ΕΡ&Σ
Π Α Ν Ε Π Ι Σ Τ Η Μ Ι Ο Θ Ε Σ Σ Α Λ Ι Α Σ
ΠΟΛΥΤΕΧΝΙΚΗ ΣΧΟΛΗ
ΤΜΗΜΑ ΜΗΧΑΝΟΛΟΓΩΝ ΜΗΧΑΝΙΚΩΝ
LFM&T
ΕΡΓΑΣΤΗΡΙΟ ΡΕΥΣΤΟΜΗΧΑΝΙΚΗΣ & ΣΤΡΟΒΙΛΟΜΗΧΑΝΩΝ
Α.Δ. Παπαθανασίου, Αν.Καθ.
Λεωφ. Αθηνών-Πεδίον Άρεως-38334 Βόλος - Τηλ. 24210 74094/11- email: fluids@mie.uth.gr - web: http://www.mie.uth.gr/labs/fluids
PROBLEM 9-157
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential
energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
Analysis (a) For this problem, we
350°C
25°C
use the properties of air from EES
5
Combustion
software. Remember that for an ideal
Heat
chamber
gas, enthalpy is a function of
exchanger
temperature only whereas entropy is
3
2 1.2 MPa
functions of both temperature and
4 500°C
pressure.
Sat. vap.
200°C
Turbine
Compress.
Process 1-2: Compression
T1  30 C 
 h1  303 .60 kJ/kg
T1  30 C

s1  5.7159 kJ/kg  K
P1  100 kPa 
1
100 kPa
30°C
P2  1200 kPa

h2 s  617 .37 kJ/kg
s 2  s1  5.7159 kJ/kg.K 
C 
h2 s  h1
617 .37  303 .60

 0.82 

 h2  686 .24 kJ/kg
h2  h1
h2  303 .60
Process 3-4: Expansion
T4  500 C 
 h4  792 .62 kJ/kg
T 
h3  h4
h  792 .62

 0.82  3
h3  h4 s
h3  h4 s
We cannot find the enthalpy at state 3 directly. We find h3 = 1404.7 kJ/kg, T3 =
1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error
approach.
Also, T5  350 C  h5  631 .44 kJ/kg
The inlet water is compressed liquid at 25ºC and at the saturation pressure of steam at
200ºC (1555 kPa). This is not available in the tables but we can obtain it in EES. The
alternative is to use saturated liquid enthalpy at the given temperature.
Tw1  25 C

hw1  106 .27 kJ/kg
P1  1555 kPa 
Tw2  200 C
hw2  2792 .0 kJ/kg
x2  1

The net work output is
wC,in  h2  h1  686 .24  303 .60  382 .64 kJ/kg
wT, out  h3  h4  1404 .7  792 .62  612 .03 kJ/kg
wnet  wT, out  wC,in  612 .03  382 .64  229 .39 kJ/kg
The mass flow rate of air is
W net
800 kJ/s

 3.487 kg/s
wnet 229 .39 kJ/kg
m a 
(b) The back work ratio is
rbw 
wC,in
wT, out

382 .64
 0.625
612 .03
The rate of heat input and the thermal efficiency are
 a (h3  h2 )  (3.487 kg/s)(1404.7  686.24)kJ/kg  2505 kW
Q in  m
 th 
W net
800 kW

 0.319

2505 kW
Qin
(c) An energy balance on the heat exchanger gives
m a (h4  h5 )  m w (hw2  hw1 )
(3.487 kg/s)(792. 62  631 .44 )kJ/kg  m w (2792.0  106 .27 )kJ/kg 
 m w  0.2093 kg/s
(d) The heat supplied to the water in the heat exchanger (process heat) and the
utilization efficiency are
Q p  m w (hw2  hw1 )  (0.2093 kg/s)(2792 .0  106 .27 )kJ/kg  562 .1 kW
u 
W net  Q p 800  562 .1

 0.544
2505 kW
Q in
PROBLEM 9-131
Assumptions 1 The air-standard
assumptions are applicable. 2 Kinetic
and potential energy changes are
negligible. 3 Air is an ideal gas with
constant specific heats.
Properties The properties of air at
500ºC = 773 K are cp = 1.093
kJ/kg·K, cv = 0.806 kJ/kg·K, R =
0.287 kJ/kg·K, and
k = 1.357
(Table A-2b).
Analysis (a) For the compressor and
the turbine:
k 1
Regenerator
6
5
100 kPa
30°C
1
2
700 kPa
260°C
Combustion
chamber
4
400°C
Compress.
871°C
3
Turbine
1.357-1
P  k
 700 kPa  1.357
T2 s  T1  2 
 303 K 
 505 .6 K

 100 kPa 
 P1 
T T
(505 .6  303 )K
 C  2s 1 
 0.881
T2  T1
(533  303 )K
( k 1) / k
P 
 100 kPa 
T4 s  T3  4 
 1144 K 
 685 .6 K

P
 700 kPa 
 3
T  T4
(1144  T4 )K
T  3

 0.85 

 T4  754 .4 K
T3  T4 s
(1144  685 .6)K
(1.357-1)/1.357
(b) The effectiveness of the regenerator is
 regen 
T5  T2
(673  533 )K

 0.632
T4  T2 (754 .4  533 )K
(c) The fuel rate and air-fuel ratio are
Q in  m f q HV c  (m f  m a )c p (T3  T5 )
m f (42 ,000 kJ/kg)(0.9 7)  (m f  12 .6)(1.093 kJ/kg.K)(1 144  673)K 
 m f  0.1613 kg/s
AF 
Also,
m a
12 .6

 78.14
m f
0.1613
m  m a  m f  12 .6  0.1613  12 .76 kg/s
 f q HV c  (0.1613 kg/s)(42,0 00 kJ/kg)(0.9 7)  6570 kW
Q in  m
(d) The net power and the back work ratio are
W C,in  m a c p (T2  T1 )  (12 .6 kg/s)(1.09 3 kJ/kg.K)(5 33  303 )K  3168 kW
WT, out  m c p (T3  T4 )  (12 .76 kg/s)(1.09 3 kJ/kg.K)(1 144  754 .4)K  5434 kW
Wnet  WT, out  WC, in  5434  3168  2267 kW
W C,in
3168 kW
rbw 

 0.583

WT, out 5434 kW
(e) The thermal efficiency is
 th 
W net 2267 kW

 0.345
6570 kW
Q in
(f) The second-law efficieny of the cycle is defined as the ratio of actual thermal
efficiency to the maximum possible thermal efficiency (Carnot efficiency). The
maximum temperature for the cycle can be taken to be the turbine inlet temperature.
That is,
 max  1 
and
 II 
T1
303 K
1
 0.735
T3
1144 K
 th
0.345

 0.469
 max 0.735
(g) The exergy efficiency for the compressor is defined as the ratio of stream exergy
difference between the inlet and exit of the compressor to the actual power input:


T
P  
X C  m a h2  h1  T0 ( s 2  s1 )  m a c p T2  T1   T0 c p ln 2  R ln 2  
T1
P1  




 533 
 700  
 (12 .6)(1.093 )(533  303 ) (303 ) (1.093 )ln 
  0.287 ln 
   2943 kW

 303 
 100  

X C 2943 kW
 II,C 

 0.929
W C,in 3168 kW
The exergy efficiency for the turbine is defined as the ratio of actual turbine power to
the stream exergy difference between the inlet and exit of the turbine:


T
P  
X T  m c p T3  T4   T0 c p ln 3  R ln 3  
T4
P4  



 1144 
 700  

 (12 .76 )(1.093 )(1144  754 .4) (303 ) (1.093 )ln 
  0.287 ln 
   5834 kW

 754.4 
 100  

W T, in 5434 kW
 II ,T 

 0.932
X T 5834 kW
An energy balance on the regenerator gives
m a c p (T5  T2 )  m c p (T4  T6 )
(12 .6)(1.093)(6 73  533 )  (12 .76 )(1.093 )(754 .4  T6 ) 
 T6  616 .2 K
The exergy efficiency for the regenerator is defined as the ratio of the exergy increase
of the cold fluid to the exergy decrease of the hot fluid:


 
T
X regen,hot  m c p T4  T6   T0 c p ln 4  0 
T6


 


 754.4   
 (12 .76 )(1.093 )( 754 .4  616 .2) (303 ) (1.093 )ln 
  0   1073 kW

 616.2   



 
T
X regen,cold  m c p T5  T2   T0 c p ln 5  0 
T2


 


 673   
 (12 .76 )(1.093 )( 673  533 ) (303 ) (1.093 )ln 
  0   954 .8 kW

 533   

X regen,cold 954 .8 kW
 II ,T 

 0.890
1073 kW
X regen,hot
The exergy of the combustion gases at the regenerator exit:


 
T
X 6  m c p T6  T0   T0 c p ln 6  0 
T0


 


 616.2   
 (12 .76 )(1.093 )( 616 .2  303 ) (303 ) (1.093 )ln 
  0   1351 kW

 303   

PROBLEM 9-134
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential
energy changes are negligible. 3 Air is an ideal gas with variable specific heats.
P
Properties The properties of air are given in Table A-17.
q23 3
Analysis (b) We treat air as an ideal gas with variable specific heats,
2
T1  300 K 
 u1  214.07 kJ/kg
q12
h1  300.19 kJ/kg
 900 K 
 u 2  674.58 kJ/kg
h2  932.93 kJ/kg
4
1
 300 kPa 
P2v 2 P1v 1
P
300 K 


 T2  2 T1  
T2
T1
P1
 100 kPa 
qout
T
Pr 4 
 100 kPa 
P4
330.9   110 .3
Pr3  
P3
 300 kPa 

 h4  1036.46 kJ/kg
q in  q12,in  q 23,in  u 2  u1   h3  h2 
 674 .58  214 .07   1395 .97  932 .93 
 923.55 kJ/kg
q out  h4  h1  1036 .46  300 .19  736.27 kJ/kg
wnet  q in  q out  923 .55  736 .27  187.28 kJ/kg
(c)
 th 
wnet 187.28 kJ/kg

 20.3%
q in
923.55 kJ/kg
3
q23
T3  130 0 K 
 u 3  1022.82 kJ/kg
h3  1395.97 kJ/kg, Pr 3  330 .9
v
2
q12
1
4
qout
s