ENSC 461
Assignment #5 (Exergy)
Assignment date:
Due date:
Problem 1:
A system undergoes a refrigeration cycle while receiving Qc by heat transfer
at temperature Tc and discharging energy QH by heat transfer at a higher
temperature TH. There are no other heat transfers.
a) Using an exergy balance, show that the network input to the cycle cannot
be zero.
b) Show that the coefficient of performance of the cycle can be expressed as:
 Tc
COP  
 TH  Tc
TH X d estroyed 


1 


 T0 QH  Qc  
where Xdestroyed is the exergy destruction and T0 is the temperature of the
surroundings.
c) Using the result of part (b), obtain an expression for the coefficient of
performance.
Problem 2:
Helium gas enters an insulated nozzle operating at steady state at 1300 K, 4
bar, and 10 m/s. At the exit, the temperature and pressure of the helium are
900 K and 1.45 bar, respectively. Determine:
a) the exit velocity in m/s
b) the isentropic nozzle efficiency
c) the rate of exergy destruction, in kJ/kg of gas flowing through the nozzle.
Assume the ideal gas model for helium and ignore the effects of gravity. Let
T0 = 20°C and P0 = 1atm.
M. Bahrami
ENSC 461 (S 11)
Assignmnet 5
1
Solution, Problem 1:
TH
QH
Wcycle = QH - QC
Refrigerator
QC
TC
Assumptions:
1) The system shown undergoes a refrigeration cycle
2) QC and QH are the only heat transfers and are in the directions of the arrows
3) Tc and TH are constant and TH > TC and the surroundings’ temperature is T0.
Analysis:
a)
An exergy balance for the cycle reads:
 T
X cycle  1  0
 Tc


T
 Qc  1  0
 TH




 QH  Wcycle  P0 
V   X destroyed  0

0 


where ∆Xcycle and ∆V are zero for a cycle. Introducing the energy balance,
QH  Wcycle  Qc
Since the work is being done on the cycle, it should be considered negative. We get:
 T
0  1  0
 Tc


T
 Qc  1  0
 TH

 1
1
 T0 

 TH Tc

 Wcycle  Qc    Wcycle   X destroyed


T
 Qc  0 Wcycle  X destroyed
TH

(1)
Solving for Xdestroyed, and setting Wcycle to zero.
M. Bahrami
ENSC 461 (S 11)
Assignmnet 5
2
 1
T
1
X destroyed  T0 
  Qc  0 Wcycle  X destroyed  0
 T


Tc  
TH 
H
0 
0
0


impossible!
0
Thus, Wcycle cannot be zero!
b) Solving Eq. (1) for COP, one finds:
COP 
Qc
Wcycle
 TH  Tc

 TH Tc
W
X

 Qc  cycle  destroyed
TH
T0

 Tc  TH X destroyed
Qc
1 
COP 
 
Wcycle  TH  Tc 
T0Wcycle
  Tc

 T T
c
  H
T X


1  H destroyed 
 T0 QH  Qc  
c) From the result of part b), COP increases as Xdestroyed 0. Thus, when Xdestroyed = 0:
 Tc
COPmax  
 TH  Tc



As expected.
Solution, Problem 2:
T
P1 = 4 bar
T1 = 1800 K
T2 = 900K
P2 = 1.45 bar
T1 = 1300K
P2 = 4 bar
V1 = 10 m/s
P2 = 1.45 bar
2
2S
Assumptions:
1) The control volume shown is at steady-state
2) No work or heat transfer occurs in the c.v. shown
3) Potential energy effects are negligible.
4) Helium is modeled as ideal gas.
5) the environment is at T0 = 20C and P0 = 1 atm.
Analysis:
Energy balance for the nozzle at steady-state reduces to:
M. Bahrami
ENSC 461 (S 11)
Assignmnet 5
3
s



 V 2  V22
0  Q cv  W cv  m h1  h2   1



2


0


  gz 

 0 
(1)
V2  V12  2h1  h2 
For monatomic gases, such as He, Ne, and Ar, c p over a wide range of temperature range is
very nearly equal to 5/2 R . Thus, h1 – h2 = cp (T1 – T2) and:
2
 1kg.m / s 2

8314 N .m 
 m




V2  10   2 2.5
1300
900

 s
 4.003 kg.K 
 1N

m
  2038
s

To determine the isentropic nozzle efficiency requires the temperature at state 2s. With k =
1.667, we have:
P
T2 s  T1  2
 P1



k 1
k
0.667
 1.45  1.667
 1300
 866.2 K

 4 
Then with Eq. (1)
V2 s  V12  2h1  h2 s   V12  2c p T1  T2 s   2122
m
s
The isentropic nozzle efficiency is:
2
 nozzle
V22 / 2  2038 
 2

  0.922
V2 s / 2  2122 
92.2% 
The exergy destruction given by:

X destruction

m

 S gen
 T0  
 m





No heat transfer and work transfer occurs in the nozzle, thus an entropy balance gives:

X destruction

m


T
P 
T
P
 T0 s 2  s1   T0 c p ln 2  R ln 2   T0 R 2.5 ln 2  ln 2
T1
P1 
T1
P1





 kJ 
 8.314 kJ 
900
1.45 
 2.5 ln
 ln
 293
  58.1 
1300
4 
 kg 
 4.003 kg.K 
M. Bahrami
ENSC 461 (S 11)
Assignmnet 5
4