kg × 0.287
kJ
kg K × 736 K/ 2 m3 = 613 kPa
BA

5.112
Find the specific heat transfer for the helium in Problem 4.62
A helium gas is heated at constant volume from a state of 100 kPa, 300 K to 500 K. A
following process expands the gas at constant pressure to three times the initial
volume. What is the specific work in the combined process?
Solution :
C.V. Helium. This is a control mass.
Energy Eq.5.11: u3 - u1 = 1q3 - 1w3
V
P
3
1
2
P
P
1
2
Use ideal gas approximation for helium.
State 1: T, P => v1 = RT1/P1
State 2: V2 = V1 => P2 = P1 (T2/T1)
State 3: P3 = P2 => V3 = 3V2; T3 = T2 v3/v2 = 500 × 3 = 1500 K
1w3 = 1w2 + 2w3 = 2w3 = P3(v3 - v2) = R(T3 - T2)
= 2.0771(1500 - 500) = 2077 kJ/kg
The heat transfer is from the energy equation
1q3 = u3 - u1 + 1w3 = Cvo (T3 – T1) + 1w3
= 3.116 (1500 – 300) + 2077 = 5816 kJ/kg

5.113
A rigid insulated tank is separated into two rooms by a stiff plate. Room A of 0.5 m3
contains air at 250 kPa, 300 K and room B of 1 m3 has air at 500 kPa, 1000 K. The
plate is removed and the air comes to a uniform state without any heat transfer. Find
the final pressure and temperature.
Solution:
C.V. Total tank. Control mass of constant volume.
Mass and volume: m2 = mA + mB; V = VA + VB = 1.5 m3
Energy Eq.: U2 – U1 = m2 u2 – mAuA1 – mBuB1 = Q – W = 0
Process Eq.: V = constant ⇒ W = 0; Insulated ⇒ Q = 0
Ideal gas at 1: mA = PA1VA/RTA1 = 250 × 0.5/(0.287 × 300) = 1.452 kg
u A1= 214.364 kJ/kg from Table A.7
Ideal gas at 2: mB = PB1VB/RT B1= 500 × 1/(0.287 × 1000) = 1.742 kg
u B1= 759.189 kJ/kg from Table A.7
m2 = mA + mB = 3.194 kg
u2 =
mAuA1 + mBuB1
m2
=
1.452 × 214.364 + 1.742 × 759.189
3.194 = 511.51 kJ/kg
=> Table A.7.1: T2 = 698.6 K
P2 = m2 RT2 /V = 3.194 × 0.287 × 698.6/1.5 = 426.9 kPa
A B
cb

5.114
A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide
at 500 kPa, 400°C. It is cooled to 40°C, at which point the pressure is 300 kPa.
Calculate the heat transfer for the process.
Solution:
C.V. The carbon dioxide, which is a control mass.
Energy Eq.: m (u2 − u1) = 1Q2 - 1W2
Process Eq.: P = A + BV (linear spring)
1W2 = ⌡⌠PdV =
1
2(P1 + P2)(V2 - V1)
Equation of state: PV = mRT (ideal gas)
State 1: V1 = mRT1/P1 = 2 × 0.18892 × 673.15 /500 = 0.5087 m3
State 2: V2 = mRT2/P2 = 2 × 0.18892 × 313.15 /300 = 0.3944 m3
1W2 =
1
2(500 + 300)(0.3944 - 0.5087) = -45.72 kJ
To evaluate u2 - u1 we will use the specific heat at the average temperature.
From Figure 5.11: Cpo(Tavg) = 45/44 = 1.023 ⇒ Cvo = 0.83 = Cpo - R
For comparison the value from Table A.5 at 300 K is Cvo = 0.653 kJ/kg K
1Q2 = m(u2 - u1) + 1W2 = mCvo(T2 - T1) + 1W2
= 2 × 0.83(40 - 400) - 45.72 = -643.3 kJ
CO2
P
v
2
1
Remark: We could also have used the ideal gas table in A.8 to get u2 - u1.

5.115
A piston/cylinder has 0.5 kg air at 2000 kPa, 1000 K as shown. The cylinder has stops
so Vmin = 0.03 m3. The air now cools to 400 K by heat transfer to the ambient. Find
the final volume and pressure of the air (does it hit the stops?) and the work and heat
transfer in the process.
We recognize this is a possible two-step process, one of constant P and one of
constant V. This behavior is dictated by the construction of the device.
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: P = constant = F/A = P1 if V > Vmin
V = constant = V1a = Vmin if P < P1
State 1: (P, T) V1 = mRT1/P1 = 0.5 × 0.287 × 1000/2000 = 0.07175 m3
The only possible P-V combinations for this system is shown in the diagram so both
state 1 and 2 must be on the two lines. For state 2 we need to know if it is on the
horizontal P line segment or the vertical V segment. Let us check state 1a:
State 1a: P1a = P1, V1a = Vmin
Ideal gas so T1a = T1
V1a
V1
= 1000 ×
0.03
0.07175 = 418 K
We see that T2 < T1a and state 2 must have V2 = V1a = Vmin = 0.03 m3.
P2 = P1×
T2
T1
×
V1
V2
= 2000 ×
400
1000 ×
0.07175
0.03 = 1913.3 kPa
The work is the area under the process curve in the P-V diagram
1W2 = ⌡⌠
1
2 P dV = P1 (V1a – V1) = 2000 kPa (0.03 – 0.07175) m3 = – 83.5 kJ
Now the heat transfer is found from the energy equation, u’s from Table A.7.1,
1Q2 = m(u2 - u1) + 1W2 = 0.5 (286.49 - 759.19) – 83.5 = -319.85 kJ
V
P
1
2
1a
P
P
2
1
V
T
1
2
1a
T
T
2
1a
T
1

5.116
A piston/cylinder contains 1.5 kg of air at 300 K and 150 kPa. It is now heated up in a
two-step process. First constant volume to 1000 K (state 2) and then followed by a
constant pressure process to 1500 K, state 3. Find the heat transfer for the process.
Solution:
C.V. Helium. This is a control mass.
Energy Eq.5.11: U3 - U1 = 1Q3 - 1W3
V
P
3
1
2
P
P
1
2
Use ideal gas approximation for air.
State 1: T, P => V1 = mRT1/P1 = 1.5×0.287×300/150 = 0.861 m3
State 2: V2 = V1 => P2 = P1 (T2/T1) = 150×1000/300 = 500 kPa
State 3: P3 = P2 => V3 = V2 (T3/T2) = 0.861×1500/1000 = 1.2915 m3
1W3 = 1W2 + 2W3 = 2W3 = P3(V3 - V2)
= 500 kPa (1.2915 - 0.861) m3 = 215.3 kJ
The heat transfer is from the energy equation and we will use Table A.7 for u
1Q3 = U3 – U1 + 1W3 = m (u3 – u1) + 1W3
= 1.5 (1205.25 – 214.36) + 215.3 = 1701.6 kJ
Comment: We used Table A.7 due to the large temperature differences.

5.117
Air in a rigid tank is at 100 kPa, 300 K with a volume of 0.75 m3. The tank is heated
400 K, state 2. Now one side of the tank acts as a piston letting the air expand slowly
at constant temperature to state 3 with a volume of 1.5 m3. Find the pressures at states
2 and 3, Find the total work and total heat transfer.
State 1: m = P1V1 / RT1 =
100 × 0.75
0.287 × 300
kPa m3
kJ/kg = 0.871 kg
Process 1 to 2: Constant volume heating, dV = 0 => 1W2 = 0
P2 = P1 T2 / T1 = 100 × 400 / 300 = 133.3 kPa
Process 2 to 3: Isothermal expansion, dT = 0 => u3 = u2 and
P3 = P2 V2 / V3 = 133.3 × 0.75 / 1.5 = 66.67 kPa
2W3 = ⌡⌠
2
3 P dV = P2V2 ln (
V3
V2
) = 133.3 × 0.75 ln(2) = 69.3 kJ
The overall process:
1W3 = 1W2 + 2W3 = 2W3 = 69.3 kJ
1Q3 = m(u3 – u1) + 1W3 = m Cv (T3 – T1) + 1W3
= 0.871 × 0.717 (400 – 300) + 69.3 = 131.8 kJ

5.118
Water at 100 kPa, 400 K is heated electrically adding 700 kJ/kg in a constant pressure
process. Find the final temperature using
a) The water tables B.1 b) The ideal gas tables A.8
c) Constant specific heat from A.5
Solution :
Energy Eq.5.11: u2 - u1 = 1q2 - 1w2
Process: P = constant => 1w2 = P ( v2 - v1 )
Substitute this into the energy equation to get
1q2 = h2 - h1
Table B.1:
h1 ≅ 2675.46 +
126.85 - 99.62
150 - 99.62 × (2776.38 –2675.46) = 2730.0 kJ/kg
h2 = h1 + 1q2 = 2730 + 700 = 3430 kJ/kg
T2 = 400 + ( 500 – 400 ) ×
3430 - 3278.11
3488.09 - 3278.11 = 472.3°C
Table A.8:
h2 = h1 + 1q2 = 742.4 + 700 = 1442.4 kJ/kg
T2 = 700 + (750 – 700 ) ×
1442.4 - 1338.56
1443.43 - 1338.56 = 749.5 K = 476.3°C
Table A.5
h2 - h1 ≅ Cpo ( T2 - T1 )
T2 = T1 + 1q2 / Cpo = 400 + 700 / 1.872 = 773.9K = 500.8°C

5.119
Air in a piston/cylinder at 200 kPa, 600 K, is expanded in a constant-pressure process
to twice the initial volume (state 2), shown in Fig. P5.101. The piston is then locked
with a pin and heat is transferred to a final temperature of 600 K. Find P, T, and h for
states 2 and 3, and find the work and heat transfer in both processes.
Solution:
C.V. Air. Control mass m2 = m3 = m1
Energy Eq.5.11: u2 - u1 = 1q2 - 1w2 ;
Process 1 to 2: P = constant => 1w2 = ∫ P dv = P1(v2 -v1) = R(T2 -T1)
Ideal gas Pv = RT ⇒ T2 = T1v2/v1 = 2T1 = 1200 K
P2 = P1 = 200 kPa, 1w2 = RT1 = 172.2 kJ/kg
Table A.7 h2 = 1277.8 kJ/kg, h3 = h1 = 607.3 kJ/kg
1q2 = u2 - u1 + 1w2 = h2 - h1 = 1277.8 - 607.3 = 670.5 kJ/kg
Process 2→3: v3 = v2 = 2v1 ⇒ 2w3 = 0,
P3 = P2T3/T2 = P1T1/2T1 = P1/2 = 100 kPa
2q3 = u3 - u2 = 435.1 - 933.4 = -498.3 kJ/kg
Po
Air
cb
T
v
2
1 3
P
v
21
3
1200
600100
200

5.120
A spring loaded piston/cylinder contains 1.5 kg of air at 27C and 160 kPa. It is now
heated to 900 K in a process where the pressure is linear in volume to a final volume
of twice the initial volume. Plot the process in a P-v diagram and find the work and
heat transfer.
Take CV as the air.
m2 = m1 = m ; m(u2 -u1) = 1Q2 - 1W2
Process: P = A + BV => 1W2 = ∫ P dV = area = 0.5(P1 + P2)(V2 -V1)
State 1: Ideal gas. V1 = mRT1/P1 = 1.5× 0.287 × 300/160 = 0.8072 m3
Table A.7 u1 = u(300) = 214.36 kJ/kg
State 2: P2V2 = mRT2 so ratio it to the initial state properties
P2V2 /P1V1 = P22 /P1 = mRT2 /mRT1 = T2 /T1 =>
P2 = P1 (T2 /T1 )(1/2) = 160 × (900/300) × (1/2) = 240 kPa
Work is done while piston moves at linearly varying pressure, so we get
1W2 = 0.5(P1 + P2)(V2 -V1) = 0.5×(160 + 240) kPa × 0.8072 m3 = 161.4 kJ
1Q2 = m(u2 - u1) + 1W2 = 1.5×(674.824 - 214.36) + 161.4 = 852.1 kJ
P
V
W
1
2
T
V
1
2

Energy Equation: Polytropic Process

5.121
A helium gas in a piston cylinder is compressed from 100 kPa, 300 K to 200 kPa in a
polytropic process with n = 1.5. Find the specific work and specific heat transfer.
Energy Eq.: u2 − u1 = 1q2 − 1w2
We can calculate the actual specific work from Eq.4.5 and heat transfer from the
energy equation be first finding T2 as:
Process: T2 = T1 (P2/P1)
n - 1
n
= 300 (2)
0.5
1.5
= 377.98 K
1w2 =
1
1 - n (P2v2 − P1v1) =
R
1 - n (T2 − T1)
=
2.0771
1 - 1.5 (377.98 − 300) = -323.9 kJ/kg
u2 − u1 = Cv(T2 − T1) = 242.99 kJ/kg,
Energy Eq.: 1q2 = u2 − u1 + 1w2 = −80.9 kJ/kg
Helium Table A.5
k = γ = 1.667 so n < k
Cv = 3.116 kJ/kgK,
R = 2.0771 kJ/kgK
TP n = 12
2 1
1 T (n = 1)
qn = 1.5
v s

5.122
Oxygen at 300 kPa, 100°C is in a piston/cylinder arrangement with a volume of 0.1
m3. It is now compressed in a polytropic process with exponent, n = 1.2, to a final
temperature of 200°C. Calculate the heat transfer for the process.
Solution:
Continuty: m2 = m1
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
State 1: T1 , P1 & ideal gas, small change in T, so use Table A.5
⇒ m =
P1V1
RT1
=
300 × 0.1 m3
0.25983 × 373.15 = 0.309 kg
Process: PVn = constant
1W2 =
1
1-n (P2V2 - P1V1) =
mR
1-n (T2 - T1) =
0.309 × 0.25983
1 - 1.2 (200 - 100)
= -40.2 kJ
1Q2 = m(u2 - u1) + 1W2 ≅ mCv(T2 - T1) + 1W2
= 0.3094 × 0.662 (200 - 100) - 40.2 = -19.7 kJ
P
v
2
1
T
v
2
1T
T
1
2
T = C v-0.2
P = C v-1.2

5.123
A piston cylinder contains 0.1 kg air at 300 K and 100 kPa. The air is now slowly
compressed in an isothermal (T = C) process to a final pressure of 250 kPa. Show the
process in a P-V diagram and find both the work and heat transfer in the process.
Solution :
Process: T = C & ideal gas ⇒ PV = mRT = constant
1W2 = ∫ PdV =
⌡

⌠mRT
V dV = mRT ln
V2
V1
= mRT ln
P1
P2
= 0.1 × 0.287 × 300 ln (100 / 250 ) = -7.89 kJ
since T1 = T2 ⇒ u2 = u1
The energy equation thus becomes
1Q2 = m × (u2 - u1 ) + 1W2 = 1W2 = -7.89 kJ
P
v
2
1
T
v
2 1
T = C
P = C v-1

5.124
A piston cylinder contains 0.1 kg nitrogen at 100 kPa, 27oC and it is now compressed
in a polytropic process with n = 1.25 to a pressure of 250 kPa. Find the heat transfer.
Take CV as the nitrogen. m2 = m1 = m ;
State 2: T2 = T1 (P2 / P1)
n-1
n
= 300.15 ( 250
100 )
0.25
1.25
= 360.5 K
From process eq.:
1W2 = ∫ P dV = area =
m
1−n
(P2v2 − P1v1) =
mR
1−n
(T2 − T1)
=
0.1 × 0.2968
1 - 1.25 (360.5 – 300.15) = −7.165 kJ
From energy eq.:
1Q2 = m(u2 − u1) + 1W2 = mCv(T2 − T1) + 1W2
= 0.1 × 0.745(360.5 – 300.15) – 7.165 = −2.67 kJ
T
v
P
2
1
P = C T5 -1.25P P = C v T
-0.25T = C v2
T 22
1 1T
1
v

5.125
process with n = 1.667. How much heat transfer is involved?
Solution:
C.V. Helium gas, this is a control mass.
Energy equation: m(u2 – u1) = 1Q2 – 1W2
n
= constant = P1V
n
1 = P2V
n
2
Ideal gas (A.5): m = PV/RT =
125 × 0.25
2.0771 × 350
= 0.043 kg
Solve for the volume at state 2
V2 = V1 (P1/P2)
1/n
= 0.25 ×





125
100
0.6
= 0.2852 m3
T2 = T1 P2V2/(P1V1) = 350
100 × 0.2852
125 × 0.25 = 319.4 K
Work from Eq.4.4
1W2 =
P2V2- P1 V1
1-n =
100× 0.2852 - 125× 0.25
1 - 1.667 kPa m3 = 4.09 kJ
Use specific heat from Table A.5 to evaluate u2 – u1, Cv = 3.116 kJ/kg K
1Q2 = m(u2 – u1) + 1W2 = m Cv (T2 – T1) + 1W2
= 0.043 × 3.116 × (319.4 – 350) + 4.09 = -0.01 kJ

5.126
Find the specific heat transfer for problem 4.52
Air goes through a polytropic process from 125 kPa, 325 K to 300 kPa and 500 K.
Find the polytropic exponent n and the specific work in the process.
Solution:
Process: Pvn = Const = P1v
n
1 = P2 v
n
2
Ideal gas Pv = RT so
v1 =
RT
P =
0.287 × 325
125 = 0.7462 m3/kg
v2 =
RT
P =
0.287 × 500
300 = 0.47833 m3/kg
(P2/ P1) = (v1/ v2)
n
=> ln(P2/ P1) = n ln(v1/ v2)
n = ln(P2/ P1) / ln(v1/ v2) =
ln 2.4
ln 1.56 = 1.969
The work is now from Eq.4.4 per unit mass
1w2 =
P2v2-P1v1
1-n =
R(T2 - T1)
1-n =
0.287(500 - 325)
1-1.969 = -51.8 kJ/kg
The energy equation (per unit mass) gives
1q2 = (u2 - u1) + 1w2 ≅ Cv(T2 - T1) + 1w2
= 0.717 (500 - 325) - 51.8 = 73.67 kJ/kg

5.127
A piston/cylinder has nitrogen gas at 750 K and 1500 kPa. Now it is expanded in a
polytropic process with n = 1.2 to P = 750 kPa. Find the final temperature, the specific
work and specific heat transfer in the process.
C.V. Nitrogen. This is a control mass going through a polytropic process.
Continuty: m2 = m1
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
Process: Pvn = constant
Substance ideal gas: Pv = RT
T2 = T1 (P2/P1)
n-1
n = 750





750
1500
0.2
1.2
= 750 × 0.8909 = 668 K
The work is integrated as in Eq.4.4
1w2 = ⌡⌠Pdv =
1
1 - n (P2v2 - P1v1) =
R
1 - n (T2 - T1)
=
0.2968
1 - 1.2(668 - 750) = 121.7 kJ/kg
The energy equation with values of u from Table A.8 is
1q2 = u2 - u1 + 1w2 = 502.8 - 568.45 + 121.7 = 56.0 kJ/kg
If constant specific heat is used from Table A.5
1q2 = Cv(T2 - T1) + 1w2 = 0.745(668 – 750) + 121.7 = 60.6 kJ/kg

5.128
A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527oC after
combustion and this is expanded in a polytropic process with n = 1.5 to a volume 10
times larger. Find the expansion work and heat transfer using Table A.5 heat capacity.
Take CV as the air. m2 = m1 = m ;
Energy Eq.5.11 m(u2 − u1) = 1Q2 − 1W2
State 2: P2 = P1 (v2 / v1)
-n
= 4000 ×10
-1.5
= 126.5 kPa
T2 = T1 ( P2v2 / P1v1) = (1527 + 273)
126.5 × 10
4000 = 569.3 K
From process eq.: 1W2 = ∫ P dV =
m
1−n
(P2v2 − P1v1) =
mR
1−n
(T2 − T1)
=
0.1 × 0.287
1 - 1.5 (569.3 – 1800) = 70.64 kJ
From energy eq.: 1Q2 = m(u2 − u1) + 1W2 = mCv(T2 − T1) + 1W2
= 0.1 × 0.717(569.3 – 1800) + 70.64 = −17.6 kJ
T
v
P
1
2
P = C T3 -1.5P P = C v T
-0.5T = C v1
T 11
2 2T2
v

5.129
Solve the previous problem using Table A.7
Take CV as the air. m2 = m1 = m ;
Energy Eq.5.11 m(u2 − u1) = 1Q2 − 1W2
State 2: P2 = P1 (v2 / v1)
-n
= 4000 ×10
-1.5
= 126.5 kPa
T2 = T1 ( P2v2 / P1v1) = (1527 + 273)
126.5 × 10
4000 = 569.3 K
From process eq.: 1W2 = ∫ P dV =
m
1−n
(P2v2 − P1v1) =
mR
1−n
(T2 − T1)
=
0.1 × 0.287
1 - 1.5 (569.3 – 1800) = 70.64 kJ
From energy eq.: 1Q2 = m(u2 − u1) + 1W2
= 0.1 (411.78 – 1486.33) + 70.64 = −36.8 kJ
The only place where Table A.7 comes in is for values of u1 and u2
T
v
P
1
2
P = C T3 -1.5P P = C v T
-0.5T = C v1
T 11
2 2T2
v

5.130
A piston/cylinder arrangement of initial volume 0.025 m3 contains saturated water
vapor at 180°C. The steam now expands in a polytropic process with exponent n = 1
to a final pressure of 200 kPa, while it does work against the piston. Determine the
heat transfer in this process.
Solution:
State 1: Table B.1.1 P = 1002.2 kPa, v1 = 0.19405 m3/kg, u1 = 2583.7 kJ/kg ,
m = V/v1 = 0.025/0.19405 = 0.129 kg
Process: Pv = const. = P1v1 = P2v2 ; polytropic process n = 1.
⇒ v2 = v1P1/P2 = 0.19405 × 1002.1/200 = 0.9723 m3/kg
State 2: P2, v2 ⇒ Table B.1.3 T2 ≅ 155°C , u2 = 2585 kJ/kg
1W2 = ⌡⌠PdV = P1V1 ln
v2
v1
= 1002.2 kPa × 0.025 m3 ln
0.9723
0.19405 = 40.37 kJ
1Q2 = m(u2 - u1) + 1W2 = 0.129(2585 - 2583.7) + 40.37 = 40.54 kJ
P
v
2
1
T
v
21
P = C v-1
T = C
Sat vapor line
Notice T drops, it is not an ideal gas.

5.131
A piston/cylinder in a car contains 0.2 L of air at 90 kPa, 20°C, shown in Fig. P5.131.
The air is compressed in a quasi-equilibrium polytropic process with polytropic
exponent n = 1.25 to a final volume six times smaller. Determine the final pressure,
temperature, and the heat transfer for the process.
Solution:
C.V. Air. This is a control mass going through a polytropic process.
Continuty: m2 = m1
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
Process: Pvn = const.
P1v1
n = P2v2
n ⇒ P2 = P1(v1/v2)n = 90 × 61.25
= 845.15 kPa
Substance ideal gas: Pv = RT
T2 = T1(P2v2/P1v1) = 293.15(845.15/90 × 6) = 458.8 K
P
v
2
1
P = C v
-1.25
T
v
2
1
T = C v
-0.25
m =
PV
RT =
90 × 0.2×10-3
0.287 × 293.15 = 2.14×10-4 kg
The work is integrated as in Eq.4.4
1w2 = ⌡⌠Pdv =
1
1 - n (P2v2 - P1v1) =
R
1 - n (T2 - T1)
=
0.287
1 - 1.25(458.8 - 293.15) = -190.17 kJ/kg
The energy equation with values of u from Table A.7 is
1q2 = u2 - u1 + 1w2 = 329.4 - 208.03 – 190.17 = -68.8 kJ/kg
1Q2 = m 1q2 = -0.0147 kJ (i.e a heat loss)

5.132
A piston/cylinder has 1 kg propane gas at 700 kPa, 40°C. The piston cross-sectional
area is 0.5 m2, and the total external force restraining the piston is directly
proportional to the cylinder volume squared. Heat is transferred to the propane until its
temperature reaches 700°C. Determine the final pressure inside the cylinder, the work
done by the propane, and the heat transfer during the process.
Solution:
C.V. The 1 kg of propane.
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Process: P = Pext = CV2 ⇒ PV-2 = constant, polytropic n = -2
Ideal gas: PV = mRT, and process yields
P2 = P1(T2/T1)
n
n-1 = 700





700+273.15
40+273.15
2/3
= 1490.7 kPa
The work is integrated as Eq.4.4
1W2 = ⌡⌠
1
2
PdV =
P2V2 - P1V1
1 - n =
mR(T2 - T1)
1 - n
=
1× 0.18855 × (700 – 40)
1– (–2) = 41.48 kJ
The energy equation with specific heat from Table A.5 becomes
1Q2 = m(u2 - u1) + 1W2 = mCv(T2 - T1) + 1W2
= 1 × 1.490 × (700 - 40) + 41.48 = 1024.9 kJ
P
V
2
1
P = C V
2
T
V
2
1
T = C V
3

5.133
A piston cylinder contains pure oxygen at ambient conditions 20°C, 100 kPa. The
piston is moved to a volume that is 7 times smaller than the initial volume in a
polytropic process with exponent n = 1.25. Use constant heat capacity to find the final
pressure and temperature, the specific work and the specific heat transfer.
Energy Eq.: u2 – u1 = 1q2 - 1w2
Process Eq: Pv
n
= C; P2 = P1 (v1/v2)
n
= 100 ( )7
1.25
= 1138.6 kPa
From the ideal gas law and state 2 (P, v) we get
T2 = T1( P2/P1)(v1/v2) = 293 ×
1138.6
100 × (1/7) = 476.8 K
We could also combine process eq. and gas law to give: T2 = T1 (v1/v2)
n-1
Polytropic work Eq. 4.5: 1w2 =
1
1-n (P2v2 – P1v1) =
R
1-n (T2 – T1)
1w2 =
0.2598
1 - 1.25
kJ
kg K × (476.8 – 293.2) K = –190.88 kJ/kg
1q2 = u2 – u1 + 1w2 = Cv (T2 – T1) + 1w2
= 0.662 (476.8 – 293.2) – 190.88 = –69.3 kJ/kg
on a steeper curve
than n = 1.
P
v
w
1
2
n = 1100
2P

5.134
An air pistol contains compressed air in a small cylinder, shown in Fig. P5.134.
Assume that the volume is 1 cm3, pressure is 1 MPa, and the temperature is 27°C
when armed. A bullet, m = 15 g, acts as a piston initially held by a pin (trigger); when
released, the air expands in an isothermal process (T = constant). If the air pressure is
0.1 MPa in the cylinder as the bullet leaves the gun, find
a. The final volume and the mass of air.
b. The work done by the air and work done on the atmosphere.
c. The work to the bullet and the bullet exit velocity.
Solution:
C.V. Air.
Air ideal gas: mair = P1V1/RT1 = 1000 × 10-6/(0.287 × 300) = 1.17×10-5 kg
Process: PV = const = P1V1 = P2V2 ⇒ V2 = V1P1/P2 = 10 cm3
1W2 = ⌡⌠PdV =
⌡

⌠P1V1
V dV = P1V1 ln (V2/V1) = 2.303 J
1W2,ATM = P0(V2 - V1) = 101 × (10 − 1) × 10-6 kJ = 0.909 J
Wbullet = 1W2 - 1W2,ATM = 1.394 J =
1
2 mbullet(Vexit)2
Vexit = (2Wbullet/mB)1/2 = (2 × 1.394/0.015)1/2 = 13.63 m/s

5.135
Consider a piston cylinder with 0.5 kg of R-134a as saturated vapor at -10°C. It is now
compressed to a pressure of 500 kPa in a polytropic process with n = 1.5. Find the
final volume and temperature, and determine the work done during the process.
Solution:
Take CV as the R-134a which is a control mass
Continuity: m2 = m1 = m ; Energy: m(u2 -u1) = 1Q2 - 1W2
Process: Pv1.5
= constant. Polytropic process with n = 1.5
1: (T, x) P = Psat = 201.7 kPa from Table B.5.1
v1 = 0.09921 m3/kg, u1 = 372.27 kJ/kg
2: (P, process)
v2 = v1 (P1/P2) (1/1.5)
= 0.09921× (201.7/500)0.667
= 0.05416 m3/kg
=> Table B.5.2 superheated vapor, T2 = 79°C, V2 = mv2 = 0.027 m3
u2 = 440.9 kJ/kg
Process gives P = C v (-1.5)
, which is integrated for the work term, Eq.4.4
1W2 = ∫ P dV = m(P2v2 - P1v1)/(1-1.5)
= -2×0.5× (500×0.05416 - 201.7×0.09921) = -7.07 kJ
1Q2 = m(u2 -u1) + 1W2 = 0.5(440.9 - 372.27) + (-7.07) = 27.25 kJ

Energy Equation in Rate Form

5.136
A crane use 2 kW to raise a 100 kg box 20 m. How much time does it take?
Power = W
.
= FV = mgV = mg
L
t
t =
mgL
W
. =
100 kg 9.807 m/s2 20 m
2000 W = 9.81 s

5.137
A crane lifts a load of 450 kg vertically up with a power input of 1 kW. How fast can
the crane lift the load?
Solution :
Power is force times rate of displacement
W
.
= F V = mg⋅V⋅
V =
W
.
mg =
1000
450 × 9.806
W
N = 0.227 m/s

5.138
A pot of 1.2 kg water at 20oC is put on a stove supplying 1250 W to the water. What
is the rate of temperature increase (K/s)?
Energy Equation on a rate form:
dEwater
dt =
dUwater
dt = Q
.
- W
.
= Q
.
- PV
.
Q
.
=
dUwater
dt + PV
.
=
dHwater
dt = mwaterCp
dTwater
dt
dTwater
dt = Q
.
/ mwaterCp = 1.250 / (1.2 × 4.18 ) = 0.2492 K/s

5.139
The rate of heat transfer to the surroundings from a person at rest is about 400 kJ/h.
Suppose that the ventilation system fails in an auditorium containing 100 people.
Assume the energy goes into the air of volume 1500 m3 initially at 300 K and 101
kPa. Find the rate (degrees per minute) of the air temperature change.
Solution:
Q
.
= n q
.
= 100× 400 = 40 000 kJ/h = 666.7 kJ/min
dEair
dt = Q
.
= mairCv
dTair
dt
mair = PV/RT = 101 × 1500 / 0.287 × 300 = 1759.6 kg
dTair
dt = Q
.
/mCv = 666.7 / (1759.6 × 0.717) = 0.53°C/min

5.140
A pot of water is boiling on a stove supplying 325 W to the water. What is the rate of
mass (kg/s) vaporizing assuming a constant pressure process?
To answer this we must assume all the power goes into the water and that the
process takes place at atmospheric pressure 101 kPa, so T = 100oC.
Energy equation
dQ = dE + dW = dU + PdV = dH = hfg dm
dQ
dt = hfg
dm
dt
dm
dt =
.
Q
hfg
=
325 W
2257 kJ/kg = 0.144 g/s
The volume rate of increase is
dV
dt =
dm
dt vfg = 0.144 g/s × 1.67185 m3/kg
= 0.24 × 10-3 m3/s = 0.24 L/s

5.141
A pot of 1.2 kg water at 20oC is put on a stove supplying 1250 W to the water. After
how long time can I expect it to come to a boil (100oC)?
Energy Equation on a rate form:
dEwater
dt =
dUwater
dt = Q
.
- W
.
= Q
.
- PV
.
Q
.
=
dUwater
dt + PV
.
=
dHwater
dt = mwaterCp
dTwater
dt
Integrate over time
Q = Q
.
∆t = ∆H = mwater (h2 – h1) ≈ mwaterCp (T2 – T1)
∆t = mwater (h2 – h1)/Q
.
≈ mwaterCp (T2 – T1)/Q
.
= 1.2 (419.02 – 83.94)/1.25 ≈ 1.2 × 4.18 (100 – 20)/1.25
= 321.7 s ≈ 5.5 min
Comment: Notice how close the two results are, i.e. use of constant Cp is OK.

5.142
A mass of 3 kg nitrogen gas at 2000 K, V = C, cools with 500 W. What is dT/dt?
Process: V = C 1W2= 0
dE
dt =
dU
dt = m
dU
dt = mCv
dT
dt = Q
.
– W = Q
.
= -500 W
Cv 2000 =
du
dT =
∆u
∆T
=
u2100 - u1900
2100-1900 =
1819.08 - 1621.66
200 = 0.987 kJ/kg K
dT
dt =
Q
.
mCv
=
-500 W
3 × 0.987 kJ/K
= -0.17
K
s
Remark: Specific heat from Table A.5 has Cv 300 = 0.745 kJ/kg K which is nearly
25% lower and thus would over-estimate the rate with 25%.

5.143
A computer in a closed room of volume 200 m3 dissipates energy at a rate of 10 kW.
The room has 50 kg wood, 25 kg steel and air, with all material at 300 K, 100 kPa.
Assuming all the mass heats up uniformly, how long will it take to increase the
temperature 10°C?
Solution:
C.V. Air, wood and steel. m2 = m1 ; no work
Energy Eq.5.11: U2 - U1 = 1Q2 = Q
.
∆t
The total volume is nearly all air, but we can find volume of the solids.
Vwood = m/ρ = 50/510 = 0.098 m3 ; Vsteel = 25/7820 = 0.003 m3
Vair = 200 - 0.098 - 0.003 = 199.899 m3
mair = PV/RT = 101.325 × 199.899/(0.287 × 300) = 235.25 kg
We do not have a u table for steel or wood so use heat capacity from A.3.
∆U = [mair Cv + mwood Cv + msteel Cv ]∆T
= (235.25 × 0.717 + 50 × 1.38 + 25 × 0.46) 10
= 1686.7 + 690 +115 = 2492 kJ = Q
.
× ∆t = 10 kW × ∆t
=> ∆t = 2492/10 = 249.2 sec = 4.2 minutes

5.144
A drag force on a car, with frontal area A = 2 m2, driving at 80 km/h in air at 20oC is
Fd = 0.225 A ρairV2. How much power is needed and what is the traction force?
W
.
= FV
V = 80
km
h = 80 ×
1000
3600 ms-1 = 22.22 ms-1
ρAIR =
P
RT =
101
0.287 × 293
= 1.20 kg/m3
F = 0.225 AρV2 = 0.225 × 2 × 1.2 × 22.222 = 266.61 Nd
W
.
= FV = 266.61 N × 22.22 m/s = 5924 W = 5.92 kW

5.145
A piston/cylinder of cross sectional area 0.01 m2 maintains constant pressure. It
contains 1 kg water with a quality of 5% at 150oC. If we heat so 1 g/s liquid turns into
vapor what is the rate of heat transfer needed?
Solution:
Control volume the water.
Continuity Eq.: mtot = constant = mvapor + mliq
on a rate form: m
.
tot = 0 = m
.
vapor + m
.
liq ⇒ m
.
liq = -m
.
vapor
Energy Eq.:
dE
dt =
d
dt(mvaporug + mliquf) = m
.
vapor( ug - uf) = m
.
vaporufg = Q
.
- W
.
Vvapor = mvapor vg , Vliq = mliq vf ; Vtot = Vvapor + Vliq
V
.
tot = V
.
vapor + V
.
liq = m
.
vaporvg + m
.
liqvf = m
.
vapor (vg- vf ) = m
.
vapor vfg
W
.
= PV
.
= P m
.
vapor vfg
Substitute the rate of work into the energy equation and solve for the heat transfer
Q
.
= m
.
vaporufg + W
.
= m
.
vaporufg + P m
.
vapor vfg = m
.
vapor hfg
= 0.001 × 2114.26 = 2.114 kW

5.146
A small elevator is being designed for a construction site. It is expected to carry four
75-kg workers to the top of a 100-m tall building in less than 2 min. The elevator cage
will have a counterweight to balance its mass. What is the smallest size (power)
electric motor that can drive this unit?
Solution:
m = 4 × 75 = 300 kg ; ∆Z = 100 m ; ∆t = 2 minutes
-W
.
= ∆PE
.
= mg
∆Z
∆t
=
300 × 9.807 × 100
1000 × 2 × 60 = 2.45 kW

5.147
The heaters in a spacecraft suddenly fail. Heat is lost by radiation at the rate of 100
kJ/h, and the electric instruments generate 75 kJ/h. Initially, the air is at 100 kPa, 25°C
with a volume of 10 m3. How long will it take to reach an air temperature of −20°C?
Solution:
C.V.
C.M. Air
elQ
.
radQ
.
Continuity Eq:
dM
dt = 0
Energy Eq:
dE
dt = Q
.
el - Q
.
rad
W
.
= 0
KE
.
= 0
PE
.
= 0
E
.
= U
.
= Q
.
el - Q
.
rad = Q
.
net ⇒ U2 - U1 = m(u2 - u1) = Q
.
net(t2 - t1)
Ideal gas: m =
P1V1
RT1
=
100 ×10
0.287 × 298.15 = 11.688 kg
m(u2 - u1) = m Cv0(T2 − T1) = 11.688 ×0.717 (-20 - 25) = −377.1 kJ
t2 - t1 = m (u2 - u1)/Q
.
net = (−377.1 kJ) /(-25 kJ/h) = 15.08 h

5.148
A steam generating unit heats saturated liquid water at constant pressure of 800 kPa in
a piston cylinder. If 1.5 kW of power is added by heat transfer find the rate (kg/s) of
saturated vapor that is made.
Solution:
Energy equation on a rate form aking saturated vapor from saturated liquidm
U
.
= (mu)
.
= m
.
∆u = Q
.
- W
.
= Q
.
- PV
.
= Q
.
– P m
.
∆v
Rearrange to solve for heat transfer rate
Q
.
= m
.
(∆u + ∆vP) = m
.
∆h = m
.
hfg
So now
m
.
= Q
.
/ hfg = 1500 / 2048.04 = 0.732 kg/s

5.149
As fresh poured concrete hardens, the chemical transformation releases energy at a
rate of 2 W/kg. Assume the center of a poured layer does not have any heat loss and
that it has an average heat capacity of 0.9 kJ/kg K. Find the temperature rise during 1
hour of the hardening (curing) process.
Solution:
U
.
= (mu)
.
= mCvT
.
= Q
.
= mq
.
T
.
= q
.
/Cv = 2×10-3 / 0.9
= 2.222 × 10-3 °C/sec
∆T = T
.
∆t = 2.222 × 10-3 × 3600 = 8 °C

5.150
Water is in a piston cylinder maintaining constant P at 700 kPa, quality 90% with a
volume of 0.1 m3. A heater is turned on heating the water with 2.5 kW. What is the
rate of mass (kg/s) vaporizing?
Solution:
Control volume water.
Continuity Eq.: mtot = constant = mvapor + mliq
on a rate form: m
.
tot = 0 = m
.
vapor + m
.
liq ⇒ m
.
liq = -m
.
vapor
Energy equation: U
.
= Q
.
- W
.
= m
.
vapor ufg = Q
.
- P m
.
vapor vfg
Rearrange to solve for m
.
vapor
m
.
vapor (ufg + Pvfg) = m
.
vapor hfg = Q
.
m
.
vapor = Q
.
/hfg =
2.5
2066.3
kW
kJ/kg = 0.0012 kg/s

5.151
A 500 Watt heater is used to melt 2 kg of solid ice at −10oC to liquid at +5oC at a
constant pressure of 150 kPa.
a) Find the change in the total volume of the water.
b) Find the energy the heater must provide to the water.
c) Find the time the process will take assuming uniform T in the water.
Solution:
Take CV as the 2 kg of water. m2 = m1 = m ;
Energy Eq.5.11 m(u2 − u1) = 1Q2 − 1W2
State 1: Compressed solid, take saturated solid at same temperature.
v = vi(−10) = 0.0010891 m3/kg, h = hi = −354.09 kJ/kg
State 2: Compressed liquid, take saturated liquid at same temperature
v = vf = 0.001, h = hf = 20.98 kJ/kg
Change in volume:
V2 − V1 = m(v2 − v1) = 2(0.001 − 0.0010891) = 0.000178 m3
Work is done while piston moves at constant pressure, so we get
1W2 = ∫ P dV = area = P(V2 − V1) = -150 × 0.000178 = −0.027 kJ = −27 J
1Q2 = m(u2 − u1) + 1W2 = m(h2 − h1) = 2 × [20.98 − (−354.09)] = 750 kJ
The elapsed time is found from the heat transfer and the rate of heat transfer
t = 1Q2/Q
.
= (750 kJ /500 W) × 1000 J/kJ = 1500 s = 25 min
P
T
v
V
L
S
C.P.
2
1
S+V
L+V
P C.P.
v
T
C.P.
v
12
1
2
P = const.

Problem Analysis (no numbers required)

5.152
Consider Problem 5.57 with the steel bottle as C.V. Write the process equation that is
valid until the valve opens and plot the P-v diagram for the process.
Process: constant volume process
constant mass
V = mv = C ⇒ v2 = v1
1W2 = ∫ P dV = 0
State 1: (T, x) so two-phase in Table B.3.1
v
P
1
2

5.153
Consider Problem 5.50. Take the whole room as a C.V. and write both conservation of
mass and energy equations. Write some equations for the process (two are needed) and
use those in the conservation equations. Now specify the four properties that
determines initial (2) and final state (2), do you have them all? Count unknowns and
match with equations to determine those.
Mass: m2 - m1 = 0 ; m2 = m1 = Vreactor/v1
Energy: m(u2 - u1) = 1Q2 - 1W2
Process: Room volume constant V = C 1W2 = 0
Room insulated 1Q2 = 0
Using these in the equation for mass and energy gives:
m2 = V2/v2 = m1 ; m(u2 – u1) = 0 – 0 = 0
State 1: P1, T1 so Table B.1.4 gives v1, u1 m1
State 2: P2, ?
We do not know one state 2 property and the total room volume
Energy equation then gives u2 = u1 (a state 2 property)
State 2: P2, u2 v2
Now we have the room volume as
Continuity Eq.: m2 = V/v2 = m1 so V = m1 v2

5.154
Take problem 5.61 and write the left hand side (storage change) of the conservation
equations for mass and energy. How do you write m1 and Eq. 5.5?
C.V.: Both rooms A and B in tank. BA
Continuity Eq.: m2 – mA1 – mB1 = 0 ;
Energy Eq.: m2u2 – mA1uA1 – mB1uB1 = 1Q2 – 1W2
Notice how the state 1 term split into two terms
m1 = ∫ ρ dV = ∫ (1/v) dV = VA/vA1 + VB/vB1 = mA1 + mB1
and for energy as
m1u1 = ∫ ρu dV = ∫ (u/v) dV = (uA1/vA1)VA + (uB1/vB1)VB
= mA1uA1 + mB1uB1
Formulation continues as:
Process constant total volume: Vtot = VA + VB and 1W2 = 0/
m2 = mA1 + mB1 => v2 = Vtot/m2
etc.

5.155
Consider Problem 5.70 with the final state given but that you were not told the piston
hits the stops and only told Vstop = 2 V1. Sketch the possible P-v diagram for the
process and determine which number(s) you need to uniquely place state 2 in the
diagram. There is a kink in the process curve what are the coordinates for that state?
Write an expression for the work term.
C.V. R-410a. Control mass goes through process: 1 -> 2 -> 3
As piston floats pressure is constant (1 -> 2) and the volume is constant for the
second part (2 -> 3). So we have: v3 = v2 = 2 × v1
State 2: V2 = Vstop ⇒ v2 = 2 × v1 = v3 and P2= P1 ⇒ T2 = …
State 3: Table B.4.2 (P, T) Compare P3 > P2 and T3 > T2
v3 = 0.02015 m3/kg, u3 = 248.4 kJ/kg
W = ∫ P dV = P(V2 – V1) = Pm (v2 – v1)
v
P
1 2
3
R-410a
Po
cb

5.156
Look at problem 5.115 and plot the P-v diagram for the process. Only T2 is given,
how do you determine the 2nd
property of the final state? What do you need to check
and does it have an influence on the work term?
Process: P = constant = F/A = P1 if V > Vmin
V = constant = V1a = Vmin if P < P1
State 1: (P, T) V1 = mRT1/P1 = 0.5 × 0.287 × 1000/2000 = 0.07175 m3
The only possible P-V combinations for this system are shown in the diagram so both
state 1 and 2 must be on the two lines. For state 2 we need to know if it is on the
horizontal P line segment or the vertical V segment. Let us check state 1a:
State 1a: P1a = P1, V1a = Vmin , Ideal gas so T1a = T1
V1a
V1
We see if T2 < T1a then state 2 must have V2 = V1a = Vmin = 0.03 m3. So state 2 is
known by (T2, v2) and P2 = P1×
T2
T1
×
V1
V2
If it was that T2 > T1a then we know state 2 as: T2, P2 = P1 and we then have
V2 = V1×
T2
T1
The work is the area under the process curve in the P-V diagram and so it does make a
difference where state 2 is relative to state 1a. For the part of the process that proceeds
along the constant volume Vmin the work is zero there is only work when the volume
changes.
1W2 = ⌡⌠
1
2 P dV = P1 (V1a – V1)
V
P
1
2
1a
P
P
2
1
V
T
1
2
1a
T
T
2
1a
T
1

Review

5.157
Ten kilograms of water in a piston/cylinder setup with constant pressure is at 450°C
and a volume of 0.633 m3. It is now cooled to 20°C. Show the P–v diagram and find
the work and heat transfer for the process.
Solution:
Energy Eq.5.11: m(u2 - u1) = 1Q2 − 1W2
Process: P = C ⇒ 1W2 = mP(v2 -v1)
State 1: (T, v1 = 0.633/10 = 0.0633 m3/kg) Table B.1.3
P1 = 5 MPa, h1 = 3316.2 kJ/kg
State 2: (P = P = 5 MPa, 20°C) ⇒ Table B.1.4
v2 = 0.000 999 5 m3/kg ; h2 = 88.65 kJ/kg
v
P
12
v
T
1
2
5 MPa
The work from the process equation is found as
1W2 = 10 × 5000 ×(0.0009995 - 0.0633) = -3115 kJ
The heat transfer from the energy equation is
1Q2 = m(u2 - u1) + 1W2 = m(h2 - h1)
1Q2 = 10 ×(88.65 - 3316.2) = -32276 kJ

5.158
Ammonia, NH3, is contained in a sealed rigid tank at 0°C, x = 50% and is then heated
to 100°C. Find the final state P2, u2 and the specific work and heat transfer.
Solution:
Continuity Eq.: m2 = m1 ;
Energy Eq.5.11: E2 - E1 = 1Q2 ; ( )1W2 = 0/
Process: V2 = V1 ⇒ v2 = v1 = 0.001566 + 0.5 × 0.28783 = 0.14538 m3/kg
Table B.2.2: v2 & T2 ⇒ between 1000 kPa and 1200 kPa
P2 = 1000 + 200
0.14538 – 0.17389
0.14347 – 0.17389 = 1187 kPa
V
P
1
2
u2 = 1490.5 + (1485.8 – 1490.5) × 0.935
= 1485.83 kJ/kg
u1 = 179.69 + 0.5 × 1138.3 = 748.84 kJ/kg
Process equation gives no displacement:1w2 = 0 ;
The energy equation then gives the heat transfer as
1q2 = u2 - u1 = 1485.83 – 748.84 = 737 kJ/kg

5.159
A piston/cylinder (Fig. P4.122) contains 1 kg of water at 20°C with a volume of 0.1
m3. Initially the piston rests on some stops with the top surface open to the
atmosphere, Po and a mass so a water pressure of 400 kPa will lift it. To what
temperature should the water be heated to lift the piston? If it is heated to saturated
vapor find the final temperature, volume and the work, 1W2.
Solution:
m2 = m1 = m ; m(u2 - u1) = 1Q2 - 1W2
V
P
1
2
H O
Po
2
1a
State 1: 20 C, v1 = V/m = 0.1/1 = 0.1 m3/kg
x = (0.1 - 0.001002)/57.789 = 0.001713
u1 = 83.94 + 0.001713 × 2318.98 = 87.92 kJ/kg
To find state 2 check on state 1a:
P = 400 kPa, v = v1 = 0.1 m3/kg
Table B.1.2: vf < v < vg = 0.4625 m3/kg
State 2 is saturated vapor at 400 kPa since state 1a is two-phase.
v2 = vg = 0.4625 m3/kg , V2 = m v2 = 0.4625 m3, u2 = ug= 2553.6 kJ/kg
Pressure is constant as volume increase beyond initial volume.
1W2 = ∫ P dV = P (V2 - V1) = Plift (V2 – V1) = 400 (0.4625 – 0.1) = 145 kJ
1Q2 = m(u2 - u1) + 1W2 = 1 (2553.6 – 87.92) + 145 = 2610.7 kJ

5.160
A piston/cylinder contains 1 kg of ammonia at 20°C with a volume of 0.1 m3, shown
in Fig. P5.160. Initially the piston rests on some stops with the top surface open to the
atmosphere, Po, so a pressure of 1400 kPa is required to lift it. To what temperature
should the ammonia be heated to lift the piston? If it is heated to saturated vapor find
the final temperature, volume, and the heat transfer.
Solution:
C.V. Ammonia which is a control mass.
m2 = m1 = m ; m(u2 -u1) = 1Q2 - 1W2
State 1: 20°C; v1 = 0.10 < vg ⇒ x1 = (0.1 – 0.001638)/0.14758 = 0.6665
u1 = uf + x1 ufg = 272.89 + 0.6665 ×1059.3 = 978.9 kJ/kg
Process: Piston starts to lift at state 1a (Plift, v1)
State 1a: 1400 kPa, v1 Table B.2.2 (superheated vapor)
Ta = 50 + (60 – 50)
0.1 – 0.09942
0.10423 – 0.09942 = 51.2 °C
1400
1200
857
P
v
1
2
1a
T
v
1
2
1a
State 2: x = 1.0, v2 = v1 => V2 = mv2 = 0.1 m3
T2 = 30 + (0.1 – 0.11049) × 5/(0.09397 – 0.11049) = 33.2 °C
u2 = 1338.7 kJ/kg; 1W2 = 0;
1Q2 = m1q2 = m(u2 – u1) = 1 (1338.7 – 978.9) = 359.8 kJ/kg

5.161
Consider the system shown in Fig. P5.161. Tank A has a volume of 100 L and contains
saturated vapor R-134a at 30°C. When the valve is cracked open, R-134a flows slowly
into cylinder B. The piston mass requires a pressure of 200 kPa in cylinder B to raise the
piston. The process ends when the pressure in tank A has fallen to 200 kPa. During this
process heat is exchanged with the surroundings such that the R-134a always remains at
30°C. Calculate the heat transfer for the process.
Solution:
C.V. The R-134a. This is a control mass.
Energy Eq.5.11: m(u2 − u1) = 1Q2 - 1W2
Process in B: If VB > 0 then P = Pfloat (piston must move)
⇒ 1W2 = ∫ Pfloat dV = Pfloatm(v2 - v1)
Work done in B against constant external force (equilibrium P in cyl. B)
State 1: 30°C, x = 1. Table B.5.1: v1 = 0.02671 m3/kg, u1 = 394.48 kJ/kg
m = V/v1 = 0.1 / 0.02671 = 3.744 kg
State 2: 30°C, 200 kPa superheated vapor Table B.5.2
v2 = 0.11889 m3/kg, u2 = 403.1 kJ/kg
1W2 = Pfloatm(v2 - v1) = 200×3.744×(0.11889 - 0.02671) = 69.02 kJ
1Q2 = m(u2 - u1) + 1W2 = 3.744 ×(403.1 - 394.48) + 69.02 = 101.3 kJ
V
P
1 2
B
B
B
A
Tank
Po
g

5.162
Water in a piston/cylinder, similar to Fig. P5.160, is at 100oC, x = 0.5 with mass 1 kg
and the piston rests on the stops. The equilibrium pressure that will float the piston is
300 kPa. The water is heated to 300oC by an electrical heater. At what temperature
would all the liquid be gone? Find the final (P,v), the work and heat transfer in the
process.
Continuty: m2 = m1 = m ; Energy: m(u2 - u1) = 1Q2 - 1W2
Process: V = constant if P < Plift otherwise P = Plift see P-v diagram.
State 1: (T,x) Table B.1.1
v1 = 0.001044 + 0.5 × 1.6719 = 0.83697 m3/kg
u1 = 418.91 + 0.5 × 2087.58 = 1462.7 kJ/kg
State 1a: (300 kPa, v = v1 > vg 300 kPa = 0.6058 m3/kg) so superheated vapor
Piston starts to move at state 1a, 1W1a = 0, u1a = 2768.82 kJ/kg
1Q1a = m(u – u) = 1 (2768.82 – 1462.7) = 1306.12 kJ
State 1b: reached before state 1a so v = v1 = vg see this in B.1.1
T1b = 120 + 5 (0.83697 – 0.8908)/(0.76953 – 0.8908) = 122.2oC
State 2: (T2 > T1a) Table B.1.3 => v2 = 0.87529, u2 = 2806.69 kJ/kg
Work is seen in the P-V diagram (when volume changes P = Plift)
1W2 = 1aW2 = P2 m(v2 - v1) = 300 × 1(0.87529 - 0.83697) = 11.5 kJ
1Q2 = 1 (2806.69 - 1462.7) + 11.5 = 1355.5 kJ
H O
Po
2
cb
P
1a 2
1b
v
100 C
o
1cb

5.163
A rigid container has two rooms filled with water, each 1 m3 separated by a wall (see
Fig. P5.61). Room A has P = 200 kPa with a quality x = 0.80. Room B has P = 2 MPa
and T = 400°C. The partition wall is removed and the water comes to a uniform state,
which after a while due to heat transfer has a temperature of 200°C. Find the final
pressure and the heat transfer in the process.
Solution:
C.V. A + B. Constant total mass and constant total volume.
Continuity: m2 – mA1– mB1= 0 ; V2= VA+ VB= 2 m3
Energy Eq.5.11: U2 – U1 = m2u2 – mA1uA1 – mA1uA1 = 1Q2 – 1W2 = 1Q2
Process: V = VA + VB = constant => 1W2 = 0
State 1A: Table B.1.2 uA1= 504.47 + 0.8 × 2025.02 = 2124.47 kJ/kg,
vA1= 0.001061 + 0.8 × 0.88467 = 0.70877 m3/kg
State 1B: Table B.1.3 u B1= 2945.2, vB1= 0.1512
mA1= 1/vA1= 1.411 kg mB1= 1/vB1= 6.614 kg
State 2: T2, v2 = V2/m 2= 2/(1.411 + 6.614) = 0.24924 m3/kg
Table B.1.3 superheated vapor. 800 kPa < P2 < 1 MPa
Interpolate to get the proper v2
P2 ≅ 800 +
0.24924-0.2608
0.20596-0.2608 × 200 = 842 kPa u2 ≅ 2628.8 kJ/kg
1Q2 = 8.025 × 2628.8 – 1.411 × 2124.47 – 6.614 × 2945.2 = - 1381 kJ
A B
Q
P
A1
v
PA1
PB1
B1
2

5.164
A piston held by a pin in an insulated cylinder, shown in Fig. P5.164, contains 2 kg
water at 100°C, quality 98%. The piston has a mass of 102 kg, with cross-sectional
area of 100 cm2, and the ambient pressure is 100 kPa. The pin is released, which
allows the piston to move. Determine the final state of the water, assuming the process
to be adiabatic.
Solution:
C.V. The water. This is a control mass.
Energy Eq.5.11: m(u2 − u1) = 1Q2 - 1W2
Process in cylinder: P = Pfloat (if piston not supported by pin)
P2 = Pfloat = P0 + mpg/A = 100 +
102 × 9.807
100×10-4 × 103
= 200 kPa
We thus need one more property for state 2 and we have one equation namely the
energy equation. From the equilibrium pressure the work becomes
1W2 = ∫ Pfloat dV = P2 m(v2 - v1)
With this work the energy equation gives per unit mass
u2 − u1 = 1q2 - 1w2 = 0 - P2(v2 - v1)
or with rearrangement to have the unknowns on the left hand side
u2 + P2v2 = h2 = u1 + P2v1
h2 = u1 + P2v1 = 2464.8 + 200 × 1.6395 = 2792.7 kJ/kg
State 2: (P2 , h2) Table B.1.3 => T2 ≅ 161.75°C

5.165
A piston/cylinder arrangement has a linear spring and the outside atmosphere acting
on the piston, shown in Fig. P5.165. It contains water at 3 MPa, 400°C with the
volume being 0.1 m3. If the piston is at the bottom, the spring exerts a force such that
a pressure of 200 kPa inside is required to balance the forces. The system now cools
until the pressure reaches 1 MPa. Find the heat transfer for the process.
Solution:
C.V. Water.
Energy Eq.5.11: m(u2 − u1) = 1Q2 - 1W2
v2 v10
200 kPa
3 MPa
1 MPa
2
1
P
V, v
v1 = 0.09936 m3/kg, u1 = 2932.8 kJ/kg
m = V/v1 = 0.1/0.09936 = 1.006 kg
Process: Linear spring so P linear in v.
P = P0 + (P1 - P0)v/v1
v2 =
(P2 - P0)v1
P1 - P0
=
(1000 - 200)0.09936
3000 - 200 = 0.02839 m3/kg
State 2: P2 , v2 ⇒ x2 = (v2 - 0.001127)/0.19332 = 0.141, T2 = 179.91°C,
u2 = 761.62 + x2 × 1821.97 = 1018.58 kJ/kg
Process => 1W2 = ⌡⌠PdV =
1
2 m(P1 + P2)(v2 - v1)
=
1
2 1.006 (3000 + 1000)(0.02839 -0.09936) = -142.79 kJ
Heat transfer from the energy equation
1Q2 = m(u2 - u1) + 1W2 = 1.006(1018.58 - 2932.8) - 142.79 = -2068.5 kJ

5.166
A piston/cylinder, shown in Fig. P5.166, contains R-410a at − 20°C, x = 20%. The
volume is 0.2 m3. It is known that Vstop = 0.4 m3, and if the piston sits at the bottom,
the spring force balances the other loads on the piston. It is now heated up to 20°C.
Find the mass of the fluid and show the P–v diagram. Find the work and heat transfer.
Solution:
C.V. R-410a, this is a control mass. Properties in Table B.4.
Continuity Eq.: m2 = m1
Energy Eq.5.11: E2 - E1 = m(u2 - u1) = 1Q2 - 1W2
Process: P = A + BV, V < 0.4 m3, A = 0 (at V = 0, P = 0)
State 1: v1 = 0.000803 + 0.2 × 0.0640 = 0.0136 m3/kg
u1 = 27.92 + 0.2 × 218.07 = 71.5 kJ/kg
m = m1 = V1/v1 = 14.706 kg
System: on line
V ≤ Vstop;
P1 = 399.6 kPa
Pstop = 2P1 = 799.2 kPa
vstop = 2v1 = 0.0272 m3/kg
P
2
2P1
State stop: (P,v) ⇒ Tstop ≅ 0°C TWO-PHASE STATE
Since T2 > Tstop ⇒ v2 = vstop = 0.0272 m3/kg
State 2: (T2 , v2) Table B.4.2: Interpolate between 1000 and 1200 kPa
P2 = 1035 kPa ; u2 = 366.5 kJ/kg
From the process curve, see also area in P-V diagram, the work is
1W2 = ⌡⌠PdV =
1
2 (P1 + Pstop)(Vstop - V1) =
1
2 (399.6 + 799.2)0.2 = 119.8 kJ
1Q2 = m(u2 - u1) + 1W2 = 14.706(266.5 – 71.5) + 119.8 = 2987.5 kJ
0
0 0.2 0.4
1
V
P1
T
stop
≅ 0°C

5.167
Consider the piston/cylinder arrangement shown in Fig. P5.167. A frictionless piston
is free to move between two sets of stops. When the piston rests on the lower stops,
the enclosed volume is 400 L. When the piston reaches the upper stops, the volume is
600 L. The cylinder initially contains water at 100 kPa, 20% quality. It is heated until
the water eventually exists as saturated vapor. The mass of the piston requires 300 kPa
pressure to move it against the outside ambient pressure. Determine the final pressure
in the cylinder, the heat transfer and the work for the overall process.
Solution:
C.V. Water. Check to see if piston reaches upper stops.
Energy Eq.5.11: m(u4 - u1) = 1Q4 − 1W4
Process: If P < 300 kPa then V = 400 L, line 2-1 and below
If P > 300 kPa then V = 600 L, line 3-4 and above
If P = 300 kPa then 400 L < V < 600 L line 2-3
These three lines are shown in the P-V diagram below and is dictated by the
motion of the piston (force balance).
State 1: v1 = 0.001043 + 0.2×1.693 = 0.33964; m = V1/v1 =
0.4
0.33964 = 1.178 kg
u1 = 417.36 + 0.2 × 2088.7 = 835.1 kJ/kg
State 3: v3 =
0.6
1.178 = 0.5095 < vG = 0.6058 at P3 = 300 kPa
⇒ Piston does reach upper stops to reach sat. vapor.
State 4: v4 = v3 = 0.5095 m3/kg = vG at P4 From Table B.1.2
=> P4 = 361 kPa, u4 = 2550.0 kJ/kg
1W4 = 1W2 + 2W3 + 3W4 = 0 + 2W3 + 0
1W4 = P2(V3 - V2) = 300 × (0.6 - 0.4) = 60 kJ
1Q4 = m(u4 - u1) + 1W4 = 1.178(2550.0 - 835.1) + 60 = 2080 kJ
Water
2
1
3
4
P4
2P = P = 3003
P1
T
v

5.168
A spherical balloon contains 2 kg of R-410a at 0°C, 30% quality. This system is heated
until the pressure in the balloon reaches 1 MPa. For this process, it can be assumed that
the pressure in the balloon is directly proportional to the balloon diameter. How does
pressure vary with volume and what is the heat transfer for the process?
Solution:
C.V. R-410a which is a control mass.
m2 = m1 = m ;
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
State 1: 0°C, x = 0.3. Table B.4.1 gives P1 = 798.7 kPa
v1 = 0.000855 + 0.3 × 0.03182 = 0.01040 m3/kg
u1 = 57.07 + 0.3 × 195.95 = 115.86 kJ/kg
Process: P ∝ D, V ∝ D3 => PV -1/3 = constant, polytropic n = -1/3.
=> V2 = mv2 = V1 ( P2 /P1 )3 = mv1 ( P2 /P1 )3
v2 = v1 ( P2 /P1 )3 = 0.01040 × (1000 / 798.7)3 = 0.02041 m3/kg
State 2: P2 = 1 MPa, process : v2 = 0.02041 → Table B.4.2, T2 = 7.25°C (sat)
vf = 0.000877, vfg = 0.02508 m3/kg, uf = 68.02, ufg = 187.18 kJ/kg
x2 = 0.7787, u2 = 213.7 kJ/kg,
1W2 = ∫ P dV = m
P2v2 - P1v1
1 - n = 2
1000 × 0.02041 - 798.7 × 0.0104
1 - (-1/3) = 18.16 kJ
1Q2 = m(u2- u1) + 1W2 = 2 (213.7 – 115.86) + 18.16 = 213.8 kJ
P
V
W
1
2
sat.
vapor
line
Notice: The R-410a is not an ideal gas at any state in this problem.
R410a

5.169
A 1 m3 tank containing air at 25oC and 500 kPa is connected through a valve to
another tank containing 4 kg of air at 60oC and 200 kPa. Now the valve is opened and
the entire system reaches thermal equilibrium with the surroundings at 20oC. Assume
oconstant specific heat at 25 C and determine the final pressure and the heat transfer.
Control volume all the air. Assume air is an ideal gas.
tinu 2 A1 – mB1 = 0
2 1 2 2 A1 A1 – mB1uB1 = 1Q2 - 1W2
1 2 0
RTA1
Con ity Eq.: m – m
Energy Eq.: U − U = m u – m u
Process Eq.: V = constant ⇒ W =
State 1:
mA1 =
PA1VA1
=
(500 kPa)(1m3)
(0.287 kJ/kgK)(298.2 K) = 5.84 kg
VB1 =
mB1RTB1
=
(4 kg)(0.287 kJ/kgK)(333.2 K
PB1 (200 kN/m )
)
2 = 1.91 m3
State 2: T = 20°C, v = V /m
m2 = mA1 + mB1 = 4 + 5.84 = 9.84 kg
P2 V2
2 2 2 2
V2 = VA1 + VB1 = 1 + 1.91 = 2.91 m3
=
m2RT2 (9.84 kg)(0.287 kJ/k
=
gK)(293.2 K)
32.91 m
Energy Eq.5.5 or 5.11:
= 284.5 kPa
= mA1(u2 – uA1) + mB1(u2 – uB1)
= mA1Cv0(T2 – TA1) + mB1Cv0 2 – TB1)
= 5.84 × 0.717 (20 – 25) + 4 × 0.717 (20 – 60) = −135.6 kJ
The air gave energy out.
1Q2 = U2 − U1 = m2u2 – mA1uA1 – mB1uB1
(T
A B

1 . 01 m 1
Process gives: v2 = v1 )1/n = 1.2101 (100/200) 1/1.3 = 0.710 m3/kg
2 = 24oC
1W2 =
m
5.170
Ammonia (2 kg) in a piston/cylinder is at 100 kPa, -20oC and is now heated in a
polytropic process with n = 1.3 to a pressure of 200 kPa. Do not use ideal gas
approximation and find T2, the work and heat transfer in the process.
Energy Eq.5.11: m(u2 − u1) = 1Q2 - 1W2
Process: Pvn = constant (n = 1.3)
State 1: Superheated vapor table B.2.2.
v = 1 21 3/kg, u = 1307.8 kJ/kg
(P1/P2
State 2: Table B.2.2 at 200 kPa interpolate: u2 = 1376.49 kJ/kg, T
1 - n (P2v2 - 1v ) =
2
P 1 1 - 1.3(200×0.71 – 100×1.2101) = -139.9 kJ
uation
= 2 (137 13
=
Heat transfer is found from the energy eq
1Q2 = m(u2 - u1) + 1W2
6.49 – 07.8) – 139.9
-2.52 kJ P2
P
1
P
v
1
2

5.171
A piston/cylinder arrangement B is connected to a 1-m3 tank A by a line and valv
shown in Fig. P5.171. Initially both contain water, with A at 100 kPa, saturated
and B at 400°C, 300 kPa, 1 m3. The valve is now opened and, the wat
e,
vapor
er in both A and
B comes to a uniform state.
a. Find the initial mass in A and B.
b. If the process results in T = 200°C, find the heat transfer and work.
2 A1 + mB1) = 0 ;
2 2 A1 A1 1 B1 1 2 1 2
B B1
2
1 2 ⌡⌠ B B B1 2 B B1 2 1 tot
State A1: Table B.1.1, x = 1
m = V /v = 0.5903 kg
State B1: Table B.1.2 sup. vapor
v = 1.0315 m /kg, uB1 = 2965.5 kJ/kg
mB1 = VB1/vB1 = 0.9695 kg
2
Solution:
C.V.: A + B. This is a control mass.
Continuity equation: m - (m
Energy: m u - m u - mB u = Q - W
System: if VB ≥ 0 piston floats ⇒ P = P = const.
if VB = 0 then P < PB1 and v = VA/mtot see P-V diagram
W = P dV = P (V - V ) = P (V - V )1
P
V
2a
P
2
vA1 = 1.694 m3/kg, uA1 = 2506.1 kJ/kg
A1 A A1
B1
3
B1
m2 = mTOT = 1.56 kg
* At (T2 , PB1) v2 = 0.7163 > va = VA/mtot = 0.641 so VB2 > 0
so now state 2: P2 = PB1 = 300 kPa, T2 = 200 °C
=> u2 = 2650.7 kJ/kg and V2 = m2 v2 = 1.56 × 0.7163 = 1.117 m3
(we could also have checked Ta at: 300 kPa, 0.641 m3/kg => T = 155 °C)
1W2 = PB1(V2 - V1) = -264.82 kJ
1Q2 = m2u2 - mA1uA1 - mB1uB1 + 1W2 = -484.7 kJ

al
r a while the bag is heated to 30oC at which time the
ssur eat transfer in the process and the elapsed
e.
Tak .
Con inu
Ene y u = – W
Pro ss
State 1: Compressed liquid P > Ps , take saturated liquid at same temperature.
v1 = vf -10 = 0.001534 m3/kg, 1 = uf = 133.96 kJ/kg
State 2: Table B.2.1 at 30 C : P < P so er ate or
347.1 kJ/kg, V2 = mv2 = 0.0132 m3
Work ssure, so we get
3206 – 0.001534) = 8.484 kJ
Heat quation
3.96) + 8.484
kJ
5 – atts
Assume the constant rate Q = dQ/dt = Q / t, so the time becomes
=
129800
5.172
A small flexible bag contains 0.1 kg ammonia at –10oC and 300 kPa. The bag materi
is such that the pressure inside varies linear with volume. The bag is left in the sun
with with an incident radiation of 75 W, loosing energy with an average 25 W to the
ambient ground and air. Afte
pre e is 1000 kPa. Find the work and h
tim
e CV as the Ammonia, constant mass
t ity Eq.: m2 = m1 = m ;
rg Eq.5.11: m(u2 − 1) 1Q2 1 2
ce : P = A + BV (linear in V)
at
u
o
sat sup he d vap
v2 = 0.13206 m3/kg, u2 = 1
is done while piston moves at increacing pre
1W2 = ½(300 + 1000)*0.1(0.1
transfer is found from the energy e
1Q2 = m(u2 – u1) + 1W2 = 0.1 (1347.1 – 13
= 121.314 + 8.484 = 129.8
.
Qnet = 7 25 = 50 W
.
net 1 2
.
t = 1Q2 / Qnet 50 = 2596 s = 43.3 min
P
v
1
T
v
1
30
C.P.
2
C.P.
2
NH3 1000
300 -10

CONTENT
SUBSECTION PROB NO.
In-Text concept questions a-k
Concept-Study guide problems 1- 10
Continuity equation and flow rates 11-19
Single flow, single-device processes
Nozzles, diffusers 20-30
Throttle flow 21-39
Turbines, expanders 40-47
Compressors, fans 48-57
Heaters, coolers 58-67
Pumps, pipe and channel flows 68-75
Multiple flow, single-device processes
Turbines, compressors, expanders 76-82
Heat exchangers 83-92
Mixing processes 93-101
Multiple devices, cycle processes 102-110
Transient processes 111-127
Review Problems 128-141
Heat Transfer Problems 142-145

6.a
A mass flow rate into a control volume requires a normal velocity component.
Why?
The tangential velocity component does not bring any substance across the
control volume surface as it flows parallel to it, the normal component of velocity
brings substance in or out of the control volume according to its sign. The normal
component must be into the control volume to bring mass in, just like when you
enter a bus (it does not help that you run parallel with the bus side).
V
Vnormal
Vtangential
6.b
Can a steady state device have boundary work?
No. Any change in size of the control volume would require either a
change in mass inside or a change in state inside, neither of which is possible in a
steady-state process.

6.c
Can you say something about changes in m
.
and V
.
through a steady flow device?
The continuity equation expresses the conservation of mass, so the total
amount of m
.
entering must be equal to the total amount leaving. For a single flow
device the mass flow rate is constant through it, so you have the same mass flow
rate across any total cross-section of the device from the inlet to the exit.
The volume flow rate is related to the mass flow rate as
V
.
= v m
.
so it can vary if the state changes (then v changes) for a constant mass flow rate.
This also means that the velocity can change (influenced by the area as V
.
= VA)
and the flow can experience an acceleration (like in a nozzle) or a deceleration (as
in a diffuser).
6.d
In a multiple device flow system, I want to determine a state property. Where
should I be looking for information—upstream or downstream?
Generally flow is affected more by what happened to it which is upstream
than what is in front of it. Only the pressure information can travel upstream and
give rise to accelerations (nozzle) or decelerations (stagnation type flow). Heat
transfer that can heat or cool a flow can not travel very fast and is easily
overpowered by the convection. If the flow velocity exceeds the speed of sound
even the pressure information can not travel upstream.

6.e
How does a nozzle or sprayhead generate kinetic energy?
By accelerating the fluid from a high pressure
towards the lower pressure, which is outside the
nozzle. The higher pressure pushes harder than
the lower pressure so there is a net force on any
mass element to accelerate it.
6.f
What is the difference between a nozzle flow and a throttle process?
In both processes a flow moves from a higher to a lower pressure. In the
nozzle the pressure drop generates kinetic energy, whereas that does not take
place in the throttle process. The pressure drop in the throttle is due to a flow
restriction and represents a loss.

6.g
If you throttle a saturated liquid what happens to the fluid state? What if this is
done to an ideal gas?
The throttle process is approximated as a constant enthalpy process.
Changing the state from saturated liquid to a lower pressure with the same h gives
a two-phase state so some of the liquid will vaporize and it becomes colder.
1 2
2
P
v
1
Th = C
h = C
If the same process happens in an ideal gas then same h gives the same
temperature (h a function of T only) at the lower pressure.

6.h
A turbine at the bottom of a dam has a flow of liquid water through it. How does
that produce power? Which terms in the energy equation are important if the CV
is the turbine only? If the CV is the turbine plus the upstream flow up to the top of
the lake, which terms in the energy equation are then important?
The water at the bottom of the dam in the turbine inlet is at a high
pressure. It runs through a nozzle generating kinetic energy as the pressure drops.
This high kinetic energy flow impacts a set of rotating blades or buckets, which
converts the kinetic energy to power on the shaft and the flow leaves at low
pressure and low velocity.
T
H
DAM
Lake
CV Turbine only.
The high P in and the low P out shows up in the (h = u + Pv) flow terms of the
energy equation giving the difference in the flow work terms Pv in and out.
CV Turbine plus upstream flow.
For this CV the pressures in and out are the same (1 atm) so the difference is in
the potential energy terms (gz) included in htot.

6.i
If you compress air the temperature goes up, why? When the hot air, high P flows
in long pipes it eventually cools to ambient T. How does that change the flow?
As the air is compressed, volume decreases so work is done on a mass
element, its energy and hence temperature goes up. If it flows at nearly
constant P and cools its density increases (v decreases) so it slows down
for same mass flow rate ( m
.
= ρAV ) and flow area.

6.j
A mixing chamber has all flows at the same P, neglecting losses. A heat
exchanger has separate flows exchanging energy, but they do not mix. Why have
both kinds?
You might allow mixing when you can use the resulting output mixture,
say it is the same substance. You may also allow it if you definitely want the
outgoing mixture, like water out of a faucet where you mix hot and cold water.
Even if it is different substances it may be desirable, say you add water to dry air
to make it more moist, typical for a winter time air-conditioning set-up.
In other cases it is different substances that flow at different pressures with
one flow heating or cooling the other flow. This could be hot combustion gases
heating a flow of water or a primary fluid flow around a nuclear reactor heating a
transfer fluid flow. Here the fluid being heated should stay pure so it does not
absorb gases or radioactive particles and becomes contaminated. Even when the
two flows have the same substance there may be a reason to keep them at separate
pressures.
1
2
3MIXING
CHAMBER
cb
1
2
3
cb
4
An open mixing chamber A closed tube in shell heat exchanger.

6.k
An initially empty cylinder is filled with air from 20oC, 100 kPa until it is full.
Assuming no heat transfer is the final temperature larger, equal to or smaller than
20oC? Does the final T depend on the size of the cylinder?
This is a transient problem with no heat transfer and no work. The balance
equations for the tank as C.V. become
Continuity Eq.: m2 – 0 = mi
Energy Eq.: m2u2 – 0 = mihi + Q – W = mihi + 0 – 0
Final state: u2 = hi = ui + Pivi & P2 = Pi
T2 > Ti and it does not depend on V

Concept-Study Guide Problems

6.1
A temperature difference drives a heat transfer. Does a similar concept apply to m
.
?
Yes. A pressure difference drives the flow. The fluid is accelerated in the
direction of a lower pressure as it is being pushed harder behind it than in front of
it. This also means a higher pressure in front can decelerate the flow to a lower
velocity which happens at a stagnation on a wall.
F = P A
1 1
F = P A
2 2

6.2
What kind of effect can be felt upstream in a flow?
Only the pressure can be felt upstream in a subsonic flow. In a supersonic
flow no information can travel upstream. The temperature information travels by
conduction and even small velocities overpowers the conduction with the
convection of energy so the temperature at a given location is mainly given by the
upstream conditions and influenced very little by the downstream conditions.
6.3
Which one of the properties (P, v, T) can be controlled in a flow? How?
Since the flow is not contained there is no direct control over the volume
and thus no control of v. The pressure can be controlled by installation of a pump
or compressor if you want to increase it or use a turbine, nozzle or valve through
which the pressure will decrease. The temperature can be controlled by heating or
cooling the flow in a heat exchanger.

6.4
Air at 500 kPa is expanded to 100 kPa in two steady flow cases. Case one is a
nozzle and case two is a turbine, the exit state is the same for both cases. What
can you say about the specific turbine work relative to the specific kinetic energy
in the exit flow of the nozzle?
For these single flow devices let us assume they are adiabatic and that the turbine
does not have any exit kinetic energy then the energy equations become:
Energy Eq.6.13 nozzle: h1+
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2
Energy Eq.6.13 turbine: h1+
1
2 V
2
1 + gZ1 = h2 + gZ2 + wT
Comparing the two we get
1
2 V
2
2 = wT so the result is that the nozzle delivers
kinetic energy of the same amount as the turbine delivers shaft work.

6.5
Pipes that carry a hot fluid like steam in a power plant, exhaust pipe for a diesel
engine in a ship etc. are often insulated. Is that to reduce the heat loss or is there
another purpose?
You definitely want to insulate pipes that carry hot steam from the boiler to the
turbines in a power plant and pipes that flows hot water from one location to
another. Even if the energy in the flow is unwanted the pipes should be insulated
for safety. Any place that people could touch a hot surface there is a risk of a burn
and that should be avoided.
Pipes for district heating with hot water.

6.6
A windmill takes a fraction of the wind kinetic energy out as power on a shaft. In
what manner does the temperature and wind velocity influence the power? Hint:
write the power as mass flow rate times specific work.
The work as a fraction f of the flow of kinetic energy becomes
W
.
= m
.
w = m
.
f
1
2 V
2
in = ρAVin f
1
2 V
2
in
so the power is proportional to the velocity cubed. The temperature enters through
the density, so assuming air as ideal gas
ρ = 1/v = P/RT
and the power is inversely proportional to temperature.
A windmill farm west of Denmark in the North Sea.

6.7
An underwater turbine extracts a fraction of the kinetic energy from the ocean
current. In what manner does the temperature and water velocity influence the
power? Hint: write the power as mass flow rate times specific work.
The work as a fraction f of the flow of kinetic energy becomes
W
.
= m
.
w = m
.
f
1
2 V
2
in = ρAVin f
1
2 V
2
in
so the power is proportional to the velocity cubed. The temperature enters through
the density, so assuming water is incompressible density is constant and the
power does not vary with the temperature.
A proposed underwater tidal flow turbine farm. Each turbine is for 1 MW with a
diameter of 11.5 m mounted on the seafloor.

6.8
A liquid water turbine in the bottom of a dam takes energy out as power on a
shaft. Which term(s) in the energy equation are changing and important?
The water at the bottom of the dam in the turbine inlet is at a high
pressure. In a standard turbine it runs through blade passages like a propeller
shown below to the left. In this case the inlet high pressure is used directly to
generate the force on the moving blades.
For the Pelton turbine shown below it runs through a nozzle generating
kinetic energy as the pressure drops. The high kinetic energy flow impacts a set of
rotating blades or buckets and converts the kinetic energy to power on the shaft so
the flow leaves at low pressure and low velocity.
A Pelton turbine above and insert on left.
A blade type turbine to the left. The
pressure drop and flow rate determine
which one is best for the given case.

6.9
You blow a balloon up with air. What kind of work terms including flow work do
you see in that case? Where is the energy stored?
As the balloon is blown up mass flow in has flow work associated with it. Also as
the balloon grows there is a boundary work done by the inside gas and a smaller
boundary work from the outside of the balloon to the atmosphere. The difference
between the latter two work terms goes into stretching the balloon material and
thus becomes internal energy (or you may call that potential energy) of the
balloon material. The work term to the atmosphere is stored in the atmosphere
and the part of the flow work that stays in the gas is stored as the gas internal
energy.

6.10
A storage tank for natural gas (NG) has a top dome that can move up or down as
gas is added or subtracted from the tank maintaining 110 kPa, 290 K inside. A
pipeline at 110 kPa, 290 K now supplies some NG to the tank. Does it change
state during the filling process? What happens to the flow work?
As the pressure inside the storage tank is the same as in the pipeline the state does
not change. However the tank volume goes up and work is done on the moving
boundary at the 110 kPa so this work equals the flow work. The net effect is the
flow work goes into raising the dome.

Continuity equation and flow rates

6.11
Carbon dioxide at 200 kPa, 10oC flows at 1 kg/s in a 0.25 m2 cross sectional area
pipe. Find the velocity and the volume flow rate.
m
.
= AV/v = V
.
/v
v = RT/P = 0.1889 kJ/kg-K × 283.15 K/ 200 kPa = 0.267 m3/kg
V
.
= v m
.
= 0.267 × 1 = 0.267 m3/s
V = V
.
/ A = 0.267 (m3/s) / 0.25 m2 = 1.07 m/s

6.12
Air at 35°C, 105 kPa, flows in a 100 mm × 150 mm rectangular duct in a heating
system. The volumetric flow rate is 0.015 m3/s. What is the velocity of the air
flowing in the duct and what is the mass flow rate?
Solution:
Assume a constant velocity across the duct area with
A = 100 × 150 ×10-6 m2 = 0.015 m2
and the volumetric flow rate from Eq.6.3,
V
.
= m
.
v = AV
V =
V
.
A =
0.015 m3/s
0.015 m2 = 1.0 m/s
Ideal gas so note:
v =
RT
P =
0.287 × 308.2
105 = 0.8424 m3/kg
m
.
=
V
.
v =
0.015
0.8424 = 0.0178 kg/s

6.13
An empty bathtub has its drain closed and is being filled with water from the
faucet at a rate of 10 kg/min. After 10 minutes the drain is opened and 4 kg/min
flows out and at the same time the inlet flow is reduced to 2 kg/min. Plot the mass
of the water in the bathtub versus time and determine the time from the very
beginning when the tub will be empty.
Solution:
During the first 10 minutes we have
dmcv
dt = m
.
i = 10 kg/min , ∆m = m
.
∆t1 = 10 × 10 = 100 kg
So we end up with 100 kg after 10 min. For the remaining period we have
dmcv
dt = m
.
i - m
.
e= 2 – 4 = -2 kg/min
∆m2 = m
.
net ∆t2 ∆t2 =
∆m
m
.
net
= -100/-2 = 50 min.
So it will take an additional 50 min. to empty
∆ttot = ∆t1 + ∆t2 = 10 + 50 = 60 min.
1010 20
100
m
-2
0
10
0 0
t
min
0
m
.
t
min
kg

6.14
Saturated vapor R-134a leaves the evaporator in a heat pump system at 10°C,
with a steady mass flow rate of 0.1 kg/s. What is the smallest diameter tubing that
can be used at this location if the velocity of the refrigerant is not to exceed 7
m/s?
Solution:
Mass flow rate Eq.6.3: m
.
= V
.
/v = AV/v
Exit state Table B.5.1: (T = 10°C, x =1) => v = vg = 0.04945 m3/kg
The minimum area is associated with the maximum velocity for given m
.
AMIN =
m
.
vg
VMAX
=
0.1 kg/s × 0.04945 m3/kg
7 m/s = 0.000706 m2 =
π
4 D
2
MIN
DMIN = 0.03 m = 30 mm
cb
Exit

6.15
A boiler receives a constant flow of 5000 kg/h liquid water at 5 MPa, 20°C and it
heats the flow such that the exit state is 450°C with a pressure of 4.5 MPa.
Determine the necessary minimum pipe flow area in both the inlet and exit pipe(s)
if there should be no velocities larger than 20 m/s.
Solution:
Mass flow rate from Eq.6.3, both V ≤ 20 m/s
m
.
i = m
.
e = (AV/v) i = (AV/v) e = 5000
1
3600 kg/s
Table B.1.4 vi = 0.001 m3/kg,
Table B.1.3 ve = (0.08003 + 0.00633)/2 = 0.07166 m3/kg,
Ai ≥ vi m
.
/Vi = 0.001×
5000
3600 / 20 = 6.94 × 10-5 m2 = 0.69 cm2
Ae ≥ ve m
.
/Ve = 0.07166 ×
5000
3600 / 20 = 4.98 × 10-3 m2 = 50 cm2
Inlet
liquid
i e
Q
Q
boiler
Super
heatervapor
cb
Exit
Superheated vapor

6.16
A hot air home heating system takes 0.25 m3/s air at 100 kPa, 17oC into a furnace
and heats it to 52oC and delivers the flow to a square duct 0.2 m by 0.2 m at 110
kPa. What is the velocity in the duct?
Solution:
The inflate flow is given by a m
.
i
Continuity Eq.: m
.
i = V
.
i / vi = m
.
e = AeVe/ve
Ideal gas: vi =
RTi
Pi
=
0.287 × 290
100 = 0.8323
m3
kg
ve =
RTe
Pe
=
0.287 × (52 + 273)
110
= 0.8479 m3/ kg
m
.
i = V
.
i/vi = 0.25/0.8323 = 0.30 kg/s
Ve = m
.
ve/ Ae =
0.3 × 0.8479
0.2 × 0.2
m3/s
m2 = 6.36 m/s

6.17
A flat channel of depth 1 m has a fully developed laminar flow of air at Po, To
with a velocity profile as: V = 4Vc (1 – x/H)x/H, where Vc is the velocity on the
center-line and x is the distance across the channel as shown in Fig. P6.17. Find
the total mass flow rate and the average velocity both as functions of Vc and H.
m
.
= AV/v = V
.
/v
Since the velocity is distributed we need to integrate over the area. From Eq.6.2
V
.
= ⌡⌠ Vlocal dA = ⌡⌠ V(x) W dx
where W is the depth. Substituting the velocity we get
V
.
= ⌡⌠ 4Vc (x/H) (1 - x/H) W dx
= 4 Vc WH ⌡⌠
0
1
z (1 - z) dz (z = x/H)
= 4 Vc WH (
1
2 z2 -
1
3 z3 |1
0 =
2
3 Vc WH =
2
3 Vc A
V = V
.
/A =
2
3 Vc
m
.
= V
.
/v =
2
3 Vc WH/v

6.18
Nitrogen gas flowing in a 50-mm diameter pipe at 15°C, 200 kPa, at the rate of
0.05 kg/s, encounters a partially closed valve. If there is a pressure drop of 30 kPa
across the valve and essentially no temperature change, what are the velocities
upstream and downstream of the valve?
Solution:
Same inlet and exit area: A =
π
4 (0.050)2 = 0.001963 m2
Ideal gas: vi =
RTi
Pi
=
0.2968 × 288.2
200 = 0.4277 m3/kg
From Eq.6.3,
Vi =
m
.
vi
A =
0.05 × 0.4277
0.001963 = 10.9 m/s
Ideal gas: ve =
RTe
Pe
=
0.2968 × 288.2
170 = 0.5032 m3/kg
Ve =
m
.
ve
A =
0.05 × 0.5032
0.001963 = 12.8 m/s

6.19
A household fan of diameter 0.75 m takes air in at 98 kPa, 22oC and delivers it at
105 kPa, 23oC with a velocity of 1.5 m/s. What are the mass flow rate (kg/s), the
inlet velocity and the outgoing volume flow rate in m3/s?
Solution:
Continuity Eq. m
.
i = m
.
e = AV/ v
Ideal gas v = RT/P
Area : A =
π
4 D 2 =
π
4× 0.752 = 0.442 m2
V
.
e = AVe = 0.442 ×1.5 = 0.6627 m3/s
ve =
RTe
Pe
=
0.287 × (23 + 273)
105 = 0.8091 m3/kg
m
.
i = V
.
e/ve = 0.6627/0.8091 = 0.819 kg/s
AVi /vi = m
.
i = AVe / ve
Vi = Ve × (vi / ve) = Ve ×
RTi
Pive
= 1.5 ×
0.287 × (22 + 273)
98 × 0.8091
= 1.6 m/s

Single flow single device processes
Nozzles, diffusers

6.20
Liquid water at 15oC flows out of a nozzle straight up 15 m. What is nozzle Vexit?
Energy Eq.6.13: hexit +
1
2 V
2
exit + gHexit = h2 +
1
2 V
2
2 + gH2
If the water can flow 15 m up it has specific potential energy of gH2 which must
equal the specific kinetic energy out of the nozzle V
2
exit/2. The water does not
change P or T so h is the same.
V
2
exit/2 = g(H2 – Hexit) = gH =>
Vexit = 2gH = 2 × 9.807 × 15 m2/s2 = 17.15 m/s

6.21
Nitrogen gas flows into a convergent nozzle at 200 kPa, 400 K and very low
velocity. It flows out of the nozzle at 100 kPa, 330 K. If the nozzle is insulated
find the exit velocity.
Solution:
C.V. Nozzle steady state one inlet and exit flow, insulated so it is adiabatic.
Inlet
Low V
Exit
Hi V
Hi P, A
Low P, Acb
Energy Eq.6.13: h1 + ∅ = h2 +
1
2 V
2
2
V
2
2 = 2 ( h1 - h2 ) ≅ 2 CPN2
(T1 – T2 ) = 2 × 1.042 (400 – 330)
= 145.88 kJ/kg = 145 880 J/kg
⇒ V2 = 381.9 m/s

6.22
A nozzle receives 0.1 kg/s steam at 1 MPa, 400oC with negligible kinetic energy.
The exit is at 500 kPa, 350oC and the flow is adiabatic. Find the nozzle exit
velocity and the exit area.
Solution:
Energy Eq.6.13: h1+
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2
Process: Z1 = Z2
State 1: V1 = 0 , Table B.1.3 h1 = 3263.88 kJ/kg
Then from the energy equation
1
2 V
2
2 = h1 – h2 = 3263.88 – 3167.65 = 96.23 kJ/kg
V2 = 2(h1 - h2) = 2 × 96.23 × 1000 = 438.7 m/s
The mass flow rate from Eq.6.3
m
.
= ρAV = AV/v
A = m
.
v/V = 0.1 × 0.57012 / 438.7 = 0.00013 m2
= 1.3 cm2
Inlet
Low V
Exit
Hi V
Hi P, A
Low P, Acb

6.23
In a jet engine a flow of air at 1000 K, 200 kPa and 30 m/s enters a nozzle, as
shown in Fig. P6.23, where the air exits at 850 K, 90 kPa. What is the exit
velocity assuming no heat loss?
Solution:
C.V. nozzle. No work, no heat transfer
Continuity m
.
i = m
.
e = m
.
Energy : m
.
(hi + ½Vi
2) = m
.
(he+ ½Ve
2)
Due to high T take h from table A.7.1
½Ve
2 = ½ Vi
2 + hi - he
=
1
2000 (30)2 + 1046.22 – 877.4
= 0.45 + 168.82 = 169.27 kJ/kg
Ve = (2000 × 169.27)1/2 = 581.8 m/s

6.24
In a jet engine a flow of air at 1000 K, 200 kPa and 40 m/s enters a nozzle where
the air exits at 500 m/s, 90 kPa. What is the exit temperature assuming no heat
loss?
Solution:
C.V. nozzle, no work, no heat transfer
Continuity m
.
i = m
.
e = m
.
Energy : m
.
(hi + ½Vi
2) = m
.
(he+ ½Ve
2)
Due to the high T we take the h value from Table A.7.1
he = hi + ½ Vi
2 - ½Ve
2
= 1046.22 + 0.5 × (402 – 5002) (1/1000)
= 1046.22 – 124.2 = 922.02 kJ/kg
Interpolation in Table A.7.1
Te = 850 + 50
922.02 - 877.4
933.15 - 877.4 = 890 K
40 m/s
200 kPa
500 m/s
90 kPa

6.25
Superheated vapor ammonia enters an insulated nozzle at 20°C, 800 kPa, shown
in Fig. P6.25, with a low velocity and at the steady rate of 0.01 kg/s. The
ammonia exits at 300 kPa with a velocity of 450 m/s. Determine the temperature
(or quality, if saturated) and the exit area of the nozzle.
Solution:
C.V. Nozzle, steady state, 1 inlet and 1 exit flow, insulated so no heat transfer.
Energy Eq.6.13: q + hi + V
2
i /2 = he + V
2
e/2,
Process: q = 0, Vi = 0
Table B.2.2: hi = 1464.9 = he + 4502/(2×1000) ⇒ he = 1363.6 kJ/kg
Table B.2.1: Pe = 300 kPa Sat. state at −9.2°C :
he = 1363.6 = 138.0 + xe × 1293.8,
=> xe = 0.947, ve = 0.001536 + xe × 0.4064 = 0.3864 m3/kg
Ae = m
.
eve/Ve = 0.01 × 0.3864 / 450 = 8.56 × 10-6 m2
Inlet
Low V
Exit
Hi V
Hi P, A
Low P, Acb

6.26
Air flows into a diffuser at 300 m/s, 300 K and 100 kPa. At the exit the velocity is
very small but the pressure is high. Find the exit temperature assuming zero heat
transfer.
Solution:
Energy Eq.: h1 +
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2
Process: Z1 = Z2 and V2 = 0
h2 = h1 +
1
2 V
2
1
T2 = T1 +
1
2 × (V
2
1 / Cp)
= 300 +
1
2 × 3002 / (1.004 × 1000) = 344.8K
Inlet
Low V
Exit
Hi V
Hi P, ALow P, A

6.27
A sluice gate dams water up 5 m. There is a small hole at the bottom of the gate
so liquid water at 20oC comes out of a 1 cm diameter hole. Neglect any changes
in internal energy and find the exit velocity and mass flow rate.
Solution:
Energy Eq.6.13: h1+
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2
Process: h1 = h2 both at P = 1 atm
V1 = 0 Z1 = Z2 + 5 m
Water 5 m
1
2 V
2
2 = g (Z1 − Z2)
V2 = 2g(Z1 - Z2) = 2 × 9.806 × 5 = 9.902 m/s
m
.
= ρΑV = AV/v =
π
4 D2 × (V2/v)
=
π
4 × (0.01)2 × (9.902 / 0.001002) = 0.776 kg/s

6.28
A diffuser, shown in Fig. P6.28, has air entering at 100 kPa, 300 K, with a
velocity of 200 m/s. The inlet cross-sectional area of the diffuser is 100 mm2. At
the exit, the area is 860 mm2, and the exit velocity is 20 m/s. Determine the exit
pressure and temperature of the air.
Solution:
Continuity Eq.6.3: m
.
i = AiVi/vi = m
.
e = AeVe/ve,
Energy Eq.(per unit mass flow)6.13: hi +
1
2Vi
2 = he +
1
2Ve
2
he - hi =
1
2 ×2002/1000 −
1
2 ×202/1000 = 19.8 kJ/kg
Te = Ti + (he - hi)/Cp = 300 + 19.8/1.004 = 319.72 K
Now use the continuity equation and the ideal gas law
ve = vi





AeVe
AiVi
= (RTi/Pi)





AeVe
AiVi
= RTe/Pe
Pe = Pi





Te
Ti 




AiVi
AeVe
= 100





319.72
300 




100 × 200
860 × 20
= 123.92 kPa
Inlet
Low V
Exit
Hi V
Hi P, ALow P, A

6.29
A diffuser receives an ideal gas flow at 100 kPa, 300 K with a velocity of 250 m/s
and the exit velocity is 25 m/s. Determine the exit temperature if the gas is argon,
helium or nitrogen.
Solution:
C.V. Diffuser: m
.
i = m
.
e & assume no heat transfer ⇒
Energy Eq.6.13: hi +
1
2 V
2
i =
1
2 V
2
e + he ⇒ he = hi +
1
2 V
2
i -
1
2V
2
e
he – hi ≈ Cp ( Te – Ti ) =
1
2 ( V
2
i - V
2
e ) =
1
2 ( 2502 – 252 )
= 30937.5 J/kg = 30.938 kJ/kg
Specific heats for ideal gases are from table A.5
Argon Cp = 0.52 kJ/kg K; ∆T =
30.938
0.52 = 59.5 Te = 359.5 K
Helium Cp = 5.913 kJ/kg K; ∆T =
30.938
5.193 = 5.96 Te = 306 K
Nitrogen Cp = 1.042 kJ/kg K; ∆T =
30.938
1.042 = 29.7 Te = 330 K
Inlet
Low V
Exit
Hi V
Hi P, ALow P, A

6.30
The front of a jet engine acts as a diffuser receiving air at 900 km/h, -5°C, 50 kPa,
bringing it to 80 m/s relative to the engine before entering the compressor. If the
flow area is reduced to 80% of the inlet area find the temperature and pressure in
the compressor inlet.
Solution:
C.V. Diffuser, Steady state, 1 inlet, 1 exit flow, no q, no w.
.
i = m
.
e = (AV/v)
Energy Eq.6.12: m
.
( hi +
1
2 V
2
i ) = m
.
(
1
2 V
2
e + he )
he – hi = Cp ( Te – Ti ) =
1
2 V
2
i -
1
2 V
2
e =
1
2 (900 × 1000
3600 )2
−
1
2 (80)2
= 28050 J/kg = 28.05 kJ/kg
∆T = 28.05/1.004 = 27.9 ⇒ Te = −5 + 27.9 = 22.9°C
Now use the continuity eq.:
AiVi /vi = AeVe /ve ⇒ ve = vi





AeVe
AiVi
ve = vi ×
0.8 × 80
1 × 250
= vi × 0.256
Ideal gas: Pv = RT => ve = RTe/Pe = RT i × 0.256/Pi
Pe = Pi (Te/T i)/0.256 = 50 × 296/268 × 0.256 = 215.7 kPa
Fan

Throttle flow

6.31
Carbon dioxide used as a natural refrigerant flows out of a cooler at 10 MPa,
40oC after which it is throttled to 1.4 MPa. Find the state (T, x) for the exit flow.
C.V. Throttle. Steady state,
Process with: q = w = 0; and Vi = Ve, Zi = Ze
Energy Eq.6.13: hi = he,
Inlet state: Table B.3.2 hi = 200.14 kJ/kg
Exit state: Table B.3.1 since h < hg = 323.87 kJ/kg
Interpolate: T = -30.6oC, hf = 19.2, hfg = 304.67 kJ/kg
x = (200.14 – 19.2) / 304.67 = 0.594

6.32
R-134a at 30oC, 800 kPa is throttled so it becomes cold at –10oC. What is exit P?
State 1 is slightly compressed liquid so
Table B.5.1: h = hf = 241.79 kJ/kg
At the lower temperature it becomes two-phase since the throttle flow has
constant h and at –10oC: hg = 392.28 kJ/kg
1 2
2
P
v
1
Th = C
h = C

6.33
Helium is throttled from 1.2 MPa, 20°C, to a pressure of 100 kPa. The diameter of
the exit pipe is so much larger than the inlet pipe that the inlet and exit velocities
are equal. Find the exit temperature of the helium and the ratio of the pipe
diameters.
Solution:
C.V. Throttle. Steady state,
Process with: q = w = 0; and Vi = Ve, Zi = Ze
Energy Eq.6.13: hi = he, Ideal gas => Ti = Te = 20°C
m
.
=
AV
RT/P But m
.
, V, T are constant => PiAi = PeAe
⇒
De
Di
=





Pi
Pe
1/2
=





1.2
0.1
1/2
= 3.464

6.34
Saturated vapor R-134a at 500 kPa is throttled to 200 kPa in a steady flow
through a valve. The kinetic energy in the inlet and exit flow is the same. What is
the exit temperature?
Solution:
Steady throttle flow
Continuity m
.
i = m
.
e = m
.
Energy Eq.6.13: h1 +
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2
Process: Z1 = Z2 and V2 = V1
⇒ h2 = h1 = 407.45 kJ/kg from Table B.5.2
State 2: P2 & h2 ⇒ superheated vapor
Interpolate between 0oC and 10oC in table B.5.2 in the 200 kPa subtable
T2 = 0 + 10
407.45 – 400.91
409.5 – 400.91 = 7.6oC
i e
cb
e
T
v
i
500 kPa
200
h = C

6.35
Saturated liquid R-410a at 25oC is throttled to 400 kPa in your refrigerator. What
is the exit temperature? Find the percent increase in the volume flow rate.
Solution:
Steady throttle flow. Assume no heat transfer and no change in kinetic or
potential energy.
he = hi = hf 25oC = 97.59 kJ/kg = hf e + xe hfg e at 400 kPa
From table B.3.1 we get Te = Tsat ( 400 kPa ) = -20oC
xe =
he – hf e
hfg e
=
97.59 – 28.24
243.65 = 0.28463
ve = vf + xe vfg = 0.000803 + xe 0.064 = 0.01902 m3/kg
vi = vf 25oC = 0.000944 m3/kg
V
.
= m
.
v so the ratio becomes
V
.
e
V
.
i
=
m
.
ve
m
.
vi
=
ve
vi
=
0.01902
0.000994 = 20.23
So the increase is 19.23 times or 1923 %
i e
cb
e
T
v
i
h = C

6.36
Carbon dioxide is throttled from 20oC, 2 MPa to 800 kPa. Find the exit
temperature assuming ideal gas behavior and repeat for real-gas behavior.
C.V. Throttle (valve, restriction), Steady flow, 1 inlet and exit, no q, w
Energy Eq.6.13: hi = he
Ideal gas: same h gives Ti = Te = 20oC
Real gas :


hi = he = 368.42 kJ/kg
Pe = 0.8 MPa
Table B.3.2
Te = 5.3°C ( = 278 K)

6.37
Liquid water at 180oC, 2000 kPa is throttled into a flash evaporator chamber
having a pressure of 500 kPa. Neglect any change in the kinetic energy. What is
the fraction of liquid and vapor in the chamber?
Solution:
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2
⇒ h2 = h1 = 763.71 kJ/kg from Table B.1.4
State 2: P2 & h2 ⇒ 2 – phase
h2 = hf + x2 hfg
x2 = (h2 - hf ) / hfg=
763.71 – 640.21
2108.47 = 0.0586
Fraction of Vapor: x2 = 0.0586 (5.86 %)
Liquid: 1 - x2 = 0.941 (94.1 %)
Two-phase out of the
valve. The liquid drops
to the bottom.

6.38
R-134a is throttled in a line flowing at 25oC, 750 kPa with negligible kinetic
energy to a pressure of 165 kPa. Find the exit temperature and the ratio of exit
pipe diameter to that of the inlet pipe (Dex/Din) so the velocity stays constant.
Solution:
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2
State 1, Table B.5.1: h1 = 234.59 kJ/kg, v1 = vf = 0.000829 m3/kg
Use energy eq.: ⇒ h2 = h1 = 234.59 kJ/kg
State 2: P2 & h2 ⇒ 2 – phase and T2 = Tsat (165 kPa) = -15°C
h2 = hf + x2 hfg = 234.59 kJ/kg
x2 = (h2 - hf ) / hfg= (234.59 – 180.19) / 209 = 0.2603
v2 = vf + x2 × vfg = 0.000746 + 0.2603 × 0.11932 = 0.0318 m3/kg
Now the continuity equation with V2 = V1 gives, from Eq.6.3,
m
.
= ρΑV = AV/v = A1V1/v1 = (A2 V1) / v2
(A2 / A1) = v2 / v1 = (D2 / D1)
2
(D2/D1) = (v2 / v1)
0.5
= (0.0318 / 0.000829)
0.5
= 6.19

6.39
Water flowing in a line at 400 kPa, saturated vapor, is taken out through a valve
to 100 kPa. What is the temperature as it leaves the valve assuming no changes in
kinetic energy and no heat transfer?
Solution:
C.V. Valve. Steady state, single inlet and exit flow
Continuity Eq.: m
.
1 = m
.
2
Energy Eq.6.12: m
.
1h1 + Q
.
= m
.
2h2 + W
.
1 2
Process: Throttling
Small surface area: Q
.
= 0;
No shaft: W
.
= 0
Table B.1.2: h2 = h1 = 2738.6 kJ/kg ⇒ T2 = 131.1°C

Turbines, Expanders

6.40
A steam turbine has an inlet of 2 kg/s water at 1000 kPa, 350oC and velocity of
15 m/s. The exit is at 100 kPa, 150oC and very low velocity. Find the specific
work and the power produced.
Solution:
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2 + wT
Table B.1.3: h1 = 3157.65 kJ/kg, h2 = 2776.38 kJ/kg
wT = h1 +
1
2 V
2
1 – h2 = 3157.65 +
152
2000 – 2776.38 = 381.4 kJ/kg
(remember to convert m2/s2 = J/kg to kJ/kg by dividing with 1000)
W
.
T = m
.
× wT = 2 kg/s × 381.4 kJ/kg
= 762.8 kW
WT
1
2

6.41
Air at 20 m/s, 260 K, 75 kPa with 5 kg/s flows into a jet engine and it flows out at
500 m/s, 800 K, 75 kPa. What is the change (power) in flow of kinetic energy?
m
.
∆KE = m
. 1
2 (V
2
e – V
2
i )
= 5 kg/s ×
1
2 (5002 – 202) (m/s)2 1
1000 (kW/W) = 624 kW
cb

6.42
A liquid water turbine receives 2 kg/s water at 2000 kPa, 20oC and velocity of 15
m/s. The exit is at 100 kPa, 20oC and very low velocity. Find the specific work
and the power produced.
Solution:
1
2 V
2
1 + gZ1 = h2 +
1
2 V
2
2 + gZ2 + wT
State 2: Table B.1.1 h2 = 83.94 (which is at 2.3 kPa so we
should add ∆Pv = 97.7 × 0.001 to this)
wT = h1 +
1
2 V
2
1 − h2 = 85.82 + 15
2
/2000 – 83.94 = 1.99 kJ/kg
W
.
T = m
.
× wT = 2 × 1.9925 = 3.985 kW
Notice how insignificant the specific kinetic energy is.

6.43
A windmill with rotor diameter of 30 m takes 40% of the kinetic energy out as
shaft work on a day with 20oC and wind speed of 30 km/h. What power is
produced?
Solution:
Continuity Eq. m
.
i = m
.
e = m
.
Energy m
.
(hi + ½Vi
2 + gZi) = m
.
(he+ ½Ve
2 + gZe) + W
.
Process information: W
.
= m
.
½Vi
2 × 0.4
m
.
= ρAV =AVi /vi
A =
π
4 D 2 =
π
4 302 = 706.85 m2
vi = RTi/Pi =
0.287 × 293
101.3 = 0.8301 m3/kg
Vi = 30 km/h =
30 × 1000
3600 = 8.3333 m/s
m
.
= AVi /vi =
706.85 × 8.3333
0.8301 = 7096 kg/s
½ Vi
2 = ½ 8.33332 m2/s2 = 34.722 J/kg
W
.
= 0.4 m
.
½ Vi
2 = 0.4 ×7096 × 34.722 = 98 555 W
= 98.56 kW

6.44
Hoover Dam across the Colorado River dams up Lake Mead 200 m higher than
the river downstream. The electric generators driven by water-powered turbines
deliver 1300 MW of power. If the water is 17.5°C, find the minimum amount of
water running through the turbines.
Solution:
C.V.: H2O pipe + turbines,
T
H
DAM
Lake
Mead
Continuity: m
.
in = m
.
ex;
Energy Eq.6.13: (h+ V2/2 + gz)in = (h+ V2/2 + gz)ex + wT
Water states: hin ≅ hex ; vin ≅ vex
Now the specific turbine work becomes
wT = gzin - gzex = 9.807 × 200/1000 = 1.961 kJ/kg
m
.
= W
.
T/wT =
1300×103 kW
1.961 kJ/kg = 6.63 ×105 kg/s
V
.
= m
.
v = 6.63 ×105 × 0.001001 = 664 m3/s

6.45
A small expander (a turbine with heat transfer) has 0.05 kg/s helium entering at
1000 kPa, 550 K and it leaves at 250 kPa, 300 K. The power output on the shaft is
measured to 55 kW. Find the rate of heat transfer neglecting kinetic energies.
Solution:
C.V. Expander. Steady operation
Cont. m
.
i = m
.
e = m
.
Energy m
.
hi + Q
.
= m
.
he + W
.
WT
i
e
Q
.
cb
Q
.
= m
.
(he - hi) + W
.
Use heat capacity from tbl A.5: Cp He = 5.193 kJ/kg K
Q
.
= m
.
Cp (Te - Ti) + W
.
= 0.05 × 5.193 (300 - 550) + 55
= - 64.91 + 55 = -9.9 kW

6.46
A small turbine, shown in Fig. P 6.46, is operated at part load by throttling a 0.25
kg/s steam supply at 1.4 MPa, 250°C down to 1.1 MPa before it enters the turbine
and the exhaust is at 10 kPa. If the turbine produces 110 kW, find the exhaust
temperature (and quality if saturated).
Solution:
C.V. Throttle, Steady, q = 0 and w = 0. No change in kinetic or potential
energy. The energy equation then reduces to
Energy Eq.6.13: h1 = h2 = 2927.2 kJ/kg from Table B.1.3
C.V. Turbine, Steady, no heat transfer, specific work: w =
110
0.25 = 440 kJ/kg
Energy Eq.: h1 = h2 = h3 + w = 2927.2 kJ/kg
⇒ h3 = 2927.2 - 440 = 2487.2 kJ/kg
State 3: (P, h) Table B.1.2 h < hg
2487.2 = 191.83 + x3 × 2392.8
T
v
1
2
3
⇒ T = 45.8°C , x3 = 0.959

6.47
A small, high-speed turbine operating on compressed air produces a power output
of 100 W. The inlet state is 400 kPa, 50°C, and the exit state is 150 kPa, −30°C.
Assuming the velocities to be low and the process to be adiabatic, find the
required mass flow rate of air through the turbine.
Solution:
C.V. Turbine, no heat transfer, no ∆KE, no ∆PE
Energy Eq.6.13: hin = hex + wT
Ideal gas so use constant specific heat from Table A.5
wT = hin - hex ≅ Cp(Tin - Tex)
= 1.004 (kJ/kg K) [50 - (-30)] K = 80.3 kJ/kg
W
.
= m
.
wT ⇒ m
.
= W
.
/wT = 0.1/80.3 = 0.00125 kg/s
The dentist’s drill has a
small air flow and is not
really adiabatic.

Compressors, fans

6.48
A compressor in a commercial refrigerator receives R-410a at -25oC, x = 1. The
exit is at 800 kPa, 40oC. Neglect kinetic energies and find the specific work.
Solution:
C.V. Compressor, steady state, single inlet and
exit flow. For this device we also assume no
heat transfer and Zi = Ze
WC
i
e
cb
-
From Table B.4.1 : hi = 269.77 kJ/kg
From Table B.4.2 : he = 319.42 kJ/kg
Energy Eq.6.13 reduces to
wc = hi – he = (269.77 – 319.42) = -49.65 kJ/kg

6.49
A refrigerator uses the natural refrigerant carbon dioxide where the compressor
brings 0.02 kg/s from 1 MPa, -20oC to 6 MPa using 2 kW of power. Find the
compressor exit temperature.
Solution:
C.V. Compressor, steady state, single inlet and
exit flow. For this device we also assume no
heat transfer and Zi = Ze
WC
i
e
cb
-
From Table B.3.2 : hi = 342.31 kJ/kg
Energy Eq.6.13 reduces to
W
.
= m
.
wc = m
.
(hi – he) ⇒ he = hi – W
.
/m
.
he = (342.31 – (–2)/0.02) = 442.31 kJ/kg
From Table B.3.2 :
Te = 100 + 20 (442.31 – 421.69)/(445.02 – 421.69) = 117.7 oC

6.50
A compressor brings R-134a from 150 kPa, -10oC to 1200 kPa, 50oC. It is water
cooled with a heat loss estimated as 40 kW and the shaft work input is measured
to be 150 kW. How much is the mass flow rate through the compressor?
Solution:
C.V Compressor. Steady flow.
Neglect kinetic and potential energies.
Energy : m
.
hi + Q
.
= m
.
he + W
.
m
.
= (Q
.
- W
.
)/(he - hi)
1
2
Qcool
Compressor
-Wc
Look in table B.5.2
hi = 393.84 kJ/kg, he = 426.84 kJ/kg
m
.
=
-40 – (-150)
426.84 – 393.84 = 3.333 kg/s

6.51
An ordinary portable fan blows 0.2 kg/s room air with a velocity of 18 m/s (see
Fig. P6.19). What is the minimum power electric motor that can drive it? Hint:
Are there any changes in P or T?
Solution:
C.V. Fan plus space out to near stagnant inlet room air.
Energy Eq.6.13: q + hi + Vi
2/2 = he + Ve
2/2 + w
Here q ≅ 0, Vi ≅ 0 and hi = he same P and T
−w = Ve
2/2 = 182/2000 = 0.162 kJ/kg
−W
.
= −m
.
w = 0.2 kg/s × 0.162 kJ/kg = 0.032 kW

6.52
The compressor of a large gas turbine receives air from the ambient at 95 kPa,
20°C, with a low velocity. At the compressor discharge, air exits at 1.52 MPa,
430°C, with velocity of 90 m/s. The power input to the compressor is 5000 kW.
Determine the mass flow rate of air through the unit.
Solution:
C.V. Compressor, steady state, single inlet and exit flow.
Energy Eq.6.13: q + hi + Vi
2/2 = he + Ve
2/2 + w
Here we assume q ≅ 0 and Vi ≅ 0 so using constant CPo from A.5
-w = CPo(Te - Ti) + Ve
2/2 = 1.004(430 - 20) +
(90)2
2 × 1000 = 415.5 kJ/kg
Notice the kinetic energy is 1% of the work and can be neglected in most
cases. The mass flow rate is then from the power and the specific work
m
.
=
W
.
c
-w =
5000
415.5 = 12.0 kg/s

6.53
A compressor in an industrial air-conditioner compresses ammonia from a state of
saturated vapor at 150 kPa to a pressure of 800 kPa. At the exit the temperature is
measured to be 100oC and the mass flow rate is 0.5 kg/s. What is the required
motor size (kW) for this compressor?
Solution:
C.V. Compressor. Assume adiabatic and neglect kinetic energies.
Energy Eq.6.13: wC = h1 – h2
States: 1: B.2.2: h1 = 1410.9 kJ/kg
2: B.1.3 h2,AC = 1670.6 kJ/kg
Energy equation:
–wC = h2 – h1 = 1670.6 – 1410.9 = 259.7 kJ/kg
W
.
= 0.5 kg/s × 259.7 kJ/kg = 129.8 kW
P
2 ac
1
v

6.54
An air compressor takes in air at 100 kPa, 17°C and delivers it at 1 MPa, 600 K to
a constant-pressure cooler, which it exits at 300 K. Find the specific compressor
work and the specific heat transfer in the cooler.
Solution
C.V. air compressor q = 0
Continuity Eq.: m
.
2 = m
.
1
Energy Eq.6.13: h1 + wc = h2
1
32
Qcool
Compressor
-Wc
Compressor section Cooler section
Table A.7:
wc in = h2 - h1 = 607.02 - 290.17 = 316.85 kJ/kg
C.V. cooler w = 0/
Continuity Eq.: m
.
3 = m
.
1
Energy Eq.6.13: h2 = qout + h3
qout = h2 - h3 = 607.02 - 300.19 = 306.83 kJ/kg

6.55
An exhaust fan in a building should be able to move 2.5 kg/s air at 98 kPa, 20oC
through a 0.4 m diameter vent hole. How high a velocity must it generate and how
much power is required to do that?
Solution:
C.V. Fan and vent hole. Steady state with uniform velocity out.
Continuity Eq.: m
.
= constant = ρΑV = AV / v =AVP/RT
Ideal gas : Pv = RT, and area is A =
π
4 D2
Now the velocity is found
V = m
.
RT/(
π
4 D2P) = 2.5 × 0.287 × 293.15 / (
π
4 × 0.42 × 98) = 17.1 m/s
The kinetic energy out is
1
2 V
2
2 =
1
2 × 17.12 / 1000 = 0.146 kJ/kg
which is provided by the work (only two terms in energy equation that does
not cancel, we assume V1 = 0)
W
.
in = m
. 1
2 V
2
2 = 2.5 × 0.146 = 0.366 kW

6.56
How much power is needed to run the fan in Problem 6.19?
A household fan of diameter 0.75 m takes air in at 98 kPa, 22oC and delivers it at
105 kPa, 23oC with a velocity of 1.5 m/s. What are the mass flow rate (kg/s), the
inlet velocity and the outgoing volume flow rate in m3/s?
Solution:
Continuity Eq. m
.
i = m
.
e = AV/ v
Ideal gas v = RT/P
Area : A =
π
4 D2 =
π
4× 0.752 = 0.442 m2
V
.
e = AVe = 0.442 ×1.5 = 0.6627 m3/s
ve =
RTe
Pe
=
0.287 × 296
105 = 0.8091 m3/kg
m
.
i = V
.
e/ve = 0.6627/0.8091 = 0.819 kg/s
AVi /vi = m
.
i = AVe / ve
Vi = Ve × (vi / ve) = Ve × (RTi)/(Pive) = 1.5 ×
0.287 × (22 + 273)
98 × 0.8091
= 1.6 m/s
m
.
(hi + ½Vi
2) = m
.
(he+ ½Ve
2) +W
.
W
.
= m
.
(hi + ½Vi
2 – he – ½Ve
2 ) = m
.
[Cp (Ti-Te) + ½ Vi
2 – ½Ve
2 ]
= 0.819 [ 1.004 (-1) +
1.62 - 1.52
2000 ] = 0.819 [ -1.004 + 0.000155]
= - 0.81 kW

6.57
A compressor in an air-conditioner receives saturated vapor R-410a at 400 kPa
and brings it to 1.8 MPa, 60oC in an adiabatic compression. Find the flow rate for
a compressor work of 2 kW?
C.V. Compressor. Assume adiabatic and neglect kinetic energies.
Energy Eq.6.13: w = h1 - h2
States: 1: B.4.2 h1 = 271.9 kJ/kg
2: B.4.2 h2 = 323.92 kJ/kg
–wC = h2 – h1 = 323.92 – 271.9 = 52.02 kJ/kg
W
.
= m
.
wC m
.
= W
.
/ wC = 2 kW/ 52.02 kJ/kg = 0.038 kg/s

Heaters/Coolers

6.58
Carbon dioxide enters a steady-state, steady-flow heater at 300 kPa, 300 K, and
exits at 275 kPa, 1500 K, as shown in Fig. P6.58. Changes in kinetic and potential
energies are negligible. Calculate the required heat transfer per kilogram of
carbon dioxide flowing through the heater.
Solution:
C.V. Heater Steady state single inlet and exit flow.
Energy Eq.6.13: q + hi = he
Q
i
e
Table A.8: q = he - hi = 1614.9 – 214.4 = 1400.5 kJ/kg
[If we use Cp0 from A.5 then q ≅ 0.842(1500 - 300) = 1010.4 kJ/kg]
Too large ∆T, Tave to use Cp0 at room temperature.

6.59
A condenser (cooler) receives 0.05 kg/s R-410a at 2000 kPa, 60oC and cools it to
15oC. Assume the exit properties are as for saturated liquid same T. What cooling
capacity (kW) must the condenser have?
Solution:
C.V. R-410a condenser. Steady state single flow, heat transfer out and no
work.
Energy Eq.6.12: m
.
h1 = m
.
h2 + Q
.
out
Inlet state: Table B.4.2 h1 = 320.62 kJ/kg,
Exit state: Table B.4.1 h2 = 81.15 kJ/kg (compressed liquid)
Process: Neglect kinetic and potential energy changes.
Cooling capacity is taken as the heat transfer out i.e. positive out so
Q
.
out = m
.
( h1- h2) = 0.05 kg/s (320.62 – 81.15) kJ/kg
= 11.9735 kW = 12 kW
1 2
Qcool

6.60
Saturated liquid nitrogen at 600 kPa enters a boiler at a rate of 0.005 kg/s and
exits as saturated vapor. It then flows into a super heater also at 600 kPa where it
exits at 600 kPa, 280 K. Find the rate of heat transfer in the boiler and the super
heater.
Solution:
C.V.: boiler steady single inlet and exit flow, neglect KE, PE energies in flow
Continuity Eq.: m
.
1 = m
.
2 = m
.
3
1 2 3
Q
Q
boiler
Super
heatervapor
cb
600
P
1 2 3
v
T
1 2
3
v
Table B.6.1: h1 = -81.469 kJ/kg, h2 = 86.85 kJ/kg,
Table B.6.2: h3 = 289.05 kJ/kg
Energy Eq.6.13: qboiler = h2 – h1 = 86.85 - (- 81.469) = 168.32 kJ/kg
Q
.
boiler = m
.
1qboiler = 0.005 × 168.32 = 0.842 kW
C.V. Superheater (same approximations as for boiler)
Energy Eq.6.13: qsup heater = h3 – h2 = 289.05 – 86.85 = 202.2 kJ/kg
Q
.
sup heater = m
.
2qsup heater = 0.005 × 202.2 = 1.01 kW

6.61
The air conditioner in a house or a car has a cooler that brings atmospheric air
from 30oC to 10oC both states at 101 kPa. If the flow rate is 0.5 kg/s find the rate
of heat transfer.
Solution:
CV. Cooler. Steady state single flow with heat transfer.
Neglect changes in kinetic and potential energy and no work term.
Energy Eq.6.13: qout = hi – he
Use constant heat capacity from Table A.5 (T is around 300 K)
qout = hi − he = Cp (Ti − Te)
= 1.004
kJ
kg K × (30 – 10) K = 20.1 kJ/kg
Q
.
out = m
.
qout = 0.5 × 20.1 = 10 kW

6.62
A chiller cools liquid water for air-conditioning purposes. Assume 2.5 kg/s water
at 20oC, 100 kPa is cooled to 5oC in a chiller. How much heat transfer (kW) is
needed?
Solution:
C.V. Chiller. Steady state single flow with heat transfer. Neglect changes in
kinetic and potential energy and no work term.
Energy Eq.6.13: qout = hi – he
Properties from Table B.1.1:
hi = 83.94 kJ/kg and he = 20.98 kJ/kg
Now the energy equation gives
qout = 83.94 – 20.98 = 62.96 kJ/kg
Q
.
out = m
.
qout = 2.5 × 62.96 = 157.4 kW
Alternative property treatment since single phase and small ∆T
If we take constant heat capacity for the liquid from Table A.4
qout = hi – he ≅ Cp (Ti - Te )
= 4.18 (20 – 5) = 62.7 kJ/kg
Q
.
out = m
.
qout = 2.5 × 62.7 = 156.75 kW
1 2
Qcool

6.63
Carbon dioxide used as a natural refrigerant flows through a cooler at 10 MPa,
which is supercritical so no condensation occurs. The inlet is at 200oC and the
exit is at 40oC. Find the specific heat transfer.
C.V. Cooler. Steady state single flow with heat transfer. Neglect changes in
kinetic and potential energy and no work term.
Energy Eq.6.13: 0 = hi – he + q
Properties from Table B.3.2:
hi = 519.49 kJ/kg and he = 200.14 kJ/kg
Now the energy equation gives
q = 200.14 – 519.49 = –319.35 kJ/kg

6.64
A flow of liquid glycerine flows around an engine, cooling it as it absorbs energy.
The glycerine enters the engine at 60oC and receives 19 kW of heat transfer.
What is the required mass flow rate if the glycerine should come out at maximum
95oC?
Solution:
C.V. Liquid flow (glycerine is the coolant), steady flow. no work.
Energy Eq.: m
.
hi + Q
.
= m
.
he
m
.
= Q
.
/( he - hi) =
Q
.
Cgly (Te - Ti)
From table A.4 Cgly = 2.42 kJ/kg-K
m
.
=
19
2.42 (95 – 60) = 0.224 kg/s
Exhaust flow
Air intake filter
Coolant flow
Atm.
airShaft
Fan
power
Radiator
cb

6.65
In a steam generator, compressed liquid water at 10 MPa, 30°C, enters a 30-mm
diameter tube at the rate of 3 L/s. Steam at 9 MPa, 400°C exits the tube. Find the
rate of heat transfer to the water.
Solution:
C.V. Steam generator. Steady state single inlet and exit flow.
Constant diameter tube: Ai = Ae =
π
4 (0.03)2 = 0.0007068 m2
Table B.1.4 m
.
= V
.
i/vi = 0.003/0.0010003 = 3.0 kg/s
Vi = V
.
i/Ai = 0.003/0.0007068 = 4.24 m/s
Exit state properties from Table B.1.3
Ve = Vi × ve/vi = 4.24 × 0.02993/0.0010003 = 126.86 m/s
The energy equation Eq.6.12 is solved for the heat transfer as
Q
.
= m
.



(he - hi) + ( )Ve
2 - Vi
2 /2
= 3.0






3117.8 - 134.86 +
126.862 - 4.242
2 × 1000 = 8973 kW
Typically hot
combustion
gas in
Steam exit
cb
liquid water in
gas
out

6.66
In a boiler you vaporize some liquid water at 100 kPa flowing at 1 m/s. What is
the velocity of the saturated vapor at 100 kPa if the pipe size is the same? Can the
flow then be constant P?
The continuity equation with average values is written
m
.
i = m
.
e = m
.
= ρAV = AV/v = AVi/vi = AVe/ve
From Table B.1.2 at 100 kPa we get
vf = 0.001043 m3/kg; vg = 1.694 m3/kg
Ve = Vi ve/vi = 1
1.694
0.001043 = 1624 m/s
To accelerate the flow up to that speed you need a large force ( ∆PA ) so a
large pressure drop is needed.
Pi cb
Pe < Pi

6.67
A cryogenic fluid as liquid nitrogen at 90 K, 400 kPa flows into a probe used in
cryogenic surgery. In the return line the nitrogen is then at 160 K, 400 kPa. Find
the specific heat transfer to the nitrogen. If the return line has a cross sectional
area 100 times larger than the inlet line what is the ratio of the return velocity to
the inlet velocity?
Solution:
C.V line with nitrogen. No kinetic or potential energy changes
Continuity Eq.: m
.
= constant = m
.
e = m
.
i = AeVe/ve = AiVi/vi
Energy Eq.6.13: q = he − hi
State i, Table B.6.1: hi = -95.58 kJ/kg, vi = 0.001343 m3/kg
State e, Table B.6.2: he = 162.96 kJ/kg, ve = 0.11647 m3/kg
q = he − hi = 162.96 – (-95.58) = 258.5 kJ/kg
From the continuity equation
Ve/Vi = Ai/Ae (ve/vi) =
1
100
0.11647
0.001343 = 0.867

Pumps, pipe and channel flows

6.68
A steam pipe for a 300-m tall building receives superheated steam at 200 kPa at
ground level. At the top floor the pressure is 125 kPa and the heat loss in the pipe
is 110 kJ/kg. What should the inlet temperature be so that no water will condense
inside the pipe?
Solution:
C.V. Pipe from 0 to 300 m, no ∆KE, steady state, single inlet and exit flow.
Neglect any changes in kinetic energy.
Energy Eq.6.13: q + hi = he + gZe
No condensation means: Table B.1.2, he = hg at 125 kPa = 2685.4 kJ/kg
hi = he + gZe - q = 2685.4 +
9.807 × 300
1000 - (-110) = 2810.1 kJ/kg
At 200 kPa: T ~ 170oC Table B.1.3

6.69
A small stream with 20oC water runs out over a cliff creating a 100 m tall
waterfall. Estimate the downstream temperature when you neglect the horizontal
flow velocities upstream and downstream from the waterfall. How fast was the
water dropping just before it splashed into the pool at the bottom of the waterfall?
Solution:
CV. Waterfall, steady state. Assume no Q
.
nor W
.
Energy Eq.6.13: h +
1
2V2 + gZ = const.
State 1: At the top zero velocity Z1 = 100 m
State 2: At the bottom just before impact, Z2 = 0
State 3: At the bottom after impact in the pool.
h1 + 0 + gZ1 = h2 +
1
2 V
2
2 + 0 = h3 + 0 + 0
Properties: h1 ≅ h2 same T, P
=>
1
2 V
2
2 = gZ1
V2 = 2gZ1 = 2 × 9.806 × 100 = 44.3 m/s
Energy equation from state 1 to state 3
h3 = h1 + gZ1
use ∆h = Cp ∆T with value from Table A.4 (liquid water)
T3 = T1 + gZ1 / Cp
= 20 + 9.806 × 100 /4180 = 20.23 °C

6.70
An irrigation pump takes water from a river at 10oC, 100 kPa and pumps it up to
an open canal, where it flows out 100 m higher at 10oC. The pipe diameter in and
out of the pump is 0.1 m and the motor driving the unit is 5 hp. What is the flow
rate neglecting kinetic energy and losses?
C.V. Pump plus pipe to canal.
This is a single steady flow device with atmospheric pressure in and out.
Energy Eq.: 0 = hi + 0 + gZi + wp in – he – 0 – gZe
wp in = g(Ze – Zi) = 9.81 m/s2 × 100 (m/s)2 = 0.981 kJ/kg
W
.
p in = 5 hp = 5 hp × 0.746 kW/hp = 3.73 kW
Find the flow rate from the work
m
.
= W
.
p in / wp in = 3.73 / 0.981 = 3.8 kg/s

6.71
Consider a water pump that receives liquid water at 15oC, 100 kPa and delivers it
to a same diameter short pipe having a nozzle with exit diameter of 1 cm (0.01 m)
to the atmosphere 100 kPa. Neglect the kinetic energy in the pipes and assume
constant u for the water. Find the exit velocity and the mass flow rate if the pump
draws a power of 1 kW.
Solution:
Continuity Eq.: m
.
i = m
.
e = AV/v ; A =
π
4 D
2
e =
π
4 × 0.012 = 7.854× 10−5 m2
1
2V
2
i + gZi = he +
1
2V
2
e + gZe + w
Properties: hi = ui + Pivi = he = ue + Peve ; Pi = Pe ; vi = ve
w = −
1
2 V
2
e
⇒ −W
.
= m
.
(
1
2 V
2
e ) = A ×
1
2 V
3
e /ve
Ve = (
−2 W
.
ve
A )
1/3
= ( 2 × 1000 × 0.001001
7.854×10 −5 )
1/3
= 29.43 m/s
m
.
= AVe/ve = 7.854× 10 −5 × 29.43 / 0.001001 = 2.31 kg/s
Small water pump for a washing
machine. Notice the very simple
impeller.

6.72
A cutting tool uses a nozzle that generates a high speed jet of liquid water.
Assume an exit velocity of 500 m/s of 20oC liquid water with a jet diameter of 2
mm (0.002 m). How much mass flow rate is this? What size (power) pump is
needed to generate this from a steady supply of 20oC liquid water at 200 kPa?
Solution:
C.V. Nozzle. Steady state, single flow.
Continuity equation with a uniform velocity across A
m
.
= AV/v =
π
4 D2 V / v =
π
4 0.0022 × 500 / 0.001002 = 1.568 kg/s
Assume Zi = Ze = Ø, ue = ui and Vi = 0 Pe = 100 kPa (atmospheric)
Energy Eq.6.13: hi + Ø + Ø = he +
1
2V
2
e + Ø + w
w = hi − he −
1
2V
2
e = ui − ue + Pi vi − Pe ve −
1
2V
2
e
= (Pi - Pe) vi −
1
2V
2
e
= 0.001002 m3/kg × (200 – 100) kPa – 0.5 ×
5002 m2/s2
1000 J/kJ
= 0.1002 – 125 ≅ −125 kJ/kg
W
.
= m
.
w = 1.568 kg/s (–125 kJ/kg) = −196 kW

6.73
A small water pump is used in an irrigation system. The pump takes water in from
a river at 10oC, 100 kPa at a rate of 5 kg/s. The exit line enters a pipe that goes up
to an elevation 20 m above the pump and river, where the water runs into an open
channel. Assume the process is adiabatic and that the water stays at 10oC. Find
the required pump work.
Solution:
C.V. pump + pipe. Steady state , 1 inlet, 1 exit flow. Assume same velocity in
and out, no heat transfer.
Continuity Eq.: m
.
in = m
.
ex = m
.
Energy Eq.6.12:
m
.
(hin + (1/2)Vin
2 + gzin) =
m
.
(hex + (1/2) Vex
2 + gzex) + W
.
States: hin = hex same (T, P)
i
e
H
cb
W
.
= m
.
g(zin - zex) = 5 × 9.807 × (0 - 20)/1000 = −0.98 kW
I.E. 0.98 kW required input

6.74
The main waterline into a tall building has a pressure of 600 kPa at 5 m below
ground level. A pump brings the pressure up so the water can be delivered at 200
kPa at the top floor 150 m above ground level. Assume a flow rate of 10 kg/s
liquid water at 10oC and neglect any difference in kinetic energy and internal
energy u. Find the pump work.
Solution:
C.V. Pipe from inlet at -5 m up to exit at +150 m, 200 kPa.
1
2Vi
2 + gZi = he +
1
2Ve
2 + gZe + w
With the same u the difference in h’s are the Pv terms
w = hi - he +
1
2 (Vi
2 - Ve
2) + g (Zi- Ze)
= Pivi – Peve + g (Zi – Ze)
= 600 × 0.001 – 200 × 0.001 + 9.806 × (-5 - 150)/1000
= 0.4 – 1.52 = -1.12 kJ/kg
W
.
= m
.
w = 10 × (-1.12) = -11.2 kW

6.75
A pipe flows water at 15oC from one building to another. In the winter time the
pipe loses an estimated 500 W of heat transfer. What is the minimum required
mass flow rate that will ensure that the water does not freeze (i.e. reach 0oC)?
Solution:
Energy Eq.: m
.
hi + Q
.
= m
.
he
Assume saturated liquid at given T from table B.1.1
m
.
=
Q
.
he - hi
=
-500 × 10-3
0 - 62.98 =
0.5
62.98 = 0.007 94 kg/s
1 2
-Q
.

Multiple flow single device processes
Turbines, Compressors, Expanders

6.76
A steam turbine receives steam from two boilers. One flow is 5 kg/s at 3 MPa,
700°C and the other flow is 15 kg/s at 800 kPa, 500°C. The exit state is 10 kPa,
with a quality of 96%. Find the total power out of the adiabatic turbine.
Solution:
C.V. whole turbine steady, 2 inlets, 1 exit, no heat transfer Q
.
= 0
Continuity Eq.6.9: m.
1 + m.
2 = m.
3 = 5 + 15 = 20 kg/s
Energy Eq.6.10: m.
1h1 + m.
2h2 = m.
3h3 + W
.
T
Table B.1.3: h1 = 3911.7 kJ/kg,
h2 = 3480.6 kJ/kg
Table B.1.2: h3 = 191.8 + 0.96 × 2392.8
= 2488.9 kJ/kg
WT
1
2
3
W
.
T = 5 × 3911.7 + 15 × 3480.6 – 20 × 2488.9 = 21 990 kW = 22 MW

6.77
A compressor receives 0.05 kg/s R-410a at 200 kPa, -20oC and 0.1 kg/s R-410a at
400 kPa, 0oC. The exit flow is at 1000 kPa, 60oC as shown in Fig. P6.77. Assume
it is adiabatic, neglect kinetic energies and find the required power input.
C.V. whole compressor steady, 2 inlets, 1 exit, no heat transfer Q
.
= 0
Continuity Eq.6.9: m.
1 + m.
2 = m.
3 = 0.05 + 0.1 = 0.15 kg/s
Energy Eq.6.10: m.
1h1 + m.
2h2 − W
.
C = m.
3h3
Table B.3.2: h1 = 278.72 kJ/kg,
h2 = 290.42 kJ/kg
Table B.3.2: h3 = 335.75 kJ/kg
-WC
3
2
1
W
.
C = 0.05 × 278.72 + 0.1 × 290.42 – 0.15 × 335.75 = −7.385 kW
Power input is 7.4 kW

6.78
Two steady flows of air enters a control volume, shown in Fig. P6.78. One is
0.025 kg/s flow at 350 kPa, 150°C, state 1, and the other enters at 450 kPa, 15°C,
state 2. A single flow of air exits at 100 kPa, −40°C, state 3. The control volume
rejects 1 kW heat to the surroundings and produces 4 kW of power. Neglect
kinetic energies and determine the mass flow rate at state 2.
Solution:
C.V. Steady device with two inlet and one
exit flows, we neglect kinetic energies. Notice
here the Q is rejected so it goes out.
1
2
3Engine
Q
.
loss
W
.
.
1 + m
.
2 = m
.
3 = 0.025 + m
.
2
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3h3 + W
.
CV + Q
.
loss
Substitute the work and heat transfer into the energy equation and use
constant heat capacity
0.025 × 1.004 × 423.2 + m
.
2 × 1.004 × 288.2
= (0.025 + m
.
2) 1.004 × 233.2 + 4.0 + 1.0
Now solve for m
.
2.
m
.
2 =
4.0 + 1.0 + 0.025 × 1.004 × (233.2 – 423.2)
1.004 (288.2 - 233.2) = 0.0042 kg/s

6.79
A steam turbine receives water at 15 MPa, 600°C at a rate of 100 kg/s, shown in
Fig. P6.79. In the middle section 20 kg/s is withdrawn at 2 MPa, 350°C, and the
rest exits the turbine at 75 kPa, and 95% quality. Assuming no heat transfer and
no changes in kinetic energy, find the total turbine power output.
Solution:
C.V. Turbine Steady state, 1 inlet and 2 exit flows.
.
1 = m
.
2 + m
.
3 ; => m
.
3 = m
.
1 - m
.
2 = 80 kg/s
Energy Eq.6.10: m
.
1h1 = W
.
T + m
.
2h2 + m
.
3h3
Table B.1.3 h1 = 3582.3 kJ/kg,
h2 = 3137 kJ/kg
Table B.1.2 : h3 = hf + x3hfg = 384.3 + 0.95×2278.6
= 2549.1 kJ/kg
WT
1
2
3
From the energy equation, Eq.6.10
=> W
.
T = m
.
1h1 − m
.
2h2 − m
.
3h3 = 91.565 MW

6.80
Cogeneration is often used where a steam supply is needed for industrial process
energy. Assume a supply of 5 kg/s steam at 0.5 MPa is needed. Rather than
generating this from a pump and boiler, the setup in Fig. P6.80 is used so the
supply is extracted from the high-pressure turbine. Find the power the turbine
now cogenerates in this process.
Solution:
C.V. Turbine, steady state, 1 inlet and 2 exit flows, assume adiabatic, Q
.
CV = 0
.
1 = m
.
2 + m
.
3
Energy Eq.6.10: Q
.
CV + m
.
1h1 = m
.
2h2 + m
.
3h3 + W
.
T ;
Supply state 1: 20 kg/s at 10 MPa, 500°C
Process steam 2: 5 kg/s, 0.5 MPa, 155°C,
Exit state 3: 20 kPa, x = 0.9
Table B.1.3: h1 = 3373.7, h2 = 2755.9 kJ/kg,
Table B.1.2: h3 = 251.4 + 0.9 × 2358.3
= 2373.9 kJ/kg
WT
1
2
3
HP LP
W
.
T = 20 × 3373.7 − 5 × 2755.9 − 15 × 2373.9 = 18.084 MW

6.81
A compressor receives 0.1 kg/s R-134a at 150 kPa, -10oC and delivers it at 1000
kPa, 40oC. The power input is measured to be 3 kW. The compressor has heat
transfer to air at 100 kPa coming in at 20oC and leaving at 25oC. How much is the
mass flow rate of air?
Solution:
C.V. Compressor, steady state, single inlet and exit
flow. For this device we also have an air flow
outside the compressor housing no changes in
kenetic or potential energy.
WC
1
2
cb
-
3
4
Air
Air
Continuity Eq.: m
.
2 = m
.
1
Energy Eq. 6.12: m
.
1h1 + W
.
in + m
.
airh3 = m
.
2h2 + m
.
airh4
Ideal gas for air and constant heat capacity: h4 - h3 ~ Cp air (T4 –T3)
m
.
air = [m
.
1 (h1 –h2) + W
.
in ] / Cp air (T4 –T3)
=
0.1 ( 393.84 – 420.25) + 3
1.004 (25-20) =
0.359
5
= 0.0715 kg/s

6.82
A large expansion engine has two low velocity flows of water entering. High
pressure steam enters at point 1 with 2.0 kg/s at 2 MPa, 500°C and 0.5 kg/s
cooling water at 120 kPa, 30°C enters at point 2. A single flow exits at point 3
with 150 kPa, 80% quality, through a 0.15 m diameter exhaust pipe. There is a
heat loss of 300 kW. Find the exhaust velocity and the power output of the
engine.
Solution:
C.V. : Engine (Steady state)
Constant rates of flow, Q
.
loss and W
.
State 1: Table B.1.3: h1 = 3467.6 kJ/kg
State 2: Table B.1.1: h2 = 125.77 kJ/kg
h3 = 467.1 + 0.8 × 2226.5 = 2248.3 kJ/kg
1
2
3
Engine
Q
.
loss
W
.
v3 = 0.00105 + 0.8 × 1.15825 = 0.92765 m3/kg
.
1+ m
.
2 = m
.
3 = 2 + 0.5= 2.5 kg/s = (AV/v) = (π/4)D2V/v
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3(h3 + 0.5 V2) + Q
.
loss + W
.
V = m
.
3v3 / [
π
4 D2] = 2.5 × 0.92765 / (0.7854 × 0.152 ) = 131.2 m/s
0.5 V2 = 0.5 × 131.22 /1000 = 8.6 kJ/kg ( remember units factor 1000)
W
.
= 2 ×3467.6 + 0.5 ×125.77 – 2.5 (2248.3 + 8.6) – 300 = 1056 kW

Heat Exchangers

6.83
A condenser (heat exchanger) brings 1 kg/s water flow at 10 kPa from 300°C to
saturated liquid at 10 kPa, as shown in Fig. P6.84. The cooling is done by lake
water at 20°C that returns to the lake at 30°C. For an insulated condenser, find the
flow rate of cooling water.
Solution:
C.V. Heat exchanger
Energy Eq.6.10: m
.
coolh20 + m
.
H2Oh300 = m
.
coolh30 + m
.
H2Ohf, 10 kPa
300°C
30°C 20°C
m
.
cool
1 kg/s
sat. liq.
Table B.1.1: h20 = 83.96 kJ/kg , h30 = 125.79 kJ/kg
Table B.1.3: h300, 10kPa = 3076.5 kJ/kg, B.1.2: hf, 10 kPa = 191.83 kJ/kg
m
.
cool = m
.
H2O
h300 - hf, 10kPa
h30 - h20
= 1 ×
3076.5 - 191.83
125.79 - 83.96 = 69 kg/s

6.84
In a co-flowing (same direction) heat exchanger 1 kg/s air at 500 K flows into one
channel and 2 kg/s air flows into the neighboring channel at 300 K. If it is
infinitely long what is the exit temperature? Sketch the variation of T in the two
flows.
C.V. mixing section (no W
.
, Q
.
)
Continuity Eq.: m
.
1 = m
.
3 and m
.
2 = m
.
4
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
1h3 + m
.
2h4
Same exit T: h3 = h4 = [m
.
1h1 + m
.
2h2] / [m
.
1 + m
.
2]
Using conctant specific heat
T3 = T4 =
m
.
1
m
.
1 + m
.
2
T1 +
m
.
2
m
.
1 + m
.
2
T2 =
1
3 × 500 +
2
3 × 300 = 367 K
x
cb
3
4
1
2
T
x300
500
1T
2T

6.85
A heat exchanger, shown in Fig. P6.85, is used to cool an air flow from 800 K to
360 K, both states at 1 MPa. The coolant is a water flow at 15°C, 0.1 MPa. If the
water leaves as saturated vapor, find the ratio of the flow rates m
.
H2O/m
.
air
Solution:
C.V. Heat exchanger, steady flow 1 inlet
and 1 exit for air and water each. The
two flows exchange energy with no heat
transfer to/from the outside. 3 water
1 air
4
2
Continuity Eqs.: Each line has a constant flow rate through it.
Energy Eq.6.10: m
.
airh1 + m
.
H2Oh3 = m
.
airh2 + m
.
H2Oh4
Process: Each line has a constant pressure.
Air states, Table A.7.1: h1 = 822.20 kJ/kg, h2 = 360.86 kJ/kg
Water states, Table B.1.1: h3 = 62.98 kJ/kg (at 15°C),
Table B.1.2: h4 = 2675.5 kJ/kg (at 100 kPa)
m
.
H2O/m
.
air =
h1 - h2
h 4 - h3
=
822.20 - 360.86
2675.5 - 62.99 = 0.1766

6.86
Air at 600 K flows with 3 kg/s into a heat exchanger and out at 100oC. How much
(kg/s) water coming in at 100 kPa, 20oC can the air heat to the boiling point?
C.V. Total heat exchanger. The flows are not mixed so the two flowrates are
constant through the device. No external heat transfer and no work.
Energy Eq.6.10: m
.
airhair in + m
.
waterhwater in = m
.
airhair out + m
.
waterhwater out
m
.
air[hair in - hair out] = m
.
water[hwater out – hwater in]
Table B.1.2: hwater out – hwater in = 2675.46 – 83.94 = 2591.5 kJ/kg
Table A.7.1: hair in - hair out = 607.32 – 374.14 = 233.18 kJ/kg
Solve for the flow rate of water from the energy equation
m
.
water = m
.
air
hair in - hair out
hwater out - hwater in
= 3 ×
233.18
2591.5 = 0.27 kg/s
Air
in cb
Air
out

6.87
An automotive radiator has glycerine at 95oC enter and return at 55oC as shown
in Fig. P6.87. Air flows in at 20oC and leaves at 25oC. If the radiator should
transfer 25 kW what is the mass flow rate of the glycerine and what is the volume
flow rate of air in at 100 kPa?
Solution:
If we take a control volume around the whole radiator then there is no external
heat transfer - it is all between the glycerin and the air. So we take a control
volume around each flow separately.
Glycerine: m
.
hi + (-Q
.
) = m
.
he
Table A.4: m
.
gly =
-Q
.
he - hi
=
-Q
.
Cgly(Te-Ti) =
-25
2.42(55 - 95) = 0.258 kg/s
Air m
.
hi+ Q
.
= m
.
he
Table A.5: m
.
air =
Q
.
he - hi
=
Q
.
Cair(Te-Ti) =
25
1.004(25 - 20) = 4.98 kg/s
V
.
= m
.
vi ; vi =
RTi
Pi
=
0.287 × 293
100 = 0.8409 m3/kg
V
.
air = m
.
vi = 4.98 × 0.8409 = 4.19 m3/s
Exhaust flow
Air intake filter
Coolant flow 55 C
Atm.
airShaft
power
95 C
o
o
cb

6.88
A superheater brings 2.5 kg/s saturated water vapor at 2 MPa to 450oC. The
energy is provided by hot air at 1200 K flowing outside the steam tube in the
opposite direction as the water, which is a counter flowing heat exchanger. Find
the smallest possible mass flow rate of the air so the air exit temperature is 20oC
larger than the incoming water temperature (so it can heat it).
Solution:
C.V. Superheater. Steady state with no
external Q
.
or any W
.
the two flows
exchanges energy inside the box. Neglect
kinetic and potential energies at all
states.
3 water
1 air
4
2
Energy Eq.6.10: m
.
H2O h3 + m
.
air h1 = m
.
H2O h4 + m
.
air h2
Process: Constant pressure in each line.
State 1: Table B.1.2 T3 = 212.42°C, h3 = 2799.51 kJ/kg
State 3: Table A.7 h1 = 1277.81 kJ/kg
State 4: T2 = T3 + 20 = 232.42°C = 505.57 K
A.7 : h2 = 503.36 +
5.57
20 (523.98 – 503.36) = 509.1 kJ/kg
From the energy equation we get
m
.
air / m
.
H2O = (h4 - h3)/(h1 - h2)
= 2.5 (3357.48 – 2799.51) / (1277.81 – 509.1) = 1.815 kg/s

6.89
A cooler in an air conditioner brings 0.5 kg/s air at 35oC to 5oC, both at 101 kPa
and it then mix the output with a flow of 0.25 kg/s air at 20oC, 101 kPa sending
the combined flow into a duct. Find the total heat transfer in the cooler and the
temperature in the duct flow.
Solution:
1 2
3
4
Qcool
Cooler section Mixing section
C.V. Cooler section (no W
.
)
Energy Eq.6.12: m
.
h1 = m
.
h2 + Q
.
cool
Q
.
cool = m
.
(h1 - h2) = m
.
Cp (T1 - T2) = 0.5 × 1.004 × (35-5) = 15.06 kW
C.V. mixing section (no W
.
, Q
.
)
Continuity Eq.: m
.
2 + m
.
3 = m
.
4
Energy Eq.6.10: m
.
2h2 + m
.
3h3 = m
.
4h4
m
.
4 = m
.
2 + m
.
3 = 0.5 + 0.25 = 0.75 kg/s
m
.
4h4 = (m
.
2 + m
.
3)h4 = m
.
2h2 + m
.
3h3
m
.
2 (h4 - h2) + m
.
3 (h4 - h3) = Ø
m
.
2 Cp (T4 - T2) + m
.
3 Cp (T4 - T3) = Ø
T4 = (m
.
2 / m
.
4) T2 + (m
.
3 / m
.
4) T3 = 5(0.5/0.75) + 20(0.25/0.75) = 10°C

6.90
Steam at 500 kPa, 300oC is used to heat cold water at 15oC to 75oC for domestic
hot water supply. How much steam per kg liquid water is needed if the steam
should not condense?
Solution:
C.V. Each line separately. No work but there is heat transfer out of the steam flow
and into the liquid water flow.
Water line energy Eq.: m
.
liqhi + Q
.
= m
.
liqhe ⇒ Q
.
= m
.
liq(he – hi)
For the liquid water look in Table B.1.1
∆hliq = he – hi = 313.91 – 62.98 = 250.93 kJ/kg
( ≅ Cp ∆T = 4.18 (75 – 15) = 250.8 kJ/kg )
Steam line energy has the same heat transfer but it goes out
Steam Energy Eq.: m
.
steamhi = Q
.
+ m
.
steamhe ⇒ Q
.
= m
.
steam(hi – he)
For the steam look in Table B.1.3 at 500 kPa
∆hsteam = hi – he = 3064.2 – 2748.67 = 315.53 kJ/kg
Now the heat transfer for the steam is substituted into the energy equation for the
water to give
m
.
steam / m
.
liq = ∆hliq / ∆hsteam =
250.93
315.53 = 0.795
cb
Steam
inSteam
out
Cold water in
Hot water out

6.91
A two fluid heat exchanger has 2 kg/s liquid ammonia at 20oC, 1003 kPa entering
state 3 and exiting at state 4. It is heated by a flow of 1 kg/s nitrogen at 1500 K,
state 1, leaving at 600 K, state 2 similar to Fig. P6.85. Find the total rate of heat
transfer inside the heat exchanger. Sketch the temperature versus distance for the
ammonia and find state 4 (T, v) of the ammonia.
Solution:
CV: Nitrogen flow line, steady rates of flow, Q
.
out and W
.
= 0
Continiuty: m
.
1 = m
.
2 = 1 kg/s ; Energy Eq: m
.
1h1 = m
.
2h2 + Q
.
out
Tbl. A.8: h1 = 1680.7 kJ/kg; h2 = 627.24 kJ/kg
Q
.
out = m
.
1(h1 - h2) = 1 (1680.7 – 627.24) = 1053.5 kW
If Tbl A.5 is used: Cp = 1.042 kJ/kg K
Q
.
out = m
.
1 Cp (T1 - T2) = 1×1.042 (1500 - 600) = 937.8 kW
CV The whole heat exchanger: No external Q
.
, constant pressure in each line.
m
.
1h1 + m
.
3h3 = m
.
1h2 + m
.
3h4 => h4 = h3 + m
.
1(h1 - h2)/m
.
3
h4 = 274.3 + 1053.5 /2 = 801 kJ/kg < hg => 2-phase
x4 = (h4 - hf)/ hfg = (801 - 298.25) / 1165.2 = 0.43147
v4 = vf + x4 vfg = 0.001658 + 0.43147×0.12647 = 0.05623 m3/kg
T4 = T3a = 25oC This is the boiling temperature for 1003 kPa.
T
x293
298
3
3a 4
3 NH
1 N
4
2
3
2

6.92
A copper wire has been heat treated to 1000 K and is now pulled into a cooling
chamber that has 1.5 kg/s air coming in at 20oC; the air leaves the other end at
60oC. If the wire moves 0.25 kg/s copper, how hot is the copper as it comes out?
Solution:
C.V. Total chamber, no external heat transfer
Energy eq.: m
.
cu h icu + m
.
air hi air = m
.
cu he cu + m
.
air he air
m
.
cu ( he – hi )cu = m
.
air( hi – he )air
m
.
cu Ccu ( Te – Ti )cu = m
.
air Cp air( Te – Ti )air
Heat capacities from A.3 for copper and A.5 for air
( Te – Ti )cu =
m
.
airCp air
m
.
cuCcu
( Te – Ti )air =
1.5 ×1.004
0.25 × 0.42
(20 - 60) = - 573.7 K
Te = Ti – 573.7 = 1000 - 573.7 = 426.3 K
Air
Air
Cu

Mixing processes

6.93
Two air flows are combined to a single flow. Flow one is 1 m3/s at 20oC and the
other is 2 m3/s at 200oC both at 100 kPa. They mix without any heat transfer to
produce an exit flow at 100 kPa. Neglect kinetic energies and find the exit
temperature and volume flow rate.
Solution:
Cont. m
.
i = m
.
e = m
.
Energy m
.
1h1 + m
.
2h2 = m
.
3h3
= (m
.
1 + m
.
2)h3
1
2
3
Mixing section
m
.
1 (h3 -h1) + m
.
2 (h3 -h2) = 0
m
.
1Cp ( T3-T1) + m
.
2Cp( T3-T2) = 0
T3 = (m
.
i/m
.
3)/T1 + (m
.
2/m
.
3)T2
We need to find the mass flow rates
v1 = RT1/P1 = (0.287 × 293)/100 = 0.8409 m3/kg
v2 = RT2/P2 = (0.287 × 473)/100 = 1.3575 m3/kg
m
.
1 =
V
.
1
v1
=
1
0.8409 = 1.1892
kg
s , m
.
2 =
V
.
2
v2
=
2
1.3575 = 1.4733
kg
s
m
.
3 = m
.
1+ m
.
2 = 2.6625 kg/s
T3 =
1.1892
2.6625 × 20 +
1.4733
2.6625 × 200 = 119.6o C
v3 =
RT3
P3
=
0.287 (119.6 + 273)
100 = 1.1268 m3/kg
V
.
3 = m
.
3 v3 = 2.6625 × 1.1268 = 3.0 m3/s

6.94
A de-superheater has a flow of ammonia 1.5 kg/s at 1000 kPa, 100oC which is
mixed with another flow of ammonia at 25oC and quality 25% in an adiabatic
mixing chamber. Find the flow rate of the second flow so the outgoing ammonia
is saturated vapor at 1000 kPa.
C.V. Desuperheater.
No external Q
.
or W
.
1
2
3MIXING
CHAMBER
cb
.
1 + m
.
2 = m
.
3
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3h3 = (m
.
1+ m
.
2)h3
State 2: Table B.2.1 h2 = 298.25 + 0.25 ×1165.2 = 589.55 kJ/kg
m
.
2 = m
.
1×
h1 - h3
h3 - h2
= 1.5 ×
1664.3 – 1463.4
1463.4 – 589.55 = 0.345 kg/s
T
v
12 3
1000 kPa
2
P
13
v

6.95
An open feedwater heater in a powerplant heats 4 kg/s water at 45oC, 100 kPa by
mixing it with steam from the turbine at 100 kPa, 250oC. Assume the exit flow is
saturated liquid at the given pressure and find the mass flow rate from the turbine.
Solution:
C.V. Feedwater heater.
No external Q
.
or W
.
1
2
3MIXING
CHAMBER
cb
.
1 + m
.
2 = m
.
3
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3h3 = (m
.
1+ m
.
2)h3
State 1: Table B.1.1 h = hf = 188.42 kJ/kg at 45oC
State 3: Table B.1.2 h3 = hf = 417.44 kJ/kg at 100 kPa
m
.
2 = m
.
1×
h1 - h3
h3 - h2
= 4 ×
188.42 – 417.44
417.44 – 2974.33 = 0.358 kg/s
T
v1
23
100 kPa
2
P
v
31

6.96
A flow of water at 2000 kPa, 20oC is mixed with a flow of 2 kg/s water at 2000
kPa, 180oC. What should the flowrate of the first flow be to produce an exit state
of 200 kPa and 100oC?
Solution:
C.V. Mixing chamber and valves. Steady state no heat transfer or work terms.
.
1 + m
.
2 = m
.
3
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3h3 = (m
.
1+ m
.
2)h3
1
2
3MIXING
CHAMBER
Properties Table B.1.1: h1 = 85.8 kJ/kg; h3 = 419.0 kJ/kg
Table B.1.4: h2 = 763.7 kJ/kg
m
.
1 = m
.
2×
h2 - h3
h3 - h1
= 2 ×
763.7 – 419.0
419.0 – 85.8 = 2.069 kg/s
2
P
1
3
v

6.97
A mixing chamber with heat transfer receives 2 kg/s of R-410a at 1 MPa, 40°C in
one line and 1 kg/s of R-410a at 15°C, quality 50% in a line with a valve. The
outgoing flow is at 1 MPa, 60°C. Find the rate of heat transfer to the mixing
chamber.
Solution:
C.V. Mixing chamber. Steady with 2 flows in and 1 out, heat transfer in.
1
2
3
Heater
Mixer
Q
.
2
P
v
31
.
1 + m
.
2 = m
.
3 ; => m
.
3 = 2 + 1 = 3 kg/s
Energy Eq.6.10: m
.
1h1 + m
.
2h2 + Q
.
= m
.
3h3
Properties: Table B.4.2: h1 = 316.05 kJ/kg, h3 = 335.75 kJ/kg
Table B.4.1: h2 = 81.15 + 0.5 × 201.64 = 181.97 kJ/kg
Energy equation then gives the heat transfer as
Q
.
= 3 × 335.75 – 2 × 316.05 – 1 × 181.97 = 193.18 kW

6.98
An insulated mixing chamber receives 2 kg/s R-134a at 1 MPa, 100°C in a line
with low velocity. Another line with R-134a as saturated liquid 60°C flows
through a valve to the mixing chamber at 1 MPa after the valve, as shown in Fig.
P6.97. The exit flow is saturated vapor at 1 MPa flowing at 20 m/s. Find the flow
rate for the second line.
Solution:
C.V. Mixing chamber. Steady state, 2 inlets and 1 exit flow.
Insulated q = 0, No shaft or boundary motion w = 0.
.
1 + m
.
2 = m
.
3 ;
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3( h3 +
1
2 V
2
3 )
m
.
2 (h2 – h3 –
1
2 V
2
3 ) = m
.
1 ( h3 +
1
2 V
2
3 – h1 )
1: Table B.5.2: 1 MPa, 100°C, h1 = 483.36 kJ/kg
2: Table B.5.1: x = ∅, 60°C, h2 = 287.79 kJ/kg
3: Table B.5.1: x = 1, 1 MPa, 20 m/s, h3 = 419.54 kJ/kg
Now solve the energy equation for m
.
2
m
.
2 = 2 × [419.54 +
1
2 202 ×
1
1000 – 483.36] / [287.79 – 419.54 –
1
2
202
1000]
= 2 × [–63.82 + 0.2] / [–131.75 – 0.2] = 0.964 kg/s
Notice how kinetic energy was insignificant.
1
2
3MIXING
CHAMBER
cb
2
P
v
3 1

6.99
To keep a jet engine cool some intake air bypasses the combustion chamber.
Assume 2 kg/s hot air at 2000 K, 500 kPa is mixed with 1.5 kg/s air 500 K, 500
kPa without any external heat transfer. Find the exit temperature by using
constant heat capacity from Table A.5.
Solution:
C.V. Mixing Section
.
1 + m
.
2 = m
.
3 => m
.
3 = 2 + 1.5 = 3.5 kg/s
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3h3
h3 = (m
.
1h1 + m
.
2h2) / m
.
3 ;
For a constant specific heat divide the equation for h3 with Cp to get
T3 =
m
.
1
m
.
3
T1 +
m
.
2
m
.
3
T2 =
2
3.5 2000 +
1.5
3.5 500 = 1357 K
1
2
3
Mixing section

6.100
Solve the previous problem using values from Table A.,7
To keep a jet engine cool some intake air bypasses the combustion chamber.
Assume 2 kg/s hot air at 2000 K, 500 kPa is mixed with 1.5 kg/s air 500 K, 500
kPa without any external heat transfer. Find the exit temperature by using values
from Table A.7.
Solution:
C.V. Mixing Section
.
1 + m
.
2 = m
.
3 => m
.
3 = 2 + 1.5 = 3.5 kg/s
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3h3
h3 = (m
.
1h1 + m
.
2h2) / m
.
3 ;
Using A.7 we look up the h at states 1 and 2 to calculate h3
h3 =
m
.
1
m
.
3
h1 +
m
.
2
m
.
3
h2 =
2
3.5 2251.58 +
1.5
3.5 503.36 = 1502 kJ/kg
Now we can backinterpolate to find at what temperature do we have that h
T3 = 1350 + 50
1502 – 1455.43
1515.27 – 1455.43 = 1389 K
This procedure is the most accurate.
1
2
3
Mixing section

6.101
Two flows are mixed to form a single flow. Flow at state 1 is 1.5 kg/s water at
400 kPa, 200oC and flow at state 2 is 500 kPa, 100oC. Which mass flow rate at
state 2 will produce an exit T3 = 150oC if the exit pressure is kept at 300 kPa?
Solution:
C.V. Mixing chamber and valves. Steady state no heat transfer or work terms.
.
1 + m
.
2 = m
.
3
Energy Eq.6.10: m
.
1h1 + m
.
2h2 = m
.
3h3 = (m
.
1+ m
.
2)h3
1
2
3MIXING
CHAMBER
2
P
v
3
1
Properties Table B.1.3: h1 = 2860.51 kJ/kg; h3 = 2760.95 kJ/kg
Table B.1.4: h2 = 419.32 kJ/kg
m
.
2 = m
.
1×
h1 - h3
h3 - h2
= 1.5 ×
2860.51 – 2760.95
2760.95 – 419.32 = 0.0638 kg/s

Multiple Devices, Cycle Processes

6.102
A flow of 5 kg/s water at 100 kPa, 20oC should be delivered as steam at 1000
kPa, 350oC to some application. Consider compressing it to 1000 kPa, 20oC and
then heat it at constant 1000 kPa to 350oC. Which devices are needed and find the
specific energy transfers in those devices.
To raise the pressure of a liquid flow requires a pump which delivers the
difference between the flow work going out and in. Constant pressure heating is a
simple heater (or heat exchanger).
Inlet state: B.1.1: h1 = 83.94 kJ/kg, v1 = 0.001002 m3/kg
State 3: h3 = 3157.65 kJ/kg
The pump delivers the difference between the flow work out and flow work in.
Pump: –wp = (P2 – P1)v1 = (1000 – 100 ) kPa × 0.001002 m3/kg
= 0.9 kJ/kg.
Heater:q = h3 – h2 = h3 – (h1 – wp)
= 3157.65 – (83.94 + 0.9) = 3072.8 kJ/kg

6.103
The following data are for a simple steam power plant as shown in Fig. P6.103.
State 1 2 3 4 5 6 7
P MPa 6.2 6.1 5.9 5.7 5.5 0.01 0.009
T °C 45 175 500 490 40
h kJ/kg - 194 744 3426 3404 - 168
State 6 has x6 = 0.92, and velocity of 200 m/s. The rate of steam flow is 25 kg/s,
with 300 kW power input to the pump. Piping diameters are 200 mm from steam
generator to the turbine and 75 mm from the condenser to the steam generator.
Determine the velocity at state 5 and the power output of the turbine.
Solution:
Turbine A5 = (π/4)(0.2)2 = 0.031 42 m2, v5 = 0.06163 m3/kg
V5 = m
.
v5/A5 = 25 kg/s × 0.061 63 m3/kg / 0.031 42 m2 = 49 m/s
h6 = 191.83 + 0.92 × 2392.8 = 2393.2 kJ/kg
wT = h5 - h6 +
1
2 ( V
2
5 - V
2
6 )
= 3404 - 2393.2 + (492 - 2002 )/(2 × 1000) = 992 kJ/kg
W
.
T = m
.
wT = 25 × 992 = 24 800 kW
Remark: Notice the kinetic energy change is small relative to enthalpy change.

6.104
For the same steam power plant as shown in Fig. P6.103 and Problem 6.103,
assume the cooling water comes from a lake at 15°C and is returned at 25°C.
Determine the rate of heat transfer in the condenser and the mass flow rate of
cooling water from the lake.
Solution:
Condenser A7 = (π/4)(0.075)2 = 0.004 418 m2, v7 = 0.001 008 m3/kg
V7 = m
.
v7/A7 = 25 × 0.001 008 / 0.004 418 = 5.7 m/s
h6 = 191.83 + 0.92 × 2392.8 = 2393.2 kJ/kg
qCOND = h7 - h6 +
1
2 ( V
2
7 - V
2
6 )
= 168 − 2393.2 + (5.72 − 2002 )/(2×1000) = −2245.2 kJ/kg
Q
.
COND = 25 × (−2245.2) = −56 130 kW
This rate of heat transfer is carried away by the cooling water so
−Q
.
COND = m
.
H2O(hout − hin)H2O = 56 130 kW
=> m
.
H2O =
56 130
104.9 - 63.0 = 1339.6 kg/s

6.105
For the same steam power plant as shown in Fig. P6.103 and Problem 6.103,
determine the rate of heat transfer in the economizer, which is a low temperature
heat exchanger. Find also the rate of heat transfer needed in the steam generator.
Solution:
Economizer A7 = πD
2
7/4 = 0.004 418 m2, v7 = 0.001 008 m3/kg
V2 = V7 = m
.
v7/A7 = 25 × 0.001 008/0.004 418 = 5.7 m/s,
V3 = (v3/v2)V2 = (0.001 118 / 0.001 008) 5.7 = 6.3 m/s ≈ V2
so kinetic energy change unimportant
qECON = h3 - h2 = 744 - 194 = 550.0 kJ/kg
Q
.
ECON = m
.
qECON = 25 (550.0) = 13 750 kW
Generator A4 = πD
2
4/4 = 0.031 42 m2, v4 = 0.060 23 m3/kg
V4 = m
.
v4/A4 = 25 × 0.060 23/0.031 42 = 47.9 m/s
qGEN = 3426 - 744 + (47.92 - 6.32)/(2×1000) = 2683 kJ/kg
Q
.
GEN = m
.
qGEN = 25 × (2683) = 67 075 kW

6.106
A somewhat simplified flow diagram for a nuclear power plant shown in Fig. 1.4
is given in Fig. P6.106. Mass flow rates and the various states in the cycle are
shown in the accompanying table. The cycle includes a number of heaters in
which heat is transferred from steam, taken out of the turbine at some
intermediate pressure, to liquid water pumped from the condenser on its way to
the steam drum. The heat exchanger in the reactor supplies 157 MW, and it may
be assumed that there is no heat transfer in the turbines.
a. Assume the moisture separator has no heat transfer between the
two turbinesections, determine the enthalpy and quality (h4, x4).
b. Determine the power output of the low-pressure turbine.
c. Determine the power output of the high-pressure turbine.
d. Find the ratio of the total power output of the two turbines to the total power
delivered by the reactor.
Solution:
HP
W
2 3
17
12
moisture
separator
WLP
9
4
5
8
a) Moisture Separator, steady state, no heat transfer, no work
Mass: m
.
3 = m
.
4 + m
.
9, Energy: m
.
3h3 = m
.
4h4 + m
.
9h9 ;
62.874 × 2517 = 58.212 × h4 + 4.662 × 558 ⇒ h4 = 2673.9 kJ/kg
h4 = 2673.9 = 566.18 + x4 × 2160.6 => x4 = 0.9755
b) Low Pressure Turbine, steady state no heat transfer
Energy Eq.: m
.
4h4 = m
.
5h5 + m
.
8h8+ W
.
CV(LP)
W
.
CV(LP) = m
.
4h4 - m
.
5h5 - m
.
8h8
= 58.212 × 2673.9 - 55.44 × 2279 - 2.772 × 2459
= 22 489 kW = 22.489 MW
c) High Pressure Turbine, steady state no heat transfer
Energy Eq.: m
.
2h2 = m
.
3h3 + m
.
12h12 + m
.
17h17 + W
.
CV(HP)
W
.
CV(HP) = m
.
2h2 - m
.
3h3 - m
.
12h12 - m
.
17h17
= 75.6 × 2765 - 62.874 × 2517 - 8.064 × 2517 - 4.662 × 2593
= 18 394 kW = 18.394 MW
d) η = (W
.
HP + W
.
LP)/Q
.
REACT = 40.883/157 = 0.26

6.107
Consider the power plant as described in the previous problem.
a. Determine the quality of the steam leaving the reactor.
b. What is the power to the pump that feeds water to the reactor?
Solution:
a) Reactor: Cont.: m
.
20 = m
.
21; Q
.
CV = 157 MW
Energy Eq.6.12: Q
.
CV + m
.
20h20 = m
.
21h21
157 000 + 1386 × 1221 = 1386 × h21
h21 = 1334.3 = 1282.4 + x21 × 1458.3
=> x21 = 0.0349
21
19
20
Q
b) C.V. Reactor feedwater pump
Cont. m
.
19 = m
.
20 Energy Eq.6.12: m
.
19h19 = m
.
19h20 + W
.
Cv,P
Table B.1: h19 = h(277°C, 7240 kPa) = 1220 kJ/kg, h20 = 1221 kJ/kg
W
.
Cv,P = m
.
19(h19 - h20) = 1386(1220 - 1221) = -1386 kW

6.108
A R-410a heat pump cycle shown in Fig. P6.108 has a R-410a flow rate of 0.05
kg/s with 5 kW into the compressor. The following data are given
State 1 2 3 4 5 6
P, kPa 3100 3050 3000 420 400 390
T, °C 120 110 45 -10 -5
h, kJ/kg 377 367 134 - 280 284
Calculate the heat transfer from the compressor, the heat transfer from the R-410a
in the condenser and the heat transfer to the R-410a in the evaporator.
Solution:
CV: Compressor
Q
.
COMP = m
.
(h1 - h6) + W
.
COMP = 0.05 (377 - 284) - 5.0 = -0.35 kW
CV: Condenser
Q
.
COND = m
.
(h3 - h2) = 0.05 kg/s (134 - 367) kJ/kg = -11.65 kW
C.V. Valve:
h4 = h3 = 134 kJ/kg
CV: Evaporator
Q
.
EVAP = m
.
(h5- h4) = 0.05 kg/s (280 - 134) kJ/kg = 7.3 kW
v
P
compressor
evaporator
condenser
valve
12
3
4 5
6
Q
H
-WC
QL
.
cb
Evaporator
Condenser
3
4
56
1
2Qloss

6.109
A modern jet engine has a temperature after combustion of about 1500 K at 3200
kPa as it enters the turbine setion, see state 3 Fig. P.6.109. The compressor inlet is
80 kPa, 260 K state 1 and outlet state 2 is 3300 kPa, 780 K; the turbine outlet state
4 into the nozzle is 400 kPa, 900 K and nozzle exit state 5 at 80 kPa, 640 K.
Neglect any heat transfer and neglect kinetic energy except out of the nozzle. Find
the compressor and turbine specific work terms and the nozzle exit velocity.
Solution:
The compressor, turbine and nozzle are all steady state single flow devices
and they are adiabatic.
We will use air properties from table A.7.1:
h1 = 260.32, h2 = 800.28, h3 = 1635.80, h4 = 933.15, h5 = 649.53 kJ/kg
Energy equation for the compressor gives
wc in = h2 – h1 = 800.28 – 260.32 = 539.36 kJ/kg
Energy equation for the turbine gives
wT = h3 – h4 = 1635.80 – 933.15 = 702.65 kJ/kg
Energy equation for the nozzle gives
h4 = h5 + ½ V
2
5
½ V
2
5 = h4 - h5 = 933.15 – 649.53 = 283.62 kJ/kg
V5 = [2( h4 – h5) ] 1/2 = ( 2× 283.62 ×1000 ) 1/2 = 753 m/s
cb

6.110
A proposal is made to use a geothermal supply of hot water to operate a steam
turbine, as shown in Fig. P6.110. The high-pressure water at 1.5 MPa, 180°C, is
throttled into a flash evaporator chamber, which forms liquid and vapor at a lower
pressure of 400 kPa. The liquid is discarded while the saturated vapor feeds the
turbine and exits at 10 kPa, 90% quality. If the turbine should produce 1 MW,
find the required mass flow rate of hot geothermal water in kilograms per hour.
Solution:
Separation of phases in flash-evaporator
constant h in the valve flow so
Table B.1.3: h1 = 763.5 kJ/kg
h1 = 763.5 = 604.74 + x × 2133.8
⇒ x = 0.07439 = m
.
2/m
.
1
Table B.1.2: h2 = 2738.6 kJ/kg;
FLASH
EVAP.
H O
2
Sat. liquid
out
Sat. vapor
W
Turb
1
2
3
4
.
h3 = 191.83 + 0.9 × 2392.8 = 2345.4 kJ/kg
Energy Eq.6.12 for the turbine
W
.
= m
.
2(h2 - h3) => m
.
2 =
1000
2738.6 - 2345.4 = 2.543 kg/s
⇒ m
.
1 = m
.
2/x = 34.19 kg/s = 123 075 kg/h

Transient processes

6.111
An initially empty cylinder is filled with air from 20oC, 100 kPa until it is full.
Assuming no heat transfer is the final temperature larger, equal to or smaller than
20oC? Does the final T depend on the size of the cylinder?
This is a transient problem with no heat transfer and no work. The balance
equations for the tank as C.V. become
Continuity Eq.: m2 – 0 = mi
Energy Eq.: m2u2 – 0 = mihi + Q – W = mihi + 0 – 0
Final state: u2 = hi & P2 = Pi
T2 > Ti and it does not depend on V

6.112
An evacuated 150-L tank is connected to a line flowing air at room temperature,
25°C, and 8 MPa pressure. The valve is opened allowing air to flow into the tank
until the pressure inside is 6 MPa. At this point the valve is closed. This filling
process occurs rapidly and is essentially adiabatic. The tank is then placed in
storage where it eventually returns to room temperature. What is the final pressure?
Solution:
C.V. Tank:
Continuity Eq.6.15: mi = m2
Energy Eq.6.16: mihi = m2u2 => u2 = hi
Use constant specific heat CPo from table A.5 then energy equation:
T2 = (CP/Cv) Ti = kTi= 1.4 × 298.2 = 417.5 K
Process: constant volume cooling to T3:
P3 = P2 × T3/T2 = 6.0 × 298.15/ 417.5 = 4.29 MPa
line

6.113
A 2.5-L tank initially is empty and we want 10 g of ammonia in it. The ammonia
comes from a line with saturated vapor at 25°C. To end up with the desired amount
we cool the can while we fill it in a slow process keeping the can and content at
30°C. Find the final pressure to reach before closing the valve and the heat
transfer?
Solution:
C.V. Tank:
Continuity Eq.6.15: mi = m2
Energy Eq.6.16: m2u2 – 0 = mihi + 1Q2
State 2: 30°C, v2 = V/m2 = 0.0025/0.010 = 0.25 m3/kg
From Table B.2.2 we locate the state between 500 and 600 kPa.
P2 = 500 + (600 – 500)
0.25 – 0.28103
0.23152 – 0.28103 = 562.7 kPa
u2 = 1370 kJ/kg,
State i Table B.2.2: hi = 1463.5 kJ/kg
Now use the energy ewquation to solve for the heat transfer
1Q2 = m2u2 − mihi = m2(u2 – hi)
= 0.01 × (1370 − 1463.5 ) = −0.935 kJ
line

6.114
A tank contains 1 m3 air at 100 kPa, 300 K. A pipe flowing air at 1000 kPa, 300
K is connected to the tank and it is filled slowly to 1000 kPa. Find the heat
transfer to reach a final temperature of 300 K.
C.V. The tank volume and the compressor.
This is a transient problem (filling of tank).
Continuity Eq.6.15: m2 – m1 = min
Q
i
TANK
1 ⇒ 2
Energy Eq.6.16: m2u2 – m1u1 = 1Q2 – 1W2 + minhin
Process: Constant volume 1W2 = 0 ,
States: u1 = u2 = uin = u300 ; hin = uin + RTin
m1 = P1V1/RT1 = 100 × 1/(0.287 × 300) = 1.1614 kg
m2 = P2V2/RT2 = 1000 × 1/(0.287 × 300) = 11.6144 kg
Heat transfer from the energy equation
1Q2 = m2u2 – m1u1 – minhin = (m1 + min) u1 – m1u1 – minuin – minRTin
= m1u1 – m1u1 + minu1 – minuin – minRTin = – minRTin
= (11.6144 – 1.1614) kg × 0.287 kJ/kg-K × 300 K = –900 kJ

6.115
An initially empty canister of volume 0.2 m3 is filled with carbon dioxide from a
line at 800 kPa, 400 K. Assume the process runs until it stops by itself and it is
adiabatic. Use constant heat capacity to find the final temperature in the canister.
C.V. Canister and valve, transient process with no heat transfer or work.
Continuity Eq.6.15: m2 - m1 = min ;
Energy Eq.6.16: m2u2 – m1u1 = - min hin + 1Q2 - 1W2
Process: 1Q2 = 0, V= constant so 1W2 = 0
State 1: m1 = 0 => m2 = min
State 2: P2 = Pline = 800 kPa, one more property
Energy Eq.: u2 = hin = uin + RTin ⇒ Cvo T2 = Cvo Tin + RTin = CPoTin
T2 = (CPo/Cvo) Tin = k × Tin = 1.289 × 400 K = 515.6 K

6.116
Repeat the previous problem but use the ideal gas Tables A.8 to solve it.
C.V. Canister and valve, transient process with no heat transfer or work.
Energy Eq.6.16: m2u2 – m1u1 = - min hin + 1Q2 - 1W2
Process: 1Q2 = 0, V= constant so 1W2 = 0
State 1: m1 = 0 => m2 = min
State 2: P2 = Pline = 800 kPa, one more property
Energy Eq.: u2 = hin = 303.76 kJ/kg from A.8
back interpolate for u2: T2 = 495.9 K

6.117
An initially empty bottle is filled with water from a line at 0.8 MPa, 350oC.
Assume no heat transfer and that the bottle is closed when the pressure reaches
the line pressure. If the final mass is 0.75 kg find the final temperature and the
volume of the bottle.
Solution;
C.V. Bottle, transient process with no heat transfer or work.
Energy Eq.6.16: m2u2 – m1u1 = - min hin
State 1: m1 = 0 => m2 = min and u2 = hin
Line state: Table B.1.3: hin = 3161.68 kJ/kg
State 2: P2 = Pline = 800 kPa, u2 = 3161.68 kJ/kg from Table B.1.3
T2 = 520oC and v2 = 0.4554 m3/kg
V2 = m2v2 = 0.75 × 0.4554 = 0.342 m3
line

6.118
A 1-m3 tank contains ammonia at 150 kPa, 25°C. The tank is attached to a line
flowing ammonia at 1200 kPa, 60°C. The valve is opened, and mass flows in until
the tank is half full of liquid, by volume at 25°C. Calculate the heat transferred
from the tank during this process.
Solution:
C.V. Tank. Transient process as flow comes in.
State 1 Table B.2.2 interpolate between 20 °C and 30°C:
v1 = 0.9552 m3/kg; u1 = 1380.6 kJ/kg
m1 = V/v1 = 1/0.9552 = 1.047 kg
State 2: 0.5 m3 liquid and 0.5 m3 vapor from Table B.2.1 at 25°C
vf = 0.001658 m3/kg; vg = 0.12813 m3/kg
mLIQ2 = 0.5/0.001658 = 301.57 kg, mVAP2 = 0.5/0.12813 = 3.902 kg
m2 = 305.47 kg, x2 = mVAP2/m2 = 0.01277,
From continuity equation
mi = m2 - m1 = 304.42 kg
Table B.2.1: u2 = 296.6 + 0.01277 × 1038.4 = 309.9 kJ/kg
State inlet: Table B.2.2 hi = 1553.3 kJ/kg
Energy Eq.6.16:
QCV + mihi = m2u2 - m1u1
QCV = 305.47 × 309.9 - 1.047 × 1380.6 - 304.42 × 1553.3 = -379 636 kJ
line
Q

6.119
A 25-L tank, shown in Fig. P6.119, that is initially evacuated is connected by a
valve to an air supply line flowing air at 20°C, 800 kPa. The valve is opened, and
air flows into the tank until the pressure reaches 600 kPa.Determine the final
temperature and mass inside the tank, assuming the process is adiabatic. Develop
an expression for the relation between the line temperature and the final
temperature using constant specific heats.
Solution:
C.V. Tank:
Continuity Eq.6.15: m2 = mi
Energy Eq.6.16: m2u2 = mihi
Table A.7: u2 = hi = 293.64 kJ/kg
⇒ T2 = 410.0 K
TANK
m2 =
P2V
RT2
=
600 × 0.025
0.287 × 410 = 0.1275 kg
Assuming constant specific heat,
hi = ui + RTi = u2 , RTi = u2 - ui = Cvo(T2 - Ti)
CvoT2 = ( Cvo + R )Ti = CPoTi , T2 =





CPo
Cvo
Ti = kTi
For Ti = 293.2K & constant CPo, T2 = 1.40×293.2 = 410.5 K

6.120
A 200 liter tank initially contains water at 100 kPa and a quality of 1%. Heat is
transferred to the water thereby raising its pressure and temperature. At a pressure
of 2 MPa a safety valve opens and saturated vapor at 2 MPa flows out. The
process continues, maintaining 2 MPa inside until the quality in the tank is 90%,
then stops. Determine the total mass of water that flowed out and the total heat
transfer.
Solution:
C.V. Tank, no work but heat transfer in and flow
out. Denoting State 1: initial state, State 2: valve
opens, State 3: final state.
Continuity Eq.: m3 − m1 = − me
Energy Eq.: m3u3 − m1u1 = − mehe + 1Q3
e
Q
sat vap
cv
.
State 1 Table B.1.2: v1 = vf + x1vfg = 0.001043 + 0.01×1.69296
= 0.01797 m3/kg
u1 = uf + x1ufg = 417.33 + 0.01×2088.72 = 438.22 kJ/kg
m1 = V/v1 = 0.2 m3/(0.01797 m3/kg) = 11.13 kg
State 3 (2MPa): v3 = vf + x3vfg = 0.001177 + 0.9×0.09845 = 0.8978 m3/kg
u3 = uf + x3ufg = 906.42 + 0.9×1693.84 = 2430.88 kJ/kg
m3 = V/v3 = 0.2 m3/(0.08978 m3/kg) = 2.23 kg
Exit state (2MPa): he = hg = 2799.51 kJ/kg
Hence: me = m1 − m3 = 11.13 kg − 2.23 kg = 8.90 kg
Applying the 1st law between state 1 and state 3
1Q3 = m3u3 − m1u1 + mehe
= 2.23 × 2430.88 − 11.13 × 438.22 + 8.90 × 2799.51
= 25 459 kJ = 25.46 MJ

6.121
Helium in a steel tank is at 250 kPa, 300 K with a volume of 0.1 m3. It is used to
fill a balloon. When the tank pressure drops to 150 kPa the flow of helium stops
by itself. If all the helium still is at 300 K how big a balloon did I get? Assume
the pressure in the balloon varies linearly with volume from 100 kPa (V = 0) to
the final 150 kPa. How much heat transfer did take place?
Solution:
Take a C.V. of all the helium.
This is a control mass, the tank mass
changes density and pressure.
Energy Eq.: U2 – U1 = 1Q2 - 1W2
Process Eq.: P = 100 + CV
State 1: P1, T1, V1
State 2: P2, T2, V2 = ?
Ideal gas:
P2 V2 = mRT2 = mRT1 = P1V1
c
i
r
c
u
s
t
h
e
r
m
o
cb
V2 = V1(P1/P2) = 0.1× (250/150) = 0.16667 m3
Vbal = V2 – V1 = 0.16667 – 0.1 = 0.06667 m3
1W2 = ∫ P dV = AREA = ½ ( P1 + P2 )( V2 –V1 )
= ½( 250 + 150) × 0.06667 = 13.334 kJ
U2 – U1 = 1Q2 - 1W2 = m (u2 – u1) = mCv ( T2 –T1 ) = 0
so 1Q2 = 1W2 = 13.334 kJ
Remark: The process is transient, but you only see the flow mass if you
select the tank or the balloon as a control volume. That analysis leads to
more terms that must be elliminated between the tank control volume and
the balloon control volume.

6.122
An empty cannister of volume 1 L is filled with R-134a from a line flowing
saturated liquid R-134a at 0oC. The filling is done quickly so it is adiabatic. How
much mass of R-134a is there after filling? The cannister is placed on a storage
shelf where it slowly heats up to room temperature 20oC. What is the final
pressure?
C.V. cannister, no work and no heat transfer.
Continuity Eq.6.15: m2 = mi
Energy Eq.6.16: m2u2 – 0 = mihi = mihline
Table B.5.1: hline = 200.0 kJ/kg, Pline = 294 kPa
u2 = hline = 200 kJ/kg > uf = 199.77 kJ/kg
State 2 is two-phase P2 = Pline = 294 kPa and T2 = 0°C
x2 =
u2 - uf
ufg
=
200 – 199.77
178.24 = 0.00129
v2 = 0.000773 + x2 0.06842 = 0.000 861 m3/kg
m2 = V/v2 = 0.01/0.000861 = 11.61 kg
At 20°C: vf = 0.000 817 m3/kg < v2 so still two-phase

6.123
A nitrogen line, 300 K and 0.5 MPa, shown in Fig. P6.123, is connected to a
turbine that exhausts to a closed initially empty tank of 50 m3. The turbine
operates to a tank pressure of 0.5 MPa, at which point the temperature is 250 K.
Assuming the entire process is adiabatic, determine the turbine work.
Solution:
C.V. turbine & tank ⇒ Transient process
Conservation of mass Eq.6.15: mi = m2 ⇒ m
Energy Eq.6.16: mihi = m2u2 + WCV ; WCV = m(hi - u2)
Table B.6.2: Pi = 0.5 MPa, Ti = 300 K, Nitrogen; hi = 310.28 kJ/kg
2: P2 = 0.5 MPa, T2 = 250 K, u2 = 183.89 kJ/kg, v2 = 0.154 m3/kg
m2 = V/v2 = 50/0.154 = 324.7 kg
WCV = 324.7 (310.28 - 183.89) = 41 039 kJ = 41.04 MJ
W
Turb
1
2
TANK
.
We could with good accuracy have solved using ideal gas and Table A.5

6.124
A 750-L rigid tank, shown in Fig. P6.124, initially contains water at 250°C, 50%
liquid and 50% vapor, by volume. A valve at the bottom of the tank is opened,
and liquid is slowly withdrawn. Heat transfer takes place such that the
temperature remains constant. Find the amount of heat transfer required to the
state where half the initial mass is withdrawn.
Solution:
C.V. vessel
Continuity Eq.6.15: m2 − m1 = − me
Energy Eq.6.16: m2u2 − m1u1 = QCV − mehe
State 1: mLIQ1 =
0.375
0.001251 = 299.76 kg; mVAP1 =
0.375
0.05013 = 7.48 kg
m1u1 = 299.76 × 1080.37 + 7.48 × 2602.4 = 343 318 kJ
m1 = 307.24 kg; me = m2 = 153.62 kg
State 2: v2 =
0.75
153.62 = 0.004882 = 0.001251 + x2 × 0.04888
x2 = 0.07428 ; u2 = 1080.37 + 0.07428 × 1522 = 1193.45 kJ/kg
Exit state: he = hf = 1085.34 kJ/kg
Energy equation now gives the heat transfer as
QCV = m2u2 - m1u1 + mehe
= 153.62 × 1193.45 – 343 318 + 153.62 × 1085.34 = 6750 kJ
Vapor
Sat. liq.
out
Liquid

6.125
Consider the previous problem but let the line and valve be located in the top of
the tank. Now saturated vapor is slowly withdrawn while heat transfer keeps the
temperature inside constant. Find the heat transfer required to reach a state where
half the original mass is withdrawn.
Solution:
C.V. vessel
Continuity Eq.6.15: m2 − m1 = − me
Energy Eq.6.16: m2u2 − m1u1 = QCV − mehe
State 1: mLIQ1 =
0.375
0.001251 = 299.76 kg; mVAP1 =
0.375
0.05013 = 7.48 kg
m1u1 = 299.76 × 1080.37 + 7.48 × 2602.4 = 343 318 kJ
m1 = 307.24 kg; me = m2 = 153.62 kg
State 2: v2 =
0.75
153.62 = 0.004882 = 0.001251 + x2 × 0.04888
x2 = 0.07428 ; u2 = 1080.37 + 0.07428 × 1522 = 1193.45 kJ/kg
Exit state: he = hg = 2801.52 kJ/kg
Energy equation now gives the heat transfer as
= 153.62 × 1193.45 – 343 318 + 153.62 × 2801.52 = 270 389 kJ
Sat. vapor
out
Liquid
Vapor

6.126
A 2-m3 insulated vessel, shown in Fig. P6.126, contains saturated vapor steam at
4 MPa. A valve on the top of the tank is opened, and steam is allowed to escape.
During the process any liquid formed collects at the bottom of the vessel, so that
only saturated vapor exits. Calculate the total mass that has escaped when the
pressure inside reaches 1 MPa.
Solution:
C.V. Vessel: Mass flows out.
Continuity Eq.6.15: me = m1 - m2
Energy Eq.6.16: m2u2 - m1u1 = - (m1-m2)he or m2(he-u2) = m1(he-u1)
Average exit enthalpy he ≈ (hG1+hG2)/2 = (2801.4+2778.1)/2 = 2789.8
State 1: m1 = V/v1 = 40.177 kg, m2 = V/v2
Energy equation ⇒
2
v2
(2789.8-u2) = 40.177(2789.8-2602.3) = 7533.19
But v2 = .001 127 + .193 313 x2 and u2 = 761.7 + 1822 x2
Substituting and solving, x2 = 0.7936
⇒ m2 = V/v2 = 12.94 kg, me = 27.24 kg
Sat. vapor
out
Liquid
Vapor
cb

6.127
A 2 m tall cylinder has a small hole in the bottom. It is filled with liquid water 1
m high, on top of which is 1 m high air column at atmospheric pressure of 100
kPa. As the liquid water near the hole has a higher P than 100 kPa it runs out.
Assume a slow process with constant T. Will the flow ever stop? When?
Solution:
Pbot = Pair + ρgLf
For the air PV = mRT and the total height is H = 2 m
Pair = mRT/Vair ; Vair = A Lair = A ( H - Lf )
Pbot =
maRaTa
A(H-Lf) + ρfg Lf =
Pa1Va1
A(H-Lf) + ρf gLf =
Pa1La1
H-Lf
+ ρf gLf ≥ Po
Solve for Lf ; ρf = 1/(vf) = 1/0.0021002 = 998 kg/m3
Pa1 La1 + ρf g Lf ( H – Lf ) ≥ Po ( H – Lf )
(ρfgH + Po ) Lf – ρfgL2
f = Po H - Pa1 La1 ≥ 0
Put in numbers and solve quadratic eq.
L2
f – [ H +(Po/ρg) ] Lf +
PoH - Pa1La1
ρg
= 0
(Po/ρg) =
100 kPa m3 s3
998 ×9.807 kg m
= 10.217 m
PoH - Pa1La1
ρg
=
100 (2-1)
998×9.807
= 10.217 m
L2
f – 12.217 Lf + 10.217 = 0
Air
Water
1 m
1 m
Lf =
12.217
2 ±
12.2172
4 -
12.217
4 = 6.1085 ± 5.2055
=> 11.314 or 0.903 m so the second root is the solution
Verify
Pa2 = Pa1.
La1
H-Lf
= 100
1
2 - 0.903 = 91.158 kPa
ρgLf = 998 × 9.807 × 0.903 = 8838 Pa = 8.838 kPa
Pbot = Pa2 + ρgLf = 91.158 + 8.838 = 99.996 kPa OK

6.128
A pipe of radius R has a fully developed laminar flow of air at Po, To with a
velocity profile as: V = Vc [1 – (r/R) 2], where Vc is the velocity on the center-
line and r is the radius as shown in Fig. P6.128. Find the total mass flow rate and
the average velocity both as functions of Vc and R.
m
.
= AV/v = V
.
/v
Since the velocity is distributed we need to integrate over the area. From Eq.6.2
V
.
= ⌡⌠ Vlocal dA = ⌡⌠ V(r) 2πr dr
where W is the depth. Substituting the velocity we get
V
.
= ⌡⌠ Vc 2πr [1 - (r/R)2] dr
= 2π Vc R2
⌡⌠
0
1
z (1 - z2) dz
= 2π Vc R2 (
1
2 z2 -
1
4 z4 |1
0 =
π
2 Vc R2 =
1
2 Vc A
V = V
.
/A =
1
2 Vc
m
.
= V
.
/v =
π
2 Vc R2/v

6.129
Steam at 3 MPa, 400°C enters a turbine with a volume flow rate of 5 m3/s. An
extraction of 15% of the inlet mass flow rate exits at 600 kPa, 200°C. The rest
exits the turbine at 20 kPa with a quality of 90%, and a velocity of 20 m/s.
Determine the volume flow rate of the extraction flow and the diameter of the
final exit pipe.
Solution:
Inlet flow : m
.
i = V
.
/v = 5/0.09936 = 50.32 kg/ s (Table B.1.3)
Extraction flow : m
.
e = 0.15 m
.
i = 7.55 kg/ s; v = 0.35202 m3/kg
V
.
ex = m
.
ev = 7.55 × 0.35202 = 2.658 m3/ s
Exit flow : m
.
= 0.85 m
.
i = 42.77 kg /s
Table B.1.2 v = 0.001017 + 0.9 × 7.64835 = 6.8845 m3/kg
m
.
= AV/v ⇒ A = (π/4) D2 = m
.
v/V = 42.77 × 6.8845/20 = 14.723 m2
⇒ D = 4.33 m
WT
1
2
3
Exit flow
Extraction flowInlet
flow
HP
LP sectionsection

6.130
In a glass factory a 2 m wide sheet of glass at 1500 K comes out of the final
rollers that fix the thickness at 5 mm with a speed of 0.5 m/s. Cooling air in the
amount of 20 kg/s comes in at 17oC from a slot 2 m wide and flows parallel with
the glass. Suppose this setup is very long so the glass and air comes to nearly the
same temperature (a co-flowing heat exchanger) what is the exit temperature?
Solution:
Energy Eq.: m
.
glasshglass 1 + m
.
airhair 2 = m
.
glasshglass 3 + m
.
airhair 4
m
.
glass = ρV
.
= ρAV = 2500× 2 ×0.005× 0.5 = 12.5 kg/s
m
.
glassCglass ( T3 – T1 ) + m
.
air CPa ( T4 – T2 ) = ∅
T4 = T3 , Cglass = 0.80 kJ/kg K, CPa = 1.004 kJ/kg K
T3 =
m
.
glassCglass T1 + m
.
airCPa T2
m
.
glassCglass + m
.
airCPa
=
12.5×0.80×1500 + 20×1.004×290
12.5×0.80 + 20×1.004
= 692.3 K
We could use table A.7.1 for air, but then it will be trial and error
1
2
3
4
Air

6.131
Assume a setup similar to the previous problem but the air flows in the opposite
direction of the glass, it comes in where the glass goes out. How much air flow at
17oC is required to cool the glass to 450 K assuming the air must be at least 120
K cooler than the glass at any location?
Solution:
Energy Eq.: m
.
1h1 + m
.
4h4 = m
.
3h3 + m
.
2h2
T4 = 290 K and T3 = 450 K
m
.
glass = ρV
.
= ρAV = 2500× 2 ×0.005× 0.5 = 12.5 kg/s
T2 ≤ T1 – 120 K = 1380 K
m
.
= m
.
4 = m
.
2 = m
.
1
h1-h3
h2-h4
Let us check the limit and since T is high use table A.7.1 for air.
h4 = 290.43 kJ/kg, h2 = 1491.33 kJ/kg
m
.
= m
.
4 = m
.
2 = m
.
1
h1-h3
h2-h4
= m
.
1
Cglass(T1-T3)
h2-h4
m
.
= 12.5
0.8 (1500-450)
1491.33 – 290.43 = 8.743 kg/s
1
2
3
4
Air

6.132
Two kg of water at 500 kPa, 20oC is heated in a constant pressure process to
1700oC. Find the best estimate for the heat transfer.
Solution:
C.V. Heater; steady state 1 inlet and exit, no work term, no ∆KE, ∆PE .
Continuity Eq.: m
.
in = m
.
ex = m
.
,
Energy Eq.6.13: q + hin = hex ⇒ q = hex - hin
steam tables only go up to 1300oC so use an intermediate state at lowest
pressure (closest to ideal gas) hx(1300oC, 10 kPa) from Table B.1.3 and table
A.8 for the high T change ∆h from 1300oC to 1700oC.
hex - hin = (hex - hx) + (hx - hin)
= (4515.55 – 3416.77) + (5409.7 - 83.96) = 6424.5 kJ/kg
Q = m(hex - hin) = 2 × 6424.5 = 12 849 kJ

6.133
A 500-L insulated tank contains air at 40°C, 2 MPa. A valve on the tank is
opened, and air escapes until half the original mass is gone, at which point the
valve is closed. What is the pressure inside then?
Solution:
State 1: ideal gas m1 = P1V/RT1 =
2000 × 0.5
0.287 × 313.2 = 11.125 kg
Continuity eq.6.15: me = m1 - m2, m2 = m1/2 ⇒ me = m2 = 5.5625 kg
Energy Eq.6.16: m2u2 - m1u1 = - mehe AV
Substitute constant specific heat from table A.5 and evaluate the exit enthalpy
as the average between the beginning and the end values
5.5625×0.717 T2 - 11.125×0.717×313.2 = - 5.5625×1.004 (313.2 + T2)/2
Solving, T2 = 239.4 K
P2 =
m2RT2
V =
5.5625 × 0.287 × 239.4
0.5 = 764 kPa
cb

6.134
Three air flows all at 200 kPa are connected to the same exit duct and mix without
external heat transfer. Flow one has 1 kg/s at 400 K, flow two has 3 kg/s at 290 K
and flow three has 2 kg/s at 700 K. Neglect kinetic energies and find the volume
flow rate in the exit flow.
Solution:
Continuity Eq. m
.
1+ m
.
2 + m
.
3 = m
.
4h4
Energy Eq.: m
.
1h1 + m
.
2h2 = m
.
3h3 + m
.
4h4
V
.
4 = m
.
v4
h4 =
m
.
1
m
.
4
h1 +
m
.
2
m
.
4
h2+
m
.
3
m
.
4
h3 =
1
6 × 401.3 +
3
6 × 290.43 +
2
6 × 713.56
= 449.95 kJ/kg
T4 = 440 + 20
449.95 - 441.93
462.34 - 441.93 = 447.86 K
v4 = RT4 /P4 = 0.287×447.86/200 = 0.643 m3/kg
V
.
4 = m
.
4v4 = 6 × 0.643 = 3.858 m3/s
2
4
1
3

6.135
Consider the power plant as described in Problem 6.106.
a. Determine the temperature of the water leaving the intermediate pressure
heater, T
13
, assuming no heat transfer to the surroundings.
b. Determine the pump work, between states 13 and 16.
Solution:
a) Intermediate Pressure Heater
Energy Eq.6.10: m
.
11h11 + m
.
12h12 + m
.
15h15 = m
.
13h13 + m
.
14h14
75.6×284.6 + 8.064×2517 + 4.662×584 = 75.6×h13 + 12.726×349
h13 = 530.35 → T13 = 126.3°C
b) The high pressure pump
Energy Eq.6.12: m
.
13h13 = m
.
16h16 + W
.
Cv,P
W
.
Cv,P = m
.
13(h13 - h16) = 75.6(530.35 - 565) = -2620 kW

6.136
Consider the powerplant as described in Problem 6.106.
a. Find the power removed in the condenser by the cooling water (not shown).
b. Find the power to the condensate pump.
c. Do the energy terms balance for the low pressure heater or is there a
heat transfer not shown?
Solution:
a) Condenser:
Energy Eq.6.10: Q
.
CV + m
.
5h5 + m
.
10h10 = m
.
6h6
Q
.
CV + 55.44 × 2279 + 20.16 × 142.51 = 75.6 × 138.3
Q
.
CV = -118 765 kW = -118.77 MW
b) The condensate pump
W
.
Cv,P = m
.
6(h6 - h7) = 75.6(138.31 - 140) = -127.8 kW
c) Low pressure heater Assume no heat transfer
m
.
14h14 + m
.
8h8 + m
.
7h7 + m
.
9h9 = m
.
10h10 + m
.
11h11
LHS = 12.726×349 + 2.772×2459 + 75.6×140 + 4.662×558 = 24 443 kW
RHS = (12.726 + 2.772 + 4.662) × 142.51 + 75.6 × 284.87 = 24 409 kW
A slight imbalance, but OK.

6.137
A 1-m3, 40-kg rigid steel tank contains air at 500 kPa, and both tank and air are at
20°C. The tank is connected to a line flowing air at 2 MPa, 20°C. The valve is
opened, allowing air to flow into the tank until the pressure reaches 1.5 MPa and
is then closed. Assume the air and tank are always at the same temperature and
the final temperature is 35°C. Find the final air mass and the heat transfer.
Solution:
Control volume: Air and the steel tank.
Continuity Eq.6.15: m2 - m1 = mi
Energy Eq.6.16: (m2u2 - m1u1)AIR + mST(u2 - u1)ST = mihi + 1Q2
m1 AIR =
P1V
RT1
=
500 × 1
0.287 × 293.2 = 5.94 kg
m2 AIR =
P2V
RT2
=
1500 × 1
0.287 × 308.2 = 16.96 kg
mi = (m2 - m1)air = 16.96 - 5.94 = 11.02 kg
The energy equation now gives
1Q2 = (m2u2 - m1u1)air + mst(u2 - u1)st - mihi
= m1(u2 - u1) + mi(u2 - ui - RTi) + mstCst(T2 – T1)
≅ m1Cv(T2 – T1) + mi[Cv(T2 – Ti) - RTi] + mstCst(T2 – T1)
= 5.94 × 0.717(35 – 20) + 11.02[0.717(35 – 20) – 0.287× 293.2]
+ 40 × 0.46(35 – 20)
= 63.885 – 808.795 + 276
= – 468.9 kJ

6.138
A steam engine based on a turbine is shown in Fig. P6.138. The boiler tank has a
volume of 100 L and initially contains saturated liquid with a very small amount
of vapor at 100 kPa. Heat is now added by the burner, and the pressure regulator
does not open before the boiler pressure reaches 700 kPa, which it keeps constant.
The saturated vapor enters the turbine at 700 kPa and is discharged to the
atmosphere as saturated vapor at 100 kPa. The burner is turned off when no more
liquid is present in the boiler. Find the total turbine work and the total heat
transfer to the boiler for this process.
Solution:
C.V. Boiler tank. Heat transfer, no work and flow out.
Continuity Eq.6.15: m2 - m1 = − me
Energy Eq.6.16: m2u2 - m1u1 = QCV - mehe
State 1: Table B.1.1, 100 kPa => v1 = 0.001 043, u1 = 417.36 kJ/kg
=> m1 = V/v1 = 0.1/0.001 043 = 95.877 kg
State 2: Table B.1.1, 700 kPa => v2 = vg = 0.2729, u2 = 2572.5 kJ/kg
=> m2 = V/vg = 0.1/0.2729 = 0.366 kg,
Exit state: Table B.1.1, 700 kPa => he = 2763.5 kJ/kg
From continuity eq.: me = m1 - m2 = 95.511 kg
= 0.366 × 2572.5 - 95.877 × 417.36 + 95.511 × 2763.5
= 224 871 kJ = 224.9 MJ
C.V. Turbine, steady state, inlet state is boiler tank exit state.
Turbine exit state: Table B.1.1, 100 kPa => he = 2675.5 kJ/kg
Wturb = me (hin- hex) = 95.511 × (2763.5 - 2675.5) = 8405 kJ
W
cb

6.139
An insulated spring-loaded piston/cylinder, shown in Fig. P6.139, is connected to
an air line flowing air at 600 kPa, 700 K by a valve. Initially the cylinder is empty
and the spring force is zero. The valve is then opened until the cylinder pressure
reaches 300 kPa. By noting that u2 = uline + CV(T2 − Tline) and hline − uline =
RTline find an expression for T2 as a function of P2, Po, Tline. With P = 100 kPa,
find T2.
Solution:
C.V. Air in cylinder, insulated so 1Q2 = 0
Continuity Eq.6.15: m2 - m1 = min
Energy Eq.6.16: m2u2 - m1u1 = minhline - 1W2
m1 = 0 ⇒ min = m2 ; m2u2 = m2hline -
1
2 (P0 + P2)m2v2
⇒ u2 +
1
2 (P0 + P2)v2 = hline
Use constant specific heat in the energy equation
Cv(T2 - Tline) + uline +
1
2 (P0 + P2)RT2/P2 = hline






Cv +
1
2
P0 + P2
P2
R T2 = (R + Cv)Tline
with #'s: T2 =
R + Cv
2
3 R + Cv
Tline ; Cv/R = 1/(k-1) , k = 1.4
T2 =
k - 1 + 1
2
3 k -
2
3 + 1
Tline =
3k
2k + 1 Tline = 1.105 Tline = 773.7 K

6.140
A mass-loaded piston/cylinder, shown in Fig. P6.140, containing air is at 300 kPa,
17°C with a volume of 0.25 m3, while at the stops V = 1 m3. An air line, 500 kPa,
600 K, is connected by a valve that is then opened until a final inside pressure of
400 kPa is reached, at which point T = 350 K. Find the air mass that enters, the
work, and heat transfer.
Solution:
C.V. Cylinder volume.
Continuity Eq.6.15: m2 - m1 = min
Energy Eq.6.16: m2u2 - m1u1 = minhline + QCV - 1W2
Process: P1 is constant to stops, then constant V to state 2 at P2
State 1: P1, T1 m1 =
P1V
RT1
=
300 × 0.25
0.287 × 290.2 = 0.90 kg
State 2:
Open to P2 = 400 kPa, T2 = 350 K
m2 =
400 × 1
0.287 × 350 = 3.982 kg
mi = 3.982 - 0.90 = 3.082 kg
AIR
Only work while constant P
1W2 = P1(V2 - V1) = 300(1 - 0.25) = 225 kJ
Energy Eq.: QCV + mihi = m2u2 - m1u1 + 1W2
QCV = 3.982 × 0.717 × 350 - 0.90 × 0.717 × 290.2 + 225
- 3.082 × 1.004 × 600 = -819.2 kJ
We could also have used the air tables A.7.1 for the u’s and hi.

6.141
A 2-m3 storage tank contains 95% liquid and 5% vapor by volume of liquified
natural gas (LNG) at 160 K, as shown in Fig. P6.62. It may be assumed that LNG
has the same properties as pure methane. Heat is transferred to the tank and
saturated vapor at 160 K flows into a steady flow heater which it leaves at 300 K.
The process continues until all the liquid in the storage tank is gone. Calculate the
total amount of heat transfer to the tank and the total amount of heat transferred to
the heater.
Solution:
CV: Tank, flow out, transient.
Continuity Eq.: m2 - m1 = -me
Energy Eq.:
QTank = m2u2 - m1u1 + mehe
At 160 K, from Table B.7:
LIQUID Q
tank
Q
heaterVAPOR
mf = Vf /vf =
0.95 × 2
0.00297 = 639.73 kg , mg = Vg/vg =
0.05 × 2
0.03935 = 2.541 kg
m1 = 642.271 kg, m2 = V/vg2 = 2/0.03935 = 50.826 kg
m1u1 = 639.73(-106.35) + 2.541(207.7) = -67507 kJ
me = m1 - m2 = 591.445 kg
QTank = 50.826 × 207.7 - (-67 507) + 591.445 × 270.3
= +237 931 kJ
CV: Heater, steady flow, P = PG 160 K = 1593 kPa
QHeater = me Tank(he - hi)Heater
= 591.445(612.9 - 270.3) = 202 629 kJ

Heat transfer problems
6.142
Liquid water at 80oC flows with 0.2 kg/s inside a square duct, side 2 cm insulated
with a 1 cm thick layer of foam k = 0.1 W/m K. If the outside foam surface is at
25oC how much has the water temperature dropped for 10 m length of duct?
Neglect the duct material and any corner effects (A = 4sL).
Solution:
Conduction heat transfer
Q
.
out = kA
dT
dx = k 4 sL
∆T
∆x
= 0.1× 4×0.02×10×(80-25)/0.01 = 440 W
Energy equation: m
.
1h1= m
.
he + Q
.
out
he– hi = -Q
.
/m
.
= - (440/0.2) = -2200 J/kg = - 2.2 kJ/kg
he= hi -2.2 kJ/kg = 334.88 – 2.2 = 332.68 kJ/kg
Te = 80 –
2.2
334.88-313.91 5 = 79.48oC
∆T = 0.52oC
We could also have used he– hi = Cp∆T
s
Lcb

6.143
Saturated liquid carbon dioxide at 2500 kPa flows at 2 kg/s inside a 10 cm outer
diameter steel pipe and outside of the pipe is a flow of air at 22oC with a
convection coefficient of h = 150 W/m2 K. Neglect any ∆T in the steel and any
inside convection h and find the length of pipe needed to bring the carbon dioxide
to saturated vapor.
Solution:
Energy Eq. water: Q
.
= m
.
(he – hi) = m
.
hfg
Table B.3.1: hfg = 263.65 kJ/kg, T = Tsat = -12oC and
The energy is transferred by heat transfer so
Q
.
= h A ∆T = h πD L ∆T
Equate the two expressions for the heat transfer and solve for the length L
L =
Q
.
h πD ∆T
=
m
.
hfg
h πD ∆T
=
2 × 263.65 × 1000
150 × π × 0.1 ×(22 - (-12))
= 329 m
L

6.144
A counter-flowing heat exchanger conserves energy by heating cold outside fresh
air at 10oC with the outgoing combustion gas (air) at 100oC. Assume both flows
are 1 kg/s and the temperature difference between the flows at any point is 50oC.
What is the incoming fresh air temperature after the heat exchanger? What is the
equivalent (single) convective heat transfer coefficient between the flows if the
interface area is 2 m2?
Solution:
The outside fresh air is heated up to T4 = 50oC (100 – 50), the heat transfer
needed is
Q
.
= m
.
∆h = m
.
Cp∆T = 1 kg/s × 1.004
kJ
kg K × (50 – 10) K = 40 kW
This heat transfer takes place with a temperature difference of 50oC throughout
Q
.
= h A ∆T ⇒ h =
Q
.
A∆T
=
40 000
2 × 50
W
m2 K
= 400
W
m2 K
Often the flows may
be concentric as a
smaller pipe inside a
larger pipe.
Wall
1 2
34
Heat transfer
Hot gas
fresh air

6.145
A flow of 1000 K, 100 kPa air with 0.5 kg/s in a furnace flows over a steel plate
of surface temperature 400 K. The flow is such that the convective heat transfer
coefficient is h = 125 W/m2 K. How much of a surface area does the air have to
flow over to exit with a temperature of 800 K? How about 600 K?
Solution:
Convection heat transfer
Q
.
= hA ∆T
Inlet: ∆Ti = 1000 - 400 = 600 K
a)
Exit: ∆Te = 800 - 400 = 400 K,
so we can use an average of ∆T = 500 K for heat transfer
Q
.
= m
.
a (hi – he) = 0.5(1046.22 – 822.2) =112 kW
A =
Q
.
h ∆T
=
112 × 1000
125 × 500
= 1.79 m2
b)
Q
.
= m
.
a (hi – he) = 0.5 (1046.22 – 607.32) = 219.45 kW
Exit: ∆Tout = 600 - 400 = 200 K,
so we have an average of ∆T = 400 K for heat transfer
A =
Q
.
h∆T
=
219.45 × 1000
125 × 400
= 4.39 m2
400 K
air1000 K T
exit
Q
from air to steel

CONTENT CHAPTER 7
SUBSECTION PROB NO.
In-Text concept questions a-g
Heat engines and refrigerators 15-36
Second law and processes 37-43
Carnot cycles and absolute temperature 44-77
Finite ∆T heat transfer 78-91
Ideal gas Carnot cycles 92-95
review problems 96-113

7.a
Electrical appliances (TV, stereo) use electric power as input. What happens to
the power? Are those heat engines? What does the second law say about those
devices?
Most electric appliances such as TV, VCR, stereo and clocks dissipate
power in electrical circuits into internal energy (they get warm) some power goes
into light and some power into mechanical energy. The light is absorbed by the
room walls, furniture etc. and the mechanical energy is dissipated by friction so
all the power eventually ends up as internal energy in the room mass of air and
other substances.
These are not heat engines, just the opposite happens, namely electrical
power is turned into internal energy and redistributed by heat transfer. These are
irreversible processes.

7.b
Geothermal underground hot water or steam can be used to generate electric
power. Does that violate the second law?
No.
Since the earth is not uniform we consider the hot water or steam supply
as coming from one energy source (the high T) and we must reject heat to
a low temperature reservoir as the ocean, a lake or the atmosphere which
is another energy reservoir.
Iceland uses a
significant amount
of steam to heat
buildings and to
generate
electricity.

7.c
A windmill produces power on a shaft taking kinetic energy out of the wind. Is it
a heat engine? Is it a perpetual machine? Explain.
Since the wind is generated by a complex
system driven by solar heat input and
radiation out to space it is a kind of heat
engine.
Within our lifetime it looks like it is perpetual. However with a different
time scale the climate will change, the sun will grow to engulf the earth as it
burns out of fuel. There is a storage effect and a non-uniform distribution of states
in the system that drives this.

7.d
heat engines and heat pumps (refrigerators) are energy conversion devices
altering amounts of energy transfer between Q and W. Which conversion
direction (Q → W or W → Q) is limited and which is unlimited according to the
second law.
The work output of a heat engine is limited (Q to W).
You can transform W to Q unlimited (a heat pump that does not work well
or you may think about heat generated by friction).
7.e
Ice cubes in a glass of liquid water will eventually melt and all the water approach
room temperature. Is this a reversible process? Why?
There is heat transfer from the warmer ambient
to the water as long as there is a temperature
difference. Eventually the temperatures
approach each other and there is no more heat
transfer. This is irreversible, as we cannot
make ice-cubes out of the water unless we run
a refrigerator and that requires a work from the
surroundings, which does not leave the
surroundings unchanged.

7.f
Does a process become more or less reversible with respect to heat transfer if it is
fast rather than slow? Hint: Recall from Chapter 4 that Q
.
= CA ∆T.
If the higher heat transfer rate is caused by a larger ∆T then the process is more
irreversible so as the process would be slower due to a lower ∆T then it
approaches a reversible process. If the rate of heat transfer is altered due to the
factor CA with the same ∆T then it is irreversible to the same degree.

7.g
If you generated hydrogen from, say, solar power, which of these would be more
efficient: (1) transport it and then burn it in an engine or (2) convert the solar
power to electricity and transport that? What else would you need to know in
order to give a definite answer?
Case (1):
First there is a certain efficiency when converting solar power to hydrogen. Then
the transport and packaging of hydrogen has some energy expenditures associated
with it. The hydrogen could be compressed to a high pressure (typically 70 MPa)
which is expensive in terms of work input and then stored in a tank. One
alternative would be to cool it down to become a liquid to have a much smaller
volume but the temperature at which this happens is very low so the cooling and
continued cooling under transport requires a significant work input also. Certain
materials like metal-hydrides, boron salt slurries and nano-carbon fibers allows
storage of hydrogen at more modest pressures and are all currently under
investigation as other alternative storage methods. After the hydrogen is
transported to an engine then the engine efficiency determines the work output.
Case (2):
If the solar power is located where there is access to electrical transmission lines
then it can be used in solar panels, solar heating of water or other substance to run
a heat engine cycle like a power plant to generate electricity. To make new
transmission lines is costly and has an impact on the environment that must be
considered.
You also need to look at the time of day/year at which the power is required and
when it is available. The end use also presents some limitations like if the power
should be used for a car then the energy must be stored temporarily like in a
battery.

7.1
Two heat engines operate between the same two energy reservoirs and both
receives the same QH. One engine is reversible and the other is not. What can you
say about the two QL’s?
The reversible heat engine can produce more work (has a higher efficiency) than
the irreversible heat engine and due to the energy conservation it then gives out a
smaller QL compared to the irreversible heat engine.
Wrev = QH - QL rev > Wirrev = QH - QL irrev
QL rev < QL irrev

7.2
Compare two domestic heat pumps (A and B) running with the same work input.
If A is better than B which one heats the house most?
The statement that A is better means it has a higher COP and since
QH A = COPA W > QH B = COPB W
it can thus provide more heat to the house. The higher heat comes from the higher
QL it is able to draw in.

7.3
Suppose we forget the model for heat transfer as Q
.
= CA ∆T, can we draw some
information about direction of Q from the second law?
One of the classical statements of the second law is the Clausius statement saying
that you cannot have heat transfer from a lower temperature domain to a higher
temperature domain without work input.
The opposite, namely a transfer of heat from a high temperature domain towards a
lower temperature domain can happen (which is a heat engine with zero
efficiency).

7.4
A combination of two heat engines is shown in Fig. P7.4. Find the overall thermal
efficiency as a function of the two individual efficiencies.
The overall efficiency
ηTH = W
.
net / Q
.
H = (W
.
1 + W
.
2) / Q
.
H = η1 + W
.
2 / Q
.
H
For the second heat engine and the energy Eq. for the first heat engine
W
.
2 = η2 Q
.
M = η2 (1 – η1) Q
.
H
so the final result is
ηTH = η1 + η2 (1 – η1)

7.5
Compare two heat engines receiving the same Q, one at 1200 K and the other at
1800 K; they both reject heat at 500 K. Which one is better?
The maximum efficiency for the engines are given by the Carnot heat
engine efficiency as
ηTH = W
.
net / Q
.
H = 1 –
TL
TH
Since they have the same low temperature the one with the highest TH will have a
higher efficiency and thus presumably better.

7.6
A car engine takes atmospheric air in at 20oC, no fuel, and exhausts the air at –
20oC producing work in the process. What do the first and the second laws say
about that?
Energy Eq.: W = QH − QL = change in energy of air. OK
2nd
law: Exchange energy with only one reservoir. NOT OK.
This is a violation of the statement of Kelvin-Planck.
Remark: You cannot create and maintain your own energy reservoir.

7.7
A combination of two refrigerator cycles is shown in Fig. P7.7. Find the overall
COP as a function of COP1 and COP2.
The overall COP becomes
COP = β =
Q
.
L
W
.
tot
=
Q
.
L
W
.
1
W
.
1
W
.
tot
= COP1
W
.
1
W
.
tot
= COP1
1
1 + W
.
2/W
.
1
where we used W
.
tot = W
.
1 + W
.
2. Use definition of COP2 and energy equation for
refrigerator 1 to eliminate Q
.
M and we have
W
.
2 = Q
.
M / COP2 = (W
.
1 + Q
.
L) / COP2
so then
W
.
2 / W
.
1 = (1 + Q
.
L/W
.
1) / COP2 = (1 + COP1) / COP2
Finally substitute into the first equation and rearrange a little to get
COP = β =
COP1 COP2
COP1 + COP2 + 1

7.8
After you have returned from a car trip the car engine has cooled down and is thus
back to the state in which it started. What happened to all the energy released in
the burning of the gasoline? What happened to all the work the engine gave out?
Solution:
All the energy from the fuel generates heat and work out of the engine. The heat
is directly dissipated in the atmosphere and the work is turned into kinetic energy
and internal energy by all the frictional forces (wind resistance, rolling resistance,
brake action). Eventually the kinetic energy is lost by braking the car so in the end
all the energy is absorbed by the environment increasing its internal energy.

7.9
Does a reversible heat engine burning coal (which, in practice, cannot be done
reversibly) have impacts on our world other than depletion of the coal reserve?
Solution:
When you burn coal you form carbon dioxide CO2 which is a greenhouse gas. It
absorbs energy over a wide spectrum of wavelengths and thus traps energy in the
atmosphere that otherwise would go out into space.
Coal from various locations also has sulfur and other substances like heavy metals
in it. The sulfur generates sulfuric acid (resulting in acid rain) in the atmosphere
and can damage the forests.

7.10
If the efficiency of a power plant goes up as the low temperature drops, why do
power plants not just reject energy at say –40oC?
In order to reject heat the ambient must be at the low temperature. Only if
we moved the plant to the North Pole would we see such a low T.
Remark: You cannot create and maintain your own energy reservoir.

7.11
If the efficiency of a power plant goes up as the low temperature drops why not
let the heat rejection go to a refrigerator at, say, –10oC instead of ambient 20oC?
The refrigerator must pump the heat up to 20oC to reject it to the ambient. The
refrigerator must then have a work input that will exactly offset the increased
work output of the power plant, if they are both ideal. As we can not build ideal
devices the actual refrigerator will require more work than the power plant will
produce extra.

7.12
A coal-fired power plant operates with a high T of 600oC whereas a jet engine has
about 1400 K. Does that mean we should replace all power plants with jet
engines?
The thermal efficiency is limited by the Carnot heat engine efficiency.
That is, the low temperature is also important. Here the power plant has a
much lower T in the condenser than the jet engine has in the exhaust flow so the
jet engine does not necessarily have a higher efficiency than the power plant.
Gas-turbines are used in power plants where they can cover peak power
demands needed for shorter time periods and their high temperature exhaust can
be used to boil additional water for the steam cycle.
WT
Q
H
QL
.
WP, in
from coal
to ambient

7.13
A heat transfer requires a temperature difference, see chapter 4, to push the Q
.
.
What implications do that have for a real heat engine? A refrigerator?
This means that there are temperature differences between the source of
energy and the working substance so TH is smaller than the source temperature.
This lowers the maximum possible efficiency. As heat is rejected the working
substance must have a higher temperature TL than the ambient receiving the Q
.
L,
which lowers the efficiency further.
For a refrigerator the high temperature must be higher than the ambient to
which the Q
.
H is moved. Likewise the low temperature must be lower than the
cold space temperature in order to have heat transfer from the cold space to the
cycle substance. So the net effect is the cycle temperature difference is larger than
the reservoir temperature difference and thus the COP is lower than that estimated
from the cold space and ambient temperatures.

7.14
Hot combustion gases (air) at 1500 K are used as heat source in a heat engine
where the gas is cooled to 750 K and the ambient is at 300 K. This is not a
constant T source. How does that affect the efficiency?
Solution:
If the efficiency is written as
ηTH = W
.
net / Q
.
H = 1 –
TL
TH
then TH is somewhere between 1500 K
and 750 K and it is not a linear average.
HQ
W
L
Q
TL
HE
1 2
cb
After studying chapter 8 and 9 we can solve this problem and find the
proper average high temperature based on properties at states 1 and 2.

Heat Engines and Refrigerators

7.15
A gasoline engine produces 20 hp using 35 kW of heat transfer from burning fuel.
What is its thermal efficiency and how much power is rejected to the ambient?
Conversion Table A.1: 20 hp = 20 × 0.7457 kW = 14.91 kW
Efficiency: ηTH = W
.
out/Q
.
H =
14.91
35 = 0.43
Energy equation: Q
.
L = Q
.
H - W
.
out = 35 – 14.91 = 20.1 kW
Q
.
H
⇒
Q
.
L
⇒
W
.
out
⇒

7.16
Calculate the thermal efficiency of the steam power plant cycle described in
Example 6.9.
Solution:
From solution to Example 6.9,
wnet = wt + wp = 640.7 – 4
= 636.7 kJ/kg
qH = qb = 2831 kJ/kg
ηTH = wnet/qH =
636.7
2831 = 0.225
WT
Q
H
QL
.
WP, in
Q1 2
Notice we cannot write wnet = qH − qL as there is an extra heat transfer 1Q
.
2
as a loss in the line. This needs to be accounted for in the overall energy
equation.

7.17
A refrigerator removes 1.5 kJ from the cold space using 1 kJ work input. How
much energy goes into the kitchen and what is its coefficient of performance?
C.V. Refrigerator. The energy QH goes into the kitchen air.
Energy Eq.: QH = W + QL = 1 + 1.5 = 2.5 kJ
COP: β =
QL
W = 1.5 / 1 = 1.5
The back side of
the refrigerator
has a black grille
that heats the
kitchen air. Other
models have that
at the bottom
with a fan to
drive the air over
it.
1
2
Air in, 3
Air out, 4

7.18
Calculate the coefficient of performance of the R-134a refrigerator given in
Example 6.10.
Solution:
From the definition
β = Q. L/W. IN =
14.54
5 = 2.91
Notice we cannot write W. IN = Q. H - Q. L
as there is a small Q. in the compressor.
This needs to be accounted for in the
overall energy equation.
QH
-W
QL
.Evaporator
Condenser
Q
.
loss

7.19
A coal fired power plant has an efficiency of 35% and produces net 500 MW of
electricity. Coal releases 25 000 kJ/kg as it burns so how much coal is used per
hour?
From the definition of the thermal efficiency and the energy release by the
combustion called heating value HV we get
W
.
= η Q
.
H = η· m
.
·HV
then
m
.
=
W
.
η × HV
=
500 MW
0.35 × 25000 kJ/kg =
500 × 1000 kJ/s
0.35 × 25000 kJ/kg
= 57.14 kg/s = 205 714 kg/h

7.20
Assume we have a refrigerator operating at steady state using 500 W of electric
power with a COP of 2.5. What is the net effect on the kitchen air?
Take a C.V. around the whole kitchen. The only energy term that crosses
the control surface is the work input W
.
apart from energy exchanged with the
kitchen surroundings. That is the kitchen is being heated with a rate of W
.
.
Remark: The two heat transfer rates are both internal to the kitchen. Q
.
H goes into
the kitchen air and Q
.
L actually leaks from the kitchen into the refrigerated space,
which is the reason we need to drive it out again.

7.21
A room is heated with a 1500 W electric heater. How much power can be saved if
a heat pump with a COP of 2.0 is used instead?
Assume the heat pump has to deliver 1500 W as the Q
.
H.
Heat pump: β′ = Q
.
H/W
.
IN
W
.
IN = Q
.
H/β′ =
1500
2 = 750 W
So the heat pump requires an input of 750 W thus saving the difference
W
.
saved = 1500 W – 750 W = 750 W
HQ
W
LQ
TL
HP
Room
incb

7.22
An air-conditioner discards 5.1 kW to the ambient with a power input of 1.5 kW.
Find the rate of cooling and the coefficient of performance.
Solution:
In this case Q
.
H = 5.1 kW goes to the ambient so
Energy Eq. : Q
.
L = Q
.
H – W
.
= 5.1 – 1.5 = 3.6 kW
βREFRIG =
Q
.
L
W
.
=
3.6
1.5 = 2.4
HQ = 5.1 kW
W = 1.5 kW
LQ
TL
Tamb
REF

7.23
Calculate the thermal efficiency of the steam power plant cycle described in
Problem 6.103.
From solution to Problem 6.103,
Turbine A5 = (π/4)(0.2)2 = 0.03142 m2
V5 = m
.
v5/A5 = 25 × 0.06163 / 0.03142 = 49 m/s
h6 = 191.83 + 0.92 × 2392.8 = 2393.2 kJ/kg
wT = 3404 - 2393.2 - (2002 - 492)/(2 × 1000) = 992 kJ/kg
W
.
T = m
.
wT = 25 × 992 = 24 800 kW
W
.
NET = 24800 - 300 = 24 500 kW
From the solution to Problem 6.105
Economizer A7 = πD
2
7/4 = 0.004 418 m2, v7 = 0.001 008 m3/kg
V2 = V7 = m
.
v/A7 = 25 × 0.001008 / 0.004418 = 5.7 m/s,
V3 = (v3/v2)V2 = (0.001 118 / 0.001 008) 5.7 = 6.3 m/s ≈ V2
so kinetic energy change is unimportant
qECON = h3 - h2 = 744 - 194 = 550.0 kJ/kg
Q
.
ECON = m
.
qECON = 25 (550.0) = 13 750 kW
Generator A4 = πD
2
4/4 = 0.031 42 m2, v4 = 0.060 23 m3/kg
V4 = m
.
v4/A4 = 25 × 0.060 23/0.031 42 = 47.9 m/s
qGEN = 3426 - 744 + (47.92 - 6.32)/(2×1000) = 2683 kJ/kg
Q
.
GEN = 25 × (2683) = 67 075 kW
The total added heat transfer is
Q
.
H = 13 758 + 67 075 = 80 833 kW
⇒ ηTH = W
.
NET/Q
.
H =
24500
80833 = 0.303

7.24
A window air-conditioner unit is placed on a laboratory bench and tested in
cooling mode using 750 W of electric power with a COP of 1.75. What is the
cooling power capacity and what is the net effect on the laboratory?
Definition of COP: β = Q
.
L / W
.
Cooling capacity: Q
.
L = β W
.
= 1.75 × 750 = 1313 W
For steady state operation the Q
.
L comes from the laboratory and Q
.
H goes
to the laboratory giving a net to the lab of W
.
= Q
.
H - Q
.
L = 750 W, that is
heating it.

7.25
A water cooler for drinking water should cool 25 L/h water from 18oC to 10oC
using a small refrigeration unit with a COP of 2.5. Find the rate of cooling
required and the power input to the unit.
The mass flow rate is
m
.
= ρV
.
=
25 × 10-3
0.001002
1
3600 kg/s = 6.93 g/s
Energy equation for heat exchanger
Q
.
L = m
.
(h1 − h2) = m
.
CP (T1 − T2)
Q
W
L
TH
HQ
REF
1 2
cb
= 6.93 × 10-3 × 4.18 × (18 – 10) = 0.2318 kW
β = COP = Q
.
L / W
.
W
.
= Q
.
L / β = 0.2318 / 2.5 = 0.093 kW
Comment: The unit does not operate continuously.

7.26
A farmer runs a heat pump with a 2 kW motor. It should keep a chicken hatchery
at 30oC, which loses energy at a rate of 10 kW to the colder ambient Tamb. What
is the minimum coefficient of performance that will be acceptable for the heat
pump?
Solution:
Power input: W
.
= 2 kW
Energy Eq. for hatchery: Q
.
H = Q
.
Loss = 10 kW
Definition of COP: β = COP =
Q
.
H
W
.
=
10
2 = 5
QleakQ QHL
W = 2 kW
HP
cb

7.27
Calculate the coefficient of performance of the R-410a heat pump cycle described
in Problem 6.108.
Solution:
From solution to Problem 6.108,
CV: Condenser
Q
.
COND = m
.
(h3 − h2)
= 0.05 kg/s (134 − 367) kJ/kg
= −11.65 kW
Then with the work as -W
.
IN = 5.0 kW we
have Q
.
H = − Q
.
COND
Q
H
-WC
QL
.
cb
Evaporator
Condenser
3
4
56
1
2Q loss
Heat pump: β′ = Q
.
H/W
.
IN =
11.65
5.0 = 2.33

7.28
A power plant generates 150 MW of electrical power. It uses a supply of 1000
MW from a geothermal source and rejects energy to the atmosphere. Find the
power to the air and how much air should be flowed to the cooling tower (kg/s) if
its temperature cannot be increased more than 10oC.
Solution:
C.V. Total power plant.
Energy equation gives the amount of heat rejection to the atmosphere as
Q
.
L= Q
.
H - W
.
= 1000 – 150 = 850 MW
The energy equation for the air flow that absorbs the energy is
Q
.
L = m
.
air ∆h = m
.
air Cp ∆T
m
.
air =
Q
.
L
Cp∆T
=
850 × 1000
1.004 × 10
= 84 661 kg/s
Probably too large to make, so some cooling by liquid water or evaporative
cooling should be used.
HQ
W
TL
L
Q
HE
Air
cb

7.29
A water cooler for drinking water should cool 25 L/h water from 18oC to 10oC
while the water reservoir also gains 60 W from heat transfer. Assume a small
refrigeration unit with a COP of 2.5 does the cooling. Find the total rate of
cooling required and the power input to the unit.
The mass flow rate is
m
.
= ρV
.
=
25 × 10-3
0.001002
1
3600 kg/s = 6.93 g/s
Q
.
L = m
.
(h1 − h2) + Q
.
H TR
= m
.
CP (T1 − T2) + Q
.
H TR
Q
W
L
TH
HQ
REF
1 2
cb
H.TRQ
= 6.93 × 10-3 × 4.18 × (18 – 10) kW + 60 W
= 291.8 W
β = COP = Q
.
L / W
.
W
.
= Q
.
L / β = 291.8 / 2.5 = 116.7 W
Comment: The unit does not operate continuously.

7.30
A car engine delivers 25 hp to the driveshaft with a thermal efficiency of 30%.
The fuel has a heating value of 40 000 kJ/kg. Find the rate of fuel consumption
and the combined power rejected through the radiator and exhaust.
Solution:
Heating value (HV): Q
.
H = m
.
·HV
From the definition of the thermal efficiency
W
.
= η Q
.
H = η· m
.
·HV
m
.
=
W
.
η·HV
=
25 × 0.7355
0.3 × 40 000
= 0.00153 kg/s = 1.53 g/s
Conversion of power from hp to kW in Table A.1.
Q
.
L = Q
.
H - W
.
= (W
.
/η −W
.
) = (
1
η
−1 )W
.
= (
1
0.3 – 1) 25 × 0.7355 = 42.9 kW
Exhaust flow
Air intake filter
Shaft
Fan
power
Fuel line
cb

7.31
R-410a enters the evaporator (the cold heat exchanger) in an A/C unit at -20oC, x
= 28% and leaves at -20oC, x = 1. The COP of the refrigerator is 1.5 and the mass
flow rate is 0.003 kg/s. Find the net work input to the cycle.
Q
.
L = m
.
(h2 − h1) = m
.
[hg − (hf + x1 hfg)]
= m
.
[hfg − x1 hfg] = m
.
(1 – x1)hfg
Q L
1 2
cb
= 0.003 kg/s × 0.72 × 243.65 kJ/kg = 0.5263 kW
β = COP = Q
.
L / W
.
W
.
= Q
.
L / β = 0.5263 / 1.5 = 0.35 kW

7.32
For each of the cases below determine if the heat engine satisfies the first law
(energy equation) and if it violates the second law.
a. Q
.
H = 6 kW, Q
.
L = 4 kW, W
.
= 2 kW
b. Q
.
H = 6 kW, Q
.
L = 0 kW, W
.
= 6 kW
c. Q
.
H = 6 kW, Q
.
L = 2 kW, W
.
= 5 kW
d. Q
.
H = 6 kW, Q
.
L = 6 kW, W
.
= 0 kW
Solution:
1st
. law 2nd
law
a Yes Yes (possible)
b Yes No, impossible Kelvin - Planck
c No Yes, but energy not conserved
d Yes Yes (Irreversible Q
.
over ∆T)
HQ
W
LQ
TL
TH
HE
cb

7.33
For each of the cases in problem 7.32 determine if a heat pump satisfies the first
law (energy equation) and if it violates the second law.
a. Q
.
H = 6 kW, Q
.
L = 4 kW, W
.
= 2 kW
b. Q
.
H = 6 kW, Q
.
L = 0 kW, W
.
= 6 kW
c. Q
.
H = 6 kW, Q
.
L = 2 kW, W
.
= 5 kW
d. Q
.
H = 6 kW, Q
.
L = 6 kW, W
.
= 0 kW
Solution:
1st
. law 2nd
law
a Satisfied Does not violate
b Satisfied Does not violate
c Violated Does not violate, but 1st
law
d Satisfied Does violate, Clausius
HQ
W
LQ
TL
TH
HP
cb

7.34
A large stationary diesel engine produces 15 MW with a thermal efficiency of
40%. The exhaust gas, which we assume is air, flows out at 800 K and the intake
air is 290 K. How large a mass flow rate is that, assuming this is the only way we
reject heat? Can the exhaust flow energy be used?
Heat engine: Q
.
H = W
.
out/ηTH =
15
0.4 = 37.5 MW
Energy equation: Q
.
L = Q
.
H - W
.
out = 37.5 – 15 = 22.5 kW
Exhaust flow: Q
.
L = m
.
air(h800 - h290)
m
.
air =
Q
.
L
h800 - h290
=
22.5 × 1000
822.2 - 290.43 = 42.3 kg/s
The flow of hot gases can be used to heat a building or it can be used to
heat water in a steam power plant since that operates at lower temperatures.

7.35
In a steam power plant 1 MW is added in the boiler, 0.58 MW is taken out in the
condenser and the pump work is 0.02 MW. Find the plant thermal efficiency. If
everything could be reversed find the coefficient of performance as a refrigerator.
Solution:
WT
Q
H
QL
.
WP, in
CV. Total plant:
Energy Eq.:
Q
.
H + W
.
P,in = W
.
T + Q
.
L
W
.
T = 1 + 0.02 – 0.58 = 0.44 MW
ηTH =
W
.
T – W
.
P,in
Q
.
H
=
440 – 20
1000 = 0.42
β =
Q
.
L
W
.
T – W
.
P,in
=
580
440 – 20 = 1.38

7.36
Calculate the amount of work input a refrigerator needs to make ice cubes out of a
tray of 0.25 kg liquid water at 10oC. Assume the refrigerator has β = 3.5 and a
motor-compressor of 750 W. How much time does it take if this is the only
cooling load?
C.V. Water in tray. We neglect tray mass.
Energy Eq.: m(u2 − u1) = 1Q2 − 1W2
Process : P = constant = Po
1W2 = ∫ P dV = Pom(v2 − v1)
1Q2 = m(u2 − u1) + 1W2 = m(h2 − h1)
Tbl. B.1.1 : h1 = 41.99 kJ/kg, Tbl. B.1.5 : h2 = - 333.6 kJ/kg
1Q2 = 0.25(-333.4 – 41.99 ) = - 93.848 kJ
Consider now refrigerator
β = QL/W
W = QL/β = - 1Q2/ β = 93.848/3.5 = 26.81 kJ
For the motor to transfer that amount of energy the time is found as
W = ∫ W
.
dt = W
.
∆t
∆t = W/W
.
= (26.81 × 1000)/750 = 35.75 s
Comment: We neglected a baseload of the refrigerator so not all the 750 W are
available to make ice, also our coefficient of performance is very optimistic and
finally the heat transfer is a transient process. All this means that it will take much
more time to make ice-cubes.

Second Law and Processes

7.37
Prove that a cyclic device that violates the Kelvin–Planck statement of the second
law also violates the Clausius statement of the second law.
Solution: Proof very similar to the proof in section 7.2.
H.E. violating Kelvin receives QH from
TH and produces net W = QH.
This W input to H.P. receiving QL from TL.
H.P. discharges QH + QL to TH .
Net Q to TH is : -QH + QH + QL = QL.
H.E. + H.P. together transfers QL from TL
to TH with no W thus violates Clausius.
Q
W
HE HP
H
QL
QH +QL
T
H
TL
C.V. Total

7.38
Assume a cyclic machine that exchanges 6 kW with a 250oC reservoir and has
a. Q
.
L = 0 kW, W
.
= 6 kW
b. Q
.
L = 6 kW, W
.
= 0 kW
and Q
.
L is exchanged with a 30oC ambient. What can you say about the
processes in the two cases a and b if the machine is a heat engine? Repeat the
question for the case of a heat pump.
Solution:
Heat engine
a. Since Q
.
L = 0 impossible Kelvin – Planck
b. Possible, irreversible, ηeng = 0
Ηeat pump
a. Possible, irreversible (like an electric heater)
b. Impossible, β → ∞, Clausius

7.39
Discuss the factors that would make the power plant cycle described in Problem
6.103 an irreversible cycle.
Solution:
General discussion, but here are a few of the most significant factors.
1. Combustion process that generates the hot source of energy.
2. Heat transfer over finite temperature difference in boiler.
3. Flow resistance and friction in turbine results in less work out.
4. Flow friction and heat loss to/from ambient in all pipes.
5. Heat transfer over finite temperature difference in condenser.

7.40
Discuss the factors that would make the heat pump described in Problem 6.108 an
irreversible cycle.
Solution:
General discussion but here are a few of the most significant factors.
1. Unwanted heat transfer in the compressor.
2. Pressure loss (back flow leak) in compressor
3. Heat transfer and pressure drop in line 1 => 2.
4. Pressure drop in all lines.
5. Throttle process 3 => 4.

7.41
Consider the four cases of a heat engine in problem 7.32 and determine if any of
those are perpetual machines of the first or second kind.
a. Q
.
H = 6 kW, Q
.
L = 4 kW, W
.
= 2 kW
b. Q
.
H = 6 kW, Q
.
L = 0 kW, W
.
= 6 kW
c. Q
.
H = 6 kW, Q
.
L = 2 kW, W
.
= 5 kW
d. Q
.
H = 6 kW, Q
.
L = 6 kW, W
.
= 0 kW
HQ
W
LQ
TL
TH
HE
cb
Solution:
1st
. law 2nd
law
b Yes No, impossible Kelvin - Planck
Perpetual machine second kind
It violates the 2nd
law converts all Q
.
to W
.
c No Yes, but energy not conserved
Perpetual machine first kind
It generates energy inside
d Yes Yes (Irreversible Q
.
over ∆T)

7.42
Consider a heat engine and heat pump connected as shown in figure P7.42.
Assume TH1 = TH2 > Tamb and determine for each of the three cases if the setup
satisfy the first law and/or violates the 2nd
law.
Q
.
H1 Q
.
L1 W
.
1 Q
.
H2 Q
.
L2 W
.
2
a 6 4 2 3 2 1
b 6 4 2 5 4 1
c 3 2 1 4 3 1
Solution:
1st
. law 2nd
law
b Yes No, combine Kelvin - Planck
c Yes No, combination clausius

7.43
The water in a shallow pond heats up during the day and cools down during the
night. Heat transfer by radiation, conduction and convection with the ambient
thus cycles the water temperature. Is such a cyclic process reversible or
irreversible?
Solution:
All the heat transfer takes place over a finite ∆T and thus all the heat
transfer processes are irreversible.
Conduction and convection have ∆T in the water, which is internally
irreversible and ∆T outside the water which is externally irreversible. The
radiation is absorbed or given out at the water temperature thus internally
(for absorption) and externally (for emission) irreversible.

Carnot Cycles and Absolute Temperature

7.44
Calculate the thermal efficiency of a Carnot cycle heat engine operating between
reservoirs at 300oC and 45oC. Compare the result to that of Problem 7.16.
Solution:
ηTH = Wnet / QH = 1 –
TL
TH
= 1 –
45 + 273
300 + 273 = 0.445 (Carnot)
η7.16 = 0.225 (efficiency about ½ of the Carnot)

7.45
A Carnot cycle heat engine has an efficiency of 40%. If the high temperature is
raised 10% what is the new efficiency keeping the same low temperature?
Solution:
ηTH = Wnet / QH = 1 –
TL
TH
= 0.4
TL
TH
= 0.6
so if TH is raised 10% the new ratio becomes
TL
TH new
= 0.6 /1.1 = 0.5454 ηTH new = 1 – 0.5454 = 0.45

7.46
Find the power output and the low T heat rejection rate for a Carnot cycle heat
engine that receives 6 kW at 250oC and rejects heat at 30oC as in Problem 7.38.
Solution:
From the definition of the absolute temperature Eq. 7.8
ηcarnot = 1 –
TL
TH
= 1 –
303
523 = 0.42
Definition of the heat engine efficiency gives the work as
W
.
= η Q
.
H = 0.42 × 6 = 2.52 kW
Apply the energy equation
Q
.
L = Q
.
H - W
.
= 6 – 2.52 = 3.48 kW

7.47
Consider the setup with two stacked (temperature wise) heat engines as in Fig.
P7.4. Let TH = 900 K, TM = 600 K and TL = 300 K. Find the two heat engine
efficiencies and the combined overall efficiency assuming Carnot cycles.
The individual efficiencies
η1 = 1 –
TM
TH
= 1 –
600
900 = 0.333
η2 = 1 –
TL
TM
= 1 –
300
600 = 0.5
The overall efficiency
ηTH = W
.
net / Q
.
H = (W
.
1 + W
.
2) / Q
.
H = η1 + W
.
2 / Q
.
H
W
.
2 = η2 Q
.
M = η2 (1 – η1) Q
.
H
ηTH = η1 + η2 (1 – η1) = 0.333 + 0.5(1 – 0.333) = 0.667
Comment: It matches a single heat engine ηTH = 1 –
TL
TH
= 1 –
300
900 =
2
3

7.48
At a few places where the air is very cold in the winter, like –30oC it is possible
to find a temperature of 13oC down below ground. What efficiency will a heat
engine have operating between these two thermal reservoirs?
Solution:
ηTH = 1 –
TL
TH
The ground becomes the hot source and
the atmosphere becomes the cold side of
the heat engine
ηTH= 1 –
273 – 30
273 + 13 = 1 –
243
286 = 0.15
This is low because of the modest
temperature difference.

7.49
Find the maximum coefficient of performance for the refrigerator in your kitchen,
assuming it runs in a Carnot cycle.
Solution:
The refrigerator coefficient of performance is
β = QL/W = QL/(QH - QL) = TL/(TH - TL)
Assuming TL ~ 0°C, TH ~ 35°C,
β ≤
273.15
35 - 0 = 7.8
Actual working fluid temperatures must be such that
TL < Trefrigerator and TH > Troom
A refrigerator does not operate in a
Carnot cycle. The actual vapor
compression cycle is examined in
Chapter 11.

7.50
A refrigerator should remove 500 kJ from some food. Assume the refrigerator
works in a Carnot cycle between –10oC and 45oC with a motor-compressor of
500 W. How much time does it take if this is the only cooling load?
Assume Carnot cycle refrigerator
β =
Q
.
L
W
.
= Q
.
L / (Q
.
H - Q
.
L ) ≅
TL
TH - TL
=
273 - 10
45 - (-10) = 4.785
This gives the relation between the low T heat transfer and the work as
Q
.
L =
Q
t = 4.785 W
.
t =
Q
β W
.
=
500 × 1000
4.785 × 500
= 209 s

7.51
A car engine burns 5 kg fuel (equivalent to addition of QH) at 1500 K and rejects
energy to the radiator and the exhaust at an average temperature of 750 K. If the
fuel provides 40 000 kJ/kg what is the maximum amount of work the engine can
provide?
Solution:
A heat engine QH = m qfuel = 5 × 40 000 = 200 000 kJ
Assume a Carnot efficiency (maximum theoretical work)
η = 1 −
TL
TH
= 1 −
750
1500 = 0.5
W = η QH = 100 000 kJ
Exhaust flow
Air intake filter
Coolant flow
Atm.
airShaft
Fan
power
Radiator

7.52
A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat
at 150°C. What is the minimum amount of work (power) input that will drive
this?
For the minimum work we assume a Carnot heat pump and Q
.
L = 5 MW.
βHP =
Q
.
H
W
.
in
=
TH
TH - TL
=
273.15 + 150
150 - 85
= 6.51
βREF = βHP - 1 =
Q
.
L
W
.
in
= 5.51
Now we can solve for the work
W
.
in = Q
.
L/βREF = 5/5.51 = 0.907 MW

7.53
An air-conditioner provides 1 kg/s of air at 15°C cooled from outside atmospheric
air at 35°C. Estimate the amount of power needed to operate the air-conditioner.
Clearly state all assumptions made.
Solution:
Consider the cooling of air which needs a heat transfer as
Q
.
air = m
.
∆h ≅ m
.
Cp∆T = 1 kg/s × 1.004 kJ/kg K × 20 K = 20 kW
Assume Carnot cycle refrigerator
β =
Q
.
L
W
.
= Q
.
L / (Q
.
H - Q
.
L ) ≅
TL
TH - TL
=
273 + 15
35 - 15 = 14.4
W
.
= Q
.
L / β =
20.0
14.4 = 1.39 kW
This estimate is the theoretical maximum
performance. To do the required heat
transfer TL ≅ 5°C and TH = 45°C are
more likely; secondly
β < βcarnot
HQ
W
L
Q
REF
35 C 15 C
cb oo

7.54
A cyclic machine, shown in Fig. P7.54, receives 325 kJ from a 1000 K energy
reservoir. It rejects 125 kJ to a 400 K energy reservoir and the cycle produces 200
kJ of work as output. Is this cycle reversible, irreversible, or impossible?
Solution:
ηCarnot = 1 −
TL
TH
= 1 −
400
1000 = 0.6
ηeng =
W
QH
=
200
325 = 0.615 > ηCarnot
This is impossible.
HQ = 325 kJ
W = 200 kJ
LQ = 125 kJ
T = 1000 KH
HE
cb
T = 400 KL

7.55
A sales person selling refrigerators and deep freezers will guarantee a minimum
coefficient of performance of 4.5 year round. How would you evaluate that? Are
they all the same?
Solution:
Assume a high temperature of 35°C. If a freezer compartment is included
TL ~ -20°C (deep freezer) and fluid temperature is then TL ~ -30°C
βdeep freezer ≤
TL
TH - TL
=
273.15 - 30
35 - (-30) = 3.74
A hot summer day may require a higher TH to push QH out into the room, so
even lower β.
Claim is possible for a refrigerator, but not for a deep freezer.

7.56
A temperature of about 0.01 K can be achieved by magnetic cooling. In this
process a strong magnetic field is imposed on a paramagnetic salt, maintained at 1
K by transfer of energy to liquid helium boiling at low pressure. The salt is then
thermally isolated from the helium, the magnetic field is removed, and the salt
temperature drops. Assume that 1 mJ is removed at an average temperature of 0.1
K to the helium by a Carnot-cycle heat pump. Find the work input to the heat
pump and the coefficient of performance with an ambient at 300 K.
Solution:
β = Q
.
L/W
.
IN =
TL
TH - TL
=
0.1
299.9 = 0.00033
W
.
IN =
1×10-3
0.00033 = 3 J
Remark: This is an extremely large temperature difference for a heat pump.

7-57
The lowest temperature that has been achieved is about 1 × 10−6
K. To achieve
this an additional stage of cooling is required beyond that described in the
previous problem, namely nuclear cooling. This process is similar to magnetic
cooling, but it involves the magnetic moment associated with the nucleus rather
than that associated with certain ions in the paramagnetic salt. Suppose that 10 µJ
is to be removed from a specimen at an average temperature of 10−5
K (ten micro-
joules is about the potential energy loss of a pin dropping 3 mm). Find the work
input to a Carnot heat pump and its coefficient of performance to do this
assuming the ambient is at 300 K.
Solution:
QL = 10 µJ = 10×10-6 J at TL = 10-5 K
⇒ QH = QL ×
TH
TL
= 10×10-6 ×
300
10-5 = 300 J
Win = QH - QL = 300 - 10×10-6 ≅ 300 J
β =
QL
Win
=
10×10-6
300 = 3.33×10-8

7.58
An inventor has developed a refrigeration unit that maintains the cold space at
−10°C, while operating in a 25°C room. A coefficient of performance of 8.5 is
claimed. How do you evaluate this?
Solution:
βCarnot =
QL
Win
=
TL
TH - TL
=
263.15
25 - (-10) = 7.52
8.5 > βCarnot ⇒ impossible claim
HQ
W
LQ
T = -10CL
T = 25C
H
REF

7.59
Calculate the amount of work input a refrigerator needs to make ice cubes out of a
tray of 0.25 kg liquid water at 10oC. Assume the refrigerator works in a Carnot
cycle between –8oC and 35oC with a motor-compressor of 750 W. How much
time does it take if this is the only cooling load?
Solution:
C.V. Water in tray. We neglect tray mass.
Energy Eq.: m(u2 − u1) = 1Q2 − 1W2
Process : P = constant + Po
1W2 = ∫ P dV = Pom(v2 − v1)
1Q2 = m(u2 − u1) + 1W2 = m(h2 − h1)
Tbl. B.1.1 : h1 = 41.99 kJ/kg, Tbl. B.1.5 : h2 = - 333.6 kJ/kg
1Q2 = 0.25(-333.4 – 41.99 ) = - 93.848 kJ
Consider now refrigerator
β =
QL
W =
QL
QH - QL
=
TL
TH - TL
=
273 - 8
35 - (-8) = 6.16
W =
QL
β
= -
1Q2
β
=
93.848
6.16 = 15.24 kJ
For the motor to transfer that amount of energy the time is found as
W = ∫ W
.
dt = W
.
∆t
∆t =
W
W
.
=
15.24 ×1000
750 = 20.3 s
Comment: We neglected a baseload of the refrigerator so not all the 750 W are
available to make ice, also our coefficient of performance is very optimistic and
finally the heat transfer is a transient process. All this means that it will take much
more time to make ice-cubes.

7.60
A heat pump receives energy from a source at 80oC and delivers energy to a
boiler that operates at 350 kPa. The boiler input is saturated liquid water and the
exit is saturated vapor both at 350 kPa. The heat pump is driven by a 2.5 MW
motor and has a COP that is 60% of a Carnot heat pump COP. What is the
maximum mass flow rate of water the system can deliver?
TH = Tsat = 138.88oC = 412 K, hfg = 2148.1 kJ/kg
βHP Carnot =
Q
.
H
W
.
in
=
TH
TH - TL
=
412
138.88 - 80
= 7
βHP ac = 0.6 × 7 = 4.2 = Q
.
H/W
.
in
Q
.
H = 4.2 W
.
in = 4.2 × 2.5 MW = 10.5 MW = m
.
hfg
m
.
= Q
.
H / hfg = 10 500 kW / 2148.1 kJ/kg = 4.89 kg/s

7.61
A household freezer operates in a room at 20°C. Heat must be transferred from
the cold space at a rate of 2 kW to maintain its temperature at −30°C. What is the
theoretically smallest (power) motor required to operate this freezer?
Solution:
Assume a Carnot cycle between TL = -30°C and
TH = 20°C:
β =
Q
.
L
W
.
in
=
TL
TH - TL
=
273.15 - 30
20 - (-30)
= 4.86
W
.
in = Q
.
L/β = 2/4.86 = 0.41 kW
This is the theoretical minimum power input.
Any actual machine requires a larger input.
HQ
W
LQ
TL
Tamb
REF
2 kW

7.62
We propose to heat a house in the winter with a heat pump. The house is to be
maintained at 20°C at all times. When the ambient temperature outside drops to
−10°C, the rate at which heat is lost from the house is estimated to be 25 kW.
What is the minimum electrical power required to drive the heat pump?
Solution:
Minimum power if we
assume a Carnot cycle
Q
.
H = Q
.
leak = 25 kW
QleakQ QHL
W
HP
β′ =
Q
.
H
W
.
IN
=
TH
TH-TL
=
293.2
30 = 9.773 ⇒ W
.
IN =
25
9.773 = 2.56 kW

7.63
A certain solar-energy collector produces a maximum temperature of 100°C. The
energy is used in a cyclic heat engine that operates in a 10°C environment. What
is the maximum thermal efficiency? What is it, if the collector is redesigned to
focus the incoming light to produce a maximum temperature of 300°C?
Solution:
For TH = 100°C = 373.2 K & TL = 283.2 K
ηth max =
TH - TL
TH
=
90
373.2 = 0.241
For TH = 300°C = 573.2 K & TL = 283.2 K
ηth max =
TH - TL
TH
=
290
573.2 = 0.506

7.64
Helium has the lowest normal boiling point of any of the elements at 4.2 K. At
this temperature the enthalpy of evaporation is 83.3 kJ/kmol. A Carnot
refrigeration cycle is analyzed for the production of 1 kmol of liquid helium at 4.2
K from saturated vapor at the same temperature. What is the work input to the
refrigerator and the coefficient of performance for the cycle with an ambient at
300 K?
Solution:
For the Carnot cycle the ratio of the heat transfers is the ratio of temperatures
QL = n h
_
fg = 1 kmol × 83.3 kJ/kmol = 83.3 kJ
QH = QL ×
TH
TL
= 83.3 ×
300
4.2 = 5950 kJ
WIN = QH - QL = 5950 - 83.3 = 5886.7 kJ
β =
QL
WIN
=
83.3
5886.7 = 0.0142 [ =
TL
TH - TL
]

7.65
A thermal storage is made with a rock (granite) bed of 2 m3 which is heated to
400 K using solar energy. A heat engine receives a QH from the bed and rejects
heat to the ambient at 290 K. The rock bed therefore cools down and as it reaches
290 K the process stops. Find the energy the rock bed can give out. What is the
heat engine efficiency at the beginning of the process and what is it at the end of
the process?
Solution:
Assume the whole setup is reversible and that the heat engine operates in a
Carnot cycle. The total change in the energy of the rock bed is
u2 - u1 = q = C ∆T = 0.89 (400 - 290) = 97.9 kJ/kg
m = ρV = 2750 × 2 = 5500 kg , Q = mq = 5500 × 97.9 = 538 450 kJ
To get the efficiency use the CARNOT cycle result as
η = 1 - To/TH = 1 - 290/400 = 0.275 at the beginning of process
η = 1 - To/TH = 1 - 290/290 = 0.0 at the end of process
W
HE
Q QH L

7.66
In a cryogenic experiment you need to keep a container at −125°C although it
gains 100 W due to heat transfer. What is the smallest motor you would need for a
heat pump absorbing heat from the container and rejecting heat to the room at
20°C?
Solution:
We do not know the actual device so find the work for a Carnot cycle
βREF = Q
.
L / W
.
=
TL
TH - TL
=
148.15
20 - (-125) = 1.022
=> W
.
= Q
.
L/ βREF = 100/1.022 = 97.8 W

7.67
It is proposed to build a 1000-MW electric power plant with steam as the working
fluid. The condensers are to be cooled with river water (see Fig. P7.67). The
maximum steam temperature is 550°C, and the pressure in the condensers will be
10 kPa. Estimate the temperature rise of the river downstream from the power
plant.
Solution:
W
.
NET = 106 kW, TH = 550°C = 823.3 K
PCOND = 10 kPa → TL = TG (P = 10 kPa) = 45.8°C = 319 K
ηTH CARNOT =
TH - TL
TH
=
823.2 - 319
823.2 = 0.6125
⇒ Q
.
L MIN= 106





1 - 0.6125
0.6125 = 0.6327 × 106 kW
But m
.
H2O =
60 × 8 × 10/60
0.001 = 80 000 kg/s having an energy flow of
Q
.
L MIN = m
.
H2O ∆h = m
.
H2O CP LIQ H2O ∆TH2O MIN
⇒ ∆TH2O MIN =
Q
.
L MIN
m
.
H2OCP LIQ H2O
=
0.6327×106
80000 × 4.184
= 1.9°C

7.68
Repeat the previous problem using a more realistic thermal efficiency of 35%.
W
.
NET = 106 kW = ηTH ac Q
.
H, ηTH ac = 0.35
⇒ Q
.
L = Q
.
H - W
.
NET = W
.
NET /ηTH ac - W
.
NET = W
.
NET(1/ηTH ac – 1)
= 106 kW





1 - 0.35
0.35 = 1.857 × 106 kW
But m
.
H2O =
60 × 8 × 10/60
0.001 = 80 000 kg/s having an energy flow of
Q
.
L = m
.
H2O ∆h = m
.
H2O CP LIQ H2O ∆TH2O
⇒ ∆TH2O =
Q
.
L
m
.
H2OCP LIQ H2O
=
1.857 × 106
80 000 × 4.18
= 5.6°C

7.69
A steel bottle V = 0.1 m3 contains R-134a at 20°C, 200 kPa. It is placed in a deep
freezer where it is cooled to -20°C. The deep freezer sits in a room with ambient
temperature of 20°C and has an inside temperature of -20°C. Find the amount of
energy the freezer must remove from the R-134a and the extra amount of work
input to the freezer to do the process.
Solution:
C.V. R-134a out to the -20 °C space.
Energy equation: m(u2 − u1) = 1Q2 − 1W2
Process : V = Const => v2 = v1 => 1W2 = 0
Table B.5.2: v1 = 0.11436 m3/kg, u1 = 395.27 kJ/kg
m = V/ v1 = 0.87443 kg
State 2: v2 = v1 < vg = 0.14649 Table B.5.1 => 2 phase
=> x2 =
v - vf
vfg
=
0.11436 - 0.000738
0.14576 = 0.77957
u2 = 173.65 + 0.77957 × 192.85 = 323.99 kJ/kg
1Q2 = m(u2 − u1) = - 62.334 kJ
Consider the freezer and assume Carnot cycle
β =
QL
W =
QL
QH - QL
=
TL
TH - TL
=
273 - 20
20 - (-20) = 6.33
Win = QL / β = 62.334 / 6.33 = 9.85 kJ
-20 Co
R
134a

7.70
Sixty kilograms per hour of water runs through a heat exchanger, entering as
saturated liquid at 200 kPa and leaving as saturated vapor. The heat is supplied by
a Carnot heat pump operating from a low-temperature reservoir at 16°C. Find the
rate of work into the heat pump.
Solution:
C.V. Heat exchanger
m
.
1 = m
.
2 ; m
.
1h1 + Q
.
H = m
.
1h2
Table B.1.2: h1 = 504.7 kJ/kg,
h2 = 2706.7 kJ/kg
TH = Tsat(P) = 120.93 +273.15
= 394.08 K
Q
.
H =
60
3600(2706.7 - 504.7) = 36.7 kW
HQ
W
L
Q
TL
HP
1 2
cb
Assume a Carnot heat pump.
β′ = Q
.
H/W
.
= TH / (TH − TL) = 394.08 / 104.93 = 3.76
W
.
= Q
.
H/β′ = 36.7/3.76 = 9.76 kW

7.71
A heat engine has a solar collector receiving 0.2 kW per square meter inside
which a transfer media is heated to 450 K. The collected energy powers a heat
engine which rejects heat at 40oC. If the heat engine should deliver 2.5 kW what
is the minimum size (area) solar collector?
Solution:
TH = 450 K TL = 40oC = 313.15 K
ηHE = 1 −
TL
TH
= 1 -
313.15
450 = 0.304
W
.
= η Q
.
H => Q
.
H =
W
.
η
=
2.5
0.304 = 8.224 kW
Q
.
H = 0.2 A => A =
Q
.
H
0.2 = 41 m2

7.72
Liquid sodium leaves a nuclear reactor at 800°C and is used as the energy source
in a steam power plant. The condenser cooling water comes from a cooling tower
at 15°C. Determine the maximum thermal efficiency of the power plant. Is it
misleading to use the temperatures given to calculate this value?
Solution:
LIQ Na
800oC
REACTOR
ENERGY
TO H O2
COND.
COOLING
TOWER
ENERGY
FROM
STEAM
POWER
PLANT
TH = 800°C = 1073.2 K, TL = 15°C = 288.2 K
ηTH MAX =
TH - TL
TH
=
1073.2 - 288.2
1073.2 = 0.731
It might be misleading to use 800°C as the value for TH, since there is not a
supply of energy available at a constant temperature of 800°C (liquid Na is
cooled to a lower temperature in the heat exchanger).
⇒ The Na cannot be used to boil H2O at 800°C.
Similarly, the H2O leaves the cooling tower and enters the condenser at 15°C,
and leaves the condenser at some higher temperature.
⇒ The water does not provide for condensing steam at a
constant temperature of 15°C.

7.73
A power plant with a thermal efficiency of 40% is located on a river similar to
Fig. P7.67. With a total river mass flow rate of 1 × 105 kg/s at 15oC find the
maximum power production allowed if the river water should not be heated more
than 1 degree.
The maximum heating allowed determines the maximum Q
.
L as
Q
.
L = m
.
H2O ∆h = m
.
H2O CP LIQ H2O ∆TH2O
= 1 × 105 kg/s × 4.18 kJ/kg-K × 1 K = 418 MW
= W
.
NET(1/ηTH ac – 1)
W
.
NET = Q
.
L / (1/ηTH ac – 1) = Q
.
L
ηTH ac
1 - ηTH ac
= 418 MW ×
0.4
1 - 0.4 = 279 MW

7.74
A heat pump is driven by the work output of a heat engine as shown in figure
P7.74. If we assume ideal devices find the ratio of the total power Q
.
L1 + Q
.
H2 that
heats the house to the power from the hot energy source Q
.
H1 in terms of the
temperatures.
βHP = Q
.
H2/W
.
= Q
.
H2/(Q
.
H2- Q
.
L2) =
Troom
Troom-Tamb
W
.
= ηHE . Q
.
H1 = (1-
Troom
TH
) Q
.
H1
W
.
= Q
.
H2/βHP =
Troom
Troom-Tamb
Q
.
H2
Q
.
L1= Q
.
H1- W
.
= [1-1 +
Troom
TH
] Q
.
H1
Q
.
H2 + Q
.
L1
Q
.
H1
= 1-1 +
Troom
TH
+
1-
Troom
TH
Troom-Tamb
Troom
=
Troom
TH
+
Troom- T2
room/TH
Troom-Tamb
= Troom [
1
TH
+
1 -
Troom
TH
Troom - Tamb
] =
Troom
TH
[1 +
TH - Troom
Troom - Tamb
]
=
Troom
TH
[
TH − Tamb
Troom − Tamb
]
W
L1Q
TH
H1Q
House Troom
HE
L2Q
H.P.
H2Q
Tamb

7.75
A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a
refrigerator) besides powering the car and other auxiliary equipment. On a hot
(35oC) summer day the A/C takes outside air in and cools it to 5oC sending it into
a duct using 2 kW of power input and it is assumed to be half as good as a Carnot
refrigeration unit. Find the rate of fuel (kW) being burned extra just to drive the
A/C unit and its COP. Find the flow rate of cold air the A/C unit can provide.
W
.
extra = η Q
.
H extra Q
.
H extra = W
.
extra / η = 2 kW / 0.33 = 6 kW
β =
QL
WIN
= 0.5 βCarnot = 0.5
TL
TH - TL
= 0.5
5 + 273.15
35 - 5 = 4.636
Q
.
L = β W
.
= 4.636 × 2 kW = 9.272 kW = m
.
air CP air ∆Tair
m
.
air = Q
.
L / [CP air ∆Tair ] =
9.272 kW
1.004 kJ/kg-K × (35 - 5) K
= 0.308 kg/s

7.76
Two different fuels can be used in a heat engine operating between the fuel
burning temperature and a low temperature of 350 K. Fuel A burns at 2200 K
delivering 30 000 kJ/kg and costs $1.50/kg. Fuel B burns at 1200 K, delivering
40 000 kJ/kg and costs $1.30/kg. Which fuel will you buy and why?
Solution:
Fuel A: ηTH,A = 1 −
TL
TH
= 1 -
350
2200 = 0.84
WA = ηTH,A × QA = 0.84 × 30 000 = 25 200 kJ/kg
WA/$A = 25 200/1.5 = 16 800 kJ/$
Fuel B: ηTH,B = 1 −
TL
TH
= 1 -
350
1200 = 0.708
WB = ηTH,B × QB = 0.708 × 40 000 = 28 320 kJ/kg
WB/$B = 28 320/1.3 = 21 785 kJ/$
Select fuel B for more work per dollar though it has a lower thermal efficiency.

7.77
A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat
at 150°C. Suppose the actual heat pump has a COP of 2.5 how much power is
required to drive the unit. For the same COP how high a high temperature would
a Carnot heat pump have assuming the same low T?
This is an actual COP for the heat pump as
βHP = COP = Q
.
H/W
.
in = 2.5 ⇒ Q
.
L/W
.
in = 1.5
W
.
in = Q
.
L/ 1.5 = 5 / 1.5 = 3.333 MW
The Carnot heat pump has a COP given by the temperatures as
βHP = Q
.
H/W
.
in =
TH
TH - TL
= 2.5 ⇒ TH = 2.5 TH – 2.5 TL
⇒ TH =
2.5
1.5 TL =
5
3 (85 + 273.15) = 597 K

Finite ∆T Heat Transfer

7.78
The ocean near Havaii has 20oC near the surface and 5oC at some depth. A power
plant based on this temperature difference is being planned. How large an
efficiency could it have? If the two heat transfer terms (QH and QL) both require a
2 degree difference to operate what is the maximum efficiency then?
Solution:
TH = 20°C = 293.2 K; TL = 5°C = 278.2 K
ηTH MAX =
TH - TL
TH
=
293.2 - 278.2
293.2 = 0.051
ηTH mod =
TH' - TL'
TH'
=
291.2 - 280.2
291.2 = 0.038
This is a very low efficiency so it has to be done on a very large scale to be
economically feasible and then it will have some environmental impact.

7.79
A refrigerator keeping 5oC inside is located in a 30oC room. It must have a high
temperature ∆T above room temperature and a low temperature ∆T below the
refrigerated space in the cycle to actually transfer the heat. For a ∆T of 0, 5 and
10oC respectively calculate the COP assuming a Carnot cycle.
Solution:
From the definition of COP and assuming Carnot cycle
β =
QL
WIN
=
TL
TH - TL
when T’s are absolute temperatures
∆T TH TH TL TL β
oC K oC K
a 0 30 303 5 278 11.1
b 5 35 308 0 273 7.8
c 10 40 313 -5 268 5.96
Notice how the COP drops significantly with the increase in ∆T.

7.80
A house is heated by a heat pump driven by an electric motor using the outside as
the low-temperature reservoir. The house loses energy directly proportional to the
temperature difference as Q
.
loss = K(TH - TL). Determine the minimum electric
power to drive the heat pump as a function of the two temperatures.
Solution:
Heat pump COP: β′ = Q
.
H/W
.
in ≤ TH/(TH - TL) ;
Heat loss must be added: Q
.
H= Q
.
loss = K(TH - TL)
Solve for required work and substitute in for β′
W
.
in = Q
.
H/β′ ≥ K(TH - TL) × (TH - TL)/TH
W
.
in ≥ K(TH - TL)2/TH
Q = K(T - T )lossQ QHL
W
HP
cb
H L

7.81
A house is heated by an electric heat pump using the outside as the low-
temperature reservoir. For several different winter outdoor temperatures, estimate
the percent savings in electricity if the house is kept at 20°C instead of 24°C.
Assume that the house is losing energy to the outside as in Eq. 7.14.
Solution:
Heat Pump Q
.
loss ∝ (TH - TL)
Max
Perf.
Q
.
H
W
.
IN
=
TH
TH - TL
=
K(TH - TL)
W
.
IN
, W
.
IN =
K(TH - TL)2
TH
A: THA
= 24°C = 297.2 K B: THB
= 20°C = 293.2 K
TL,°C W
.
INA
/K W
.
INB
/K % saving
-20 6.514 5.457 16.2 %
-10 3.890 3.070 21.1 %
0 1.938 1.364 29.6 %
10 0.659 0.341 48.3 %

7.82
A car engine operates with a thermal efficiency of 35%. Assume the air-
conditioner has a coefficient of performance of β = 3 working as a refrigerator
cooling the inside using engine shaft work to drive it. How much fuel energy
should be spend extra to remove 1 kJ from the inside?
Solution:
Car engine: W = ηeng Qfuel
Air conditioner: β =
QL
W
W = ηeng Qfuel =
QL
β
Qfuel = QL / (ηeng β) =
1
0.35 × 3
= 0.952 kJ
W
LQ
TH
HQ
TL
REF
FuelQ
H.E.
L engQ
FUEL

7.83
A refrigerator uses a power input of 2.5 kW to cool a 5°C space with the high
temperature in the cycle as 50°C. The QH is pushed to the ambient air at 35°C in a
heat exchanger where the transfer coefficient is 50 W/m2K. Find the required
minimum heat transfer area.
Solution:
W
.
= 2.5 kW = Q
.
H / βHP
Q
.
H = W
.
× βHP = 2.5 × [323 / (50 - 5)] = 17.95 kW = h A ∆T
A =
Q
.
H
h ∆T
=
17.95 × 103
50 × 15
= 23.9 m2
T
H
Q
W
TL
LQ
amb REF
= 2.5 kW
35 C 50 C

7.84
A heat pump has a coefficient of performance that is 50% of the theoretical
maximum. It maintains a house at 20°C, which leaks energy of 0.6 kW per degree
temperature difference to the ambient. For a maximum of 1.0 kW power input
find the minimum outside temperature for which the heat pump is a sufficient heat
source.
Solution:
QleakQ QHL
W = 1 kW
HP
cb
= 0.6 (TH - TL )
C.V. House. For constant 20°C the heat pump must provide Q
.
leak = 0.6 ∆T
Q
.
H = Q
.
leak = 0.6 (TH - TL ) = β′ W
.
C.V. Heat pump. Definition of the coefficient of performance and the fact that
the maximum is for a Carnot heat pump.
β′ =
Q
.
H
W
.
=
Q
.
H
Q
.
H - Q
.
L
= 0.5 β′Carnot = 0.5 ×
TH
TH - TL
Substitute into the first equation to get
0.6 (TH - TL ) = [ 0.5 × TH / (TH - TL ) ] 1 =>
(TH - TL )2 = (0.5 / 0.6) TH × 1 = 0.5 / 0.6 × 293.15 = 244.29
TH - TL = 15.63 => TL = 20 - 15.63 = 4.4 °C

7.85
Consider a room at 20oC that is cooled by an air conditioner with a COP of 3.2
using a power input of 2 kW and the outside is at 35oC. What is the constant in
the heat transfer Eq. 7.14 for the heat transfer from the outside into the room?
Q
.
L = βACW
.
= 3.2 × 2 kW = 6.4 kW = Q
.
leak in = CA ∆T
CA = Q
.
L / ∆T =
6.4 kW
(35 – 20) K = 0.427 kW/K
QleakQ QH L
W = 2 kW
AC
cb
T
L
Here:
TL = Thouse = 20oC
TH = Tamb = 35oC

7.86
A farmer runs a heat pump with a motor of 2 kW. It should keep a chicken
hatchery at 30oC which loses energy at a rate of 0.5 kW per degree difference to
the colder ambient. The heat pump has a coefficient of performance that is 50% of
a Carnot heat pump. What is the minimum ambient temperature for which the
heat pump is sufficient?
Solution:
C.V. Hatchery, steady state.
To have steady state at 30oC for the hatchery
Energy Eq.: Q
.
H= Q
.
Loss = β ACW
.
Process Eq.: Q
.
Loss= 0.5 (TH –Tamb); β AC = ½ βCARNOT
COP for the reference Carnot heat pump
β CARNOT=
Q
.
H
W
.
=
Q
.
H
Q
.
H - Q
.
L
=
TH
TH - TL
=
TH
TH - Tamb
Substitute the process equations and this β CARNOT into the energy Eq.
0.5 (TH –Tamb) = ½
TH
TH - Tamb
W
.
(TH –Tamb)2 = ½ THW
.
/0.5 = THW
.
= (273 + 30) × 2 = 606 K2
TH – Tamb= 24.62 K
Tamb= 30 – 24.62 = 5.38oC
Comment: That of course is not a very low temperature and the size of the system
is not adequate for most locations.
QleakQ QHL
W = 2 kW
HP
cb

7.87
An air conditioner cools a house at TL = 20°C with a maximum of 1.2 kW power
input. The house gains 0.6 kW per degree temperature difference to the ambient
and the refrigeration COP is β = 0.6 βCarnot. Find the maximum outside
temperature, TH, for which the air conditioner provides sufficient cooling.
Solution:
QleakQ QH L
W = 1.2 kW
HP
cb
T
L
Here:
TL = Thouse
TH = Tamb
In this setup the low temperature space is the house and the high
temperature space is the ambient. The heat pump must remove the gain or
leak heat transfer to keep it at a constant temperature.
Q
.
leak = 0.6 (Tamb - Thouse) = Q
.
L which must be removed by the heat pump.
β = Q
.
L / W
.
= 0.6 βcarnot = 0.6 Thouse / (Tamb - Thouse )
Substitute in for Q
.
L and multiply with (Tamb - Thouse)W
.
:
0.6 (Tamb - Thouse )2 = 0.6 Thouse W
.
Since Thouse = 293.15 K and W
.
= 1.2 kW it follows
(Tamb - Thouse )2 = Thouse W
.
= 293.15 × 1.2 = 351.78 K2
Solving ⇒ (Tamb - Thouse ) = 18.76 ⇒ Tamb = 311.9 K = 38.8 °C

7.88
A house is cooled by an electric heat pump using the outside as the high-
temperature reservoir. For several different summer outdoor temperatures,
estimate the percent savings in electricity if the house is kept at 25°C instead of
20°C. Assume that the house is gaining energy from the outside directly
proportional to the temperature difference as in Eq. 7.14.
Solution:
Air-conditioner (Refrigerator) Q
.
LEAK ∝ (TH - TL)
Max
Perf.
Q
.
L
W
.
IN
=
TL
TH - TL
=
K(TH - TL)
W
.
IN
, W
.
IN =
K(TH - TL)2
TL
A: TLA
= 20°C = 293.2 K B: TLB
= 25°C = 298.2 K
TH,°C W
.
INA
/K W
.
INB
/K % saving
45 2.132 1.341 37.1 %
40 1.364 0.755 44.6 %
35 0.767 0.335 56.3 %

7.89
A Carnot heat engine, shown in Fig. P7.89, receives energy from a reservoir at
Tres through a heat exchanger where the heat transferred is proportional to the
temperature difference as Q
.
H = K(Tres
- TH
). It rejects heat at a given low
temperature TL. To design the heat engine for maximum work output show that
the high temperature, TH
, in the cycle should be selected as TH = Tres
TL
Solution:
W = ηTH
QH
=
TH
- TL
TH
× K(Tres
− TH
) ; maximize W(TH
) ⇒
δW
δTH
= 0
δW
δTH
= K(Tres
− TH
)TL
TH
-2
− K(1 − TL
/TH
) = 0
⇒ TH
= Tres
TL

7.90
Consider a Carnot cycle heat engine operating in outer space. Heat can be rejected
from this engine only by thermal radiation, which is proportional to the radiator
area and the fourth power of absolute temperature, Q
.
rad ~ KAT4. Show that for a
given engine work output and given TH, the radiator area will be minimum when
the ratio TL/TH = 3/4.
Solution:
W
.
NET = Q
.
H





TH - TL
TH
= Q
.
L





TH - TL
TL
; also Q
.
L = KAT
4
L
W
.
NET
KT
4
H
=
AT
4
L
T
4
H 




TH
TL
- 1 = A













TL
TH
3
-





TL
TH
4
= const
Differentiating,
dA













TL
TH
3
-





TL
TH
4
+ A








3





TL
TH
2
- 4





TL
TH
3
d





TL
TH
= 0
dA
d(TL/TH) = - A








3





TL
TH
2
- 4





TL
TH
3
/[




TL
TH
3
-





TL
TH
4
]= 0
TL
TH
=
3
4 for min. A
Check that it is minimum and
not maximum with the 2nd
derivative > 0.

7.91
On a cold (–10oC) winter day a heat pump provides 20 kW to heat a house
maintained at 20oC and it has a COPHP of 4. How much power does the heat
pump require? The next day a winter storm brings the outside to -15oC, assuming
the same COP and the same house heat transfer coefficient for the heat loss to the
outside air. How much power does the heat pump require then?
If we look at the heat loss for the house we have
Q
.
loss = 20 kW = CA ∆T ⇒ CA =
20 kW
20 - (-10) K = 0.667 kW/K
So now with the new outdoor temperature we get
Q
.
loss = CA ∆T = 0.667 kW/K × [20 – (–15)] K = 23.3 kW
Q
.
loss = Q
.
H = COP W
.
⇒ W
.
= Q
.
loss /COP =
23.3 kW
4 = 5.83 kW

Ideal Gas Carnot Cycles

7.92
Hydrogen gas is used in a Carnot cycle having an efficiency of 60% with a low
temperature of 300 K. During the heat rejection the pressure changes from 90 kPa
to 120 kPa. Find the high and low temperature heat transfer and the net cycle
work per unit mass of hydrogen.
Solution:
As the efficiency is known, the high temperature is found as
η = 0.6 = 1 −
TL
TH
= > TH = TL /(1 - 0.6) = 750 K
Now the volume ratio needed for the heat transfer, T3 = T4 = TL, is
v3 / v4 = ( RT3 / P3 ) / ( RT4 / P4 ) = P4 / P3 = 120 / 90 = 1.333
so from Eq.7.9 we have with R = 4.1243 from Table A.5
qL = RTL ln (v3/v4 ) = 355.95 kJ/kg
Using the efficiency from Eq.7.4 then
qH = qL / (1 - 0.6) = 889.9 kJ/kg
The net work equals the net heat transfer
w = qH - qL = 533.9 kJ/kg

7.93
Carbon dioxide is used in an ideal gas refrigeration cycle, reverse of Fig. 7.24.
Heat absorption is at 250 K and heat rejection is at 325 K where the pressure
changes from 1200 kPa to 2400 kPa. Find the refrigeration COP and the specific
heat transfer at the low temperature.
The analysis is the same as for the heat engine except the signs are opposite so the
heat transfers move in the opposite direction.
β = Q
.
L / W
.
= βcarnot = TL / (TH - TL ) =
250
325 − 250
= 3.33
qH = RTH ln(v2/v1) = RTH ln (
P1
P2
) = 0.1889 × 325 ln(
2400
1200) = 42.55
kJ/kg
qL = qH TL / TH = 42.55 × 250 / 325 = 32.73 kJ/kg

7.94
An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of
1200 K and a heat rejection at 400 K. During the heat addition the volume triples.
Find the two specific heat transfers (q) in the cycle and the overall cycle
efficiency.
Solution:
The P-v diagram of the cycle is
shown to the right.
From the integration along the
process curves done in the main
text we have Eq.7.7
qH = R TH ln(v2/v1)
= 0.287 × 1200 ln(3)
= 378.4 kJ/kg
P
v
1
2
3
4
1200 K
400 K
Since it is a Carnot cycle the knowledge of the temperatures gives the cycle
efficiency as
ηTH
= 1 −
TL
TH
= 1 -
400
1200 = 0.667
from which we can get the other heat transfer from Eq.7.4
qL = qH TL / TH = 378.4 400 / 1200 = 126.1 kJ/kg

7.95
Air in a piston/cylinder goes through a Carnot cycle with the P-v diagram shown
in Fig. 7.24. The high and low temperatures are 600 K and 300 K respectively.
The heat added at the high temperature is 250 kJ/kg and the lowest pressure in the
cycle is 75 kPa. Find the specific volume and pressure after heat rejection and the
net work per unit mass.
Solution:
qH = 250 kJ/kg , TH = 600 K, TL = 300 K, P3 = 75 kPa
The states as shown in figure 7.21
1: 600 K , 2: 600 K, 3: 75 kPa, 300 K 4: 300 K
Since this is a Carnot cycle and we know the temperatures the efficiency is
η = 1 −
TL
TH
= 1 -
300
600 = 0.5
and the net work becomes
wNET = ηqH = 0.5 × 250
= 125 kJ/kg
The heat rejected is
P
v
1
2
3
4
600 K
300 K
qL = qH – wNET = 125 kJ/kg
After heat rejection is state 4. From equation 7.9
3→4 Eq.7.9 : qL = RTL ln (v3/v4)
v3 = RT3 / P3 = 0.287 × 300 / 75 = 1.148 m3/kg
v4 = v3 exp(-qL/RTL) = 1.148 exp(−125/0.287 × 300) = 0.2688 m3/kg
P4 = RT4 / v4 = 0.287 × 300 / 0.2688 = 320 kPa

7.96
At certain locations geothermal energy in undergound water is available and used
as the energy source for a power plant. Consider a supply of saturated liquid
water at 150°C. What is the maximum possible thermal efficiency of a cyclic heat
engine using this source of energy with the ambient at 20°C? Would it be better to
locate a source of saturated vapor at 150°C than use the saturated liquid at 150°C?
Solution:
TMAX = 150°C = 423.2 K = TH ; TMin = 20°C = 293.2 K = TL
ηTH MAX =
TH - TL
TH
=
130
423.2 = 0.307
Yes. Saturated vapor source at 150°C would remain at 150°C as it
condenses to liquid, providing a large energy supply at that temperature.

7.97
A rigid insulated container has two rooms separated by a membrane. Room A
contains 1 kg air at 200oC and room B has 1.5 kg air at 20oC, both rooms at 100
kPa. Consider two different cases
1) Heat transfer between A and B creates a final uniform T.
2) The membrane breaks and the air comes to a uniform state.
For both cases find the final temperature. Are the two processes reversible and
different? Explain.
Solution:
C.V. Total A+B
1) Energy Eq.: U2 - U1 = 1Q2 - 1W2 = 0 − 0 = 0
U2 - U1 = 0 = mA ( U2 - U1 )A + mB ( U2 - U1 )B
≅ mA CV (Τ2 - TA1) + mB CV (Τ2 - TB1)
⇒ Τ2 =
mA
mA + mB
TA1 +
mB
mA + mB
TB1 =
1
2.5 × 200 +
1.5
2.5 × 20
= 92oC
PA2 = PA1 × T2/ TA1 = 100 × (273 + 92) /473 = 77.2 kPa
PB2 = PB1 × T2/ TB1 = 100 × (273 + 92) /293 = 124.6 kPa
2) Same energy eq. Since ideal gas u(T) same T2 = 92oC, but now also same P2
P2 = mRT2 / V1; V1 = VA + VB
V1 = mA1RTA1/ P1 + mB1RTB1/ P1
P2 = (m2RT2 / (mA1RTA1/ P1 + m B1RTB1/ P1))
= P1 (m2T2 / (mA1TA1 + mB1TB1)) = 100
2.5 (273 + 92)
1 × 473 + 1.5 × 293
= 100 kPa
Both cases irreversible 1) Q over a finite ∆T and in 2) mixing of 2 different states
(internal u redistribution)
(Case 2) is more irreversible as the final state in 1 could drive a turbine between
the two different pressures until equal.

7.98
Consider the combination of the two heat engines as in Fig. P7.4. How should the
intermediate temperature be selected so the two heat engines have the same
efficiency assuming Carnot cycle heat engines.
Heat engine 1: ηTH 1
= 1 −
TM
TH
Heat engine 2: ηTH 2
= 1 −
TL
TM
ηTH 1
= ηTH 2
1 −
TM
TH
= 1 −
TL
TM
TM
TH
=
TL
TM
TM = TLTH

7.99
A house should be heated by a heat pump, β′ = 2.2, and maintained at 20oC at all
times. It is estimated that it looses 0.8 kW per degree the ambient is lower than
the inside. Assume an outside temperature of –10oC and find the needed power to
drive the heat pump?
Solution : Ambient TL = –10oC
Heat pump : β′ = Q
.
H/W
.
House : Q
.
H = Q
.
leak = 0.8 ( TH - TL)
W
.
= Q
.
H/β′ = Q
.
leak / β′ = 0.8 ( TH - TL) / β′
= 0.8[20 – (−10)] /2.2 = 10.91 kW
QleakQ QHL
W
HP
cb

7.100
Consider a combination of a gas turbine power plant and a steam power plant as
shown in Fig. P7.4. The gas turbine operates at higher temperatures (thus called a
topping cycle) than the steam power plant (then called a bottom cycle). Assume
both cycles have a thermal efficiency of 32%. What is the efficiency of the overall
combination assuming QL in the gas turbine equals QH to the steam power plant?
Let the gas turbine be heat engine number 1 and the steam power plant the heat
engine number 2. Then the overall efficiency
ηTH = W
.
net / Q
.
H = (W
.
1 + W
.
2) / Q
.
H = η1 + W
.
2 / Q
.
H
W
.
2 = η2 Q
.
M = η2 (1 – η1) Q
.
H
ηTH = η1 + η2 (1 – η1) = 0.32 + 0.32(1 – 0.32) = 0.538

7.101
We wish to produce refrigeration at −30°C. A reservoir, shown in Fig. P7.101, is
available at 200°C and the ambient temperature is 30°C. Thus, work can be done
by a cyclic heat engine operating between the 200°C reservoir and the ambient.
This work is used to drive the refrigerator. Determine the ratio of the heat
transferred from the 200°C reservoir to the heat transferred from the −30°C
reservoir, assuming all processes are reversible.
Solution:
Equate the work from the heat engine to the refrigerator.
QH1
W
QL1
HE
QH2
QL2
REF
T = 200 CH T = 30 Co
T = 30 Co T =- 30 CL
W = QH1





TH - T0
TH
also
W = QL2





T0 - TL
TL
QH1
QL2
=





To - TL
TL 




TH
TH - To
=





60
243.2 




473.2
170 = 0.687

7.102
A 4L jug of milk at 25°C is placed in your refrigerator where it is cooled down to
5°C. The high temperature in the Carnot refrigeration cycle is 45°C and the
properties of milk are the same as for liquid water. Find the amount of energy that
must be removed from the milk and the additional work needed to drive the
refrigerator.
Solution:
C.V milk + out to the 5 °C refrigerator space
Energy Eq.: m(u2 − u1) = 1Q2 − 1W2
Process : P = constant = 1 atm => 1W2 = Pm (v2 - v1)
State 1: Table B.1.1, v1 ≅ vf = 0.001003 m3/kg, h1 ≅ hf = 104.87 kJ/kg
m2 = m1 = V1/v1 = 0.004 / 0.001003 = 3.988 kg
State 2: Table B.1.1, h2 ≅ hf = 20.98 kJ/kg
1Q2 = m(u2 − u1) + 1W2 = m(u2 − u1) + Pm (v2 - v1) = m(h2 − h1)
1Q2 = 3.998 (20.98 - 104.87) = -3.988 × 83.89 = - 334.55 kJ
C.V. Refrigeration cycle TL = 5 °C ; TH = 45 °C, assume Carnot
Ideal : β = QL / W = QL / (QH - QL ) = TL/ (TH − TL)
= 278.15 / 40 = 6.954
W = QL / β = 334.55 / 6.954 = 48.1 kJ
MILK
cb
5 C
AIR
o
Remark: If you calculate the work term 1W2 you will find that it is very small,
the volume does not change (liquid). The heat transfer could then have been done
as m(u2 − u1) without any change in the numbers.

7.103
An air-conditioner with a power input of 1.2 kW is working as a refrigerator (β =
3) or as a heat pump (β′ = 4). It maintains an office at 20°C year round which
exchanges 0.5 kW per degree temperature difference with the atmosphere. Find
the maximum and minimum outside temperature for which this unit is sufficient.
Solution:
Analyze the unit in heat pump mode
Replacement heat transfer equals the loss: Q
.
= 0.5 (TH - Tamb)
W
.
=
Q
.
H
β′
= 0.5
TH - Tamb
4
TH - Tamb = 4
W
.
0.5 = 9.6 K
Heat pump mode: Minumum Tamb = 20 - 9.6 = 10.4 °C
The unit as a refrigerator must cool with rate: Q
.
= 0.5 (Tamb - Thouse)
W
.
=
Q
.
L
β
= 0.5 (Tamb - Thouse) / 3
Tamb - Thouse = 3
W
.
0.5 = 7.2 K
Refrigerator mode: Maximum Tamb = 20 + 7.2 = 27.2 °C

7.104
Make some assumption about the heat transfer rates to solve problem 7.62 when
the outdoor temperature is -20oC. Hint: look at the heat transfer given by Eq.7.14.
Solution:
Minimum power if we
assume a Carnot cycle
QleakQ QHL
W
HP
We assume the heat transfer coefficient stays the same
Q
.
H = Q
.
leak = 25 kW = CA ∆T = CA [20 –(-10)] CA =
5
6 kW/K
Q
.
leak new = CA ∆T =
5
6 [20 – (-20)] = 33.33 kW
β′ =
Q
.
H
W
.
IN
=
TH
TH-TL
=
293.15
40 = 7.32875 ⇒ W
.
IN =
33.333
7.32875 = 4.55 kW
Comment. Leak heat transfer increases and COP is lower when T outside drops.

7.105
Air in a rigid 1 m3 box is at 300 K, 200 kPa. It is heated to 600 K by heat transfer
from a reversible heat pump that receives energy from the ambient at 300 K
besides the work input. Use constant specific heat at 300 K. Since the coefficient
of performance changes write dQ = mair Cv dT and find dW. Integrate dW with
temperature to find the required heat pump work.
Solution:
COP: β′ =
QH
W =
QH
QH − QL
≅
TH
TH − TL
mair = P1V1 / RT1 = 200 × 1 / 0.287 × 300 = 2.322 kg
dQH = mair Cv dTH = β′ dW ≅
TH
TH − TL
dW
=> dW = mair Cv [
TH
TH − TL
] dTH
1W2 = ∫ mair Cv ( 1 -
TL
T ) dT = mair Cv ∫ ( 1 -
TL
T ) dT
= mair Cv [T2 - T1 - TL ln
T2
T1
]
= 2.322 × 0.717 [ 600 - 300 - 300 ln
600
300 ] = 153.1 kJ

7.106
A combination of a heat engine driving a heat pump (see Fig. P7.106) takes waste
energy at 50°C as a source Qw1 to the heat engine rejecting heat at 30°C. The
remainder Qw2 goes into the heat pump that delivers a QH at 150°C. If the total
waste energy is 5 MW find the rate of energy delivered at the high temperature.
Solution:
Waste supply: Q
.
w1 + Q
.
w2 = 5 MW
Heat Engine:
W
.
= η Q
.
w1 = ( 1 - TL1 / TH1 ) Q
.
w1
Heat pump:
W
.
= Q
.
H / βHP = Q
.
W2 / β′
= Q
.
w2 / [TH1 / (TH - TH1 )]
Equate the two work terms:
W
QL
Qw1
HE
Qw2
QH
HP
Waste
source
Ambient
30 C
Waste
source
HEAT
150 C
( 1 - TL1 / TH1 ) Q
.
w1 = Q
.
w2 × (TH - TH1 ) / TH1
Substitute Q
.
w1 = 5 MW - Q
.
w2
(1 - 303.15/323.15)(5 - Q
.
w2 ) = Q
.
w2 × (150 - 50) / 323.15
20 ( 5 - Q
.
w2 ) = Q
.
w2 × 100 => Q
.
w2 = 0.8333 MW
Q
.
w1 = 5 - 0.8333 = 4.1667 MW
W
.
= η Q
.
w1 = 0.06189 × 4.1667 = 0.258 MW
Q
.
H = Q
.
w2 + W
.
= 1.09 MW
(For the heat pump β′ = 423.15 / 100 = 4.23)

7.107
A heat pump heats a house in the winter and then reverses to cool it in the
summer. The interior temperature should be 20°C in the winter and 25°C in the
summer. Heat transfer through the walls and ceilings is estimated to be 2400 kJ
per hour per degree temperature difference between the inside and outside.
a. If the winter outside temperature is 0°C, what is the minimum power required
to drive the heat pump?
b.For the same power as in part (a), what is the maximum outside summer tem-
perature for which the house can be maintained at 25°C?
Solution:
a) Winter:
House is TH and ambient
is at TL
QleakQ QHL
W
HP
TH = 20°C = 293.2 K , TL = 0°C = 273.2 K and Q
.
H = 2400(20 -0) kJ/h
β′ = Q
.
H/W
.
IN = 2400 (20 - 0)/ W
.
IN =
TH
TH - TL
=
293.2
20
⇒ W
.
IN = 3275 kJ/h = 0.91 kW (For Carnot cycle)
b)
QleakQ QH L
W
HP TL
Summer:
TL = Thouse
TH = Tamb
TL = 25°C = 298.2 K, W
.
IN = 3275 kJ/h and Q
.
L = 2400(TH - 298.2) kJ/h
β =
Q
.
L
W
.
IN
=
2400(TH - 298.2)
3275 =
TL
TH - TL
=
298.2
TH - 298.2
or, (TH - 298.2)2 =
298.2 × 3275
2400 = 406.92
TH = 318.4 K = 45.2°C

7.108
A remote location without electricity operates a refrigerator with a bottle of
propane feeding a burner to create hot gases. Sketch the setup in terms of cyclic
devices and give a relation for the ratio of Q
.
L in the refrigerator to Q
.
fuel in the
burner in terms of the various reservoir temperatures.
C.V.: Heat Eng.: W
.
HE = ηHEQ
.
fuel
C.V.: Refrigerator: Q
.
L2 = β W
.
HE = β ηHE Q
.
fuel
The ratio becomes
Q
.
L2 / Q
.
fuel = β ηHE
=
TL2
TH2 − TL2
(1 −
TL1
TH1
) If Carnot devices

7.109
A furnace, shown in Fig. P7.109, can deliver heat, QH1 at TH1 and it is proposed to
use this to drive a heat engine with a rejection at Tatm instead of direct room
heating. The heat engine drives a heat pump that delivers QH2 at Troom using the
atmosphere as the cold reservoir. Find the ratio QH2/QH1 as a function of the
temperatures. Is this a better set-up than direct room heating from the furnace?
Solution:
C.V.: Heat Eng.: W
.
HE = ηQ
.
H1 where η = 1 − Tatm/TH1
C.V.: Heat Pump: W
.
HP = Q
.
H2/β′ where β′ = Trm/(Trm − Tatm)
Work from heat engine goes into heat pump so we have
Q
.
H2 = β′ W
.
HP = β′ η Q
.
H1
and we may substitute T's for β′, η. If furnace is used directly Q
.
H2 = Q
.
H1,
so if β′η > 1 this proposed setup is better. Is it? For TH1 > Tatm formula shows
that it is good for Carnot cycles. In actual devices it depends wether β′η > 1 is
obtained.

7.110
Consider the rock bed thermal storage in Problem 7.65. Use the specific heat so
you can write dQH in terms of dTrock and find the expression for dW out of the
heat engine. Integrate this expression over temperature and find the total heat
engine work output.
Solution:
The rock provides the heat QH
dQH = −dUrock = −mC dTrock
dW = ηdQH = − ( 1 − To / Trock) mC dTrock
m = ρV = 2750 × 2 = 5500 kg
1W2 = ∫ − ( 1 − To / Trock) mC dTrock = − mC [T2 − T1 − To ln
T2
T1
]
= − 5500 × 0.89 [ 290 − 400 − 290 ln
290
400 ] = 81 945 kJ

7.111
On a cold (-10oC) winter day a heat pump provides 20 kW to heat a house
maintained at 20oC and it has a COPHP of 4 using the maximum power available.
The next day a winter storm brings the outside to -15oC, assuming the same COP
and the house heat loss is to the outside air. How cold is the house then?
If we look at the heat loss for the house we have
Q
.
loss = 20 kW = CA ∆T ⇒ CA =
20 kW
20 - (-10) K = 0.667 kW/K
Q
.
loss = Q
.
H = COP W
.
⇒ W
.
= Q
.
loss /COP =
20 kW
4 = 5 kW
With the same COP and the same power input we can get the same Q
.
H = 20 kW.
Q
.
loss = CA ∆T = 0.667 kW/K × [T – (–15)] K = 20 kW
T = 20 kW / 0.667 (kW/K) – 15 oC = (30 – 15) oC = 15oC
Remark: Since Q
.
H = Q
.
loss = CA ∆T is the same the ∆T becomes the same so the
house can be kept at 30oC above the ambient. In the real system the COP drops as
the outdoor T drops unless the outside heat exchanger is buried under ground with
a constant temperature independent upon the weather.

7.112
A Carnot heat engine operating between a high TH and low TL energy reservoirs
has an efficiency given by the temperatures. Compare this to two combined heat
engines one operating between TH and an intermediate temperature TM giving out
work WA and the other operating between TM and TL giving out WB. The
combination must have the same efficiency as the single heat engine so the heat
transfer ratio QH/QL = ψ(TH,TL) = [QH/QM] [QM/QL]. The last two heat transfer
ratios can be expressed by the same function ψ() involving also the temperature
TM. Use this to show a condition the function ψ() must satisfy.
The overall heat engine is a Carnot heat engine so
Q
.
H / Q
.
L =
TH
TL
= ψ(TH,TL)
The individual heat engines
Q
.
H / Q
.
M = ψ(TH,TM) and Q
.
M / Q
.
L = ψ(TM,TL)
Since an identity is
Q
.
H / Q
.
L = [Q
.
H / Q
.
M] [Q
.
M / Q
.
L] = ψ(TH,TL)
it follows that we have
ψ(TH,TL) = ψ(TH,TM) × ψ(TM,TL)
Notice here that the product of the two functions must cancel the intermediate
temperature TM, this shows a condition the function ψ() must satisfy. The Kelvin
and Rankine temperature scales are determined by the choice of the function
ψ(TH,TL) = TH / TL = Q
.
H / Q
.
L
satisfying the above restriction.

7.113
A 10-m3 tank of air at 500 kPa, 600 K acts as the high-temperature reservoir for a
Carnot heat engine that rejects heat at 300 K. A temperature difference of 25°C
between the air tank and the Carnot cycle high temperature is needed to transfer
the heat. The heat engine runs until the air temperature has dropped to 400 K and
then stops. Assume constant specific heat capacities for air and find how much
work is given out by the heat engine.
Solution:
Q
H
W
QL
HE
AIR
300 K
TH = Tair – 25°C, TL = 300 K
mair =
P1V
RT1
=
500 × 10
0.287 × 600
= 29.04 kg
dW = ηdQH =






1 -
TL
Tair - 25 dQH
dQH = –mairdu = –mairCvdTair
W = ⌡⌠dW = –mairCv
⌡

⌠






1 –
TL
Ta – 25 dTa
= –mairCv 





Ta2 – Ta1 – TL ln
Ta2 – 25
Ta1 – 25
= –29.04 × 0.717 ×






400 – 600 – 300 ln
375
575 = 1494.3 kJ

CONTENT
SUBSECTION PROB NO.
In-Text concept questions a-j
Concept-Study Guide problems 1-16
Inequality of Clausius 17-23
Entropy of a pure substance 24-34
Reversible processes 35-66
Entropy of a liquid or solid 67-79
Entropy of ideal gases 80-103
Polytropic processes 104-116
Entropy generation 117-157
Rates or fluxes of entropy 158-168
Review 169-185
Problem solution repeated, but using the Pr and vr functions in Table
A.7.2:
97, 107, 112

8.a
Does Clausius say anything about the sign for ∫o dQ ?
No.
The total (net) heat transfer can be coming in like in a heat engine (Wout = QH
– QL) in which case it is positive. It can also be net going out like in a
refrigerator or heat pump (Win = QH – QL) in which case the sign is negative.
Finally if you look at a transmission gearbox there could be no heat transfer
(first approximation) in which case the integral is zero.
8.b
Does the statement of Clausius require a constant T for the heat transfer as in a
Carnot cycle?
No.
The statement for a cycle involves an integral of dQ/T so T can vary, which it
does during most processes in actual devices. This just means that you cannot
that easily get a closed expression for the integral.
8.c
How can you change s of a substance going through a reversible process?
From the definition of entropy
ds =
dq
T
for a reversible process. Thus only heat transfer gives a change in s,
expansion/compression involving work does not give such a contribution.
8.d
A reversible process adds heat to a substance. If T is varying does that influence
the change in s?
Yes.
Reversible: ds =
dq
T
So if T goes up it means that s changes less per unit of dq, and the opposite if
T decreases then s changes more per unit of dq.

8.e
Water at 100 kPa, 150oC receives 75 kJ/kg in a reversible process by heat
transfer. Which process changes s the most: constant T, constant v or constant P?
ds =
dq
T
Look at the constant property lines in a T-s diagram, Fig. 8.5. The constant v
line has a higher slope than the constant P line also at positive slope. Thus
both the constant P and v processes have an increase in T. As T goes up the
change in s is smaller.
The constant T (isothermal) process therefore changes s the most.
8.f
A liquid is compressed in a reversible adiabatic process. What is the change in T?
If the process is reversible then s is constant, ds =
dq
T = 0
Change in s for a liquid (an incompressible substance) is
Eq. 8.10: ds =
C
T dT
From this it follows that if ds = 0 then T is constant.
8.g
An ideal gas goes through a constant T reversible heat addition process. How do
the properties (v, u, h, s, P) change (up, down or constant)?
Ideal gas: u(T), h(T) so they are both constant
Eq. 8.11 gives: ds = dq/T + dsgen = dq/T + 0 > 0 so s goes up by q/T
Eq. 8.21 gives: ds = (R/v) dv so v increases
Eq. 8.23 gives: ds = -(R/P) dP so P decreases
21
P
v
T
s
1
2
T q

8.h
Carbon dioxide is compressed to a smaller volume in a polytropic process with n
= 1.2. How do the properties (u, h, s, P, T) change (up, down or constant)?
For carbon dioxide Table A.5 k = 1.289 so we have n < k and the
process curve can be recognized in Figure 8.18. From this we see a smaller
volume means moving to the left in the P-v diagram and thus also up.
From P-v diagram: P up, T up
From T-s diagram Since T is up then s down.
As T is up so is h and u.
TP
2
2 1
1 T (n = 1)
n = 1.2
v s

8.i
A substance has heat transfer out. Can you say anything about changes in s if the
process is reversible? If it is irreversible?
Reversible: ds =
dq
T < 0 since dq < 0
Irreversible: ds =
dq
T + dsgen = ? dq < 0 but dsgen > 0
You cannot say, ds depends on the magnitude of dq/T versus dsgen
8.j
A substance is compressed adiabatically so P and T go up. Does that change s?
If the process is reversible then s is constant, ds =
dq
T = 0
If the process is irreversible then s goes up, ds =
dq
T + dsgen = dsgen > 0

8.1
When a substance has completed a cycle, v, u, h, and s are unchanged. Did
anything happen? Explain.
Yes.
During various parts of the cycle work and heat transfer may be transferred.
That happens at different P and T. The net work out equals the net heat
transfer in (energy conservation) so dependent upon the sign it is a heat
engine or a heat pump (refrigerator). The net effect is thus a conversion of
energy from one storage location to another and it may also change nature
(some Q was changed to W or the opposite)

8.2
Assume a heat engine with a given QH. Can you say anything about QL if the
engine is reversible? If it is irreversible?
For a reversible heat engine it must be that:
∫o dQ
T = 0 =
QH
TH
−
QL
TL
or integrals if T not constant
So as TL is lower than TH then QL must be correspondingly lower than QH to
obtain the net zero integral.
For an irreversible heat engine we have
∫o dQ
T =
QH
TH
−
QL
TL
< 0
This means that QL is larger than before (given QH and the T’s). The
irreversible heat engine rejects more energy and thus gives less out as work.

8.3
CV A is the mass inside a piston/cylinder, CV B is
that plus part of the wall out to a source of 1Q2 at Ts.
Write the entropy equation for the two control
volumes assuming no change of state of the piston
mass or walls.
Fig. P8.3
Po
mp
Ts
Am
The general entropy equation for a control mass is Eq.8.37
S2 – S1 =
⌡

⌠
1
2 dQ
T + 1S2 gen
The left hand side is storage so that depends of what is inside the C.V. and the
integral is summing the dQ/T that crosses the control volume surface while the
process proceeds from 1 to 2.
C.V. A: mA (s2 – s1) =
⌡

⌠
1
2 dQ
TA
+ 1S2 gen CV A
C.V. B: mA (s2 – s1) =
⌡

⌠
1
2 dQ
Ts
+ 1S2 gen CV B
In the first equation the temperature is that of mass mA which possibly changes
from 1 to 2 whereas in the second equation it is the reservoir temperature Ts. The
two entropy generation terms are also different the second one includes the first
one plus any s generated in the walls that separate the mass mA from the reservoir
and there is a Q over a finite temperature difference. When the storage effect in
the walls are neglected the left hand sides of the two equations are equal.

8.4
Consider the previous set-up with the mass mA and the piston cylinder of mass
mp starting out at two different temperatures. After a while the temperature
becomes uniform without any external heat transfer. Write the entropy equation
storage term (S2 – S1) for the total mass.
In this case the storage effect must be summed up over all the mass inside the
control volume.
S2 – S1 = mA (s2 – s1)A + mP (s2 – s1)P
≈ mA (s2 – s1)A + mP CP ln (
T2
T1 P
)
The last equation assumed a constant specific heat for the solid material of the
piston, a common assumption. There is only a single temperature T2, but there are
two different temperatures for state 1 T1 A and T1 P. The temperature T2 would be
found from the energy equation.

8.5
Water at 100
o
C, quality 50% in a rigid box is heated to 110
o
C. How do the properties
(P, v, x, u and s) change? (increase, stay about the same, or decrease)
A fixed mass in a rigid box give a constant v process. So
P goes up (in the two-phase region P = Psat at given T)
v stays constant.
x goes up ( we get closer to the saturated vapor state see P-v diagram)
u goes up (Q in and no work)
s goes up (Q in)
P T
22
1 1v s

8.6
Liquid water at 20
o
C, 100 kPa is compressed in a piston/cylinder without any heat
transfer to a pressure of 200 kPa. How do the properties (T, v, u, and s) change?
(increase, stay about the same, or decrease)
Adiabatic dq = 0: dq = T ds = 0;
Incompressible dv = 0: dw = P dv = 0
(T, v, u, and s) they are all constant.
Only the pressure and enthalpy goes up.

8.7
A reversible process in a piston/cylinder is
shown in Fig. P8.7. Indicate the storage
change u2 - u1 and transfers 1w2 and 1q2 as
positive, zero, or negative
P T
2
u = C1
21
sv
1w2 = ∫ P dv > 0 ; 1q2 = ∫ T ds > 0
u2 - u1 > 0 from general shape of the constant u curves. Further out in the ideal
gas region the constant u curve become horizontal ( u = fct(T) only ).

8.8
A reversible process in a piston/cylinder
is shown in Fig. P8.8. Indicate the
storage change u2 - u1 and transfers 1w2
and 1q2 as positive, zero, or negative
P
v
1
2 2
1
T
s
1w2 = ∫ P dv > 0 ; 1q2 = ∫ T ds = 0
u2 - u1 = 1q2 - 1w2 < 0

8.9
Air at 290 K, 100 kPa in a rigid box is heated to 325 K. How do the properties (P, v,
u and s) change? (increase, stay about the same, or decrease)
Rigid box: v = constant, (P, u, and s) all increases.
TP
2 P2 2
P11
1
sv

8.10
Air at 20
o
C, 100 kPa is compressed in a piston/cylinder without any heat transfer to a
pressure of 200 kPa. How do the properties (T, v, u and s) change? (increase, about
the same or decrease)
T goes up,
v goes down
u goes up (work in, q = 0)
s = constant
TP
2 P22
P1 1
1
sv

8.11
Carbon dioxide is compressed to a smaller volume in a polytropic process with n
= 1.4. How do the properties (u, h, s, P, T) change (up, down or constant)?
For carbon dioxide Table A.5 k = 1.289 so we have n > k and the process
curve can be recognized in Figure 8.18. From this we see a smaller volume means
moving to the left in the P-v diagram and thus also up.
P up, T up. As T is up so is h and u.
From the T-s diagram as n > k then we move to larger T means s is up.
TP
2
2
1 T (n = 1) 1
n = 1.4
v s

8.12
Process A: Air at 300 K, 100 kPa is heated to 310 K at constant pressure.
Process B: Heat air at 1300 K to 1310 K at constant 100 kPa.
Use the table below to compare the property changes.
Property ∆A > ∆B ∆A ≈ ∆B ∆A < ∆B
a ∆ = v2 – v1 √
b ∆ = h2 – h1 √
c ∆ = s2 – s1 √
a. Ideal gas Pv = RT so v goes with absolute T
∆v = (R/P) ∆T thus the same
b. Since dh = CP dT and CP increases with T
c. At constant P: ds = (CP/T) dT
CP is only 15% higher at 1300 K compared to 300 K (see Fig. 5.11)

8.13
Why do we write ∆S or S2 – S1 whereas we write ∫dQ/T and 1S2 gen?
This is similar to the terms in the continuity equation m2 – m1 versus mi, me and
the energy equation E2 – E1 versus 1Q2, 1W2. The first part is the change in the
storage S2 – S1 of entropy in the mass and the second part is the transfer or
generation during the process from 1 to 2. The storage terms correspond to the left
hand side of the balance equation for entropy. The integral ∫dQ/T represents a
transfer of entropy across the control volume surface during the process from 1
to 2 and the 1S2 gen expresses the total amount of entropy being generated inside
the control volume and both are on the right hand side of the balance equation.

8.14
A reversible heat pump has a flux of s entering as Q
.
L/TL. What can you say about
the exit flux of s at TH?
For the entropy equation 8.3 and 8.42, the rate of storage is zero and the
generation term is zero. Thus the entropy equation becomes
0 =
Q
.
L
TL
–
Q
.
H
TH
+ 0
So:
Q
.
L
TL
=
Q
.
H
TH
= flux of s
We have the same flux of s in as out matching the result in chapter 7.

8.15
An electric baseboard heater receives 1500 W of electrical power that heats the
room air which loses the same amount out through the walls and windows. Specify
exactly where entropy is generated in that process.
a Electrical heating wire (electrical work turned into internal energy,
leaving as heat transfer).
b Heat transfer from hot wire to cooler room air, i.e. in the wire coverings
c Room air to walls
d Inside walls and windows, heat transfer over a finite ∆T
e from outside wall surface to ambient T

8.16
A 500 W electric space heater with a small fan inside heats air by blowing it over
a hot electrical wire. For each control volume: a) wire at Twire only b) all the
room air at Troom and c) total room plus the heater, specify the storage, entropy
transfer terms and entropy generation as rates (neglect any Q
.
through the room
walls or windows).
Storage Q
.
/T S
.
gen
C.V. Wire only 0
−500 W
Twire
500 W
Twire
C.V. Room air
500 W
Troom
500 W
Twire
500 W (
1
Troom
−
1
Twire
)
C.V. Total room
500 W
Troom
0
500 W
Troom
Remark: Room only receives the electrical power input of 500 W.

Inequality of Clausius

8.17
Consider the steam power plant in Example 6.9 and assume an average T in the
line between 1 and 2. Show that this cycle satisfies the inequality of Clausius.
Solution:
Show Clausius:
⌡

⌠dQ
T ≤ 0
For this problem we have three heat transfer terms:
qb = 2831 kJ/kg, qloss = 21 kJ/kg, qc = 2173.3 kJ/kg
⌡

⌠dq
T =
qb
Tb
–
qloss
Tavg 1-2
–
qc
Tc
=
2831
573 −
21
568 −
2173.3
318
= –1.93 kJ/kg K < 0 OK

8.18
A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC.
Examine each of three cases with respect to the inequality of Clausius.
a. W
.
= 6 kW b. W
.
= 0 kW c. Carnot cycle
Solution:
TH = 250 + 273 = 523 K ; TL = 30 + 273 = 303 K
Case a)
⌡


⌠
dQ
.
T =
6000
523 –
0
303 = 11.47 kW/K > 0 Impossible
b)
⌡


⌠
dQ
.
T =
6000
523 –
6000
303 = –8.33 kW/K < 0 OK
c)
⌡


⌠
dQ
.
T = 0 =
6000
523 –
Q
.
L
303 ⇒
Q
.
L =
303
523 × 6 kW = 3.476 kW
W
.
= Q
.
H – Q
.
L = 2.529 kW

8.19
Use the inequality of Clausius to show that heat transfer from a warm space
towards a colder space without work is a possible process i.e. a heat engine with
no work output.
Clausius:
⌡

⌠dQ
T ≤ 0 or
⌡


⌠
dQ
.
T ≤ 0
Take C.V. as the space separating the warm and cold space. It is the same Q that
crosses each surface (from energy equation) so
⌡

⌠dQ
T =
Q
Twarm
–
Q
Tcold
= Q (
1
Twarm
–
1
Tcold
) < 0 OK

8.20
Use the inequality of Clausius to show that heat transfer from a cold space
towards a warmer space without work is an impossible process i.e. a heat pump
with no work input.
Clausius:
⌡

⌠dQ
T ≤ 0 or
⌡


⌠
dQ
.
T ≤ 0
Take C.V. as the space separating the warm and cold space. It is the same Q that
crosses each surface (from energy equation) so
⌡

⌠dQ
T =
Q
Tcold
–
Q
Twarm
= Q (
1
Tcold
–
1
Twarm
) > 0 Impossible!

8.21
Assume the heat engine in Problem 7.32 has a high temperature of 1200 K and a
low temperature of 400 K. What does the inequality of Clausius say about each of
the four cases?
Solution:
Cases a)
⌡


⌠
dQ
.
T =
6
1200 –
4
400 = – 0.005 kW/K < 0 OK
b)
⌡


⌠
dQ
.
T =
6
1200 –
0
400 = 0.005 kW/K > 0 Impossible
c)
⌡


⌠
dQ
.
T =
6
1200 –
2
400 = 0 kW/K Possible if reversible
d)
⌡


⌠
dQ
.
T =
6
1200 –
6
400 = – 0.001 kW/K < 0 OK
HQ
W
LQ
H
HE
cb
T = 400 KL
T = 1200 K

8.22
Let the steam power plant in Problem 7.35 have 700oC in the boiler and 40oC
during the heat rejection in the condenser. Does that satisfy the inequality of
Clausius? Repeat the question for the cycle operated in reverse as a refrigerator.
Solution:
Q
.
H = 1 MW Q
.
L = 0.58 MW
⌡


⌠
dQ
.
T =
1000
973 –
580
313 = –0.82 kW/K < 0 OK
Refrigerator
⌡


⌠
dQ
.
T =
580
313 –
1000
973 = 0.82 > 0 Cannot be possible
WT
QH
QL
.
WP, in

8.23
Examine the heat engine given in Problem 7.54 to see if it satisfies the inequality
of Clausius.
Solution:
QH = 325 kJ at TH = 1000 K
QL = 125 kJ at TL = 400 K
⌡

⌠ dQ
T =
325
1000 –
125
400 = 0.0125 kJ/K > 0 Impossible
HQ = 325 kJ
W = 200 kJ
LQ = 125 kJ
T = 1000 KH
HE
cb
T = 400 KL

Entropy of a pure substance

8.24
Find the missing properties of T, P, s and x for water at:
a. P = 25 kPa, s = 7.7 kJ/kgK
b. P = 10 MPa, u = 3400 kJ/kg
c. T = 150oC, s = 7.4 kJ/kgK
Solution:
a) Table B.1.2 sf < s < sg so two-phase and T = Tsat(P) = 64.97°C
x = (s – sf)/sfg =
7.70 − 0.893
6.9383 = 0.981
h = 271.9 + 0.981 × 2346.3 = 2573.8 kJ/kg
b) Table B.1.2 u > ug => Superheated vapor Table B.1.3, x = undefined
T ≅ 682°C , s ≅ 7.1223 kJ/kg K
c) Table B.1.1 s > sg so superheated vapor. Table B.1.3, x = undefined
State is located between 100 and 200 kPa, so interpolate
P = 100 kPa + 100 kPa ×
7.40 − 7.6133
7.2795 − 7.6133
= 163.9 kPa
b
P
v
T
s
b
a aT
P
cc

8.25
Determine the missing property among P, T, s, and x for R-410a at:
a. T = –20oC, v = 0.1377 m3/kg
b. T = 20oC, v = 0.01377 m3/kg
c. P = 200 kPa, s = 1.409 kJ/kgK
a) B.4.1: v > vg = 0.0648 m3/kg =>
B.4.2 superheated vapor so x is undefined
very close to 200 kPa, s = 1.1783 kJ/kg-K
b) B.4.1: 0.000923 = vf < v < vg = 0.01758 m3/kg => Two-phase
P = Psat = 1444.2 kPa
x = (v – vf)/vfg =
0.01377 − 0.000923
0.01666 = 0.77113
s = sf + x sfg = 0.3357 + 0.77113 × 0.6627 = 0.8467 kJ/kg-K
c) Table B.4.2 at 200 kPa, s > sg so superheated vapor,
x is undefined, and we find the state at T = 60oC.
b
P
v
T
s
b a
aT
P
cc

8.26
Find the missing properties and give the phase of the ammonia, NH3.
a. T = 65°C, P = 600 kPa s = ? v = ?
b. T = 20°C, P = 100 kPa v = ? s = ? x = ?
c. T = 50°C, v = 0.1185 m3/kg s = ? x = ? P = ?
a) B.2.2 average between 60°C and 70°C
v = (0.25981 + 0.26999)/2 = 0.26435 m3/kg
s = (5.6383 + 5.7094)/2 = 5.6739 kJ/kgK
b) B.2.1: P < Psat = 857.5 kPa =>
v = 1.4153 m3/kg, s = 6.2826 kJ/kgK
c) B.2.1: v > vg = 0.06337 m3/kg =>
very close to 1200 kPa, s = 5.1497 kJ/kgK
v
P
s
T
a
b
c a
b
c

8.27
Find the entropy for the following water states and indicate each state on a T-s
diagram relative to the two-phase region.
a. 250oC, v = 0.02 m3/kg
b. 250oC, 2000 kPa
c. –2oC, 100 kPa
Solution:
a) Table B.1.1: 0.001251 = vf < v < vg = 0.05013 m3/kg => Two-phase
x =
0.02 - 0.001251
0.04887 = 0.38365
s = sf + x sfg = 2.7927 + 0.38365 × 3.2802 = 4.05 kJ/kg K
b) Table B.1.1: P < Psat = 3973 kPa => superheated vapor B.1.3
s = 6.5452 kJ/kg K
c) Table B.1.1 T < Ttripple =0.01oC so goto B.1.5
Table B.1.5: P > Psat = 0.5177 kPa so compressed solid
s = –1.2369 kJ/kg K
v
P
s
T
a b
c
a
b
c

8.28
Find the entropy for the following water states and indicate each state on a T-s
diagram relative to the two-phase region.
a. 20oC, 100 kPa
b. 20oC, 10 000 kPa
Solution:
Both states are compressed liquid states P > Psat = 2.34 kPa from B.1.1
a) Table B.1.1: s = 0.2966 kJ/kg K (saturated liquid same T)
b) Table B.1.4 s = 0.2945 kJ/kg K
v
P
s
T
a
b
a
b

8.29
Determine the missing property among P, T, s, and x for CO2 at:
a. P = 1000 kPa, v = 0.05 m3/kg
b. T = 0oC, s = 1 kJ/kgK
c. T = 60oC, s = 1.8 kJ/kgK
a) Table B.3.1 at 1004 kPa: v > vg = 0.03828 m3/kg => Superheated
vapor
B.3.2: T = 0 + 20
0.05 – 0.048
0.0524 – 0.048 = 20 × 0.4545 = 9.09oC
s = 1.5371 + (1.6025 – 1.5371) × 0.4545 = 1.5514 kJ/kg-K
b) Table B.3.1: 0.3344 = sf < s < sg = 1.1797 kJ/kg-K => Two-phase
P = Psat = 3485 kPa
x =
s - sf
sfg
=
1 - 0.3344
0.8453 = 0.7874
c) Table B.3.1: T > Tcritical = 31oC so superheated vapor or dense fluid.
Table B.3.2: Start at 1000 kPa, we see s = 1.72 is too low, so lower P.
State is between 400 and 800 kPa at 60oC, interpolate
P = 400 kPa + 400 kPa ×
1.8 – 1.9033
1.7660 – 1.9033 = 300.95 kPa

8.30
Two kg water at 120oC with a quality of 25% has its temperature raised 20oC in a
constant volume process. What are the new quality and specific entropy?
Solution:
State 1 from Table B.1.1 at 120oC
v = vf + x vfg = 0.001060 + 0.25 × 0.8908 = 0.22376 m3/kg
State 2 has same v at 140oC also from Table B.1.1
x =
v - vf
vfg
=
0.22376 - 0.00108
0.50777 = 0.4385
s = sf + x sfg = 1.739 + 0.4385 × 5.1908 = 4.015 kJ/kg K
P TC.P. C.P.
140 C
361.3
120
140
120 C
198.5
v v

8.31
Two kg water at 200 kPa with a quality of 25% has its temperature raised 20oC in a
constant pressure process. What is the change in entropy?
Solution:
s = sf + x sfg = 1.53 + 0.25 × 5.597 = 2.9293 kJ/kg
T2 = Tsat + 20 = 120.23 + 20 = 140.23oC
so state 2 is superheated vapor (x = undefined) from Table B.1.3
s2 = 7.1271 + (7.2795 – 7.1271)
20
150 - 120.23 = 7.2295 kJ/kgK
s2 – s1 = 7.2295 – 2.9293 = 4.3002 kJ/kgK
T
C.P.
v
200 kPa
P C.P.
v
T
200
120
140
120.2 C
140 C

8.32
Saturated liquid water at 20oC is compressed to a higher pressure with constant
temperature. Find the changes in u and s when the final pressure is
a. 500 kPa b. 2000 kPa c. 20 000 kPa
Solution:
kJ/kg kJ/kg K
B.1.1: u1 = 83.94 s1 = 0.2966
B.1.4: ua = 83.91 sa = 0.2965 ∆u = –0.03 ∆s = –0.0001
B.1.4: ub = 83.82 sb = 0.2962 ∆u = –0.12 ∆s = –0.0004
B.1.4: uc = 82.75 sc = 0.2922 ∆u = –1.19 ∆s = –0.0044
Nearly constant u and s, incompressible media
v
P
s
T
a
b
c
1
c,b,a,1

8.33
Saturated vapor water at 150oC is expanded to a lower pressure with constant
temperature. Find the changes in u and s when the final pressure is
a. 100 kPa b. 50 kPa c. 10 kPa
Solution:
Table B.1.1 for the first state then B.1.3 for the a, b and c states.
kJ/kg kJ/kg K kJ/kg kJ/kg K
u1= 2559.54 s1= 6.8378
ua = 2582.75 sa = 7.6133 ∆u = 23.21 ∆s = 0.7755
ub = 2585.61 sb = 7.94 ∆u = 26.07 ∆s = 1.1022
uc = 2587.86 sc = 8.6881 ∆u = 28.32 ∆s = 1.8503
v
P
s
T
476 kPa
50
100
10a b c
1
1 a b c
Remark: You approach ideal gas as P drops so u is u(T) but s is still s(T,P).

8.34
Determine the missing property among P, T, s, x for the following states:
a. Ammonia 25oC, v = 0.10 m3/kg
b. Ammonia 1000 kPa, s = 5.2 kJ/kg K
c. R-134a 5oC, s = 1.7 kJ/kg K
d. R-134a 50oC, s = 1.9 kJ/kg K
Solution:
Table P kPa T oC s kJ/kg K x
a) B2.1 1003 25 4.1601 0.7776
b) B2.2 1000 42.53 5.2 -----
c) B5.1 350.9 5 1.7 0.96598
d) B5.2 232.3 50 1.9 -----
a) x = (0.1 – 0.001658)/0.12647 = 0.7776
s = sf + x sfg = 1.121 + x × 3.9083 = 4.1601 kJ/kg K
b) T = 40 + 10 × (5.2 – 5.1778)/(5.2654 – 5.1778) = 42.53oC
superheated vapor so x is undefined
c) x = (1.7 – 1.0243)/0.6995 = 0.96598
d) superheated vapor between 200 and 300 kPa
P = 200 + 100 × (1.9 – 1.9117)/(1.8755 – 1.9117) = 232.3 kPa
v
P
s
T
da b
c
a
b
cd

Reversible processes

8.35
Consider a Carnot-cycle heat engine with water as the working fluid. The heat
transfer to the water occurs at 300°C, during which process the water changes
from saturated liquid to saturated vapor. The heat is rejected from the water at
40°C. Show the cycle on a T–s diagram and find the quality of the water at the
beginning and end of the heat rejection process. Determine the net work output
per kilogram of water and the cycle thermal efficiency.
Solution:
From the definition of the Carnot cycle, two constant s and two constant T
processes.
1 2
34
40
300
T
s
From table B.1.1
State 2 is saturated vapor so
s3 = s2 = 5.7044 kJ/kg K
= 0.5724 + x3(7.6845)
x3 = 0.6678
State 1 is saturated liquid so
s4 = s1 = 3.2533 kJ/kg K = 0.5724 + x4(7.6845)
x4 = 0.3489
ηTH =
wNET
qH
=
TH – TL
TH
=
260
573.2 = 0.4536
qH = TH(s2 – s1) = 573.2 K (5.7044 – 3.2533) kJ/kg K = 1405.0 kJ/kg
wNET = ηTH × qH = 637.3 kJ/kg

8.36
A piston cylinder compresses R-410a at 200 kPa, -20oC to a pressure of
1200 kPa in a reversible adiabatic process. Find the final temperature and the
specific compression work?
C.V. The R-410a
Energy Eq.: u2 – u1 = 1q2 – 1w2 = – 1w2
Entropy Eq.: s2 – s1 = ⌡⌠ dq/T + 0 = 0 ⇒ s2 = s1
State 1: Sup. vapor u1 = 251.18 kJ/kg, s1 = 1.1783 kJ/kgK
State 2: (P2, s2) u2 = 300.05 kJ/kg (same interpolation as for T),
T2 = 60 + 20
1.1783 – 1.1747
1.2331 – 1.1747 = 61.2oC
1w2 = u1 – u2 = 251.18 – 300.05 = –48.87 kJ/kg
Process line shown by CATT3 program in T-s diagram

8.37
In a Carnot engine with ammonia as the working fluid, the high temperature is
60°C and as QH is received, the ammonia changes from saturated liquid to
saturated vapor. The ammonia pressure at the low temperature is 190 kPa. Find
TL, the cycle thermal efficiency, the heat added per kilogram, and the entropy, s,
at the beginning of the heat rejection process.
Solution:
T
s
1 2
34
Constant T ⇒ constant P from 1 to 2, Table B.2.1
qH = ∫ Tds = T (s2 – s1) = T sfg
= h2 – h1 = hfg = 997.0 kJ/kg
States 3 & 4 are two-phase, Table B.2.1
⇒ TL = T3 = T4 = Tsat(P) = –20°C
ηcycle = 1 –
TL
TH
= 1 –
253.2
333.2 = 0.24
Table B.2.1: s3 = s2 = sg(60°C) = 4.6577 kJ/kg K

8.38
Water is used as the working fluid in a Carnot cycle heat engine, where it changes
from saturated liquid to saturated vapor at 200°C as heat is added. Heat is rejected
in a constant pressure process (also constant T) at 20 kPa. The heat engine powers
a Carnot cycle refrigerator that operates between –15°C and +20°C. Find the heat
added to the water per kg water. How much heat should be added to the water in
the heat engine so the refrigerator can remove 1 kJ from the cold space?
Solution:
Carnot cycle heat engine:
T
s
1 2
34
Constant T ⇒ constant P from 1 to 2, Table B.2.1
qH = ∫ Tds = T (s2 – s1) = T sfg = hfg
= 473.15 (4.1014) = 1940 kJ/kg
States 3 & 4 are two-phase, Table B.2.1
⇒ TL = T3 = T4 = Tsat(P) = 60.06oC
Carnot cycle refrigerator (TL and TH are different from above):
βref =
QL
W =
TL
TH – TL
=
273 – 15
20 – (–15) =
258
35 = 7.37
W =
QL
β
=
1
7.37 = 0.136 kJ
The needed work comes from the heat engine
W = ηHE QH H2O ; ηHE = 1 –
TL
TH
= 1 –
333
473 = 0.296
QH H2O =
W
ηHE
=
0.136
0.296 = 0.46 kJ

8.39
Water at 200 kPa, x = 1.0 is compressed in a piston/cylinder to 1 MPa, 250°C in a
reversible process. Find the sign for the work and the sign for the heat transfer.
Solution:
The process is not specified, but the beginning and end states are
State 1: Table B.1.1:
v1 = 0.8857 m3/kg; u1 = 2529.5 kJ/kg; s1 = 7.1271 kJ/kg K
Reversible process: dw = P dv dq = T ds
v2 < v1 => 1w2 = ∫ P dv < 0
s2 < s1 => 1q2 = ∫ T ds < 0
2
1
P
v
T
s
1
2

8.40
Water at 200 kPa, x = 1.0 is compressed in a piston/cylinder to 1 MPa, 350oC in a
reversible process. Find the sign for the work and the sign for the heat transfer.
Solution:
1w2 = ∫ P dv so sign dv
1q2 = ∫ T ds so sign ds
B1.2 v1 = 0.88573 m3/kg s1 = 7.1271 kJ/kg K
B1.3 v2 = 0.28247 m3/kg s2 = 7.301 kJ/kg K
dv < 0 => w is negative
ds > 0 => q is positive
2
1
P
v
T
s
1
2

8.41
R-410a at 1 MPa and 60oC is expanded in a piston cylinder to 500 kPa, 20oC in a
reversible process. Find the sign for both the work and the heat transfer for this
process.
Solution:
1w2 = ∫ P dv so sign dv
1q2 = ∫ T ds so sign ds
B.4.2 v1 = 0.03470 m3/kg s1 = 1.2019 kJ/kg K
B.4.2 v2 = 0.06231 m3/kg s2 = 1.1803 kJ/kg K
dv > 0 => w is positive
ds < 0 => q is negative
2
1
P
v
T
s
1
2

8.42
A piston cylinder maintaining constant pressure contains 0.1 kg saturated liquid
water at 100°C. It is now boiled to become saturated vapor in a reversible process.
Find the work term and then the heat transfer from the energy equation. Find the
heat transfer from the entropy equation, is it the same?
Energy Eq.: m(u2 - u1) = 1Q2 - 1W2
Entropy Eq.: m(s2 - s1) = ∫
dq
T + 0 =
1Q2
T
Process: P = C => 1W2 = m P(v2 - v1) = m P vfg
= 0.1 kg × 101.3 kPa × 1.67185 m3/kg = 16.936 kJ
1Q2 = m(u2 - u1) + 1W2 = m ufg + 1W2
= 0.1 × 2087.58 + 16.936 = 225.7 kJ
or = m(h2− h1) = m hfg = 0.1 × 2257.03 = 225.7 kJ
From the entropy equation we can get
1Q2 = mT(s2 − s1) = m T sfg = 0.1 × 373.15 × 6.048 = 225.68 kJ
So they are equal to within round off errors.

8.43
Consider a Carnot-cycle heat pump with R-410a as the working fluid. Heat is
rejected from the R-410a at 40°C, during which process the R-410a changes from
saturated vapor to saturated liquid. The heat is transferred to the R-410a at -5°C.
a. Show the cycle on a T–s diagram.
b. Find the quality of the R-410a at the beginning and end of the isothermal heat
addition process at -5°C.
c. Determine the coefficient of performance for the cycle.
Solution:
a)
1
23
4
40
-5
T
s
b) From Table B.4.1, state 3 is
saturated liquid
s4 = s3 = 0.4473 kJ/kg K
= 0.1989 + x4(0.8477)
=> x4 = 0.293
State 2 is saturated vapor so from Table B.4.1
s1 = s2 = 0.9552 kJ/kg K = 0.1989 + x1(0.8477)
=> x1 = 0.892
c) β′ =
qH
wIN
=
TH
TH – TL
=
313.2
45 = 6.96

8.44
Do Problem 8.43 using refrigerant R-134a instead of R-410a.
Consider a Carnot-cycle heat pump with R-410a as the working fluid. Heat is
rejected from the R-410a at 40°C, during which process the R-410a changes from
saturated vapor to saturated liquid. The heat is transferred to the R-410a at -5°C.
a. Show the cycle on a T–s diagram.
b. Find the quality of the R-410a at the beginning and end of the isothermal heat
addition process at -5°C.
c. Determine the coefficient of performance for the cycle.
Solution:
a)
1
23
4
40
-5
T
s
b) From Table B.5.1, state 3 is
saturated liquid
s4 = s3 = 1.1909 kJ/kg K
= 0.9755 + x4(0.7534)
=> x4 = 0.2859
State 2 is saturated vapor so from Table B.5.1
s1 = s2 = 1.7123 kJ/kg K = 0.9755 + x1(0.7534)
=> x1 = 0.978
c) β′ =
qH
wIN
=
TH
TH – TL
=
313.2
45 = 6.96

8.45
One kilogram of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a
reversible isobaric process to 140°C. Find the work and heat transfer for this
process.
Solution:
Control mass.
m(u2 - u1) = 1Q2 - 1W2
Process: P = constant
⇒ 1W2 = mP(v2 - v1)
2
1
P
v
T
s
1 2
1W2 = 1 × 1000(0.1955 - 0.145) = 50.5 kJ
1Q2 = m(u2 - u1) + 1W2 = 1 × (1566.7 - 1391.3) + 50.5 = 225.9 kJ

8.46
A piston cylinder contains 0.25 kg of R-134a at 100 kPa. It will be compressed in
an adiabatic reversible process to 400 kPa and should be 70oC. What should the
initial temperature be?
C.V. R-134a which is a control mass.
Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T = 0
State 2: s2 = s1 = 1.9051 kJ/kgK
Work backwards from state 2 to state 1
State 1: 100 kPa & s1 => T1 = 26.4°C
P
v
1
2 2
1
T
s

8.47
Compression and heat transfer brings carbon dioxide in a piston/cylinder from
1400 kPa, 20oC to saturated vapor in an isothermal process. Find the specific heat
transfer and the specific work.
Solution:
m = constant
Energy Eq.5.11: u2 − u1 = 1q2 − 1w2
Entropy Eq.8.3: s2 - s1= ∫ dq/T = 1q2 /T
Process: T = C and assume reversible ⇒ 1q2 = T (s2 - s1)
u1 = 259.18 kJ/kg,
s1 = 1.0057 kJ/kg K
u2 = 258.16 kJ/kg,
s2 = 0.9984 kJ/kg K
P
v
1
2
T
2 1
T
s
1q2 = (273 + 20) × (0.9984 – 1.0057) = –2.14 kJ/kg
1w2 = 1q2 + u1 – u2 = –2.14 + 259.18 – 258.16
= –1.12 kJ/kg

8.48
One kilogram of carbon dioxide in a piston/cylinder at 120°C, 1400 kPa, shown in
Fig. P8.48, is expanded to 800 kPa in a reversible adiabatic process. Find the
work and heat transfer.
Solution:
Control mass: Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2
Entropy Eq.8.3: m(s2 - s1) = ∫ dQ/T ( = since reversible)
Process: 1Q2 = 0 ⇒ s2 = s1
State 1: (P, T) Table B.3.2, u1 = 398.83 kJ/kg, s1 = 1.8093 kJ/kg K
State 2: P2 , s2 ⇒ superheated vapor, u2 = 368.52 kJ/kg
1W2 = 1 × (398.83 - 368.52) = 30.3 kJ
2
1
P
v
T
s
1
2

8.49
A cylinder fitted with a piston contains ammonia at 50°C, 20% quality with a
volume of 1 L. The ammonia expands slowly, and during this process heat is
transferred to maintain a constant temperature. The process continues until all the
liquid is gone. Determine the work and heat transfer for this process.
Solution:
C.V. Ammonia in the cylinder.
1 2
T
s
50 C
NH3
o
Table B.2.1: T1 = 50°C, x1 = 0.20, V1 = 1 L
v1 = 0.001777 + 0.2 ×0.06159 = 0.014095 m3/kg
s1 = 1.5121 + 0.2 × 3.2493 = 2.1620 kJ/kg K
m = V1/v1 = 0.001/0.014095 = 0.071 kg
v2 = vg = 0.06336 m3/kg,
s2 = sg = 4.7613 kJ/kg K
Process: T = constant to x2 = 1.0, P = constant = 2.033 MPa
From the constant pressure process
1W2 = ⌡⌠PdV = Pm(v2 - v1) = 2033 × 0.071 × (0.06336 - 0.014095) = 7.11 kJ
From the second law Eq.8.3 with constant T
1Q2 = ⌡⌠TdS = Tm(s2 - s1) = 323.2 × 0.071(4.7613 - 2.1620) = 59.65 kJ
or 1Q2 = m(u2 - u1) + 1W2 = m(h2 - h1)
h1 = 421.48 + 0.2 × 1050.01 = 631.48 kJ/kg, h2 = 1471.49 kJ/kg
1Q2 = 0.071(1471.49 - 631.48) = 59.65 kJ

8.50
Water in a piston/cylinder at 400oC, 2000 kPa is expanded in a reversible
adiabatic process. The specific work is measured to be 415.72 kJ/kg out. Find the
final P and T and show the P-v and the T-s diagram for the process.
Solution:
C.V. Water, which is a control mass. Adiabatic so: 1q2 = 0
Energy Eq.5.11: u2 − u1 = 1q2 − 1w2 = -1w2
Entropy Eq.8.3: s2 - s1= ∫ dq/T = 0 (= since reversible)
State 1: Table B.1.3 u1 = 2945.21 kJ/kg; s1 = 7.127 kJ/kg K
State 2: (s, u): u2 = u1 - 1w2 = 2945.21 – 415.72 = 2529.49 kJ/kg
=> sat. vapor 200 kPa, T = 120.23°C
v
P
s
T
2
1
1
2

8.51
A piston cylinder has R-134a at –20oC, 100 kPa which is compressed to 500 kPa
in a reversible adiabatic process. Find the final temperature and the specific work.
Solution:
C.V. R-134a, Control mass of unknown size, adiabatic 1q2 = 0
Energy Eq.5.11: u2 − u1 = 1q2 − 1w2 = - 1w2
Entropy Eq.8.3: s2 − s1 = ∫ dq/T
Process: Adiabatic and reversible => s2 = s1
State 1: (T, P) B.5.2 u1 = 367.36 kJ/kg, s1 = 1.7665 kJ/kg K
State 2: (P, s) B.5.2 P2 = 500 kPa, s2 = s1 = 1.7665 kJ/kg K
very close at 30oC u2 = 398.99 kJ/kg
1w2 = u2 - u1 = 367.36 – 398.99 = -31.63 kJ/kg
2
1
P
v
T
s
1
2

8.52
A piston/cylinder has 2 kg water at 1000 kPa, 250°C which is now cooled with a
constant loading on the piston. This isobaric process ends when the water has
reached a state of saturated liquid. Find the work and heat transfer and sketch the
process in both a P-v and a T-s diagram.
Solution:
C.V. H2O
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T
Process: P = C => W = ∫ P dV = P(V2 − V1)
State 1: B.1.3 v1= 0.23268 m3/kg, s1= 6.9246 kJ/kg K, u1 = 2709.91 kJ/kg
State 2: B.1.2 v2 = 0.001127 m3/kg, s2 = 2.1386 kJ/kg K, u2 = 761.67 kJ/kg
1W2 = m P (v2 − v1) = 2 × 1000 (0.001127 – 0.23268) = -463.1 kJ
1Q2 = m(u2 − u1) + 1W2 = 2 (761.67 – 2709.91) – 463.1 = -4359.6 kJ
2
1P
v
T
s
12

8.53
One kilogram of water at 300°C expands against a piston in a cylinder until it
reaches ambient pressure, 100 kPa, at which point the water has a quality of
90.2%. It may be assumed that the expansion is reversible and adiabatic. What
was the initial pressure in the cylinder and how much work is done by the water?
Solution:
C.V. Water. Process: Rev., Q = 0
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 = − 1W2
Process: Adiabatic Q = 0 and reversible => s2 = s1
State 2: P2 = 100 kPa, x2 = 0.902 from Table B.1.2
s2 = 1.3026 + 0.902 × 6.0568 = 6.7658 kJ/kg K
u2 = 417.36 + 0.902 × 2088.7 = 2301.4 kJ/kg
State 1 At T1 = 300°C, s1 = 6.7658 Find it in Table B.1.3
⇒ P1 = 2000 kPa, u1 = 2772.6 kJ/kg
1W2 = m(u1 - u2) = 1(2772.6 – 2301.4) = 471.2 kJ
v
P
s
T
2
1
1
2
T1

8.54
Water at 1000 kPa, 250°C is brought to saturated vapor in a rigid container,
shown in Fig. P8.54. Find the final T and the specific heat transfer in this
isometric process.
Solution:
Energy Eq.5.11: u2 − u1 = 1q2 - 1w2
Process: v = constant => 1w2 = 0
State 1: (T, P) Table B.1.3 u1 = 2709.91 kJ/kg, v1 = 0.23268 m3/kg
State 2: x = 1 and v2 = v1 so from Table B.1.1 we see P2 ≅ 800 kPa
T2 = 170 + 5 × (0.23268 – 0.24283)/(0.2168 – 0.24283)
= 170 + 5 × 0.38993 = 171.95°C
u2 = 2576.46 + 0.38993 × (2580.19 – 2576.46) = 2577.9 kJ/kg
1q2 = u2 − u1 = 2577.9 – 2709.91 = −132 kJ/kg
v
P
s
T
2
1
1
2
v = C
Notice to get 1q2 = ∫ T ds we must know the function T(s) which we do
not readily have for this process.

8.55
Estimate the specific heat transfer from the area in the T-s diagram and compare it
to the correct value for the states and process in Problem 8.54.
Solution:
Energy Eq.5.11: u2 − u1 = 1q2 - 1w2
Process: v = constant => 1w2 = 0
State 1: (T, P) Table B.1.3 u1 = 2709.91 kJ/kg, v1 = 0.23268 m3/kg,
s1 = 6.9246 kJ/kg K
State 2: x = 1 and v2 = v1 so from Table B.1.1 we see P2 ≅ 800 kPa
T2 = 170 + 5 × (0.23268 – 0.24283)/(0.2168 – 0.24283)
= 170 + 5 × 0.38993 = 171.95°C
u2 = 2576.46 + 0.38993 × (2580.19 – 2576.46) = 2577.9 kJ/kg
s2 = 6.6663 + 0.38993 (6.6256 – 6.6663) = 6.6504 kJ/kg K
1q2 actual = u2 − u1 = 2577.9 – 2709.91 = −132 kJ/kg
Assume a linear variation of T versus s.
1q2 = ∫ T ds = area ≅
1
2 (T1 + T2)(s2 − s1)
=
1
2 (171.95 + (2 × 273.15) + 250)(6.6504 – 6.9246)
= -132.74 kJ/kg
very close i.e. the v = C curve is close to a straight line in the T-s diagram.
Look at the constant v curves in Fig. E.1. In the two-phase region they curve
slightly and more so in the region above the critical point.
v
P
s
T
2
1
1
2
v = C

8.56
An insulated cylinder fitted with a piston contains 0.1 kg of water at 100°C, 90%
quality. The piston is moved, compressing the water until it reaches a pressure of
1.2 MPa. How much work is required in the process?
Solution:
C.V. Water in cylinder.
Energy Eq.5.11: 1Q2 = 0 = m(u2 - u1) + 1W2
Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T = 0 (assume reversible)
State 1: 100°C, x1 = 0.90:
Table B.1.1,
s1 = 1.3068 + 0.90×6.048
= 6.7500 kJ/kg K
P
v
1
2
2
1
T
s
u1 = 418.91 + 0.9 × 2087.58 = 2297.7 kJ/kg
State 2: Given by (P, s) B.1.3


s2 = s1 = 6.7500
P2 = 1.2 MPa ⇒


 T2 = 232.3°C
u2 = 2672.9
1W2 = -m(u2 – u1) = -0.1(2672.9 - 2297.7) = -37.5 kJ

8.57
A closed tank, V = 10 L, containing 5 kg of water initially at 25°C, is heated to
175°C by a heat pump that is receiving heat from the surroundings at 25°C.
Assume that this process is reversible. Find the heat transfer to the water and the
change in entropy.
Solution:
C.V.: Water from state 1 to state 2.
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
Process: constant volume (reversible isometric) so 1W2 = 0
State 1: v1 = V/m = 0.002 from Table B.1.1
x1 = (0.002 - 0.001003)/43.358 = 0.000023
u1 = 104.86 + 0.000023 × 2304.9 = 104.93 kJ/kg
s1 = 0.3673 + 0.000023 × 8.1905 = 0.36759 kJ/kg K
Continuity eq. (same mass) and V = C fixes v2
State 2: T2, v2 = v1 so from Table B.1.1
x2 = (0.002 - 0.001121)/0.21568 = 0.004075
u2 = 740.16 + 0.004075 × 1840.03 = 747.67 kJ/kg
s2 = 2.0909 + 0.004075 × 4.5347 = 2.1094 kJ/kg K
Energy eq. has W = 0, thus provides heat transfer as
1Q2 = m(u2 - u1) = 3213.7 kJ
The entropy change becomes
m(s2 - s1) = 5(2.1094 – 0.36759) = 8.709 kJ/K
P
v
1
2
T
2
1
T
s
Notice we do not perform the integration ∫ dQ/T to find change in s as the
equation for the dQ as a function of T is not known.

8.58
A piston/cylinder has 2 kg of R-410a at 60°C, 100 kPa which is compressed to
1000 kPa. The process happens so slowly that the temperature is constant. Find
the heat transfer and work for the process assuming it to be reversible.
Solution:
CV : R-410a Control Mass
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 ;
Process: T = constant and assume reversible process
1: (T,P), Table B.4.2: v1 = 0.37833 m3/kg, u1 = 309.4 kJ/kg,
s1 = 1.4910 kJ/kg K
2: (T,P), Table B.4.2: v2 = 0.03470 m3/kg, u2 = 301.04 kJ/kg,
s2 = 1.2019 kJ/kg K
21
P
v
T
s
1
2
T
From the entropy equation (2nd law)
1Q2 = mT(s2 − s1) = 2 × 333.15 (1.2019 − 1.4910) = −192.63 kJ
1W2 = 1Q2 − m(u2 − u1) = −192.63 − 2(301.04 − 309.4) = −175.9 kJ

8.59
A heavily-insulated cylinder fitted with a frictionless piston, as shown in Fig.
P8.59 contains ammonia at 5°C, 92.9% quality, at which point the volume is 200
L. The external force on the piston is now increased slowly, compressing the
ammonia until its temperature reaches 50°C. How much work is done by the
ammonia during this process?
Solution:
C.V. ammonia in cylinder, insulated so assume adiabatic Q = 0.
Cont.Eq.: m2 = m1 = m ;
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
State 1: T1 = 5o
C, x1 = 0.929, V1 = 200 L = 0.2 m3
Table B.2.1 saturated vapor, P1 = Pg = 515.9 kPa
v1 = vf + x1vfg = 0.001583 + 0.929 × 0.2414 = 0.2258 m3/kg,
u1 = uf + x1ufg = 202.8 + 0.929 × 1119.2 = 1242.5 kJ/kg
s1 = sf + x1sfg = 0.7951 + 0.929 × 4.44715 = 4.9491 kJ/kg K,
m1 = V1/v1 = 0.2 / 0.2258 = 0.886 kg
Process: 1 2 Adiabatic 1Q2 = 0 & Reversible => s1 = s2
State 2: T2 = 50o
C, s2 = s1 = 4.9491 kJ/kg K
superheated vapor, interpolate in Table B.2.2 =>
P2 = 1600 kPa, u2 = 1364.9 kJ/kg
v
P
s
T
2
11
2
T2
Energy equation gives the work as
1W2 = m(u1 - u2) = 0.886 ( 1242.5 – 1364.9) = −108.4 kJ

8.60
A heavily insulated cylinder/piston contains ammonia at 1200 kPa, 60°C. The
piston is moved, expanding the ammonia in a reversible process until the
temperature is −20°C. During the process 600 kJ of work is given out by the
ammonia. What was the initial volume of the cylinder?
C.V. ammonia. Control mass with no heat transfer.
State 1: Table B.2.2 v1 = 0.1238 m3/kg, s1 = 5.2357 kJ/kg K
u1 = h - Pv = 1553.3 - 1200×0.1238 = 1404.9 kJ/kg
Entropy Eq.: m(s2 − s1) = ∫ dQ/T + 1S2 gen
Process: reversible (1S2 gen = 0) and adiabatic (dQ = 0) => s2 = s1
P
v
1
2 2
1
T
s
State 2: T2, s2 ⇒ x2 = (5.2357 - 0.3657)/5.2498 = 0.928
u2 = 88.76 + 0.928×1210.7 = 1211.95 kJ/kg
1Q2 = 0 = m(u2 - u1) + 1W2 = m(1211.95 - 1404.9) + 600
⇒ m = 3.110 kg
V1 = mv1 = 3.11 × 0.1238 = 0.385 m3

8.61
Water at 1000 kPa, 250°C is brought to saturated vapor in a piston/cylinder with
an isothermal process. Find the specific work and heat transfer. Estimate the
specific work from the area in the P-v diagram and compare it to the correct
value.
Solution:
Energy Eq.:5.11 m(u2 − u1) = 1Q2 − 1W2
Process: T = constant, reversible
State 2: (T, x) Table B.1.1 P2 = 3973 kPa
v2 = 0.05013 m3/kg, u2 = 2602.37 kJ/kg, s2 = 6.0729 kJ/kg K
P
v
1
2
T
2 1
T
s
From the entropy equation
1q2 = ∫ T ds = T(s2 − s1) = (250 + 273) (6.0729 – 6.9246) = -445.6 kJ/kg
1w2 = 1q2 + u1 − u2 = -445.6 + 2709.91 – 2602.37 = -338 kJ/kg
Estimation of the work term from the area in the P-v diagram
1w2 area ≅
1
2 (P1+P2)(v2 − v1) =
1
2(1000 + 3973)(0.05013 – 0.23268)
= –454 kJ/kg
Not extremely accurate estimate; P-v curve not linear more like Pv = constant
as curve has positive curvature the linear variation over-estimates area.

8.62
A rigid, insulated vessel contains superheated vapor steam at 3 MPa, 400°C. A
valve on the vessel is opened, allowing steam to escape. The overall process is
irreversible, but the steam remaining inside the vessel goes through a reversible
adiabatic expansion. Determine the fraction of steam that has escaped, when the
final state inside is saturated vapor.
C.V.: steam remaining inside tank. Rev. & Adiabatic (inside only)
Cont.Eq.: m2 = m1 = m ; Energy Eq.: m(u2 − u1) = 1Q2 − 1W2
Entropy Eq.: m(s2 − s1) = ∫ dQ/T + 1S2 gen
P
v
1
2 2
1
T
s
C.V. m2
Rev ( 1S2 gen = 0) Adiabatic ( Q = 0) => s2 = s1 = 6.9212 = sG at T2
⇒ T2 = 141°C, v2 = vg at T2
= 0.4972 m3/kg
me
m1
=
m1-m2
m1
= 1 -
m2
m1
= 1 -
v1
v2
= 1 -
0.09936
0.4972 = 0.80

8.63
A cylinder containing R-134a at 10°C, 150 kPa, has an initial volume of 20 L. A
piston compresses the R-134a in a reversible, isothermal process until it reaches
the saturated vapor state. Calculate the required work and heat transfer to
accomplish this process.
Solution:
C.V. R-134a.
Cont.Eq.: m2 = m1 = m ;
Energy Eq.:5.11 m(u2 − u1) = 1Q2 − 1W2
Process: T = constant, reversible
State 1: (T, P) Table B.5.2 u1 = 388.36 kJ/kg, s1 = 1.822 kJ/kg K
m = V/v1 = 0.02/0.148283 = 0.1349 kg
State 2: (10°C, sat. vapor)
Table B.5.1
u2 = 383.67 kJ/kg,
s2 = 1.7218 kJ/kg K
P
v
1
2
T
2 1
T
s
As T is constant we can find Q by integration as
1Q2 = ⌡⌠Tds = mT(s2 - s1) = 0.1349 × 283.15 × (1.7218 - 1.822) = -3.83 kJ
The work is then from the energy equation
1W2 = m(u1 - u2) + 1Q2 = 0.1349 × (388.36 - 383.67) - 3.83 = -3.197 kJ

8.64
an adiabatic process. Find the final T and the specific work. Estimate the specific
work from the area in the P-v diagram and compare it to the correct value.
Solution:
C.V. Water, which is a control mass with unknown size.
Energy Eq.5.11: u2 – u1 = 0 – 1w2
Entropy Eq.8.3: s2 – s1 = ∫ dq/T = 0
Process: Adiabatic 1q2 = 0 and reversible as used above
State 1: Table B.1.3 v1 = 0.23268 m3/kg, u1 = 2709.91 kJ/kg,
s1 = 6.9246 kJ/kg K
State 2: Table B.1.1 x = 1 and s2 = s1 = 6.9246 kJ/kg K
=> T2 ≅ 140.56°C, P2 ≅ 367.34 kPa, v2 = 0.50187 m3/kg,
u2 ≅ 2550.56 kJ/kg
1w2 = u1 – u2 = 2709.91 – 2550.56 = 159.35 kJ/kg
Now estimate the work term from the area in the P-v diagram
1w2 ≅
1
2 (P1 + P2)(v2 − v1)
=
1
2 (1000 + 367.34)(0.50187 – 0.23268)
= 184 kJ/kg
The s = constant curve is not a straight line in the the P-v diagram, notice the
straight line overestimates the area slightly.
v
P
s
T
2
1
1
2

8.65
A piston/cylinder contains 2 kg water at 200°C, 10 MPa. The piston is slowly
moved to expand the water in an isothermal process to a pressure of 200 kPa. Any
heat transfer takes place with an ambient at 200°C and the whole process may be
assumed reversible. Sketch the process in a P-V diagram and calculate both the
heat transfer and the total work.
Solution:
C.V. Water.
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T = 1Q2 / T
Process: T = C and reversible as used in entropy equation
State 1: Table B.1.4 : v1 = 0.001148 m3/kg, u1 = 844.49 kJ/kg,
s1 = 2.3178 kJ/kg K,
V1 = mv1 = 0.0023 m3
State 2: Table B.1.3 : v2 = 1.08034 m3/kg, u2 = 2654.4 kJ/kg
s2 = 7.5066 kJ/kg K
V2 = mv2 = 2.1607 m3,
P
v
1
2
T
21
T
s
From the entropy equation and the process equation
1Q2 = mT(s2 − s1) = 2 × 473.15 (7.5066 - 2.3178) = 4910 kJ
1W2 = 1Q2 - m(u2 - u1) = 1290.3 kJ

8.66
an isobaric process. Find the specific work and heat transfer. Estimate the specific
heat transfer from the area in the T-s diagram and compare it to the correct value.
Solution:
C.V. H2O
Energy Eq.5.11: u2 − u1 = 1q2 − 1w2
Process: P = C => w = ∫ P dv = P(v2 − v1)
1: B1.3 v1= 0.23268 m3/kg, s1= 6.9246 kJ/kgK, u1 = 2709.91 kJ/kg
2: B1.3 v2 = 0.19444 m3/kg, s2 = 6.5864 kJ/kg K, u2 = 2583.64 kJ/kg,
T2 = 179.91°C
1w2 = P (v2 − v1) = 1000 (0.1944 – 0.23268) = -38.28 kJ/kg
1q2 = u2 − u1 + 1w2 = 2583.64 – 2709.91 – 38.28 = -164.55 kJ/kg
Now estimate the heat transfer from the T-s diagram.
1q2 = ∫ T ds = AREA ≅
1
2 (T1 + T2)(s2 − s1)
=
1
2 (250 + 179.91 + 2 × 273.15)(6.5864 – 6.9246)
= 488.105 × (-0.3382) = -165.1 kJ/kg
very close approximation. The P = C curve in the T-s diagram is nearly a
straight line. Look at the constant P curves on Fig.E.1. Up over the critical
point they curve significantly.
2
1P
v
T
s
12

Entropy of a liquid or a solid

8.67
Two 5 kg blocks of steel, one at 250oC the other at 25oC, come in thermal
contact. Find the final temperature and the change in entropy of the steel?
C.V. Both blocks, no external heat transfer, C from Table A.3.
Energy Eq.: U2 – U1 = mA(u2 – u1)A + mB(u2 – u1)B = 0 – 0
= mAC(T2 – TA1) + mBC(T2 – TB1)
T2 =
mATA1 + mBTB1
mA + mB
=
1
2 TA1 +
1
2 TB1 = 137.5oC
Entropy Eq.8.39: S2 – S1 = mA(s2 – s1)A + mB(s2 – s1)B = 1S2 gen
S2 – S1 = mAC ln
T2
TA1
+ mBC ln
T2
TB1
= 5 × 0.46 ln
137.5 + 273.15
250 + 273.15 + 5 × 0.46 ln
137.5 + 273.15
298.15
= -0.5569 + 0.7363 = 0.1794 kJ/K
A
B
Heat transfer over a finite
temperature difference is an
irreversible process

8.68
A large slab of concrete, 5 × 8 × 0.3 m, is used as a thermal storage mass in a
solar-heated house. If the slab cools overnight from 23°C to 18°C in an 18°C
house, what is the net entropy change associated with this process?
Solution:
C.V.: Control mass concrete.
V = 5 × 8 × 0.3 = 12 m3
m = ρV = 2200 × 12 = 26 400 kg
Energy Eq.: m(u2 - u1) = 1Q2 - 1W2
Entropy Eq.: m(s2 - s1) =
1Q2
T0
+ 1S2 gen
Process: V = constant so 1W2 = 0
Use heat capacity (Table A.3) for change in u of the slab
1Q2 = mC∆T = 26400 kg × 0.88 kJ/kg-K × (-5) K = -116 160 kJ
We add all the storage changes as in Eq.8.39:
∆Sslab = m(s2 - s1) = m C ln
T2
T1
= 26400 kg × 0.88 kJ/kg-K × ln
291.2
296.2 = -395.5 kJ/K
∆SSURR =
-1Q2
T0
=
+116 160
291.2 = +398.9 kJ/K
∆SNET = -395.5 + 398.9 = +3.4 kJ/K
= m(s2 - s1) −
1Q2
T0
= 1S2 gen

8.69
A piston cylinder has constant pressure of 2000 kPa with water at 20oC. It is now
heated up to 100oC. Find the heat transfer and the entropy change using the steam
tables. Repeat the calculation using constant heat capacity and incompressibility.
Solution:
C.V. Water. Constant pressure heating.
Energy Eq.5.11: u2 - u1 = 1q2 − 1w2
Entropy Eq.8.3: s2 - s1 = 1q2 / TSOURCE + 1s2 gen
Process: P = P1 => 1w2 = P(v2 - v1)
The energy equation then gives the heat transfer as
1q2= u2 - u1 + 1w2 = h2 - h1
Steam Tables B.1.4: h1 = 85.82 kJ/kg; s1= 0.2962 kJ/kg K
h2 = 420.45 kJ/kg; s2 = 1.3053 kJ/kg K
1q2= h2 - h1= -85.82 + 420.45 = 334.63 kJ/kg
s2 - s1= 1.3053 – 0.2962 = 1.0091 kJ/kg K
Now using values from Table A.4: Liquid water Cp = 4.18 kJ/kg K
h2 - h1 ≅ Cp(T2 – T1) = 4.18 × 80 = 334.4 kJ/kg
s2 - s1 ≅ Cp ln(T2/T1) = 4.18 ln
373.15
293.15 = 1.0086 kJ/kg K
Approximations are very good

8.70
A 4 L jug of milk at 25°C is placed in your refrigerator where it is cooled down to
the refrigerators inside constant temperature of 5°C. Assume the milk has the
property of liquid water and find the entropy change of the milk.
Solution:
C.V. Jug of milk. Control mass at constant pressure.
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
Entropy Eq.8.37: m(s2 − s1) = ∫ dQ/T + 1S2 gen
State 1: Table B.1.1: v1 ≅ vf = 0.001003 m3/kg, sf = 0.3673 kJ/kg K
m = V/v = 0.004 m3/ 0.001003 (m3/kg) = 3.988 kg
State 2: Table B.1.1: s = sf = 0.0761 kJ/kg K
The change of entropy becomes
S2 − S1= m(s2 − s1) = 3.988 (0.0761 − 0.3673)
= − 1.1613 kJ/K
MILK
cb
5 C
AIR
o

8.71
A foundry form box with 25 kg of 200°C hot sand is dumped into a bucket with
50 L water at 15°C. Assuming no heat transfer with the surroundings and no
boiling away of liquid water, calculate the net entropy change for the process.
Solution:
C.V. Sand and water, constant pressure process
msand(u2 - u1)sand + mH2O(u2 - u1)H2O = -P(V2 - V1)
⇒ msand∆hsand + mH2O∆hH2O = 0
For this problem we could also have said that the work is nearly zero as the
solid sand and the liquid water will not change volume to any measurable
extent. Now we get changes in u's instead of h's. For these phases CV = CP =
C which is a consequence of the incompressibility. Now the energy equation
becomes
msandCsand∆Tsand + mH2OCH2O∆TH2O = 0
25 × 0.8×(T2 - 200) + (50×10-3/0.001001) × 4.184 × (T2 - 15) = 0
T2 = 31.2°C
∆S = msand( s2 - s1) + mH2O ( s2 - s1)
= msand Csand ln(T2/T1) + mH2O CH2O ln(T2/T1)
= 25 × 0.8 ln





304.3
473.15 + 49.95 × 4.184 ln





304.3
288.15 = 2.57 kJ/K
Box holds the sand for
form of the cast part

8.72
In a sink 5 liters of water at 70oC is combined with 1 kg aluminum pots, 1 kg of
flatware (steel) and 1 kg of glass all put in at 20oC. What is the final uniform
temperature and change in stored entropy neglecting any heat loss and work?
Entropy Eq.: S2 – S1 = ∫ dQ/T + 1S2 gen
For the water: vf = 0.001023 m3/kg, V = 5 L = 0.005 m3; m = V/v = 4.8876 kg
For the liquid and the metal masses we will use the specific heats (Tbl A.3, A.4)
so
∑miCv i = 4.8876 × 4.18 + 1 × 0.9 + 1 × 0.46 + 1 × 0.8 = 22.59 kJ/K
Energy Eq.: 22.59 T2 = 4.8876 × 4.18 ×70 + (1× 0.9 + 1× 0.46 + 1× 0.8) × 20
= 1430.11 + 43.2
T2 = 65.2oC
S2 – S1 = ∑ mi(s2 – s1)i = ∑ mi Ci ln
T2
Ti1
= 4.8876 × 4.18 × ln
65.22 + 273.15
70 + 273.15
+ 1 × (0.9 + 0.46 + 0.8) ln
65.22 + 273.15
20 + 273.15
= -0.28659 + 0.30986 = 0.02327 kJ/K

8.73
A 5-kg steel container is cured at 500oC. An amount of liquid water at 15oC, 100
kPa is added to the container so a final uniform temperature of the steel and the
water becomes 75oC. Neglect any water that might evaporate during the process
and any air in the container. How much water should be added and how much was
the entropy changed?
CV. The steel and the water no external heat transfer nor any work.
Energy Eq.: mH2O( u2 − u1)H2O + mst( u2 − u1) = 0
mH2o( 313.87 – 62.98) + mstC ( T2 - T1) = 0
mH2O × 250.89 + 5 × 0.46 × (75 - 500) = 0
mH2O = 977.5/250.89 = 3.896 kg
Entropy Eq. 8.37: mH2O ( s2-s1) + mst( s2 - s1) = ∅ + 1S2 gen
S2 − S1 = 3.896 (1.0154 – 0.2245) + 5 × 0.46 ln
75 + 273
773
= 3.0813 – 1.8356 = 1.246 kJ/K
cb

8.74
A pan in an autoshop contains 5 L of engine oil at 20oC, 100 kPa. Now 2 L of hot
100oC oil is mixed into the pan. Neglect any work term and find the final
temperature and the entropy change.
Solution:
Since we have no information about the oil density, we assume the same for
both from Table A.4: ρ = 885 kg/m3
Energy Eq.: m2u2 – mAuA – mBuB ≅ 0 – 0
∆u ≅ Cv∆T so same Cv = 1.9 kJ/kg K for all oil states.
T2 =
mA
m2
TA +
mB
m2
TB =
5
7 × 20 +
2
7 × 100 = 42.868oC = 316.02 K
S2 - S1 = m2s2 − mAsA – mBsB = mA(s2 – sA) + mB(s2 – sB)
= 0.005 × 885 × 1.9 ln
316.02
293.15 + 0.002 × 885 × 1.9 ln
316.02
373.15
= 0.6316 – 0.5588 = + 0.0728 kJ/K
Entropy generation is the total change in S, recall Eq.8.39, no external Q
Oils shown before
mixed to final
uniform state.

8.75
A computer CPU chip consists of 50 g silicon, 20 g copper, 50 g polyvinyl
chloride (plastic). It heats from 15oC to 70oC as the computer is turned on. How
much did the entropy increase?
C.V. CPU chip. The process has electrical work input and no heat transfer.
Entropy Eq.: S2 - S1 = ∑mi(s2 - s1)i = ∫ dQ/T + 1S2 gen = 1S2 gen
For the solid masses we will use the specific heats, Table A.3, and they all have
the same temperature so
∑mi(s2 - s1)i = ∑miCi ln(T2 / T1)i = ln (T2/T1) ∑miCi
∑miCi = 0.05 × 0.7 + 0.02 × 0.42 + 0.05 × 0.96 = 0.0914 kJ/K
S2 - S1 = 0.0914 × ln (343.15 / 288.15) = 0.016 kJ/K

8.76
A 12 kg steel container has 0.2 kg superheated water vapor at 1000 kPa, both at
200oC. The total mass is now cooled to ambient temperature 30oC. How much
heat transfer was taken out and what is the steel-water entropy change?
Solution:
C.V.: Steel and the water, control mass of constant volume.
Energy Eq.5.11: U2 - U1 = 1Q2 - 1W2
Process: V = constant => 1W2 = 0
State 1: H20 Table B.1.3: u1 = 2621.9 kJ/kg, v1 = 0.20596 m3/kg,
s1 = 6.6939 kJ/kg K
State 2: H20: T2 , v2 = v1 => from Table B.1.1
x2 =
v - vf
vfg
=
0.20596 – 0.001004
32.8922 = 0.006231
u2 = 125.77 + x2 × 2290.81 = 140.04 kJ/kg
s2 = 0.4369 + x2 × 8.0164 = 0.48685 kJ/kg K
1Q2 = m(u2 − u1) = msteelCsteel (T2 – T1 ) + mH2O (u2 - u1) H2O
= 12 × 0.46 (30 – 200) + 0.2 (140.04 –2621.9)
= -1434.8 kJ
Entropy changes from Eq.8.11 and the water tables
S2 - S1 = m2 s2 - m1s1 = msteelCsteel ln (
T2
T1
) + mH2O (s2- s1)H2O
=12 × 0.46 ln (
303.15
473.15 ) + 0.2 (0.48685 – 6.6939)
= –2.4574 – 1.2414
= –3.699 kJ/K

8.77
Two kg of liquid lead initially at 500°C are poured into a form. It then cools at
constant pressure down to room temperature of 20°C as heat is transferred to the
room. The melting point of lead is 327°C and the enthalpy change between the
phases, hif , is 24.6 kJ/kg. The specific heats are in Tables A.3 and A.4. Calculate
the net entropy change for this process.
Solution:
C.V. Lead, constant pressure process
mPb(u2 - u1)Pb = 1Q2 - P(V2 - V1)
We need to find changes in enthalpy (u + Pv) for each phase separately and
then add the enthalpy change for the phase change.
Consider the process in several steps:
Cooling liquid to the melting temperature
Solidification of the liquid to solid
Cooling of the solid to the final temperature
1Q2 = mPb(h2 - h1) = mPb(h2 - h327,sol - hif + h327,f - h500)
= 2 × (0.138 × (20 - 327) - 24.6 + 0.155 × (327 - 500))
= -84.732 - 49.2 - 53.63 = -187.56 kJ
∆SCV = mPb[Cp solln(T2/600) - (hif/600) + CP liqln(600/T1)]
= 2 × [0.138 ln
293.15
600 -
24.6
600 + 0.155 ln
600
773.15 ] = -0.358 kJ/K
∆SSUR = -1Q2/T0 = 187.56/293.15 = 0.64 kJ/K
The net entropy change from Eq.8.39 is equivalent to total entropy generation
∆Snet = ∆SCV + ∆SSUR = 0.282 kJ/K

8.78
Find the total work the heat engine can give out as it receives energy from the
rock bed as described in Problem 7.65 (see Fig.P 8.78). Hint: write the entropy
balance equation for the control volume that is the combination of the rock bed
and the heat engine.
Solution:
To get the work we must integrate over the process or do the 2nd
law for a
control volume around the whole setup out to T0
C.V. Heat engine plus rock bed out to T0. W and QL goes out.
W
Q
H Q
L
HE
C.V.
Energy Eq.5.11: (U2 − U1)rock = – QL – W
Entropy Eq.8.3: (S2 − S1)rock = −
QL
T0
= mC ln (
T2
T1
)
= 5500 × 0.89 ln
290
400 = −1574.15 kJ/K
QL = −T0 (S2 − S1)rock = −290 (−1574.15) = 456 504 kJ
The energy drop of the rock −(U2 − U1)rock equals QH into heat engine
(U2−U1)rock = mC (T2−T1) = 5500 ×0.89 (290 − 400) = −538 450 kJ
W = −(U2 − U1)rock − QL = 538450 − 456504 = 81 946 kJ

8.79
A 5-kg aluminum radiator holds 2 kg of liquid R-134a at -10C. The setup is
brought indoors and heated with 220 kJ. Find the final temperature and the
change in entropy of all the mass.
Solution:
C.V. The aluminum radiator and the R-134a.
Energy Eq.5.11: m2u2 – m1u1 = 1Q2 – 0
Process: No change in volume so no work as used above.
The energy equation now becomes (summing over the mass)
mal (u2 - u1)al + mR134a (u2 - u1)R134a = 1Q2
Use specific heat from Table A.3 and A.4
malCal (T2 - T1) + m R134aC R134a ln (T2 - T1) = 1Q2
T2 - T1 = 1Q2 / [malCal + m R134aC R134a ]
= 220 / [5 × 0.9 + 2 × 1.43] = 29.89oC
T2 = -10 + 29.89 = 19.89oC
Entropy change for solid (A.3) and liquid (A.4) from Eq.8.11
S2 − S1 = mal (s2 − s1)al + mR134a (s2 − s1)R134a
= malCal ln (T2/T1) + m R134aC R134a ln (T2/T1)
= (5 × 0.9 + 2 × 1.43) ln
(19.89 + 273.15)
-10 + 273.15
= 0.792 kJ/K
Q1 2

Entropy of ideal gases

8.80
Air inside a rigid tank is heated from 300 to 350 K. Find the entropy increase s2 -
s1? What if it is from 1300 to 1350 K?
Process: V = C 1W2 = Ø
Entropy change from Eq.8.17:
a) s2 − s1 = Cvo ln (
T2
T1
) = 0.717 ln





350
300 = 0.1105 kJ/kgK
b) s2 − s1 = Cvo ln (
T2
T1
) = 0.717 ln





1350
1300 = 0.02706 kJ/kgK
From A.7:
case a) Cv ≈ ∆u / ∆T = 36/50 = 0.72 kJ/kg K , see A.5
case b) Cv ≈ ∆u / ∆T = 45.2/50 = 0.904 kJ/kg K (25 % higher)
so result should have been 0.0341 kJ/kgK

8.81
A piston/cylinder setup contains air at 100 kPa, 400 K which is compressed to a
final pressure of 1000 kPa. Consider two different processes (i) a reversible
adiabatic process and (ii) a reversible isothermal process. Show both processes in
P-v and a T-s diagram. Find the final temperature and the specific work for both
processes.
Solution:
C.V. Air, control mass of unknown size and mass.
Energy Eq.5.11: u2 – u1 = 1q2 – 1w2
Entropy Eq.8.14: s2 – s1 = ∫ dq/T + 1s2 gen
Process: Reversible 1s2 gen = 0
i) dq = 0 so 1q2 = 0
ii) T = C so ∫ dq/T = 1q2/T
i) For this process the entropy equation reduces to:
s2 – s1 = 0 + 0 so we have constant s, an isentropic process.
The relation for an ideal gas, constant s and k becomes Eq.8.32
T2 = T1( P2 / P1)
k-1
k
= 400





1000
100
0.4
1.4
= 400 × 10
0.28575
= 772 K
From the energy equation we get the work term
1w2 = u1 – u2 = Cv(T1 – T2) = 0.717(400 – 772) = -266.7 kJ/kg
ii) For this process T2 = T1 so since ideal gas we get
u2 = u1 also s°
T2
= s°
T1
=> Energy Eq.: 1w2 = 1q2
Now from the entropy equation we solve for 1q2
1w2 = 1q2 = T(s2 – s1) = T[s°
T2
– s°
T1
– R ln
P2
P1
] = −RT ln
P2
P1
= − 0.287 × 400 ln 10 = −264 kJ/kg
P
v
1
2ii
2ii
1
T
s
1
P
P
2
2i
2i

8.82
Prove that the two relations for changes in s, Eqs.8.16 and 8.17 are equivalent
once we assume constant specific heat. Hint: recall the relation for specific heat in
Eq. 5.27.
From Eq. 5.27: Cpo = Cvo + R
Start with Eq.8.16: s2 – s1 = Cpo ln
T2
T1
– R ln
P2
P1
Now substitute Eq.5.27 to get:
s2 – s1 = (Cvo + R) ln
T2
T1
– R ln
P2
P1
= Cvo ln
T2
T1
– R [ ln
P2
P1
– ln
T2
T1
]
= Cvo ln
T2
T1
– R ln (
P2
P1
T1
T2
)
Use the ideal gas law Pv = RT for both states to get the ratio
P2v2
RT2
=
P1v1
RT1
P2
P1
T1
T2
=
v1
v2
so then we get to Eq. 8.17 as
s2 – s1 = Cvo ln
T2
T1
– R ln
v1
v2
= Cvo ln
T2
T1
+ ln
v2
v1

8.83
Assume an ideal gas with constant specific heats. Show the functions T(s, P = C)
and T(s, v = C) mathematically and sketch them in a T-s diagram.
From Eq 8.17 when P2 = P1
s2 – s1 = Cpo ln
T2
T1
so we can solve for T2 as a function of s2 as
T2 = T1 exp [
s2 – s1
Cpo
]
and it is an exponential function. For the other function use Eq.8.16 with v2 = v1
s2 – s1 = Cvo ln
T2
T1
so we can solve for T2 as a function of s2 as
T2 = T1 exp [
s2 – s1
Cvo
]
which is also an exponential function. Since Cpo is larger than Cvo the slope of
the first function is lower than the slope of the second function.
1
T
s
1
v
P
1

8.84
Water at 150°C, 400 kPa, is brought to 1200°C in a constant pressure process.
Find the change in the specific entropy, using a) the steam tables, b) the ideal gas
water Table A.8, and c) the specific heat from A.5.
Solution:
a)
State 1: Table B.1.3 Superheated vapor s1 = 6.9299 kJ/kgK
State 2: Table B.1.3 s2 = 9.7059 kJ/kgK
s2 - s1 = 9.7059 - 6.9299 = 2.776 kJ/kgK
b)
Table A.8 at 423.15 K: s
o
T1 = 11.13891 kJ/kgK
Table A.8 at 1473.15 K: s
o
T2 = 13.86383 kJ/kgK
s2 - s1 = s
o
T2 – s
o
T1 – R ln
P2
P1
= s
o
T2 – s
o
T1
s
o
T2 – s
o
T1 = 13.86383 - 11.13891 = 2.72492 kJ/kgK
c) Table A.5: Cpo = 1.872 kJ/kgK
s2 - s1 ≈ Cpo ln (
T2
T1
) = 1.872 ln (
1473.15
423.15 ) = 2.3352 kJ/kgK
Notice how the average slope
from 150°C to 1200°C is higher
than the one at 25°C ( = Cpo)
h
h
1200 Slope at
25°C
150
h T
25 1200150

8.85
R-410a at 400 kPa is brought from 20oC to 120oC in a constant pressure process.
Evaluate the change in specific entropy using Table B.4 and using ideal gas with
Cp = 0.81 kJ/kgK.
Table B.4.2 s1 = 1.2108 kJ/kg-K, s2 = 1.4788 kJ/kg-K
s2 – s1 = 1.4788 – 1.2108 = 0.268 kJ/kg-K
Eq. 8.17: s2 - s1 ≈ Cpo ln
T2
T1
= 0.81 ln
393.15
293.15 = 0.238 kJ/kg-K
Two explanations for the difference are as the average temperature is higher than
25oC we could expect a higher value of the specific heat and secondly it is not an
ideal gas (if you calculate Z = Pv/RT = 0.94).

8.86
R-410a at 300 kPa, 20oC is brought to 500 kPa, 200oC in a constant volume
process. Evaluate the change in specific entropy using Table B.4 and using ideal
gas with Cv = 0.695 kJ/kgK.
Table B.4.2 s1 = 1.2485 kJ/kg-K, v1 = 0.10720 m3/kg
s2 = 1.6413 kJ/kg-K, v2 = 0.10714 m3/kg (very close to v1)
s2 – s1 = 1.6413 – 1.2485 = 0.3928 kJ/kg-K
Eq. 8.16: s2 - s1 ≈ Cvo ln
T2
T1
= 0.695 ln
473.15
293.15 = 0.333 kJ/kg-K
Two explanations for the difference are as the average temperature is higher than
25oC we could expect a higher value of the specific heat and secondly it is not an
ideal gas (if you calculate Z = Pv/RT = 0.96).

8.87
A mass of 1 kg of air contained in a cylinder at 1.5 MPa, 1000 K, expands in a
reversible isothermal process to a volume 10 times larger. Calculate the heat
transfer during the process and the change of entropy of the air.
Solution:
C.V. Air, control mass.
Energy Eq. 5.11: m(u2 - u1) = 1Q2 - 1W2 = 0
Process: T = constant so with ideal gas => u2 = u1
P
v
1
2
21
T
s
1P
P
2
From the process equation and ideal gas law
PV = mRT = constant
we can calculate the work term as in Eq.4.5
1Q2 = 1W2 = ⌡⌠PdV = P1V1 ln (V2/V1) = mRT1 ln (V2/V1)
= 1 × 0.287 × 1000 ln (10) = 660.84 kJ
The change of entropy from Eq.8.3 is
∆Sair = m(s2 - s1) = 1Q2/T = 660.84/1000 = 0.661 kJ/K
If instead we use Eq.8.17 we would get
∆Sair = m(s2 - s1) = m(Cvo ln
T2
T1
+ R ln
v2
v1
)
= 1 [ 0 + 0.287 ln(10) ] = 0.661 kJ/K
consistent with the above result.

8.88
Consider a small air pistol with a cylinder volume of 1 cm3 at 250 kPa, 27°C. The
bullet acts as a piston initially held by a trigger. The bullet is released so the air
expands in an adiabatic process. If the pressure should be 100 kPa as the bullet
leaves the cylinder find the final volume and the work done by the air.
Solution:
C.V. Air. Assume a reversible, adiabatic process.
Energy Eq.5.11: u2 - u1 = 0 − 1w2 ;
Entropy Eq.8.37: s2 - s1 = ∫ dq/T + 1s2 gen = 0/
State 1: (T1,P1) State 2: (P2, ?)
So we realize that one piece of information is needed to get state 2.
Process: Adiabatic 1q2 = 0 Reversible 1s2 gen = 0
With these two terms zero we have a zero for the entropy change. So this is a
constant s (isentropic) expansion process giving s2 = s1. From Eq.8.23
T2 = T1( P2 / P1)
k-1
k
= 300





100
250
0.4
1.4
= 300 × 0.4
0.28575
= 230.9 K
The ideal gas law PV = mRT at both states leads to
V2 = V1 P1 T2/P2 T1 = 1 × 250 × 230.9/100 × 300 = 1.92 cm3
The work term is from Eq.8.29 or Eq.4.4 with polytropic exponent n = k
1W2 =
1
1 - k (P2V2 - P1V1) =
1
1 - 1.4 (100 × 1.92 - 250 × 1) ×10-6
= 0.145 J

8.89
Oxygen gas in a piston cylinder at 300 K, 100 kPa with volume 0.1 m3 is
compressed in a reversible adiabatic process to a final temperature of 700 K. Find
the final pressure and volume using Table A.5.
Solution:
Energy Eq.5.11: u2 - u1 = 0 − 1w2 ;
Entropy Eq.8.37: s2 - s1 = ∫ dq/T + 1s2 gen = 0
Properties: Table A.5: k = 1.393
constant s (isentropic) expansion process. From Eq.8.23
P2 = P1( T2 / T1)
k
k-1
= 2015 kPa
Using the ideal gas law to eliminate P from this equation leads to Eq.8.24
V2 = V1( T2 / T1)
1
1-k
= 0.1 ×





700
300
1
1−1.393
= 0.0116 m3
P
v
1
1
T
s
1
P
P
222

8.90
Oxygen gas in a piston cylinder at 300 K, 100 kPa with volume 0.1 m3 is
compressed in a reversible adiabatic process to a final temperature of 700 K. Find
the final pressure and volume using constant heat capacity from Table A.8.
Solution:
Energy Eq.5.11: u2 - u1 = 0 − 1w2 ;
constant s (isentropic) expansion process. From Eq.8.19
s
o
T2 – s
o
T1 = R ln
P2
P1
Properties: Table A.8: s
o
T1 = 6.4168, s
o
T2 = 7.2336 kJ/kg K
P2
P1
= exp [(s
o
T2 – s
o
T1)/R] = exp(
7.2336 – 6.4168
0.2598 ) = 23.1955
P2 = 100 × 23.1955 = 2320 kPa
Ideal gas law: P1V1 = mRT1 and P2V2 = mRT2
Take the ratio of these so mR drops out to give
V2 = V1 × (T2 / T1) × (P1 / P2) = 0.1 × (
700
300) × (
100
2320) = 0.01 m3
P
v
1
1
T
s
1
P
P
222

8.91
A rigid tank contains 1 kg methane at 500 K, 1500 kPa. It is now cooled down to
300 K. Find the heat transfer and the change in entropy using ideal gas.
Ideal gas, constant volume so there is no work.
Energy Eq. 5.11: m(u2 - u1) = 1Q2 - 1W2 = 1Q2
Use specific heat from Table A.5
u2 - u1 = Cv (T2 – T1) = 1.736 (300 – 500) = –347.2 kJ/kg
1Q2 = m(u2 - u1) = 1 (–347.2) = –347.2 kJ
The change in s for an ideal gas, Eqs.8.16-17 and v2 = v1 gives
m(s2 - s1) = m [ Cvo ln
T2
T1
+ R ln
v2
v1
] = m Cvo ln
T2
T1
= 1 × 1.736 ln
300
500 = – 0.8868 kJ/K
P
v
1
1
T
s
1P
P
2
22

8.92
Consider a Carnot-cycle heat pump having 1 kg of nitrogen gas in a
cylinder/piston arrangement. This heat pump operates between reservoirs at 300
K and 400 K. At the beginning of the low-temperature heat addition, the pressure
is 1 MPa. During this process the volume triples. Analyze each of the four
processes in the cycle and determine
a. The pressure, volume, and temperature at each point
b. The work and heat transfer for each process
Solution:
T
s
1
34
2
N2
T1 = T2 = 300 K, T3 = T4 = 400 K,
P1 = 1 MPa, V2 = 3 × V1
a) P2V2 = P1V1 => P2 = P1/3 = 0.3333 MPa
V1 =
mRT1
P1
=
1 × 0.2968 × 300
1000 = 0.08904 m3
V2 = 0.26712 m3
P3 = P2(T3/T2)
k
k-1 = 0.3333





400
300
3.5
= 0.9123 MPa
V3 = V2 ×
P2
P3
×
T3
T2
= 0.26712 ×
0.3333
0.9123 ×
400
300 = 0.1302 m3
P4 = P1(T3/T1)
k
k-1 = 1





400
300
3.5
= 2.73707 MPa
V4 = V1 ×
P1
P4
×
T4
T1
= 0.08904 ×
1
2.737 ×
400
300 = 0.04337 m3
b) 1W2 = 1Q2 = mRT1 ln (P1/P2)
= 1 × 0.2968 × 300 ln(1/0.333) = 97.82 kJ
3W4 = 3Q4 = mRT3 ln(P3/P4)
= 1 × 0.2968 × 400 ln(0.9123/2.737) = -130.43 kJ
2W3 = -mCV0(T3 - T2) = -1 × 0.7448(400 - 300) = -74.48 kJ
4W1 = -mCV0(T1 - T4) = -1 × 0.7448(300 - 400) = +74.48 kJ
2Q3 = 0, 4Q1 = 0

8.93
A hydrogen gas in a piston/cylinder assembly at 280 K, 100 kPa with a volume of
0.1 m3 is now compressed to a volume of 0.01 m3 in a reversible adiabatic
process. What is the new temperature and how much work is required?
Entropy Eq.8.37: s2 - s1 = ∫ dq/T + 1s2 gen = 0 + 0 = 0
For an isentropic process and constant heat capacity we have Eq.8.24
From Eq.8.24: T2 = T1( v2 / v1)
1-k
= 280 (0.01/0.1)
1-1.409
= 718 K
From Eq.8.39: 1w2 =
R
1-k (T – T) =
4.1243
1-1.409 (718 – 280) = -4417 kJ/kg

8.94
A handheld pump for a bicycle has a volume of 25 cm3 when fully extended. You
now press the plunger (piston) in while holding your thumb over the exit hole so that
an air pressure of 300 kPa is obtained. The outside atmosphere is at P0, T0. Consider
two cases: (1) it is done quickly (∼1 s), and (2) it is done very slowly (∼1 h).
a. State assumptions about the process for each case.
b. Find the final volume and temperature for both cases.
Solution:
C.V. Air in pump. Assume that both cases result in a reversible process.
State 1: P0, T0 State 2: 300 kPa, ?
One piece of information must resolve the ? for a state 2 property.
Case I) Quickly means no time for heat transfer
Q = 0, so a reversible adiabatic compression.
u2 - u1 = -1w2 ; s2 - s1 = ∫ dq/T + 1s2 gen = 0/
With constant s and constant heat capacity we use Eq.8.23
T2 = T1( P2 / P1)
k-1
k
= 298





300
101.325
0.4
1.4
= 405.3 K
Use ideal gas law PV = mRT at both states so ratio gives
=> V2 = P1V1T2/T1P2 = 11.48 cm3
Case II) Slowly, time for heat transfer so T = constant = T0.
The process is then a reversible isothermal compression.
T2 = T0 = 298 K => V2 = V1P1/P2 = 8.44 cm3
P
v
1
2ii
2ii
1
T
s
1
P
P
2
2i
2i

8.95
An insulated piston/cylinder setup contains carbon dioxide gas at 400 kPa, 300 K
which is then compressed to 3 MPa in a reversible adiabatic process. Calculate
the final temperature and the specific work using a) ideal gas tables A.8 and b)
using constant specific heats A.5.
Solution:
C.V. CO2, a control mass undergoing a reversible, adiabatic process.
Energy Eq.5.11: u2 - u1 = 0 − 1w2 ;
State 1: (300 K, 400 kPa) State 2: (3000 kPa, ?)
With two terms zero in the entropy equation we have a zero for the entropy
change. So this is a constant s (isentropic) expansion process, s2 = s1.
a) Table A.8 for CO2 and Eq.8.19
s2 - s1 = 0 = s
o
T2 – s
o
T1 − R ln(P2/P1)
s
o
T2 = s
o
T1 + R ln
P2
P1
= 4.8631 + 0.1889 ln
3000
400 = 5.2437 kJ/kg-K
Now interpolate in A.8 to find T2
T2 = 450 + 50 (5.2437 – 5.2325)/(5.3375 – 5.2325) = 455.3 K
1w2 = -(u2 - u1) = –(271 – 157.7) = –113.3 kJ/kg
b) Table A.5: k = 1.289, CVo = 0.653 kJ/kg K and now Eq.8.23
T2 = T1





P2
P1
k-1
k = 300





3000
400
0.224
= 471.1 K
1w2 = -CVo(T2-T1) = -0.653 (471.1 - 300) = -111.7 kJ/kg

8.96
Extend the previous problem to also solve using a constant specific heat at an
average temperature from A.6 and resolve using Table B.3.
C.V. CO2, a control mass undergoing a reversible, adiabatic process.
Energy Eq.5.11: u2 - u1 = 0 − 1w2 ;
State 1: (300 K, 400 kPa) State 2: (3000 kPa, ?)
With two terms zero in the entropy equation we have a zero for the entropy
change. So this is a constant s (isentropic) expansion process, s2 = s1.
Find a heat capacity at an average temperature from Table A.6.
Estimate T2 ~ 500 K giving TAVE ~ 400 K => θ = 0.4
CPo = 0.45 + 1.67 ×0.4 – 1.27 ×0.42 + 0.39 ×0.43 = 0.9398 kJ/kg K
CVo = CPo – R = 0.9398 – 0.1889 = 0.7509, k = CPo/CVo = 1.2516
Eq.8.32: T2 = T1





P2
P1
k-1
k = 300





3000
400
0.201
= 449.8 K
1w2 = –CVo(T2–T1) = –0.7509 (449.8 – 300) = –112.5 kJ/kg
From Table B.3.2: s1 = 1.8102 kJ/kg-K, u1 = 336.2 kJ/kg
T2 = 186.7oC = 459.8 K, u2 = 446.9 kJ/kg
1w2 = – (u2 – u1) = –(446.9 – 336.2) = –110.7 kJ/kg

8.97
A piston/cylinder, shown in Fig. P8.97, contains air at 1380 K, 15 MPa, with V1 =
10 cm3, Acyl = 5 cm2. The piston is released, and just before the piston exits the
end of the cylinder the pressure inside is 200 kPa. If the cylinder is insulated,
what is its length? How much work is done by the air inside?
Solution:
C.V. Air, Cylinder is insulated so adiabatic, Q = 0.
Continuity Eq.: m2 = m1 = m,
Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 = - 1W2
Entropy Eq.8.37: m(s2 - s1) = ∫ dQ/T + 1S2 gen = 0 + 1S2 gen
State 1: (T1, P1) State 2: (P2, ?)
So one piece of information is needed for the ?, assume reversible process.
1S2 gen = 0 => s2 - s1 = 0
State 1: Table A.7: u1 = 1095.2 kJ/kg, s
o
T1 = 8.5115 kJ/kg K
m = P1V1/RT1 =
15000 × 10×10-6
0.287 × 1380
= 0.000379 kg
State 2: P2 and from Entropy eq.: s2 = s1 so from Eq.8.19
s°
T2
= s°
T1
+ R ln
P2
P1
= 8.5115 + 0.287 ln(
200
15000) = 7.2724 kJ/kg K
Now interpolate in Table A.7 to get T2
T2 = 440 + 20 (7.2724 – 7.25607)/(7.30142 – 7.25607) = 447.2 K
u2 = 315.64 + (330.31 – 315.64) 0.36 = 320.92 kJ/kg
V2 = V1
T2 P1
T1P2
=
10 × 447.2 × 15000
1380 × 200
= 243 cm3
⇒ L2 = V2 /Acyl = 243/5 = 48.6 cm
1w2 = u1 - u2 = 774.3 kJ/kg, 1W2 = m1w2 = 0.2935 kJ

8.98
Argon in a light bulb is at 90 kPa and 20oC when it is turned on and electric input
now heats it to 60oC. Find the entropy increase of the argon gas.
Solution:
C.V. Argon gas. Neglect any heat transfer.
Energy Eq.5.11: m(u2 - u1) = 1W2 electrical in
Entropy Eq.8.37: s2 - s1 = ∫ dq/T + 1s2 gen = 1s2 gen
Process: v = constant and ideal gas => P2/ P1 = T2/T1
Evaluate changes in s from Eq.8.16 or 8.17
s2 - s1 = Cp ln (T2/T1) – R ln (P2/ P1) Eq.8.16
= Cp ln (T2/T1) – R ln (T2/ T1) = Cv ln(T2/T1) Eq.8.17
= 0.312 kJ/kg-K × ln [
60 + 273
20 + 273 ] = 0.04 kJ/kg K
cb
Since there was no heat transfer but work input all the change in s is
generated by the process (irreversible conversion of W to internal energy)

8.99
We wish to obtain a supply of cold helium gas by applying the following
technique. Helium contained in a cylinder at ambient conditions, 100 kPa, 20°C,
is compressed in a reversible isothermal process to 600 kPa, after which the gas is
expanded back to 100 kPa in a reversible adiabatic process.
a. Show the process on a T–s diagram.
b. Calculate the final temperature and the net work per kilogram of helium.
Solution:
a)
1
T
2
=
P
2 1 3
1
2
3
32
T
s
s = s
T
P = P
2
P
v
1
600
100 3
b) The adiabatic reversible expansion gives constant s from the entropy equation
Eq.8.37. With ideal gas and constant specific heat this gives relation in
Eq.8.23
T3 = T2(P3/P2)
k-1
k = 293.15 (100/600)0.4 = 143.15 K
The net work is summed up over the two processes. The isothermal process
has work as Eq.8.31
1w2 = -RT1 ln(P2/P1) = -2.0771 × 293.15 × ln(600/100) = -1091.0 kJ/kg
The adiabatic process has a work term from energy equation with no q
2w3 = CVo(T2-T3) = 3.116 (293.15 - 143.15) = +467.4 kJ/kg
The net work is the sum
wNET = -1091.0 + 467.4 = -623.6 kJ/kg

8.100
A 1-m3 insulated, rigid tank contains air at 800 kPa, 25°C. A valve on the tank is
opened, and the pressure inside quickly drops to 150 kPa, at which point the valve
is closed. Assuming that the air remaining inside has undergone a reversible
adiabatic expansion, calculate the mass withdrawn during the process.
Solution:
C.V.: Air remaining inside tank, m2.
Cont.Eq.: m2 = m ;
Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2
Entropy Eq.8.37: m(s2 − s1) = ∫ dQ/T + 1S2 gen
Process: adiabatic 1Q2 = 0 and reversible 1S2 gen = 0
P
v
1
2 2
1
T
s
C.V. m2
Entropy eq. then gives s2 = s1 and ideal gas gives the relation in Eq.8.23
T2 = T1(P2/P1)
k-1
k = 298.2(150/800)0.286 = 184.8 K
m1 = P1V/RT1 = (800 kPa × 1 m3)/(0.287 kJ/kgK × 298.2 K) = 9.35 kg
m2 = P2V/RT2 = (150 kPa × 1 m3)/(0.287 kJ/kgK × 184.8 K) = 2.83 kg
me = m1 - m2 = 6.52 kg

8.101
Two rigid tanks, shown in Fig. P8.101, each contain 10 kg N2 gas at 1000 K, 500
kPa. They are now thermally connected to a reversible heat pump, which heats
one and cools the other with no heat transfer to the surroundings. When one tank
is heated to 1500 K the process stops. Find the final (P, T) in both tanks and the
work input to the heat pump, assuming constant heat capacities.
Solution:
Control volume of hot tank B,
Process = constant volume & mass so no work
Energy equation Eq.5.11 and specific heat in Eq.5.20 gives
U2 - U1 ≅ mCv(T2 - T1) = 1Q2 = 10 × 0.7448 × 500 = 3724 kJ
P2 = P1T2/T1 = 1.5(P1) = 750 kPa
H.P.
W
HE
1 3Q
1 2QA
1 -> 3
B
1 -> 2
State: 1 = initial,
2 = final hot
3 = final cold
To fix temperature in cold tank, C.V.: total
For this CV only WHP cross the control surface no heat transfer. The entropy
equation Eq.8.37 for a reversible process becomes
(S2 - S1)tot = 0 = mhot (s2 - s1) + mcold (s3 - s1)
Use specific heats to evaluate the changes in s from Eq.8.25 and division by m
Cp,hot ln(T2 / T1) − R ln(P2 / P1) + Cp,cold ln(T3 / T1) − R ln(P3 / P1) = 0/
P3 = P1T3/T1 and P2 = P1T2/T1
Now everything is in terms of T and Cp = Cv + R, so
Cv,hot ln(T2/T1) + Cv,cold ln(T3/T1) = 0
same Cv: T3 = T1(T1/T2) = 667 K, P3 = 333 kPa
Qcold = - 1Q3 = m Cv (T3 - T1) = -2480 kJ,
WHP = 1Q2 + Qcold = 1Q2 - 1Q3 = 1244 kJ

8.102
A hydrogen gas in a piston/cylinder assembly at 300 K, 100 kPa with a volume of
0.1 m3 is now slowly compressed to a volume of 0.01 m3 while cooling it in a
reversible isothermal process. What is the final pressure, the heat transfer and the
change in entropy?
Solution:
C.V. Hydrogen, control mass.
Energy Eq. 5.11: m(u2 - u1) = 1Q2 - 1W2
Process: T = constant so with ideal gas => u2 = u1
P
v
1
2
2 1
T
s
1
P
P
2
From the process equation (T2 = T1) and ideal gas law we get
P2 = P1 (V1/V2) = 10 P1 = 1000 kPa
we can calculate the work term as in Eq.4.5
1Q2 = 1W2 = ⌡⌠PdV = P1V1 ln (V2/V1)
= 100 × 0.1 × ln (1/10) = -23.0 kJ
The change of entropy from the entropy equation Eq.8.3 is
m(s2 - s1) = 1Q2/T1 =
-23
300 = -0.07667 kJ/K
If instead we use Eq.8.17 we would get
∆S = m(s2 - s1) = m(Cvo ln
T2
T1
+ R ln
v2
v1
) = m R ln
v2
v1
= (P1V1/T1) ln (
v2
v1
) = 1Q2/T1
consistent with the above result.

8.103
A rigid tank contains 4 kg air at 200oC, 4 MPa which acts as the hot energy
reservoir for a heat engine with its cold side at 20oC shown in Fig. P.8.103. Heat
transfer to the heat engine cools the air down in a reversible process to a final 20o
C and then stops. Find the final air pressure and the work output of the heat
engine.
W
QH Q L
HE
C.V. total
Air Ambient
Energy Eq.5.11: m(u2 − u1) = 1Q2 - 1W2
Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T (T not constant)
Process: v = constant ⇒ 1W2 = 0
State 2: T2 and v2 = v1,
P2 = P1T2/T1 = 4000 × 293.15/473.15 = 2478.3 kPa
QH = −1Q2 = −m(u2 − u1) = –m Cvo (T2 – T1)
= –4 × 0.717 (293.15 – 473.15) = 516.2 kJ
Take now CV total as the air plus heat engine out to ambient
Entropy Eq.8.3: m(s2 − s1) = −
QL
Tamb
⇒
QL = −mTamb (s2 − s1) = −mTamb Cvo ln
T2
T1
= − 4 × 293.15 × 0.717 ln(293.15/473.15) = 402.5 kJ
Now the CV heat engine can give the engine work from the energy equation
Energy H.E.: WHE = QH − QL = 516.2 – 402.5 = 113.7 kJ
Notice to get 1q2 = ∫ T ds we must know the function T(s) which we do not
readily have for this process.
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