Evaluating entropy changes
 2 Q 
S 2  S1   

 1 T  int
rev
• Since entropy is a property, the entropy change between two specified states
does not depend on the process (reversible or irreversible) that occurs
between the two states.
• However, to evaluate entropy change by the above definition a reversible
process has to be imagined connecting the initial and final states and
integration has to be carried along its path.
Using entropy change and thermodynamic temperature
to graphically interpret the heat supplied in reversible
processes
Q 

  dS
 T  rev
  Q rev  TdS
 W rev  PdV
P
T
1
2
S
V
2
Q rev   TdS
1
2
W rev 

1
P dV
Showing a Carnot cycle on a T-s diagram
Isotherms: horizontal lines
Isentropes:vertical lines
T
2
1
Application:
Carnot engine
Qnet,in
4
3
s
Isobars and isochores on the T-s
diagram for an ideal gas (Sec. 7.9)
s  s1 ~ c
avg
v
s  s1 ~ c
avg
p
T 
 v 
ln    R ln  
 T1 
 v1 
 s  s1 
Isochores:
T
~
T
exp
Isochores:
 avg 
1
 cv

T 
 P 
ln    R ln  
 T1 
 P1 
ss
1
Isobars: T ~ T exp 
1
 c avg
 p
T
v
P
s
P T
P1
T1
v1

 vT T

1

T-s diagram of a cycle: try yourself
• The Brayton cycle used in a gas turbine
engine consists of following steps:
– isentropic compression (1-2).
– isobaric heat addition (2-3).
– isentropic expansion (3-4).
– isobaric heat rejection (4-1).
Plot this process on a T-s and P-v diagram.
Use of the Tds relations in calculating
entropy changes
On an intensive (per unit mass) basis:
dS 
dU
P dV

T
dS 
First TdS relation
T
dH

V dP
T
T
ds 
du

T
Second TdS relation
ds 
dh
T
Pdv
T

vdP
T
Integrating between initial (1) and the final state (2):
2
s 2  s1 

1
2
s 2  s1 

1
du
2

T
dh
T

1
2


1
P dv
T
vdP
T
Even though the Tds equations
are derived using an internally
reversible process, the change
in entropy during an irreversible
process occuring between the
same two equilibrium states
can be
also calculated using the
integrals on the left.
Entropy change of liquids/solids (Sec. 7.8)
•Liquid/solid can be approximated as incompressible substances (v=constant): i.e.
specific volume ( or density) remains constant even when other properties change
(dv=0) Internal energy of a solid/liquid is a function of temperature
alonecp=cv=c (one specific heat)
d u  cd T
2
s 2  s1 

1
du
T
2


1
P dv
T
2
s 2  s1 

1
from First Tds
relation
cdT
T
2
c
1
dT
T
T 
c ln  2 
 T1 
if temperature variation of
specific heat (c)can be
neglected.
For solids/liquids, the isentropic process is also isothermal.
Second law for a closed system undergoing a
process
Clausius inequality:
b: boundary
Tb=T for
an internally
reversible
process
0
Tb
1

Q
Q 
 T  T 
b
1
2
 int
2
2

> for irreversible process 1-2
= internally reversible process 1-2
1
2
S 2  S1 
˜
Q

1
Q
Tb
Q
Tb
0
rev
 S1  S 2  0
Second law for an isolated system
2
 S 1 2  S 2  S 1 

1
Q
T
Also, across the boundaries of an isolated system no energy
(heat/work) is transferred (by definition)
 S 1 2
iso lated
0
The universe (any system + its surroundings) can be considered an isolated
system. “The total entropy of the universe is increasing.”
Second law for an adiabatic process
undergone by a closed system
2
 S 1 2  S 2  S 1 

1
Q
T
Second law for a system undergoing adiabatic (irreversible/reversible) process
(Q=0)
 S 1 2  0  S 2  S 1
The entropy balance for closed systems
2
S 2  S1 

1
entropy
change
Q
Tb
 S gen
entropy
generation
within the
system
entropy
transfer
accompanying
( S gen  0)
heat transfer
• Sgen>0 if irreversibilities present inside the system
• Sgen=0 for no irreversibilities inside the system
• The value of Sgen is a measure of the extent of
irreversibilities within the system. More irreversibilities
Sgen ↑
Entropy generation (Sgen>0) within the system
is due to irreversible processes within the
system boundary
Irreversible processes occurring within the system
boundary result in entropy generation.
Examples:
• Friction (solid-solid, solid-fluid) may be present
between parts of the system.
• Hot and cold zones may be present within the
system which may be interacting irreversibly
through heat transfer.
• Non-quasi-equilibrium compression/expansion
occuring between parts of the system.
• Other: mixing between substances having
different chemical composition, chemical reaction.
Example of entropy generation due to an
irreversible process in an isolated system

Remove
separator
Initial state (1)
Pf,Tf, (m1+m2)
Final state (2)
Objective:
• To show that the process is irreversible.
Isolated system has Q=0
2
S 2  S1 

1
Q
Tb
To show a process is irreversible
 S gen
) to show Sgen>0
 to show S 2  S 1  0
Example (contd.): The final state of
the process
P1,V1, T, m1
Pf,Tf, (m1+m2)
P2,V2, T, m2
High pressure
Low pressure
Final temperature:
( m 1  m 2 ) C v T f  m 1C v T  m 2 C v T  T f  T
• From the first law Uf=Ui.
• At the final state:
.
Pf 
( m1  m 2 ) RT
V1  V 2

 m1  m 2  RT
m1 RT / P1  m 2 RT / P2

 m1  m 2 
m1 / P1  m 2 / P2
Example of entropy generation due to an
irreversible process in an isolated system
P1,,,V1, T, m1
P2,V2, T, m2
High pressure
Low pressure
Entropy balance
Pf,Tf, (m1+m2)
( m 1  m 2 ) C v T f  m 1C v T  m 2 C v T  T f  T
 S isolated  S f  S i  S gen
 S isolated  S f  ( S 1  S 2 )  ( m1  m 2 ) s f  ( m1 s1  m 2 s 2 )  m1 ( s f  s1 )  m 2 ( s f  s 2 )
Entropy change of the system is the entropy change of its parts
Example of entropy generation due to an
irreversible process in an isolated system
P1,,,V1, T, m1,
Pf,Tf, (m1+m2)
P2,V2, T, m2
Second law (entropy balance)
 S isolated  m1 ( s f  s1 )  m 2 ( s f  s1 )
 p 
 pf 
T 
1
s f  s1  C p ln    R ln 

  R ln 

p
p
T 
 1 
 f 
Similarly
s f  s 2  R ln
p2
pf
from integration of Tds=dh-vdP
Example of entropy generation due to an
irreversible process in an isolated system
P1,,,V1, T, m1,
Pf,T, (m1+m2)
P2,V2, T, m2
S gen  S 2  S 1  m 1 ( s f  s1 )  m 2 ( s f  s 2 )
m
m
 p 
 p 
p1 1 p 2 2
1
2
 m1 R ln 
  m 2 R ln 
  R ln
(m m )
 p 
 p 
pf 1 2
f
f




m
S gen  R ln
p1 1 p 2
m2

 m1  m 2  


m
/
P

m
/
P
2
2 
 1 1
( m1  m 2 )
1
S gen  0  process is irreversible

1

m1
m2
m m
  p1 p 2  1 2
 ( m 1  m 2 ) R ln 
 m1  m 2 

m / P  m / P
2
2
 1 1
GM  HM







Calculating work done in a reversible
steady flow process
• Reversible cycles develop the maximum work.
• Reversible cycles consist of reversible processes
through each device.
• At every stage of the internally reversible process:
First law for control volumes
– qrev-wrev=dh +d(ke)+d(pe)
–  q rev  Tds
– T d s  d h  vd P
• Combining: wrev=-vdP-d(ke)-d(pe)
• Integrating:
2
w rev    vdP 
1
v 2  v1
2
2
2
 g ( z 2  z1 )
Graphical representation of reversible
steady flow work on a p-v diagram
if changes in KE and PE
can be neglected (e.g. in
turbines, compressor and
pumps, but not in nozzles)
2
w rev    v d P
1
Reversible steady flow work for a pump
(PUMP HANDLES LIQUIDS)
w
pum p
rev
2
   vdP
1
Since specific volume of a liquid is nearly independent of
pressure:
v~v ~v
1
pum p
w rev
  v1 ( p 2  p1 )
2
 w isen
pum p
(since pump is an adiabatic device)
Alternatively
pum p
w rev
 w isen
pum p
 h1 ( P1 ,  )  h2 ( P2 , s 2  s1 )  h1  h2 s
similar to compressor
In steam power plants, how big is “turbine
work out” compared to “pump work in”?
v
w
tu rb in e
rev
4
   vd P
3
w
pum p
rev
2
   vdP
1
 v ( P2  P1 )
Specific volume of vapors are orders of magnitude larger than the
specific volume of liquid (e.g. vf' 10-3 m3/kg vg' 2 m3/kg at 100 kPa))
w
pum p
rev
 w
com pre ss or
re v
Gas turbine pumps consume a
significant part of work developed
in turbines to run compressor