ENTROPY

1. The T-dS
Equations & Diagrams
Meeting 9
Section 4-3 & 4-4

2. Second Law of
Thermodynamics
Entropy is a non-conserved property!
 








2
1 A
1
2
T
Q
S
S
S


This can be viewed as a mathematical
statement of the second law (for a closed
system).

3. Entropy
 







2
1 rev
int
1
2
T
q
s
s

R
lbm
Btu
or
K
kg
kJ


Units are
s = S/m : intensive property

4. The Entropy Change Between
Two Specific States
The entropy
change
between two
specific states
is the same
whether the
process is
reversible or
irreversible

5. We can write entropy change as an
equality by adding a new term:
s
P
T
Q
S
2
1 A






 

 
entropy
change
entropy
transfer due
to heat
transfer
entropy
generation
or
production
Ps  0 ; not a property

6. Entropy generation
• Ps  0 is an actual irreversible
process.
• Ps = 0 is a reversible process.
• Ps  0 is an impossible process.

7. Entropy Production
• Ps quantifies irreversibilities. The
larger the irreversibilities, the greater
the value of the entropy production,
Ps.
• A reversible process will have no
entropy production, Ps = 0.
• Ps depends upon the process, and
thus it is not a property.

8. Entropy transfer and
production
Ps
T
Q
S
S
2
1
1
2 





 

 
Entropy change
of system as it
goes from 1 to 2;
can be + – or zero
depending on the
other two terms.
Entropy transfer
into/out of the
system via heat
transfer; can be
+ – or zero,
depending on Q
and its direction
Entropy production:
> 0 when internal
irrerversibilities
are present;
= 0 when no int irr
are present;
< 0 never.

9. Entropy transfer and
production



















 










 







 








s
P
T
Q
and
negative
is
Q
if
:
0
0
s
P
and
0
Q
if
:
0
s
P
T
Q
and
negative
is
Q
if
:
0
positive
always
is
s
P
because
s
P
T
Q
,
if
;
or
be
could
Q
:
0
S
S
2
1
2
1
2
1
1
2

10. Entropy Increase
–Entropy change in a general system
(ΔSsys) may be negative (due to heat
transfer out of the system)
–Entropy production cannot be negative








process
Impossible
0
process
Reversible
0
process
le
Irreversib
0
Ps

11. Entropy Transfer
• Entropy change is caused by heat
transfer, mass flow, and
irreversibilities.
• Heat transfer to a system increases
the entropy, and heat transfer from a
system decreases it.
• The effect of irreversibilities is
always to increase the entropy.

12. Some Remarks about Entropy
• Process can occur in a certain direction only,
not in any direction such that Ps  0.
• Entropy is a non-conserved property, and there
is no such thing as conservation of entropy
principle. The entropy of the universe is
continuously increasing.
• The performance of engineering systems is
degraded by the presence of irreversibilities,
and entropy generation is a measure of the
magnitudes of the irreversibilities present
during that process.

13. The Tds Equations

14. In previous slides, we
developed a new
property, entropy
 







2
1 rev
int
1
2
T
q
s
s

R
lbm
Btu
or
K
kg
kJ
are
Units



15. The entropy of a pure substance
is determined from the tables,
just as for any other property
The
Entropy
Change
of
a
Pure
Substance

16. It’s just like the other
properties we’ve
encountered in the tables
)
s
s
(
x
s
s f
g
f 


)
T
(
s
)
p
,
T
(
s f

It is tabulated just like u, v, and h.
Also,
And, for compressed or subcooled liquids,

17. T-s Diagram for Water

19. TEAMPLAY
• Use the tables in your book
• Find the entropy of water at 50 kPa and
500°C. Specify the units.
• Find the entropy of water at 100°C and a
quality of 50%. Specify the units.
• Find the entropy of water at 1 MPa and
120°C. Specify the units.

20. T-s diagram

 Pdv
w
Work was the area under the curve.
Recall that the P-v diagram was very
important in first law analysis, and that

21. For a T-s diagram
rev
int
T
Q
dS 







TdS
Q 



2
1
TdS
Q
Rearrange:
Integrate:

22. Heat
Transfer
for
Internally
Reversible
Processes
•
d
On a T-S diagram, the area under
the process curve represents the
heat transfer for internally
reversible processes

23. Derivation of Tds
equations:
For a simple closed system:
The work is given by:
Substituting gives:
dU
W
Q 


PdV
W 

PdV
dU
Q 



24. More derivation….
For a reversible process:
Make the substitution for Q in the energy
equation:
PdV
+
dU
=
TdS
Or on a per unit mass basis:
Pdv
+
du
=
Tds
Q
TdS 


25. Entropy is a property. The Tds expression
that we just derived expresses entropy in
terms of other properties. The properties
are independent of path….We can use the
Tds equation we just derived to calculate
the entropy change between any two
states:
Tds Equations
Pdv
du
Tds 


26. Starting with enthalpy h = u + Pv, it is possible
to develop a second Tds equation:
Tds Equations
vdP
dh
Tds 

vdP
Tds
vdP
Pdv
du
)
Pv
u
(
d
dh








27. Schematic of an h-s
Diagram for Water

28. Tds equations
• These two Tds relations have many uses
in thermodynamics and serve as the
starting point in developing entropy-
change relations for processes.

29. Entropy change of an
Pure substance
•The entropy-change and isentropic
relations for a process can be
summarized as follows:
1.Pure substances:
Any process: Δs = s2 - s1 [kJ/(kg-K)]
Isentropic process: s2 = s1

30. Let’s look at the entropy change
for an incompressible
substance:
dT
T
)
T
(
C
ds 
We start with the first Tds equation:
For incompressible substances, v  const, so dv = 0.
We also know that Cv(T) = C(T), so we can write:
Pdv
dT
)
T
(
C
Tds v 


31. Entropy change of an
incompressible substance
dT
T
)
T
(
C
s
s
2
1
T
T
1
2 


1
2
1
2
T
T
ln
C
s
s 

Integrating
If the specific heat does not vary with temperature:

32. Entropy Change for
Incompressible Substance
•The entropy-change and isentropic
relations for a process can be
summarized as follows:
2.Incompressible substances:
Any process:
Isentropic process: T2 = T1
1
2
av
1
2
T
T
ln
C
s
s 


33. Sample Problem
Aluminum at 100oC is placed in a large,
insulated tank having 10 kg of water at a
temperature of 30oC. If the mass of the
aluminum is 0.5 kg, find the final temperature
of the aluminum and water, the entropy of the
aluminum and the water, and the total entropy
of the universe because of this process.

34. Draw Diagram
water
AL
Insulated
wall
Closed system including aluminum and water
Constant volume, adiabatic, no work done

35. Conservation of Energy
Apply the first law
AL
W
sys U
U
U
W
Q 






0
)
T
T
(
C
m
)
T
T
(
C
m AL
1
2
AL
AL
W
1
2
W
W 



But (T2)W = (T2)AL = T2 at equilibrium
AL
AL
W
W
AL
1
AL
AL
W
1
W
W
2
C
m
C
m
)
T
(
C
m
)
T
(
C
m
T





36. Solve for Temperature
mW = 10 kg, CW = 4.177 kJ/kg.K
mAL = 0.5 kg, CAL = 0.941 kJ/kg.K
)
K
kg
/
kJ
941
.
0
)(
kg
5
.
0
(
)
K
kg
kJ
177
.
4
)(
kg
10
(
)
K
373
)(
K
kg
kJ
941
.
0
)(
kg
5
.
0
(
)
K
303
)(
K
kg
kJ
177
.
4
)(
kg
10
(
T2







T2 = 303.8 K

37. Entropy Transfer
Entropy change for water and aluminum
K
kJ
1101
.
0
)
K
303
K
8
.
303
ln(
)
K
kg
kJ
177
.
4
)(
kg
10
(
T
T
ln
C
m
S
W
,
1
2
W
W
W





K
kJ
0966
.
0
K
373
K
8
.
303
ln
)
K
kg
kJ
941
.
0
)(
kg
5
.
0
(
T
T
ln
C
m
S
AL
,
1
2
AL
AL
AL







38. Entropy Generation
Entropy production of the universe
0
K
kJ
0135
.
0
K
kJ
0966
.
0
K
kJ
1101
.
0
S
S
S
S AL
W
total
gen










Sgen > 0 : irreversible process

39. The Entropy Change
of an Ideal Gas

40. Let’s calculate the change
in entropy for an ideal gas
RT/P
v
and
dT
C
dh p 

dP
P
RT
dT
C
Tds p 

Start with 2nd Tds equation
Remember dh and v for an ideal gas?
Substituting:
vdP
dh
Tds 


41. Change in entropy
for an ideal gas
P
dP
R
T
dT
C
ds p 

Dividing through by T,
Don’t forget, Cp = Cp(T) ….. a function of
temperature!

42. Entropy change of
an ideal gas
1
2
T
T
p
1
2
P
P
ln
R
T
dT
)
T
(
C
s
s
2
1


 

2
1
T
T
p
T
dT
C
Integrating yields
To evaluate entropy change, we’ll have
to evaluate the integral:

43. Entropy change of an ideal gas
• Similarly it can be shown from
Tds = du + Pdv
that
1
2
T
T
v
1
2
v
v
R
T
dT
T
C
s
s
2
1
ln
)
( 

 

44. Entropy change of an ideal
gas for constant specific
heats: Approximation
• Now, if the temperature range is so limited
that Cp  constant (and Cv  constant),
1
2
p
2
1
p
T
T
ln
C
T
dT
C 

1
2
1
2
p
1
2
P
P
ln
R
T
T
ln
C
s
s 



45. Entropy change of an ideal gas
for constant specific heats:
Approximation
• Note
Tds = du + Pdv
• Therefore
1
2
1
2
v
1
2
v
v
ln
R
T
T
ln
C
s
s 



46. Constant
lines
of
v
and
P
for
Ideal
gases
diagrams
T
s
v = const.
P = const.
dT/ds
P
v
v
P
P
P
v
v
ds
dT
ds
dT
C
C
C
T
ds
dT
const
P
vdP
dh
Tds
C
T
ds
dT
const
v
Pdv
du
Tds















0
0

47. Summary: Entropy change of
an Pure substance
1. Pure substances:
Any process:
Δs = s2  s1 [kJ/(kg-K)] (Table)
Isentropic process:
s2 = s1

48. Summary: Entropy Change for
Incompressible Substance
Any process:
Isentropic process: T2 = T1
1
2
av
1
2
T
T
ln
C
s
s 

2. Incompressible substances:

49. Summary: Entropy
Change for Ideal gases
3. Ideal gases:
Constant specific heats (approximate treatment):
 
 
K)
kJ/(kg
P
P
R
T
T
C
s
s
K)
kJ/(kg
v
v
R
T
T
C
s
s
1
2
1
2
av
p
1
2
1
2
1
2
av
v
1
2








ln
ln
ln
ln
,
,

50. TEAMPLAY : Air is compressed in a
piston-cylinder device from 90 kPa and 20oC to 400
kPa in a reversible isothermal process.
Determine: (a) the entropy change of air,
(b) the work done and (c) the removed heat.
90kPa
293K
400kPa
293 K
s
T
1
2
Q<0
v1
v2
293K
Air is ideal gas, R = 287 Jkg-1K-1

51. 












K
J
428
90
400
ln
287
P
P
ln
R
T
T
ln
C
s
s
1
2
1
2
p
1
2
















kg
kJ
25.4
1
90
400
ln
298
287
P
P
ln
RT
W
1
2
comp
Ideal gas isothermal compression work:
Entropy change for an ideal gas with constant heat :
Rejected heat (1st law):















kg
kJ
25.4
1
W
Q
U
W
Q comp
comp

52. • The entropy of the air decreased due to the heat
extraction. Consider the heat is rejected on the
environment at 15oC.
• Evaluate the entropy change of the environment and
the air.
•The environment is a heat reservoir. The entropy
change is s = Q/T = 125400/288 = + 435 JK-1.
•The entropy change of the system plus the
environment is therefore:
s = ssystem + senviron = - 428+ 435 = 7 JK-1.
• If the environment is at 20oC, s = 0 because both
are at the same temperature (reversible heat
transfer).

53. Sample Problem
A rigid tank contains 1 lbm of carbon
monoxide at 1 atm and 90°F. Heat is
added until the pressure reaches 1.5
atm. Compute:
(a) The heat transfer in Btu.
(b) The change in entropy in Btu/R.

54. Draw diagram:
State 1:
P = 1 atm
T = 90oF
CO: m= 1 lbm
State 2:
P = 1.5 atm
Rigid Tank => volume is constant
Heat Transfer

55. Assumptions
• CO in tank is system
• Work is zero - rigid tank
• kinetic energy changes zero
• potential energy changes zero
• CO is ideal gas
• Constant specific heats

56. Apply assumptions to
conservation of energy
equation
PE
+
KE
+
U
W
Q 




 
1
2
v T
T
mC
=
Q 
For constant specific heats, we get:
0 0 0
Need T2  How do we get it?

57. Apply ideal gas
equation of state:
2
1
2
2
1
1
mRT
mRT
V
P
V
P
 Cancel common
terms...
Solve for T2:
  R
825
R
460
90
1.0
1.5
T
P
P
T 1
1
2
2 


















58. Solve for heat
transfer
 R
550
825
R
lbm
Btu
18
.
0
)
lbm
1
(
Q 









Btu
5
.
49
Q 
Now, let’s get entropy change ...

59. For constant specific
heats:











1
2
1
2
v
1
2
v
v
Rln
T
T
ln
C
m
S
S
Since v2 = v1
0
1
2
v
1
2
T
T
ln
C
m
S
S 



















R
550
R
825
ln
R
m
lb
Btu
18
.
0
)
lbm
1
(
S
S 1
2
Btu/R
073
.
0
S
S 1
2 


60. Entropy Change in
Some Selected
Irreversible Processes

61. Some Irreversible Processes
ONE WAY PROCESSES

62. PROCESSOS IRREVERSÍVEIS: EFEITO JOULE
Sistema Isolado -> não há calor
cruzando a fronteira e todo trabalho
é transformado em energia interna:
 
lei)
(2a.
P
T
Q
S
S
lei)
(1a.
t
t
Ri
U
U
s
f
i
i
f
i
f
i
f









Todo trabalho elétrico é convertido em energia interna do
sistema. A variação de S é devido a Ps, pois Q=0.
Como S não depende do caminho,
TfdS = dU =Rit -> Ps = Rit/Tf > 0.
Trabalho elétrico convertido em energia interna aumentou
a entropia do sistema isolado. Não é possível converter a
mesma quantidade de energia interna em trabalho
elétrico.
R
V
i
Q
Sistema
Isolado
Tf

63. PROCESSOS IRREVERSÍVEIS: EFEITO JOULE
Por que não é possível converter a mesma quantidade
de energia interna em trabalho elétrico?
Do ponto de vista microscópico: o sistema deveria
resfriar para diminuir a energia interna e transforma-
la em energia elétrica . Certamente este não será o
estado mais provável de encontrá-lo portanto, esta
transformação não será espontânea!
Do ponto de vista macroscópico a entropia do
sistema isolado deveria diminuir e isto violaria a 2a.
Lei! Note que de (i) -> (f), S>0
Um processo é reversível quando S não varia entre
os estados final e inicial!

64. PROCESSOS IRREVERSÍVEIS: ATRITO
Sistema Isolado -> não há calor
cruzando a fronteira e toda energia
Potencial da mola é transformada
em energia interna:
lei)
(2a.
P
T
Q
S
S
lei)
(1a.
x
k
U
U
s
f
i
i
f
2
0
i
f







Toda energ. pot. mola é convertida em energia interna do
sistema. A variação de S é devido a Ps, pois Q=0.
Como S não depende do caminho,
TdS = dU =(1/2)kx0
2 -> Ps = (1/2)kx0
2 /T > 0 & Sf > Si.
Aumentou a entropia do sistema isolado. Não é possível
converter a mesma quantidade de energia interna em
energia potencial da mola!
k
Q
Sistema
Isolado
m
x
x0

65. PROCESSOS IRREVERSÍVEIS: QUEDA LIVRE
Sistema Isolado -> não há calor
cruzando a fronteira e toda energia
Potencial é transformada
em energia interna após choque
com água
lei)
(2a.
P
T
Q
S
S
lei)
(1a.
h
g
m
U
U
s
f
i
i
f
0
i
f








Toda energ. pot. é convertida em energia interna do
sistema. A variação de S é devido a Ps, pois Q=0.
Como S não depende do caminho,
TdS = dU =mgh0 -> Ps = mgh0 /T > 0 & Sf > Si.
Aumentou a entropia do sistema isolado. Não é possível
converter a mesma quantidade de energia interna para
elevar a bola na posição inicial h0!
Q
Sistema
Isolado
m
inicial
h0
água
final

66. PROCESSOS IRREVERSÍVEIS: DIFERENÇA TEMP.
Sistema
Isolado
Sistema Isolado -> energia interna
permanece constante (blocos com
mesma massa e calor específico
porém a Th e Tc)
lei)
(1a.
2
T
T
T
U
U
U
f
q
mix
f
q
f





S do sistema isolado é considerado como a soma de S do
sistema quente e frio. Squente diminui mas Sfrio aumenta
de modo que a variação total é maior que zero.
Troca de calor com diferença de temperatura é
irreversível. O bloco que atinge Tmix não volta
expontaneamente para Th e Tc, é necessário trabalho!
inicial
Frio Tc
final
Quente Th
2
T
T
T
c
h
mix


lei)
(2a.
T
T
pq.
0
T
T
ln
C
T
T
ln
C
T
T
ln
C
S f
q
c
h
mix
h
c
mix































67. PROCESSOS IRREVERSÍVEIS: EXPANSÃO COM DIFERENÇA DE PRESSÃO
Sistema Isolado -> na expansão para o vácuo não há
calor nem trabalho cruzando a fronteira. A energia
interna do gás permanece a mesma. T1 = T2 (1a lei)
Processo isotérmico: P1V1 = P2V2 ou P2 = 0.5P1
Variação da Entropia:
 0
1
V
2
V
MRLN
S
V
dV
MR
T
PdV
dS
PdV
dU
TdS
0
















Durante a expansão contra o vácuo a capacidade de
realizar trabalho do gás foi perdida (não havia máq. p/
extrair trabalho). A transf. inversa é irreversível pois S>0
Inicial
Sistema
isolado
AR
P =P1
V1=V
T1=T
AR
P=P2
V2 =2V
T2=T
final
Vácuo

68. Process Efficiency
A reversible process may have an increase or
decrease of Entropy but THERE IS NO ENTROPY
GENERATION.
The process efficiency is often taken as the ratio
between the actual work and the ideal
reversible work.
An actual process which does or receive work is
often compared against an ideal reversible process.

69. Isentropic Process
1
2
REV S
S
or
0
dS
0
Q 




For a reversible, adiabatic process
For an ideal gas  ds=0 and vdP = CPdT
Using v = RT/P and the fact that R = CP-CV ,
 
 























1
2
1
1
2
P
v
P
T
T
P
P
T
dT
C
P
dP
C
C
Recovers the adiabatic and reversible ideal
gas relations, Pv= const.

70. 0
P
or
0
dS
0
Q S 




Adiabatic Process: Reversible x Irreversible
T
s
P1
P2
1
2s
2
For an adiabatic process (12) S is constant or
greater than zero due to entropy generation.

71. Adiabatic Process: Reversible x Irreversible
T
s
P1
P2
1
2s
2
S is constant or greater than zero due to entropy
generation independent if it is an adiabatic
expansion or compression.
T
s
P2
P1
1
2s
2
Expansion, W > 0
Compression, W < 0

72. Available Work in an Adiabatic Process:
Reversible x Irreversible
• The first law states for an adiabatic process
that W = U2 – U1 where 1 and 2 represent the
initial and final states.
• For compression, T2S < T2, since U ~ to T,
then the reversible process requires less work
than the irreversible, i.e., WREV < W .
• For expansion, T2S > T2, since U ~ to T, then
the reversible process delivers more work
than the irreversible, i.e., WREV > W .

73. Available Work in a Process:
Reversible x Irreversible Paths
The 1st and 2nd law
combined result  Irr
Q
TdS
W
dU
Q REAL








REAL
W
Irr
PdV 



0
Irr
W
W
W
Irr
W REAL
REV
REAL
REV 











eliminating dQ 
but PdV is a reversible work mode, then
or 1
0
W
W
1
REV
REAL





the work delivered in a reversible process is
always equal or greater than the one in a
irreversible process.

74. Available Work in a Process:
Reversible x Irreversible Path
Either for compression or expansion, the
irreversibilites do:
1. increase the system entropy due to the
entropy generation by the irreversibilities,
2. a fraction of the available work is spent to
overcome the irreversibilites which in turn
increase the internal energy,

75. What For Are the Reversible Process?
• They are useful for establishing references
between actual and ‘ideal’ processess.
• The process efficiency defined as the ratio
of the work delivered by an actual and an
reversible process compares how close they
are.
reversible
actual
W
W


• It must not be confused the process
efficiency with the thermal eff. of heat
engines! The later operates in a cycle.

76. TEAMPLAY
Um carro com uma potência de 90
kW tem uma eficiência térmica de
28%. Determine a taxa de
consumo de combustível se o
poder calorífico do mesmo é de
44.000 kJ/kg.

77. Recommended
Exercises
4-18 4-19 4-23 4-25
4.26 4-34

78. TEAMPLAY
A 50-kg iron block and a 20-kg copper
block, both initially at 80oC, are
dropped into a large lake at 15oC.
Thermal equilibrium is established
after a while as a result of heat
transfer between the blocks and the
lake water. Determine the total
entropy generation for this process.