SSL12 Entropy

ENTROPY
Lecture 10

Keith Vaugh BEng (AERO) MEng

Reference text: Chapter 8 - Fundamentals of Thermal-Fluid Sciences, 3rd Edition
Yunus A. Cengel, Robert H. Turner, John M. Cimbala
McGraw-Hill, 2008

OBJECTIVES
Apply the second law of thermodynamics to
processes.
}

OBJECTIVES
processes.
}
Deﬁne a new property called entropy to quantify
the second-law effects.

OBJECTIVES
processes.
}

Establish the increase of entropy principle.

OBJECTIVES
processes.
}

Establish the increase of entropy principle.

Calculate the entropy changes that take place
during processes for pure substances*,
incompressible substances, and ideal gases.

* we’re limiting our studies to pure substances (2010)

ENTROPY
Entropy is a somewhat abstract property and it is
difﬁcult to give a physical description of without
considering the microscopic state of the system
}

ENTROPY
}
The second law of thermodynamics often leads to
expressions that involve inequalities.

ENTROPY
}

An irreversible heat engine is less efﬁcient than a
reversible one operating between the same two
thermal reservoirs

ENTROPY
}

An irreversible heat engine is less efﬁcient than a
reversible one operating between the same two
thermal reservoirs

Another important inequality that has major
consequences in thermodynamics is Clausius
inequality

The system considered in the
development of the Clausius inequality.

Clausius inequality is expressed as;
δQ
— ≤0
∫T
the validity of the Clausius inequality is
demonstrated on pages 298/9 of the reference text.


δQ
— ≤0
∫T

Clausius realized he had discovered a new
thermodynamic property and named it entropy
with designation of S. It is deﬁned as;

 δQ 
dS = 
 T  int, rev
 ( kJ K )


δQ
— ≤0
∫T

Clausius realized he had discovered a new
thermodynamic property and named it entropy
with designation of S. It is deﬁned as;

 δQ 
dS = 
 T  int, rev
 ( kJ K )
The entropy change of a system during a
process can be determined by integrating
The system considered in the between the initial and ﬁnal states
2  δQ 
ΔS = S2 − S1 = ∫ 1

 T  int, rev
 ( kJ K )

The entropy change between two speciﬁed
states is the same whether the process is
reversible or irreversible

A quantity whose cyclic integral is zero
(i.e. a property like volume)

 δQ 
—T  int, rev = 0
∫ 




 δQ 
∫ 


The net change in volume
(a property) during a
cycle is always zero


 δQ 
∫ 


The net change in volume
(a property) during a
cycle is always zero

Entropy is an extensive property of a system

A special case: Internally reversible
isothermal heat transfer processes


2  δQ 
ΔS = ∫
1

 T  int, rev



2  δQ 
ΔS = ∫
1

 T  int, rev


2  δQ 
= ∫
1 T 
 0 int, rev


2  δQ 
ΔS = ∫
1

 T  int, rev


2  δQ 
= ∫
1 T 
 0 int, rev

1 2
=
T0 ∫ (δ Q )
1 int, rev


2  δQ 
ΔS = ∫
1

 T  int, rev


2  δQ 
= ∫
1 T 
 0 int, rev

1 2
=
T0 ∫ (δ Q )
1 int, rev

ΔS =
Q
T0 } This equation is particularly useful
for determining the entropy changes
of thermal energy reservoirs

EXAMPLE 8 - 1
A piston cylinder device contains liquid vapor
mixture at 300 K. During a constant pressure process,
}
750 kJ of heat is transferred to the water. As a result,
part of the liquid in the cylinder vaporizes. Determine
the entropy change of the water during this process.

SOLUTION
Heat is transferred to a liquid vapor mixture
of water in a piston cylinder device at
constant pressure. The entropy change of
water is to be determined

SOLUTION
Heat is transferred to a liquid vapor mixture
of water in a piston cylinder device at
constant pressure. The entropy change of
water is to be determined

ASSUMPTIONS
No irreversibility's occur within the system
boundaries during the process.

SOLUTION ANALYSIS
Heat is transferred to a liquid vapor mixture We take the entire water (liquid + vapor) in the
of water in a piston cylinder device at cylinder as the system. This is a closed system
constant pressure. The entropy change of since no mass crosses the system boundary during
water is to be determined the process. We note that the temperature of the
system remains constant at 300 K since the
ASSUMPTIONS temperature of a pure substance remains constant
at the saturation value during a phase change
process at constant pressure.

SOLUTION ANALYSIS
Heat is transferred to a liquid vapor mixture We take the entire water (liquid + vapor) in the
of water in a piston cylinder device at cylinder as the system. This is a closed system
constant pressure. The entropy change of since no mass crosses the system boundary during
water is to be determined the process. We note that the temperature of the
system remains constant at 300 K since the
ASSUMPTIONS temperature of a pure substance remains constant
at the saturation value during a phase change
process at constant pressure.

The system undergoes an internally reversible,
isothermal process and thus its entropy change can
be determined directly

Q 750kJ
ΔS = = = 2.5 kJ K
Tsys 300K

NOTE - Entropy change of the system is positive
as expected since heat transfer is to the system

The increase of entropy principle

A cycle composed of a reversible and an
irreversible process

δQ 2 δQ 1 δQ 
— ≤ 0 → ∫1 T + ∫2  T  int, rev ≤ 0
∫T  


— ≤ 0 → ∫1 T + ∫2  T  int, rev ≤ 0
∫T  

2 δQ 2 δQ
∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫
1 T 1 T


— ≤ 0 → ∫1 T + ∫2  T  int, rev ≤ 0
∫T  

2 δQ 2 δQ
∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫
1 T 1 T

δQ The equality holds for an internally
dS ≥ reversible process and the inequality
T for an irreversible process


— ≤ 0 → ∫1 T + ∫2  T  int, rev ≤ 0
∫T  

2 δQ 2 δQ
∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫
1 T 1 T


2 δQ
ΔSsys = S2 − S1 = ∫ + Sgen
1 T

— ≤ 0 → ∫1 T + ∫2  T  int, rev ≤ 0
∫T  

2 δQ 2 δQ
∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫
1 T 1 T


2 δQ
ΔSsys = S2 − S1 = ∫ + Sgen
1 T
irreversible process Sgen = ΔStotal = ΔSsys + ΔSsurr ≥ 0

Some entropy is generated or created during an irreversible process
and this generation is due entirely to the presence of irreversibility's.

A system and its surrounding form an
isolated system

The entropy change of an isolated system is
the sum of the entropy changes of its
components, and is never less than zero.

A system and its surrounding form an
isolated system


A system and its surrounding form an ΔSisolated ≥ 0
isolated system


isolated system
Sgen = ΔStotal = ΔSsys + ΔSsurr ≥ 0


isolated system
Sgen = ΔStotal = ΔSsys + ΔSsurr ≥ 0

> Irreversible process
The increase of 
Sgen = Reversible process
entropy principle < Impossible process


A comment on entropy

The entropy change of a system can
be negative, but the entropy
generation cannot.

Processes can occur in a certain direction only,
not in any direction. A process must proceed in
the direction that complies with the increase of
entropy principle, that is, Sgen ≥ 0. A process that
violates this principle is impossible.

generation cannot.


Entropy is a non-conserved property, and there
is no such thing as the conservation of entropy
principle. Entropy is conserved during the
idealized reversible processes only and increases
during all actual processes.

generation cannot.


Entropy is a non-conserved property, and there
is no such thing as the conservation of entropy
principle. Entropy is conserved during the
idealized reversible processes only and increases
during all actual processes.

The performance of engineering systems is
The entropy change of a system can degraded by the presence of irreversibility's, and
generation cannot.
entropy generation is a measure of the
magnitudes of the irreversibility's during that
process. It is also used to establish criteria for
the performance of engineering devices.

Entropy change of pure substances
Entropy is a property and therefore its value for
a system is ﬁxed provided that state of that
system is also ﬁxed.

Schematic of the T-s
diagram for water.

Entropy change
ΔS = mΔS = m ( s2 − s1 ) ( kJ K )
The entropy of a pure substance is
determined from the tables (like other
properties).

QUESTION 8 - 39
A well insulated rigid tank contains 2kg of a saturated
liquid vapor mixture of water at 100 kPa. Initially
three quarters of the mass is in the liquid phase. An
}
electric resistance heater placed in the tank is now
turned on and kept on until all the liquid in the tank is
vaporized. Determine the entropy change of the
steam during this process.

SOLUTION
An insulated rigid tank contains a
saturated liquid-vapor mixture of
water at a speciﬁed pressure. An
electric heater inside is turned on
and kept on until all the liquid
vaporized. The entropy change of
the water during this process is to
be determined.

SOLUTION
An insulated rigid tank contains a
be determined.

H2O
2 kg
100 kPa

We

SOLUTION ANALYSIS
An insulated rigid tank contains a From the steam tables (A-4 through A-6)
be determined.

H2O
2 kg
100 kPa

We

SOLUTION ANALYSIS
saturated liquid-vapor mixture of P1  v1 = v f + x1v fg
water at a speciﬁed pressure. An 
electric heater inside is turned on x1  s1 = s f + x1s fg
be determined.

H2O
2 kg
100 kPa

We

SOLUTION ANALYSIS
vaporized. The entropy change of v1 = v f + x1v fg
the water during this process is to v1 = 0.001 + ( 0.25 ) (1.6941 − 0.001)
be determined. 3
v1 = 0.4243 m kg

H2O
2 kg
100 kPa

We

SOLUTION ANALYSIS
be determined. 3
v1 = 0.4243 m kg

s1 = s f + x1s fg
H2O s1 = 1.3028 + ( 0.25 ) ( 6.0562 )
2 kg
100 kPa s1 = 2.8168 kJ kg ⋅ K

We

SOLUTION ANALYSIS
be determined. 3
v1 = 0.4243 m kg

s1 = s f + x1s fg
H2O s1 = 1.3028 + ( 0.25 ) ( 6.0562 )
2 kg
100 kPa s1 = 2.8168 kJ kg ⋅ K

v2 = v1 
We  s2 = 6.8649 kJ kg ⋅ K
sat. vapor 
Then the entropy change of the steam becomes

SOLUTION ANALYSIS
be determined. 3
v1 = 0.4243 m kg

s1 = s f + x1s fg
H2O s1 = 1.3028 + ( 0.25 ) ( 6.0562 )
2 kg
100 kPa s1 = 2.8168 kJ kg ⋅ K

v2 = v1 
We  s2 = 6.8649 kJ kg ⋅ K
sat. vapor 
Then the entropy change of the steam becomes

ΔS = m ( s2 − s1 ) = ( 2kg ) ( 6.8649 − 2.8168 ) kJ kg ⋅ K
ΔS = 8.10 kJ K

Isentropic processes
A process during which the entropy
remains constant is called an isentropic
process

During an internally reversible, adiabatic
(isentropic) process, the entropy remains
constant.

process

The isentropic process appears
as a vertical line segment on a
T-s diagram.

constant.

process

The isentropic process appears
as a vertical line segment on a
T-s diagram.

Δs = 0 or s 2 − s1 ( kJ
kg ⋅K)
constant.

QUESTION 8 - 44
An insulated piston-cylinder device contains 0.05 m3
of saturated refrigerant 134a vapor at 0.8 MPa
pressure. The refrigerant is now allowed to expand in
}
a reversible manner until the pressure drops to 0.4
MPa. Determine;
a. The ﬁnal temperature in the cylinder
b. The work done by the refrigerant

SOLUTION
An insulated cylinder is initially
filled with saturated R-134a vapor
at a specified pressure. The
refrigerant expands in a reversible
manner until the pressure drops to
a specified value. The final
temperature in the cylinder and the
work done by the refrigerant are to
be determined.

SOLUTION
An insulated cylinder is initially
ﬁlled with saturated R-134a vapor
be determined.

R-134a
0.8 MPa
0.05 m3

SOLUTION ASSUMPTIONS
An insulated cylinder is initially 1. The kinetic and potential energy
ﬁlled with saturated R-134a vapor changes are negligible.
be determined.

R-134a
0.8 MPa
0.05 m3

at a speciﬁed pressure. The 2. The cylinder is well-insulated and thus
refrigerant expands in a reversible heat transfer is negligible.
be determined.

R-134a
0.8 MPa
0.05 m3

manner until the pressure drops to 3. The thermal energy stored in the
a speciﬁed value. The ﬁnal cylinder itself is negligible.
be determined.

R-134a
0.8 MPa
0.05 m3

temperature in the cylinder and the 4. The process is stated to be reversible.
be determined.

R-134a
0.8 MPa
0.05 m3

be determined.
ANALYSIS
a. This is a reversible adiabatic (i.e., isentropic)
process, and thus s2 = s1. From the refrigerant
tables (Tables A-11 through A-13),

R-134a
0.8 MPa
0.05 m3

be determined.
ANALYSIS
a. This is a reversible adiabatic (i.e., isentropic)
process, and thus s2 = s1. From the refrigerant
tables (Tables A-11 through A-13),

v1 = vg@0.8 MPa = 0.02561 m3
kg
P1 = 0.8MPa 
 u1 = u g@0.8 MPa = 246.79 kJ kg
R-134a sat. vapor 
0.8 MPa s1 = sg@0.8 MPa = 0.91835 kJ kg ⋅ K
0.05 m3

V 0.05m 3
m= = m3
= 1.952kg
v1 0.025621 kg

V 0.05m 3
m= = m3
= 1.952kg
v1 0.025621 kg

s2 − s f 0.91835 − 024761
P2 = 0.4MPa  x2 = = = 0.9874
 s fg 0.67929
s2 = s1 
u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kg

V 0.05m 3
m= = m3
= 1.952kg
v1 0.025621 kg

s2 − s f 0.91835 − 024761
P2 = 0.4MPa  x2 = = = 0.9874
 s fg 0.67929
s2 = s1 
u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kg

T2 = Tsat @0.4 MPa = 8.91°C

V 0.05m 3
m= = m3
= 1.952kg
v1 0.025621 kg

s2 − s f 0.91835 − 024761
P2 = 0.4MPa  x2 = = = 0.9874
 s fg 0.67929
s2 = s1 
u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kg

T2 = Tsat @0.4 MPa = 8.91°C

b. We take the contents of the cylinder as the system. This is a closed system since no
mass enters or leaves. The energy balance for this adiabatic closed system can be
expressed as

V 0.05m 3
m= = m3
= 1.952kg
v1 0.025621 kg

s2 − s f 0.91835 − 024761
P2 = 0.4MPa  x2 = = = 0.9874
 s fg 0.67929
s2 = s1 
u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kg

T2 = Tsat @0.4 MPa = 8.91°C

expressed as

Ein − Eout = ΔEsystem
14 2 4 3 123
Net energy transfer Change in internal, kinetic,
by heat, work and mass potential, etc... energies

− Wb, out = ΔU
Wb, out = m ( u1 − u2 )

V 0.05m 3
m= = m3
= 1.952kg
v1 0.025621 kg

s2 − s f 0.91835 − 024761
P2 = 0.4MPa  x2 = = = 0.9874
 s fg 0.67929
s2 = s1 
u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kg

T2 = Tsat @0.4 MPa = 8.91°C

expressed as

Ein − Eout = ΔEsystem Substituting, the work done during this isentropic
14 2 4 3 123 process is determined to be
Net energy transfer Change in internal, kinetic,
by heat, work and mass potential, etc... energies

− Wb, out = ΔU Wb, out = m ( u1 − u2 )
Wb, out = m ( u1 − u2 ) = (1.952kg ) ( 246.79 − 232.91) kJ kg = 27.09kJ

Property diagrams involving entropy

On a T-S diagram, the area under the process curve
represents the heat transfer for internally reversible
processes.

δ Qint, rev = TdS

processes.

δ Qint, rev = TdS
2
Qint, rev = ∫ T dS
1

processes.

δ Qint, rev = TdS
2
1

δ qint, rev = Tds

processes.

δ Qint, rev = TdS
2
1

δ qint, rev = Tds
2
qint, rev = ∫ T ds
1

processes.

δ Qint, rev = TdS
2
1

δ qint, rev = Tds
2
1

Qint, rev = T0 ΔS

processes.

δ Qint, rev = TdS
2
1

δ qint, rev = Tds
2
1

Qint, rev = T0 ΔS
qint, rev = T0 Δs

processes.


For adiabatic steady-ﬂow devices,
the vertical distance ∆h on an h-s
diagram is a measure of work, and
the horizontal distance ∆s is a
measure of irreversibility's.
Mollier diagram: The h-s diagram

Entropy
Clausius inequality
The Increase of entropy principle
 Example 8-1
Entropy change of pure substances
 Question 8-39
 Question 8-44
What is entropy?
The T ds relations
Entropy change of liquids and solids
The entropy change of ideal gases
Reversible steady-flow work
Minimizing the compressor work
Isentropic efficiencies of steady-flow devices
Entropy balance

SSL12 Entropy

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