Solution cv stage 1 air steady flow process adibatic

Solution:
C.V.: Stage 1 air, Steady flow
Process: adibatic: q = 0, reversible: s
gen
= 0
Energy Eq.6.13: -w
C1
= h
2
² h
1
, Entropy Eq.9.8: s
2
= s
1
Assume constant C
P0
= 1.004 from A.5 and isentropic leads to Eq.8.32
T
2
= T
1
(P
2
/P
1
)
k-1
k
= 300(2000/100)
0.286
= 706.7 K
w
C1
= h
1
- h
2
= C
P0
(T
1
- T
2
) = 1.004(300 – 706.7) = -408.3 kJ/kg
C.V. Intercooler, no work and no changes in kinetic or potential energy.
q
23
= h
3
- h
2
= C
P0
(T
3
- T
2
) = 1.004(340 – 706.7) = -368.2 kJ/kg
C.V. Stage 2. Analysis the same as stage 1. So from Eq.8.32
T
4
= T
3
(P
4
/P
3
) = 340(15.74/2)
0.286
= 613.4 K
w
C2
= h
3
- h
4
= C
P0
(T
3
- T
4
) = 1.004(340 – 613.4) = -274.5 kJ/kg
Same flow rate through both stages so the total work is the sum of the two
w
comp
= w
C1
+ w
C2
= –408.3 – 274.5 = – 682.8 kJ/kg
For no intercooler (P
2
= 15.74 MPa) same analysis as stage 1. So Eq.8.32
T
2
= 300(15740/100)
0.286
= 1274.9 K
w
comp
= 1.004(300 – 1274.9) = – 978.8 kJ/kg
C1
C2
1
2 3
4
2
-W
·
Q
·
1
-W
·
cooler
inter-

9-24
9.22 A heat-powered portable air compressor consists of three components: (a) an
adiabatic compressor; (b) a constant pressure heater (heat supplied from an outside
source); and (c) an adiabatic turbine. Ambient air enters the compressor at 100 kPa,
300 K, and is compressed to 600 kPa. All of the power from the turbine goes into
the compressor, and the turbine exhaust is the supply of compressed air. If this
pressure is required to be 200 kPa, what must the temperature be at the exit of the
heater?
Solution:
Heater
C
T
q
H
1
2 3
4
P
2
= 600 kPa, P
4
= 200 kPa
Adiabatic and reversible compressor:
Process: q = 0 and s
gen
= 0
Energy Eq.6.13: h ² w
c
= h
2
Entropy Eq.9.8: s
2
= s
1
For constant specific heat the isentropic relation becomes Eq.8.32
T
2
= T
1
©
¨
§
¹
¸
·
P
2
P
1
k-1
k
= 300(6)
0.286
= 500.8 K
² w
c
= C
P
0
(T
2
- T
1
) = 1.004(500.8 ² 300) = 201.5 kJ/kg
Adiabatic and reversible turbine: q = 0 and s
gen
= 0
Energy Eq.6.13: h
3
= w
T
+ h
4
; Entropy Eq.9.8: s
4
= s
3
For constant specific heat the isentropic relation becomes Eq.8.32
T
4
= T
3
(P
4
/P
3
)
k-1
k
= T
3
(200/600)
0.286
= 0.7304 T
3
Energy Eq. for shaft: ² w
c
= w
T
= C
P
0
(T
3
² T
4
)
201.5 = 1.004 T
3
(1 ² 0.7304) => T
3
= 744.4 K
2
1
v
T
s
1
300
100 kPa
3
P
2 3
4
200 kPa
600 kPa
4

9-25
9.23 A certain industrial process requires a steady 0.5 kg/s supply of compressed air at
500 kPa, at a maximum temperature of 30 q C. This air is to be supplied by
installing a compressor and aftercooler. Local ambient conditions are 100 kPa,
20 q C. Using an reversible compressor, determine the power required to drive the
compressor and the rate of heat rejection in the aftercooler.
Solution:
Air Table A.5: R = 0.287 kJ/kg-K, C
p
= 1.004 kJ/kg K, k = 1.4
State 1: T
1
= T
o
= 20
o
C, P
1
= P
o
= 100 kPa, m
.
= 0.5 kg/s
State 2: P
2
= P
3
= 500 kPa
State 3: T
3
= 30
o
C, P
3
= 500 kPa
Compressor: Assume Isentropic (adiabatic q = 0 and reversible s
gen
= 0 )
From entropy equation Eq.9.8 this gives constant s which is expressed for an
ideal gas in Eq.8.32
T
2
= T
1
(
P
2
/P
1
)
k-1
k
, T
2
= 464.6 K
1
st
Law Eq.6.13: q
c
+ h
1
= h
2
+ w
c
; q
c
= 0,
assume constant specific heat from Table A.5
w
c
= C
p
(T
1
- T
2
) = -172.0 kJ/kg
W
.
C
= m
.
w
C
= -86 kW
Aftercooler:
1
st
Law Eq.6.13: q + h
2
= h
3
+ w; w = 0,
assume constant specific heat
q = C
p
(T
3
- T
2
) = -205 kJ/kg, Q
.
= m
.
q = -102.5 kW
Compressor
cooler
Q
-W
C
1
2
3
.
.
cooler
after-

9-26
Steady state irreversible processes
9.24 Analyze the steam turbine described in Problem 6.44. Is it possible?