Chem E 260HW #9
C&B: 9.91, 9.94, 9.102E , 9.103 and 10.15E, 10.17, 10.23 ,10.34
9-91 A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. The air temperature at the turbine exit, the net work output, and the 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 variable specific heats.
Properties The properties of air are given in Table A-17.
Analysis (a) Noting that process 1-2s is isentropic,
Th
Pr1
1
11 5546
= ⏐ →⏐==
310 K310.24 kJ /kg.
( )( )
( )
( )( )( )
kJ/kg 89.167
19.69292.123082.092.1230
K 3.680 and kJ/kg 692.1990.252.2078
1
2.207
kJ/kg 1230.92K 1160
kJ/kg 646.775.0
24.31058.56224.310
K 557.25 and kJ/kg 562.5844.125546.18
433443
43
443
4
33
1212
12
12
221
2
34
3
12
=
−−=
−−=⏐ →⏐−
−=
==⏐ →⏐=⎟⎠
⎞⎜⎝
⎛==
=
=⏐ →⏐=
=−
+=
−+=⏐ →⏐
−
−=
==⏐ →⏐===
sTs
T
ssrr
r
C
ssC
ssrr
hhhhhh
hh
ThPP
PP
P
hT
hhhh
hh
hh
ThPP
PP
ηη
ηη
Thus,
T4 = 770.1 K
(b)
kJ/kg 105.3=−=−=
=−=−=
=−=−=
92.4782.584
/kJ kg 478.9224.31016.789
/kJ kg 584.27.64692.1230
outinout,net
14out
23in
qqwhhqhhq
s
T
1
2s
4s
3qin
qout
1160 K
310 K
2
4
(c) 18.0%===/kJ kg 584.2/kJ kg 105.3
in
out,netth q
wη
9-94 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The net work and the 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.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis Using the isentropic relations for an ideal gas,
K .1706kPa 100
kPa 2000K) 300(
0.4/1.4/)1(
1
212 =⎟
⎠
⎞⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛=
− kk
PP
TT
Similarly,
K 9.424kPa 2000
kPa 100K) 1000(
0.4/1.4/)1(
3
434 =⎟
⎠
⎞⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛=
− kk
PP
TT
Applying the first law to the constant-pressure heat addition process 2-3 produces
kJ/kg 4.295K)1.7061000)(KkJ/kg 1.005()( 2323in =−⋅=−=−= TTchhq p
Similarly,kJ/kg 5.125K)3009.424)(KkJ/kg 1.005()( 1414out =−⋅=−=−= TTchhq p
The net work production is thenkJ/kg 169.9=−=−= 5.1254.295outinnet qqw
and the thermal efficiency of this cycle is
0.575===kJ/kg 295.4
kJ/kg 169.9
in
netth q
wη
9-102E A simple ideal Brayton cycle with argon as the working fluid operates between the specified temperature and pressure limits. The rate of heat addition, the power produced, and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats.
Properties The properties of argon at room temperature are R = 0.2686 psiaft3/lbm·R (Table A-1E), cp = 0.1253 Btu/lbm·R and k = 1.667 (Table A-2Ea).
Analysis At the compressor inlet,
/lbmft 670.9psia 15
R) 540)(R/lbmftpsia 2686.0( 33
1
11 =
⋅⋅==
PRT
v
lbm/s 05.62/lbmft 670.9
ft/s) )(200ft 3(3
2
1
11 ===v
VAm&
s
T
1
2
4
3qin
qout
1000 K
300 K
s
T
1
2
4
3qin
qout
1660 R
540 R
According to the isentropic process expressions for an ideal gas,
R 1357psia 15
psia 150R) 540(
70.667/1.66/)1(
1
212 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
PP
TT
R 7.660psia 150
psia 15R) 1660(
70.667/1.66/)1(
3
434 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− kk
PP
TT
Applying the first law to the constant-pressure heat addition process 2-3 gives
Btu/s 2356=−⋅=−= R)13571660)(RBtu/lbm 1253.0(lbm/s) 05.62()( 23in TTcmQ p&&
The net power output is
Btu/s 1417=−+−⋅=
−+−=
R)13575407.6601660)(RBtu/lbm 1253.0(lbm/s) 05.62(
)( 2143net TTTTcmW p&&
The thermal efficiency of this cycle is then
0.601===Btu/s 2356
Btu/s 1417
in
netth
Q
W&
&η
9-103 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the 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.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Analysis For the isentropic compression process,
K .1527K)(10) 273( 0.4/1.4/)1(12 === − kkprTT
The heat addition is
kJ/kg 500kg/s 1
kW 500inin ===
mQ
q&
&
Applying the first law to the heat addition process,
K 1025KkJ/kg 1.005
kJ/kg 500K 1.527
)(
in23
23in
=⋅
+=+=
−=
p
p
cq
TT
TTcq
The temperature at the exit of the turbine is
K 9.53010
1K) 1025(
10.4/1.4/)1(
34 =⎟⎠
⎞⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛=
− kk
prTT
Applying the first law to the adiabatic turbine and the compressor producekJ/kg 6.496K)9.5301025)(KkJ/kg 1.005()( 43T =−⋅=−= TTcw p
s
T
1
2
4
3qin
qout273 K
kJ/kg 4.255K)2731.527)(KkJ/kg 1.005()( 12C =−⋅=−= TTcw p
The net power produced by the engine is then
kW 241.2=−=−= kJ/kg)4.2556kg/s)(496. 1()( CTnet wwmW &&
Finally the thermal efficiency is
0.482===kW 500
kW 241.2
in
netth
Q
W&
&η
10-15E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),
Btu/lbm 52.13950.102.138Btu/lbm 50.1
ftpsia 5.404
Btu 1 psia)6500)(/lbmft 01645.0(
)(
/lbmft 01645.0
Btu/lbm 02.138
inp,12
33
121inp,
3psia 6 @1
psia 6 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
Btu/lbm 4.1120)88.995)(9864.0(02.138
9864.058155.1
24739.08075.1
psia 6
RBtu/lbm 8075.1
Btu/lbm 0.1630
F1200
psia 500
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from
lbm/s 9300.0kJ 1
Btu 0.94782
/lbm1120.4)Btu(1630.0
kJ/s 500)(
43
outT,43outT, =⎟
⎠
⎞⎜⎝
⎛−
=−
=⏐ →⏐−=hh
WmhhmW
&&&&
The rates of heat addition and rejection are
Btu/s 913.6
Btu/s 1386
=−=−=
=−=−=
Btu/lbm)02.1380.4lbm/s)(112 9300.0()(
Btu/lbm)52.1390.0lbm/s)(163 9300.0()(
14out
23in
hhmQ
hhmQ
&&
&&
and the thermal efficiency of the cycle is
0.341=−=−=1386
6.91311
in
outth
Q
Q&
&η
10-17 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The power produced by the turbine and consumed by the pump are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
qin
qout
6 psia1
3
2
4
500 psia
s
T
Analysis From the steam tables (Tables A-4, A-5, and A-6),
kJ/kg 47.25505.442.251kJ/kg 05.4 mkPa 1
kJ 1 kPa)204000)(/kgm 001017.0(
)(
/kgm 001017.0
kJ/kg 42.251
inp,12
33
121inp,
3kPa 20 @1
kPa 20 @1
=+=+==
⎟⎠
⎞⎜⎝
⎛
⋅−=
−=
==
==
whh
PPw
hh
f
f
v
vv
kJ/kg 7.2513)5.2357)(9596.0(42.251
9596.00752.7
8320.06214.7
kPa 20
KkJ/kg 6214.7
kJ/kg 3.3906
C700
kPa 4000
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
==
⋅==
⎭⎬⎫
°==
fgf
fg
f
hxhhs
ssx
ssP
sh
TP
The power produced by the turbine and consumed by the pump are
kW 203
kW 69,630
===
=−=−=
kJ/kg) kg/s)(4.05 50(
kJ/kg)7.2513.3kg/s)(3906 50()(
inP,inP,
43outT,
wmW
hhmW
&&
&&
10-23 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
kJ/kg .0327707.596.271kJ/kg 5.07
mkPa 1
kJ 1kPa 255000/kgm 00102.0
/kgm 000102.0
kJ/kg 96.271
in,12
33
121in,
3kPa 25 @1
kPa 25 @1
=+=+==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
p
f
f
whh
PPw
hh
v
vv
( )( ) kJ/kg 2.22765.23458545.096.271
8545.09370.6
8932.08210.6kPa 25
KkJ/kg 8210.6
kJ/kg 2.3317
C450
MPa 5
44
44
34
4
3
3
3
3
=+=+=
=−
=−
=⎭⎬⎫
=
=
⋅=
=
⎭⎬⎫
°=
=
fgf
fg
f
hxhh
s
ssx
ss
P
s
h
T
P
The thermal efficiency is determined from
kJ/kg 2.200496.2712.2276kJ/kg 2.304003.2772.3317
14out
23in
=−=−==−=−=
hhqhhq
and
Thus,( )( )( ) 24.5%==××=
=−=−=
96.075.03407.0
3407.02.3040
2.200411
gencombthoverall
in
outth
ηηηη
ηq
q
(b) Then the required rate of coal supply becomes
and
qin
qout
20 kPa1
3
2
4
4 MPa
s
T
Qi
n
Qou
t
25 kPa1
3
2
4
5 MPa
s
T
·
·
tons/h 150.3==⎟⎟⎠
⎞⎜⎜⎝
⎛==
===
/tons s 04174.0kg 1000
ton1/kJ kg 29,300/kJ s 1,222,992
/kJ s 992,222,12453.0
/kJ s 300,000
coal
incoal
overall
netin
CQm
WQ
&&
&&
η
10-34 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the steam tables (Tables A-4, A-5, and A-6),
( )( )( )
kJ/kg .5425912.842.251
kJ/kg 8.12mkPa 1
kJ 1kPa 208000/kgm 001017.0
/kgm 700101.0
kJ/kg 42.251
in,12
33
121in,
3kPa 20 @1
kPa 20 @1
=+=+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⋅−=
−=
==
==
p
p
f
f
whh
PPw
hh
v
vv
( )( ) kJ/kg 2.23855.23579051.042.251
9051.00752.7
8320.02359.7kPa 20
KkJ/kg 2359.7
kJ/kg 2.3457
C500
MPa 3
kJ/kg 1.3105MPa 3
KkJ/kg 7266.6
kJ/kg 5.3399
C500
MPa 8
66
66
56
6
5
5
5
5
434
4
3
3
3
3
=+=+=
=−
=−
=
⎭⎬⎫
=
=
⋅=
=
⎭⎬⎫
°=
=
=⎭⎬⎫
=
=
⋅=
=
⎭⎬⎫
°=
=
fgf
fg
f
hxhh
s
ssx
ss
P
s
h
T
P
hss
P
s
h
T
P
The turbine work output and the thermal efficiency are determined from
and( ) ( )
( ) ( ) kJ/kg 0.34921.31052.345754.2595.3399
2.23852.34571.31055.3399
4523in
6543outT,
=−+−=−+−=
=−+−=−+−=
hhhhq
hhhhw kJ/kg 1366.4
Thus,38.9%===
=−=−=
kJ/kg 3492.5
kJ/kg 1358.3
kJ/kg 3.135812.84.1366
in
netth
in,,net
q
w
www poutT
η
1
5
2
6s
T
3
48 MPa
20 kPa