Chem E 260
HW #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
T
exist. 2 The air-standard assumptions are
applicable. 3 Kinetic and potential energy
3
1160
changes are negligible. 4 Air is an ideal gas
q
in
K
with variable specific heats.
2
2s
Properties The properties of air are given in Table A-17.
4
Analysis (a) Noting that process 1-2s is isentropic,
310
1
qout 4
s
s
h1  310.24 kJ / kg
K
T1  310 K


Pr1  1.5546
Pr 2 
P2
Pr  81.5546   12 .44 
 h2 s  562.58 kJ/kg and T2 s  557.25 K
P1 1
C 
h2 s  h1
h  h1

 h2  h1  2 s
h2  h1
C
562 .58  310 .24
 646.7 kJ/kg
0.75
h3  1230.92 kJ/kg
 310 .24 
T3  1160 K 

Pr3  207 .2
P4
1
Pr   207 .2  25 .90 
 h4 s  692.19 kJ/kg and T4 s  680 .3 K
P3 3  8 
h h
T  3 4 
 h4  h3   T h3  h4 s 
h3  h4 s
 1230 .92  0.82 1230 .92  692 .19 
 789.16 kJ/kg
Pr 4 
Thus,
T4 = 770.1 K
(b)
q in  h3  h2  1230 .92  646 .7  584.2 kJ/kg
q out  h4  h1  789 .16  310 .24  478.92 kJ/kg
wnet,out  q in  q out  584 .2  478 .92  105.3 kJ/kg
(c)
 th 
wnet,out
q in

105.3 kJ/kg
 18.0%
584.2 kJ/kg
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,
P
T2  T1  2
 P1



( k 1) / k
 2000 kPa 
 (300 K) 

 100 kPa 
0.4/1.4
 706 .1 K
1000
K
Similarly,
P
T4  T3  4
 P3




( k 1) / k
 100 kPa 
 (1000 K) 

 2000 kPa 
T
0.4/1.4
3
qin
2
 424 .9 K
Applying the first law to the constant-pressure
heat addition process 2-3 produces
300
K
4
1
qout
s
qin  h3  h2  c p (T3  T2 )  (1.005 kJ/kg K)(1000  706.1)K  295.4 kJ/kg
Similarly,
qout  h4  h1  c p (T4  T1 )  (1.005 kJ/kg K)(424.9  300)K  125.5 kJ/kg
The net work production is then
wnet  qin  q out  295 .4  125 .5  169.9 kJ/kg
and the thermal efficiency of this cycle is
 th 
wnet 169.9 kJ/kg

 0.575
q in
295.4 kJ/kg
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 psiaft3/lbm·R
(Table A-1E), cp = 0.1253 Btu/lbm·R and k = 1.667 (Table A-2Ea).
Analysis At the compressor inlet,
T
RT
(0.2686 psia  ft 3 /lbm  R )(540 R)
v1  1 
 9.670 ft 3 /lbm
P1
15 psia
3
1660
q
in
AV
R
(3 ft 2 )(200 ft/s)
 1 1 
m

62
.
05
lbm/s
2
3
v
1
9.670 ft /lbm
540
R
4
1
qout
s
According to the isentropic process expressions for an ideal gas,
P
T2  T1  2
 P1



( k 1) / k
P
T4  T3  4
 P3




( k 1) / k
 150 psia 

 (540 R) 
 15 psia 
0.667/1.667
 15 psia 

 (1660 R) 
 150 psia 
 1357 R
0.667/1.667
 660 .7 R
Applying the first law to the constant-pressure heat addition process 2-3 gives
Q in  m c p (T3  T2 )  (62 .05 lbm/s) (0.1253 Btu/lbm  R )(1660  1357 )R  2356 Btu/s
The net power output is
W net  m c p (T3  T4  T1  T2 )
 (62 .05 lbm/s) (0.1253 Btu/lbm  R )(1660  660 .7  540  1357 )R
 1417 Btu/s
The thermal efficiency of this cycle is then
 th 
W net 1417 Btu/s

 0.601
2356 Btu/s
Q in
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,
T2  T1 r p( k 1) / k  (273 K)(10) 0.4/1.4  527 .1 K
T
The heat addition is
qin
3
Q
500 kW
 in 
 500 kJ/kg

m
1 kg/s
qin
2
Applying the first law to the heat addition process,
q in  c p (T3  T2 )
T3  T2 
q in
500 kJ/kg
 527 .1 K 
 1025 K
cp
1.005 kJ/kg  K
273
K
4
1
The temperature at the exit of the turbine is
 1
T4  T3 
 rp





( k 1) / k
1
 (1025 K) 
 10 
0.4/1.4
 530 .9 K
Applying the first law to the adiabatic turbine and the compressor produce
wT  c p (T3  T4 )  (1.005 kJ/kg K)(1025  530.9)K  496.6 kJ/kg
qout
s
wC  c p (T2  T1 )  (1.005 kJ/kg K)(527.1  273)K  255.4 kJ/kg
The net power produced by the engine is then
 (wT  wC )  (1 kg/s)(496.6  255.4)kJ/kg  241.2kW
W net  m
Finally the thermal efficiency is
 th 
W net 241.2 kW

 0.482
500 kW
Q in
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),
h1  h f @ 6 psia  138 .02 Btu/lbm
v 1  v f @ 6 psia  0.01645 ft 3 /lbm
T
wp,in  v 1 ( P2  P1 )

1 Btu
 (0.01645 ft /lbm)(500  6)psia 
 5.404 psia  ft 3

 1.50 Btu/lbm
h2  h1  wp,in  138 .02  1.50  139 .52 Btu/lbm
3




2
500
psia
qin
3
6 psia
P3  500 psia  h3  1630 .0 Btu/lbm
1
qout

T3  1200 F  s 3  1.8075 Btu/lbm  R
s 4  s f 1.8075  0.24739
P4  6 psia  x 4 

 0.9864
s fg
1.58155

s 4  s3
 h  h  x h  138 .02  (0.9864 )(995 .88 )  1120 .4 Btu/lbm
4
f
4 fg
4
s
Knowing the power output from the turbine the mass flow rate of steam in the cycle is
determined from
W T, out  m (h3  h4 ) 
 m 
W T, out
h3  h4

500 kJ/s
 0.94782 Btu 

  0.9300 lbm/s
(1630.0  1120.4)Btu /lbm 
1 kJ

The rates of heat addition and rejection are
Q in  m (h3  h2 )  (0.9300 lbm/s)(163 0.0  139 .52 )Btu/lbm  1386 Btu/s
Q out  m (h4  h1 )  (0.9300 lbm/s)(112 0.4  138 .02 )Btu/lbm  913.6 Btu/s
and the thermal efficiency of the cycle is
 th  1 
Q out
913 .6
 1
 0.341

1386
Q
in
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.
Analysis From the steam tables (Tables A-4, A-5, and A-6),
h1  h f @ 20 kPa  251 .42 kJ/kg
v 1  v f @ 20 kPa  0.001017 m 3 /kg
T
w p,in  v 1 ( P2  P1 )
 1 kJ

 (0.001017 m /kg )( 4000  20 )kPa 

3
1 kPa  m 

 4.05 kJ/kg
h2  h1  w p,in  251 .42  4.05  255 .47 kJ/kg
3
4 MPa
3
qin
2
20 kPa
1
qout
P3  4000 kPa  h3  3906 .3 kJ/kg

T3  700 C
 s 3  7.6214 kJ/kg  K
s4  s f
7.6214  0.8320
P4  20 kPa  x 4 

 0.9596
s fg
7.0752

s 4  s3
 h  h  x h  251 .42  (0.9596 )( 2357 .5)  2513 .7 kJ/kg
4
f
4 fg
4
s
The power produced by the turbine and consumed by the pump are
W T, out  m (h3  h4 )  (50 kg/s)(3906 .3  2513 .7)kJ/kg  69,630 kW
W P,in  m wP,in  (50 kg/s)(4.05 kJ/kg)  203 kW
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),
h1  h f @ 25 kPa  271 .96 kJ/kg
v 1  v f @ 25 kPa  0.00102 0 m 3 /kg
w p ,in  v 1 P2  P1 


 1 kJ
 0.00102 m 3 /kg 5000  25 kPa 
 1 kPa  m 3

 5.07 kJ/kg
T




h2  h1  w p ,in  271 .96  5.07  277 .03 kJ/kg
P3  5 MPa  h3  3317 .2 kJ/kg

T3  450 C  s 3  6.8210 kJ/kg  K
s 4  s f 6.8210  0.8932
P4  25 kPa 

 0.8545
 x4 
s 4  s3
s fg
6.9370

3
2
5 MPa
·
Q
i
n
25 kPa
·
1
Q
ou
t
h4  h f  x 4 h fg  271 .96  0.8545 2345 .5  2276 .2 kJ/kg
The thermal efficiency is determined from
4
s
qin  h3  h2  3317 .2  277 .03  3040 .2 kJ/kg
qout  h4  h1  2276 .2  271 .96  2004 .2 kJ/kg
and
 th  1 
Thus,
q out
2004 .2
 1
 0.3407
q in
3040 .2
 overall   th  comb  gen  0.3407 0.75 0.96   24.5%
(b) Then the required rate of coal supply becomes
W net
300,000 kJ/s
Q in 

 1,222 ,992 kJ/s
 overall
0.2453
and
m coal 
Q in
1,222,992 kJ/s  1 ton 

  0.04174 tons/s  150.3 tons/h

C coal
29,300 kJ/kg  1000 kg 
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),
h1  h f @ 20 kPa  251 .42 kJ/kg
v1  v f @ 20 kPa  0.00101 7 m 3 /kg
w p ,in  v 1 P2  P1 

T
3 5

 1 kJ


 0.001017 m 3 /kg 8000  20 kPa 
3
1 kPa  m 

 8.12 kJ/kg
h2  h1  w p ,in  251 .42  8.12  259 .54 kJ/kg
P3  8 MPa  h3  3399 .5 kJ/kg

T3  500 C  s3  6.7266 kJ/kg  K
8 MPa
4
2
20 kPa
1
P4  3 MPa 
 h4  3105 .1 kJ/kg
s4  s3

P5  3 MPa  h5  3457 .2 kJ/kg

T5  500 C  s5  7.2359 kJ/kg  K
s6  s f
7.2359  0.8320

 0.9051
P6  20 kPa  x6 
s fg
7.0752

s6  s5

h6  h f  x6 h fg  251 .42  0.9051 2357 .5  2385 .2 kJ/kg
The turbine work output and the thermal efficiency are determined from
wT, out  h3  h4   h5  h6   3399 .5  3105 .1  3457 .2  2385 .2  1366.4 kJ/kg
and
q in  h3  h2   h5  h4   3399 .5  259 .54  3457 .2  3105 .1  3492 .0 kJ/kg
6
s
wnet  wT ,out  w p,in  1366 .4  8.12  1358 .3 kJ/kg
Thus,
 th 
wnet 1358.3 kJ/kg

 38.9%
q in
3492.5 kJ/kg