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Svnas 8e ISM Chapter 05 - Chemical engineering
thermodynamics practice questions
화학공학 열역학 (Kwangwoon University)
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Chapter 5
Shown at the bottom is a PV diagram with two adiabatic lines 1 → 2 and 2 → 3, assumed to intersect at
point 2. A cycle is formed by an isothermal line from 3 → 1. An engine traversing this cycle would
produce work. For the cycle U = 0, and therefore by the first law, Q + W = 0. Since W is negative, Q
must be positive, indicating that heat is absorbed by the system. The net result is therefore a complete
conversion of heat taken in by a cyclic process into work, in violation of Statement 1a of the second law
(Pg. 160). The assumption of intersecting adiabatic lines is therefore false.
W   QH
 1
TC
TH
 T 
 323 K 
-1
-1
W  1  C  QH  1 
 250 kJ s  148.8 kJ s
T
798
K


H 

QC  QH  W  250 kJ s-1  148.8 kJ s-1  101.2 kJ s-1
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 1
Chapter 5
TC
300 K
1
 0.6
TH
750 K
thermal  1 
QC
QH
 1
TC
303
 1
 0.514
TH
623


thermal  0.55 1 

 303 
TC 
  0.55 1 
  0.35
TH 
TH 

The energy balance for the over-all process is written: Q = Ut + EK + EP
Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial
temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is
negative, and by the preceding equation, so is Q. Thus heat is transferred to the surroundings.
The total entropy change of the process is: Stotal = St + Stsurr
Just as Ut for the egg is zero, so is St . Therefore,
Since Q is negative, Stotal is positive, and the process is irreversible.
By Eq. (5.9) the thermal efficiency of a Carnot engine is:
Differentiate:
Since TC/TH is less unity, the efficiency changes more rapidly with TC than with TH. So in theory it is
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Chapter 5
more effective to decrease TC. In practice, however, TC is fixed by the environment, and is not subject
to control. The practical way to increase η is to increase TH. Of course, there are limits to this too.

 1
TC
113.7 K
1
 0.625
TH
303.15 K















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Chapter 5





S  C p ln
P
T
 R ln  
T0
 P0 
T2

Q  U  n Cv dT
t
T1
Q  U t  nCv T
T 
Q
15000 J

 500 K
nCv 1.443 mol  2.5  8.31451J mol-1 K -1
S  C p ln


P 7
T
5
 R ln    R ln 2  R ln 2  R ln 2  1.733R  14.4 J mol-1 K -1
T0
2
 P0  2


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
Chapter 5
























For an ideal gas with constant heat capacities, and for the changes T1 → T2 and P1 → P2, Eq. (5.14)
can be rewritten as:
V2 = V1, then
Whence,
Since CP > CV , this demonstrates that SP > SV .
V2 = V1, then
Whence,
This demonstrates that the signs for ST and SV are opposite.
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SVNAS 8th Edition Annotated Solutions
Chapter 5
Start with the equation just preceding Eq. (5.10):
For an ideal gas PV = RT, and lnP + ln V = ln R + ln T . Therefore,
**********************
As an additional part of the problem, one could ask for the following proof, valid for constant heat
capacities. Return to the original equation and substitute dT/T = dP/P + dV/V:
As indicated in the problem statement the
basic differential equations are:
−
−
(A)
−
(B)
where QC and QH refer to the reservoirs.
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(a) With
and
Whence, d ln TC = − d ln TH
Chapter 5
, Eq. (B) becomes:
where
≡
Integration from TH0 and TC0 to TH and TC yields:
With
and
, Eq. (B) becomes
Eliminate TC by the boxed equation of Part (a) and rearrange slightly:
For infinite time, TH =TC ≡T, and the answer of Part (a) becomes:
Because /(
+ 1) − 1 = −1/(
+ 1), then:
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Chapter 5
Because TH =T , substitution of these quantities in the answer of Part (b) yields:
As indicated in the problem statement the
basic differential equations are:
−
−
(A)
−
(B)
where QC and QH refer to the reservoirs.
(a) With
, Eq. (B) becomes:
Substitute for dQH and dQC in Eq. (A):
Integrate from TC0 to TC:
For infinite time, TC = TH, and the boxed equation above becomes:
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Chapter 5
Write Eqs. (5.9) in rate form and combine to eliminate |
|
H|=
H|:
this becomes:
Differentiate, noting that the quantity in square brackets is constant:
Equating this equation to zero, leads immediately to: 4r = 3 or r = 0.75
Hence
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thermal  1 
QC
QH
 1
Chapter 5
TC
300
 1
 0.5
TH
600

 


 

Q
T
300
W
 H 1  H 1 
 1  0.2
250
QC
QC
TC

S  C p ln
P 
T2
 R ln  2 
T1
 P1 
S  C p ln
P 
T2
 R ln  2 
T1
 P1 

7 T
 P 
S  R  ln 2  ln  2    8.314  3.5ln1.5  ln 5   1.58 J mol-1 K -1
2 T

1
 P1  


7 T
 P 
S  R  ln 2  ln  2    8.314  3.5ln1.667  ln 5   1.48 J mol-1 K -1
2 T

1
 P1  



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Chapter 5
5 T
 P 
S  R  ln 2  ln  2    8.314  2.5ln 0.667  ln 0.2   4.96 J mol-1 K -1
2 T

1
 P1  


S  C p ln
P 
T2
 R ln  2 
T1
 P1 

9 T
P
S  R  ln 2  ln  2
2 T
1
 P1


-1
-1
   8.314  4.5ln 0.75  ln 0.2   2.62 J mol K


S  C p ln
P 
T2
 R ln  2 
T1
 P1 


 P 
T
S  R  3ln 2  ln  2    8.314  4ln 0.6  ln 0.2   3.61 J mol-1 K -1


T1
 P1  

This cycle is shown below. Assuming constant specific heats, the efficiency is given by:
Temperature T4 is not given and must be calculated. The following equations are used to derive
and expression for T4.
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Chapter 5
For adiabatic steps 1 to 2 and 3 to 4:
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Chapter 5
Because W =0, Eq. (2.3) here becomes:
Q=
Ut = m CV T
A necessary condition for T to be zero when Q is non-zero is that m =∞. This is the reason that
natural bodies (air and water) that serve as heat reservoirs must be massive (oceans) or continually
renewed (rivers).

T 
P 
Sad ,rev  C p ln  rev   R ln  2   0
 T1 
 P1 
T
C p ln  rev
 T1

 P2 
  R ln  

 P1 
Trev  P2 
 
T1  P1 
Trev
R
Cp
P 
 T1  2 
 P1 

R
Cp

7
 298.15  
 2
2
7
 426.46 K


T 
P 
 471.4 
7
S  C p ln  2   R ln  2   7 / 2* R ln 
 R ln    0.3506R

 298.15 
 2
 T1 
 P1 

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Chapter 5
T 
P 
Sad ,rev  C p ln  rev   R ln  2   0
 T1 
 P1 


T 
 P  
T 
 P 
S  Sad ,rev  S  C p ln  2   R ln  2    C p ln  rev   R ln  2  
 T1 
 P1   
 T1 
 P1  

T 
T
S  C p ln  2   C p ln  rev
 T1 
 T1
 T2 


  C p ln 

 Trev 
An appropriate energy balance here is: Q = Ht = 0
Applied to the process described, with T as the final temperature, this becomes:
If m1 = m2 = m,
.
Since
St is positive
Isentropic processes are not necessarily reversible and adiabatic. The term isentropic denotes a process
for which the system does not change in entropy. There are two causes for entropy changes in a
system: The process may be internally irreversible, causing the entropy to increase; heat may be
transferred between system and surroundings, causing the entropy of the system to increase or decrease.
For processes that are internally irreversible, it is possible for heat to be transferred out of the system
in an amount such that the entropy changes from the two causes exactly compensate each other. One
can imagine irreversible processes for which the state of the system is the same at the end as at the
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Chapter 5
beginning of the process. The process is then necessarily isentropic, but neither reversible nor adiabatic.
More generally, the system conditions may change in such a way that entropy changes resulting
from temperature and pressure changes compensate each other. Such a process is isentropic, but not
necessarily reversible. Expansion of gas in a piston/cylinder arrangement is a case in point. It may be
reversible and adiabatic, and hence isentropic. But the same change of state may be irreversible with
heat transfer to the surroundings. The process is still isentropic, but neither reversible nor adiabatic.
An isentropic process must be either reversible and adiabatic or irreversible and non-adiabatic.
By definition,
By inspection, one sees that for both T > T0 and T0 > T the numerators and denominators of the
above fractions have the same sign. Thus, for both cases <CP>H is positive. Similarly,
By inspection, one sees that for both T > T0 and T0 > T the numerators and denominators of the
above fractions have the same sign. Thus, for both cases <CP>S is positive. When T = T0, both the
numerators and denominators of the above fractions become zero, and the fractions are indeterminate.
Application of l’Hˆopital’s rule leads to the result: <CP>H = <CP>S = CP.
Step 1-2: Volume decreases at constant P. Heat flows out of the system. Work is done on the
system.
Step 2-3: Isothermal compression. Work is done on the system. Heat flows
out of the system.
Step 3-1: Expansion process that produces work. Heat flows into the system. Since the PT product
is constant,
Combine with (A) this becomes
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Chapter 5
Moreover
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T

S  C p
T0
Chapter 5
P
dT
 R ln  
T
 P0 
T
P
S
D
A
   B  CT  3  dT  ln  
R
T 
T
 P0 

T0
T 
P
S
C 2
D 2
 A ln    B T  T0  
T  T02 
T  T02  ln  
2
2
R
 T0 
 P0 

T1 (K)
473.15
T2 (K)
1373.15
T2
ICPS 
 T2 


C (1/K2)
D (K2)
B (1/K)
8.01E-04 0.00E+00 -1.02E+00
 RT dT  A ln  T   B T  T   2 T
Cp
T1 (K)
523.15
T2 (K)
1473.15
T2

T1
C
2
2
2
1
1
T1
ICPS 
A
5.699

A
1.213

 T12 
ICPS
6.793

D 2
T2  T1 2
2
C (1/K2)
D (K2)
B (1/K)
2.88E-02 -8.82E-06 0.00E+00




ICPS
20.234
T 
C
D 2
dT  A ln  2   B T2  T1   T22  T12 
T2  T1 2
RT
T
2
2
 1
Cp
S (J/(mol K))
56.48
S (J/(mol K))
168.23

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Chapter 5
T

S  C p
dT
T
T0
T2

Q  H  n C p dT
t
T1
Q
800000 J

 9622 K=
nR 10 mol  8.314 J mol-1 K -1
T1 (K)
473.15
T2 (K)
1374.5
T2
ICPH 
A
1.424
B
2
1
2
2

 T12 
T1
S

R
1374.5 K


Cp
R
dT
473 K
C (1/K2)
D (K2)
B (1/K)
1.44E-02 -4.39E-06 0.00E+00
 R dT  AT  T   2 T
Cp
T2
ICPH (K)
9.622E+03
 1 1
C 3
T2  T13  D   
3
 T2 T1 


C p dT
R T
473.15 K
T1 (K)
473.15
T2 (K)
1374.5
T2
ICPS 

T1
A
1.424
C (1/K2)
D (K2)
B (1/K)
1.44E-02 -4.39E-06 0.00E+00
ICPS
10.835
T 
C 2
D 2
dT  A ln  2   B T2  T1  
T2  T12 
T2  T12
RT
T
2
2
 1
Cp




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1374.5 K
S

R

Chapter 5
C p dT
 10.835
R T
473.15 K

2500000 J
Q

 20045 K=
nR 15 mol  8.314 J mol-1 K -1
T1 (K)
533.15
T2 (K)
1413.7
T2
ICPH 
R
Cp
A
1.967
dT  A T2  T1  
T1
T1 (K)
533.15
T2 (K)
1413.7
T2
ICPS 
A
1.967
 T2 
S

R
1413.7 K

Cp
R
dT
533.15 K
ICPH (K)
2.004E+04
 1 1
B 2
C 3
T2  T12 
T2  T13  D   
2
3
 T2 T1 




D (K2)
C (1/K2)
B (1/K)
3.16E-02 -9.87E-06 0.00E+00
C
2
1
1
T1

D (K2)
C (1/K2)
B (1/K)
3.16E-02 -9.87E-06 0.00E+00
 RT dT  A ln  T   B T  T   2 T
Cp
T2
2
2

 T12 
ICPS
21.307

D 2
T2  T12
2

C p dT
 21.307
R T
533.15 K

1000000 Btu
Q

 12588 R = 6993 K =
nR 40 lbmol  1.986 Btu lbmol-1 R -1
T2

Cp
R
dT
533.15 K

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T1 (K)
533.15
T2 (K)
1202.7
T2
ICPH 
R
Cp
A
1.424
dT  A T2  T1  
T1
Chapter 5
D (K2)
C (1/K2)
B (1/K)
1.44E-02 -4.39E-06 0.00E+00
ICPH (K)
6.993E+03
 1 1
B 2
C 3
T2  T12 
T2  T13  D   
2
3
 T2 T1 





T1 (K)
533.15
T2 (K)
1202.7
T2
ICPS 

T1
A
1.424
T
dT  A ln  2
RT
 T1
Cp
S

R
1202.7 K

C (1/K2)
D (K2)
B (1/K)
1.44E-02 -4.39E-06 0.00E+00
ICPS
8.244

C 2
D 2
T2  T12 
T2  T12
  B T2  T1  
2
2





C p dT
 8.244
R T
533.15 K



248.15 K
298.15 K
1 bar
5 bar
348.15 K
1 bar

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Chapter 5
The process involves three heat reservoirs: the house, a heat source; the tank, a heat source; and the
surroundings, a heat sink. Notation is as follows:
|Q| Heat transfer to the house at temperature T
|QF| Heat transfer from the furnace at TF
|Qσ | Heat transfer from the surroundings at Tσ
The first and second laws provide the two equations:
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Chapter 5
Combine these equations to eliminate |Qσ |, and solve for |QF|:
With T = 295 K , TF = 810 K, Tσ = 265 K, and |Q’| = 1000 kJ
Shown to the right is a scheme designed to accomplish this result. A Carnot heat engine operates
with the furnace as heat source and the house as heat sink. The work produced by the engine drives a
Carnot refrigerator (reverse Carnot engine) which extracts heat from the surroundings and discharges heat
to the house. Thus the heat rejected by the Carnot engine (|Q1|) and by the Carnot refrigerator (|Q2|)
together provide the heat |Q| for the house. The energy balances for the engine and refrigerator are:
|W|engine = |Q| − |Qσ1|
|W|refrig = |Qσ2| − |Q’|
Equation (5.7) may be applied to both the engine
and the refrigerator:
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Chapter 5
The process involves three heat reservoirs: the house, a heat source; the tank, a heat source; and the
surroundings, a heat sink. Notation is as follows:
|Q| Heat transfer from the tank at temperature T
|Q‘| Heat transfer from the house at T ‘
|Qσ | Heat transfer to the surroundings at Tσ
The first and second laws provide the two equations:
Combine these equations to eliminate |Qσ |, and solve for |Q|:
With T = 448.15 K , T’ = 297.15 K, Tσ = 306.15 K, and |Q’| = 1500 kJ
Shown below is a scheme designed to accomplish this result. A Carnot heat engine operates with the tank
as heat source and the surroundings as heat sink. The work produced by the engine drives a Carnot
refrigerator (reverse Carnot engine) which extracts heat |Q_| from the house and discharges heat to the
surroundings. The energy balances for the engine and refrigerator are:
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Chapter 5
|W|engine = |Q| − |Qσ1|
|W|refrig = |Qσ2| − |Q’|
Equation (5.7) may be applied to both the engine
and the refrigerator:
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SVNAS 8th Edition Annotated Solutions

Q
 Sg
T
Sg  

Chapter 5
135 J s1
 0.45 J K -1 s-1
300 K
Q
 Sg
T
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Sg  
Chapter 5
2500 J s1
 8.33 J K -1 s-1
300 K
For a closed system the first term of Eq. (5.16) is zero, and it becomes:
Where j is here redefined to refer to the system rather than to the surroundings. Nevertheless, the second
term accounts for the entropy changes of the surroundings, and can be written simply as dStsurr/dt:
Multiplication by dt and integration over finite time yields:
The general equation applicable here is Eq. (5.17):
(a)
: For a single stream flowing within the pipe and with a single heat
source in the surroundings, this becomes:
(b)
(c)
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Chapter 5
(d)
(e)
(f)
(g)

S  C p ln
P 
T2
 R ln  2 
T1
 P1 

7 T
 P 
S  R  ln 2  ln  2    8.314  3.5ln 0.742  ln 0.2   4.698 J mol-1 K -1
2 T

1
 P1  


Updated 4/5/2017
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Chapter 5


S  
j
Qj
Tj

 Sg

Wlost  T S g
S  C p ln

P 
T2
 R ln  2 
T1
 P1 


 P 
T
S  R  4ln 2  ln  2    8.314  4ln 0.8356  ln 0.4   1.643 J mol-1 K -1


T1
 P1  





Updated 4/5/2017
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S  
j
Qj
Tj
Chapter 5
 Sg

Wlost  T S g
S  C p ln

P 
T2
 R ln  2 
T1
 P1 

 11 T
 P 
 11

S  R  ln 2  ln  2    8.314  ln 0.8724  ln 0.3   3.767 J mol-1 K -1
2 T

2

1
 P1  




S  
j
Qj
Tj

 Sg
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Chapter 5

Wlost  T S g
S  C p ln

P 
T2
 R ln  2 
T1
 P1 

9 T
P
S  R  ln 2  ln  2
2 T
1
 P1


-1
-1
   8.314  4.5ln 0.7832  ln 0.2143  3.663 J mol K




S  
j
Qj
Tj

 Sg

Wlost  T S g
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Chapter 5
S  C p ln

P 
T2
 R ln  2 
T1
 P1 

5 T
 P 
S  R  ln 2  ln  2    8.314  2.5ln 0.7327  ln 0.3  3.546 J mol-1 K -1
2 T

1
 P1  




S  
j
Qj
Tj

 Sg

Wlost  T S g
The figure below indicates the direct, irreversible transfer of heat |Q| from a reservoir at T1
to a reservoir at T2. The figure on the right depicts a completely reversible process to accomplish the
same changes in the heat reservoirs at T1 and T2.
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 5
The entropy generation for the direct heat-transfer process is:
For the completely reversible process the net work produced is Wideal:
This is the work that is lost, Wlost, in the direct, irreversible transfer of heat |Q|. Therefore,
Note that a Carnot engine operating between T1 and T2 would not give the correct Wideal or Wlost, because
the heat it transfers to the reservoir at T2 is not Q.
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Sg  

Chapter 5
QH QC
1
0.55



 6.7  105 kJ/K
523.15 298.15
TH
TC
W
QH
1
TC
293.15
1
 0.5016
TH
588.15
Equation (5.14) can be written for both the reversible and irreversible processes:
Updated 4/5/2017
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Chapter 5
Since Sirrev must be greater than zero, Tirrev must be greater than Trev.
T 1 (K)
293.15
T 2 (K)
300.15
A
3.355
T2

Cp
ICPH 
dT A T2 T1 
R
B (1/K)
5.75E-04

C (1/K2)
0
B 2
T2 T12
2

B (1/K)
5.75E-04
C (1/K2)
0
D (K2)
ICPH (K)
-1.60E+03
24.55

C 3
T2 T13
3

T1
T 1 (K)
293.15
T 2 (K)
251.15
T2

A
3.355
Cp
ICPH 
dT A T2 T1 
R
T1

B 2
T2 T12
2


1
D 
T2
1
T1
D (K2)
ICPH (K)
-1.60E+03 -146.57
C 3
T2 T13
3

1
D 
T2
DH (J/mol)
204.12
DH (J/mol)
-1218.58
1
T1
Updated 4/5/2017
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T 1 (K)
293.15
T 2 (K)
303.15
A
3.355
Chapter 5
B (1/K)
5.75E-04
T2
Cp
T
ICPS 
dT A ln  2
RT
T1

B T2 T1 
C (1/K2)
0

D (K2)
-1.60E+03
C 2
T2 T12
2

C (1/K2)
0
D (K2)
-1.60E+03
ICPS
0.118

D 2
T2
T1
2
2

T1
T 1 (K)
293.15
T 2 (K)
251.15
A
3.355
B (1/K)
5.75E-04
T2
Cp
T
ICPS 
dT A ln  2
RT
T1

B T2 T1 

C 2
T2 T12
2


ICPS
-0.540
D 2
T2
T1
2
2

T1
n
pV
1*100000

 258.6 lbmol hr 1
RT 0.7302*529.67
T2
ICPH T0,T;A,B,C,D  
R
T1
Cp
dT  A T  T0  
1 1 
B 2
C
T  T02  T 3  T03  D   
2
3
 T T0 




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T
ICPS  T0,T;A,B,C,D 

T0
n
Chapter 5

T 
D  T 2  T02 
dT  A ln    B T  T0    C  2 2 


2 
RT
T T0 
 T0 


Cp
pV 101325*3000

 122629 mol hr 1  34.06 mol s 1
RT 8.314*298.15
T2
ICPH T0,T;A,B,C,D  
R
Cp
dT  A T  T0  
T1
T0(K)
298.15
T (K)
265.15
A
3.355
T
ICPS  T0,T;A,B,C,D 

T0
T0 (K)
298.15
T (K)
265.15
B (1/K)
5.75E-04
1 1 
B 2
C 3
T  T02 
T  T03  D   
2
3
 T T0 




ig
C (1/K2)
D (K2)
ICPH (K) H (J/mol)
0
-1.60E+03 -115.4
-959.4

T 
D  T 2  T02 
dT  A ln    B T  T0    C  2 2 


2 
RT
T T0 
 T0 


Cp
A
3.355
B (1/K)
5.75E-04
C (1/K2)
D (K2)
0
-1.60E+03
ICPS
-0.410
Updated 4/5/2017
Sig (J/(mol K))
-3.4099
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SVNAS 8th Edition Annotated Solutions

Chapter 5

Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of
the steam and the gas.
Calculate entropy generation per lbmol of gas:
(b)
Calculate lbs of steam generated per lbmol of gas cooled.
Use ratio to calculate ideal work of steam per lbmol of gas
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SVNAS 8th Edition Annotated Solutions
Chapter 5
Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of
the steam and the gas.
Calculate entropy generation per lbmol of gas:
(b)
Calculate lbs of steam generated per lbmol of gas cooled.
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SVNAS 8th Edition Annotated Solutions
Chapter 5
Use ratio to calculate ideal work of steam per lbmol of gas
Now place a heat engine between the ethylene and the surroundings. This would constitute a
reversible process, therefore, the total entropy generated must be zero. calculate the heat released
to the surroundings for ΔStotal = 0.
Now apply an energy balance around the heat engine to find the work produced. Note that the
heat gained by the heat engine is the heat lost by the ethylene.
The lost work is exactly equal to the work that could be produced by the heat engine.
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