9-61
9-78 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits.
The cycle is to be sketched on the T-s cycle and the isentropic efficiency of the turbine and the cycle thermal efficiency are
to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential
energy changes are negligible. 4 Air is an ideal gas with constant specific heats.
Analysis (b) For the compression process,
4
)( 12Comp
TTcmW p
K 9.772
kPa 800
kPa 100
K) 1400(
0.4/1.4
/)1(
3
4
34
kk
sP
P
TT
kW 050,126K)9.7721400)(KkJ/kg 1.005(kg/s) 200()( 43sTurb, sp TTcmW
The actual power output from the turbine is
kW 300,120300,60000,60
TurbnetTurb
CompTurbnet
WWW
WWW
The isentropic efficiency of the turbine is then
95.4%954.0
kW 050,126
kW 300,120
sTurb,
Turb
Turb W
W
4s
s
T
1
3
873 K
303 K
9-63
Brayton Cycle with Regeneration
9-81C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at
9-83C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn,
9-65
9-85 A Brayton cycle with regeneration produces 115 kW power. The rates of heat addition and rejection are to be
determined.
K 0.585K)(10) 303(0.4/1.4/)1(
12 kk
ps rTT
K 1.627
87.0
3030.585
303
)(
)( 12
12
12
12
C
s
p
sp
C
TT
TT
TTc
TTc
K 8.555
10
1
K) 1073(
10.4/1.4
/)1(
45
kk
p
sr
TT
K 0.592
)8.5551073)(93.0(1073
)(
)(
)(
5445
54
54
sT
sp
p
TTTTT
TTc
TTc
When the first law is applied to the heat exchanger, the result is
6523 TTTT
while the regenerator temperature specification gives
K 0.582100.59210
53 TT
s
T
1
2
5s
4
qin
1073 K
303 K
3
6
qout
5
2s
9-67
9-87 An ideal Brayton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work
output and the thermal efficiency of the cycle are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and
53.5%
kJ/kg 601.94
kJ/kg 322.26
in
net
th
q
w
9-68
9-88 Problem 9-87 is reconsidered. The effects of the isentropic efficiencies for the compressor and turbine and
regenerator effectiveness on net work done and the heat supplied to the cycle are to be studied. Also, the T–s diagram for the
cycle is to be plotted.
Analysis Using EES, the problem is solved as follows:
“Input data”
P[2] = Pratio*P[1]
s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2])
“T_s[2] is the isentropic value of T[2] at compressor exit”
Eta_c = w_compisen/w_comp
“compressor adiabatic efficiency, W_comp > W_compisen”
“SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0
e_in – e_out =DELTAe_cv =0 for steady flow”
h[2] + q_in_noreg = h[3]
h[3]=ENTHALPY(Air,T=T[3])
P[3]=P[2]“process 2-3 is SSSF constant pressure”
“Actual Turbine analysis:”
h[3] = w_turb + h[4]
h[4]=ENTHALPY(Air,T=T[4])
s[4]=ENTROPY(Air,T=T[4], P=P[4])
“Cycle analysis”
9-69
P[5]=P[2]
“The regenerator effectiveness gives h[5] and thus T[5] as:”
Eta_reg = (h[5]-h[2])/(h[4]-h[2])
“Energy balance on regenerator gives h[6] and thus T[6] as:”
9-70
0.7 0.75 0.8 0.85 0.9 0.95 1
10
15
20
25
30
35
40
45
t
th
c = 0.8
With regeneration
No regeneration
0.7 0.75 0.8 0.85 0.9 0.95 1
50
95
140
185
230
275
t
wnet [kJ/kg]
c = 0.8
500
550
600
650
qin
No regeneration
With regeneration
9-71
0.6 0.65 0.7 0.75 0.8 0.85 0.9
10
15
20
25
30
35
40
45
c
th
t = 0.9
With regeneration
No regeneration
0.6 0.65 0.7 0.75 0.8 0.85 0.9
75
110
145
180
215
250
c
wnet [kJ/kg]
t = 0.9
0.6 0.65 0.7 0.75 0.8 0.85 0.9
500
520
540
560
580
600
620
640
660
680
c
qin
t = 0.9
No regeneration
With regeneration
9-72
9-89 An ideal Brayton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work
output and the thermal efficiency of the cycle are to be determined.
9-73
9-90 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit,
the net work output, and the thermal efficiency are to be determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air
is an ideal gas with variable specific heats. 3 Kinetic and potential
T
9-74
9-91 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is
considered. The power delivered by this plant is to be determined for two cases.
k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-21.
263.9
kPa 95
kPa 880
1
2
P
P
1
5
6
T
3
30,000 kW
1100 K
9-79
Carnot Vapor Cycle
two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the
cycle, (2) the turbine will have to handle steam with a high moisture content which causes erosion, and (3) it is not practical
to design a compressor that will handle two phases.
the quality at the end of the heat rejection process, and the net work output are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
2
1
0.205
3961.1
3358.062168.0
4
4
fg
f
s
ss
x
20 psia
3
4
s