Solutions manual chapter 7 (1)

Borgnakke and Sonntag
CONTENT CHAPTER 7
SUBSECTION PROB NO.
In-Text concept questions a-g
Concept problems 1-14
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
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In-Text Concept Questions

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.

Concept Problems

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

Review Problems

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

Solutions manual chapter 7 (1)

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Solutions manual chapter 7 (1)