10-86
10-85 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the
bottoming cycle. The mass flow rate of air for a specified power output is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic
and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis Working around the topping cycle gives the
following results:
T6 s
P
 T5  6
 P5




( k 1) / k
 (293 K)(8) 0.4/1.4  530.8 K
T
7
1373 K
·
Qin
c p (T6 s  T5 )
h h
 C  6s 5 
h6  h5
c p (T6  T5 )

 T6  T5 
8
T6 s  T5
6
C




( k 1) / k
8s
6s
1
 (1373 K) 
8
3
320C
6 MPa
530.8  293
 293 
 572.8 K
0.85
P
T8 s  T7  8
 P7
GAS
CYCLE
9
0.4/1.4
 758.0 K
293 K
c p (T7  T8 )
h h

 T8  T7   T (T7  T8 s )
T  7 8 
h7  h8 s c p (T7  T8 s )
 1373  (0.90)(1373  758.0)
 819.5 K
5
2
1
STEAM
CYCLE 20 kPa
·
4s
Qout
4
s
T9  Tsat @ 6000 kPa  275.6C  548.6 K
Fixing the states around the bottom steam cycle yields (Tables A-4, A-5, A-6):
h1  h f @ 20 kPa  251.42 kJ/kg
v 1  v f @ 20 kPa  0.001017 m 3 /kg
wp,in  v 1 ( P2  P1 )
 1 kJ 
 (0.001017 m 3 /kg )(6000  20)kPa 

1 kPa  m 3 

 6.08 kJ/kg
h2  h1  wp,in  251.42  6.08  257.5 kJ/kg
P3  6000 kPa  h3  2953.6 kJ/kg

 s 3  6.1871 kJ/kg  K
P4  20 kPa 
 h4 s  2035.8 kJ/kg
s 4  s3

T3  320C
T 
h3  h4

h4  h3   T (h3  h4 s )
h3  h4 s
 2953.6  (0.90)(2953.6  2035.8)
 2127.6 kJ/kg
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
10-87
The net work outputs from each cycle are
wnet, gas cycle  wT, out  wC,in
 c p (T7  T8 )  c p (T6  T5 )
 (1.005 kJ/kg  K )(1373  819.5  572.7  293)K
 275.2 kJ/kg
wnet, steam cycle  wT,out  wP,in
 (h3  h4 )  wP,in
 (2953.6  2127.6)  6.08
 819.9 kJ/kg
An energy balance on the heat exchanger gives
m a c p (T8  T9 )  m w (h3 -h2 ) 
 m w 
c p (T8  T9 )
h3 -h2
m a 
(1.005)(819.5  548.6)
 0.1010m a
2953.6  257.5
That is, 1 kg of exhaust gases can heat only 0.1010 kg of water. Then, the mass flow rate of air is
m a 
W net
100,000 kJ/s

 279.3 kg/s
wnet (1 275.2  0.1010  819.9) kJ/kg air
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.