Problem As Term o
Problem As Term o
Problem As Term o
30 NOVIEMBRE 2009
917E An air-standard cycle with variable specific heats is executed in a closed system
and is composed of the following four processes:
1-2 v = constant heat addition from 14.7 psia and 80F in the amount of 300 Btu/lbm
2-3 P = constant heat addition to 3200 R
3-4 Isentropic expansion to 14.7 psia
4-1 P= constant heat rejection to initial state
(a) Show the cycle on P-v and T-s diagrams.
(b) Calculate the total heat input per unit mass.
(c) Determine the thermal efficiency.
Consideraciones:
P
Qin
57.61psi
Qin
14.7psi
1
Qout
T
3400R
3
Qin
2
Qin
Qout
340R
1
Partimos del Edo. 1 al Edo. 2 para obtener los datos para ste ltimo:
T1 = 80F = 540R
u1 = 92.04 BTU/lbm
h1 = 129.06 BTU/lbm
u2 = Q12 + u1 = 300
BTU
BTU
BTU
+ 92.04
= 392.04
lbm
lbm
lbm
V2 = V1
Para el Edo. 3
T3 = 3200 R
u3 = 630.12 BTU/lbm
h3 = 849.48 BTU/lbm
Pr3 = 1242
P2 P1
=
T2 T1
P2 =
Pr4 Pr3
=
P4
P3
Pr4 =
P1= P4 y P3= P2
Para el Edo.4
h4 = 593.176 BTU/lbm
Pr4 = 316.9165
Del balance de energa, tenemos que,
Q23 in = h3 - h2 = 849.28
BTU
BTU
BTU
- 537.103
= 312.377
lbm
lbm
lbm
BTU
BTU
BTU
+ 312.377
= 612.377
lbm
lbm
lbm
Qout = h4 - h1 = 593.176
BTU
BTU
BTU
- 129.06
= 464.116
lbm
lbm
lbm
BTU
lbm
BTU
lbm
Diagramas T-s
T
Qin
2
1200R
350R
1
Qout
De las tablas A-17 obtenemos los siguientes valores para obtener la presin
mxima del sistema, tenemos que;
T1 = 1200 K;
P1 =
Pr1 = 238
T4 = 350K;
Pr4 = 2.379
wnet
hTH
hTH = 1 +
TL
350 K
= 1+
= 0.7083 x 100 % = 70.83 % ;
TH
1200 K
Qin =
0.5 kJ
= 0.7059 kJ
0.7083
Wnet
;
wnet
s4 - s3 = H0L - j
j0.287
i
k
kJ y
300 kPa
kJ
z
z ln
= -0.1989
kg K {
150 kPa
kg K
P4
P3
kJ y
kJ
z
z H1200 K - 350 KL = 169.093
kg K {
kg
Wnet
0.5 kJ
=
= 0.002956 kg
wnet 169.093 kJ
kg
964E An ideal Ericsson engine using helium as the working fluid operates between
temperature limits of 550 and 3000 R and pressure limits of 25 and 200 psia. Assuming a
mass flow rate of 14lbm/s, determine (a) the thermal efficiency of the cycle, (b) the
heat transfer rate in the regenerator, and (c) the power delivered.
Consideraciones:
Diagramas T-s
T
.
Qin
2
3000R
550R
1
.
Qout
hTH = 1 +
TL
550 R
= 1+
= 0.8166x 100 % = 81.66 %
TH
3000 R
lbm
BTU
BTU
N J1.25
N H3000 R - 550 RL = 42 875
s
lbm R
s
Qin = J14
Hs2 - s1L = Cp ln
T2
P2
- R ln
T1
P1
sabiendo que ln
BTU
25 psi
BTU
N ln
= 1.0316
lbm R
200 psi
lbm R
lbm
BTU
BTU
N H3000 RL J1.0316
N = 43 327.6598
s
lbm R
s
BTU
BTU
N = 35 381.367
s
s
984 A gas-turbine power plant operates on the simple Brayton cycle between the
pressure limits of 100 and 1200 kPa. The working fluid is air, which enters the
compressor at 30C at a rate of 150 m3/min and leaves the turbine at 500C. Using
variable specific heats for air and assuming a compressor isentropic efficiency of 82
percent and a turbine isentropic efficiency of 88 percent, determine (a) the net power
output, (b) the back work ratio, and (c) the thermal efficiency.
T2
= 0;
T1
Consideraciones:
T
550C
3
Qin
2
30C
Pa
0k
20
1
P=
a
kP
00
1
P=
Qout
P
1200kPa
2
Qin
s=
cte
1
100kPa
s=
cte
Qout
4
V
T1 = 30C = 303K
h1 = 303.208 kJ/kg
Pr1 = 1.43556
Pr2 Pr1
=
P2
P1
Pr2 =
Para el Edo.2
T2 = 609.2815 K
h2s = 609.2815 kJ/kg
Pr2 = 17.22672
Para el Edo.4
T4 = 500C = 773 K
T3 = 1437.345 K
h3 = 1560.315 kJ/kg
Pr4 = 41.922
Pr3 = 503.064
Pr4 Pr3
=
P4
P3
Pr3 =
Hh2 s - h1L
hC
616.7749 kJ - 303.208 kJ y
i
j
kJ
kJ y
kJ
kg
kg z
i
j
z
j
z
j
z
wNET = H0.88L j1560.315
- 792.3825
z-j
= 293.3819
z
j
z
z
kg
kg { j
0.82
kg
k
k
{
kJ
kJ
wCOMP = 382.3986
kg
kg
La razn de trabajo regenerado;
kJ
wC 382.3980 kg
rbw =
=
= 0.56586
wT 675.7806 kJ
wTUB = 675.7806
kg
wNET
hTH =
=
Qin
293.3819
h3 - h4s
kJ
kg
293.3819
kJ
kg
J1560.315 kJ - 616.7749 kJ N
kg
kg
9109 Consider an ideal gas-turbine cycle with two stages of compression and two stages
of expansion. The pressure ratio across each stage of the compressor and turbine is 3.
The air enters each stage of the compressor at 300 K and each stage of the turbine at
1200 K. Determine the back work ratio and the thermal efficiency of the cycle, assuming
T
1200K
5 7
Qin
6 8
10
2
300K
1
s
Para el Edo.1
T1 = 300 K
h1 = 300.19 kJ/kg = h4
Pr1 = 1.386
Pr2 Pr1
=
P2
P1
Pr2 =
Pr1 P2
= H3L H1.386L = 4.158
P1
T5 = 1200 K
h5 = 1277.79 kJ/kg = h7
Pr5 = 238
Pr6 Pr5
=
P6
P5
Pr6 =
Pr5 P6 1
=
H238L = 79.333
P5
3
h6 = 946.3468 kJ/kg = h8
Pr6 = 79.333
wNET = wTUB - wCOMP = hT 2 Hh5 - h6L -
2 Hh2 - h1L
hC
kJ
kJ
i 411.257 kg - 300.19 kg y
j
z
kJ
kJ
kJ
i
y
j
z
z - H2L j
z
wNET = H0.80L H2L j
j1277.79
- 946.3468
z
= 268.975
j
z
j
z
j
z
kg
kg {
0.85
kg
k
k
{
j
qin = Hh5 - h4L + Hh7 - h6L = i
j1277.79
k
hTH =
wNET
=
Qin
kJ
kg
1197.9762 kJ
kg
kJ
kJ y
kJ
kJ y
kJ
i
z
z
- 411.257
z+j
j1277.79
- 946.3468
z = 1197.9762
kg
kg { k
kg
kg {
kg
268.975
9131 A gas-turbine power plant operates on the regenerative Brayton cycle between the
pressure limits of 100 and 700 kPa. Air enters the compressor at 30C at a rate of 12.6
kg/s and leaves at 260C. It is then heated in a regenerator to 400C by the hot
combustion gases leaving the turbine. A diesel fuel with a heating value of 42,000 kJ/kg
is burned in the combustion chamber with a combustion efficiency of 97 percent. The
combustion gases leave the combustion chamber at 871C and enter the turbine whose
isentropic efficiency is 85 percent. Treating combustion gases as air and using constant
specific heats at 500C, determine (a) the isentropic efficiency of the compressor, (b)
the effectiveness of the regenerator, (c) the airfuel ratio in the combustion chamber,
(d) the net power output and the back work ratio, (e) the thermal efficiency, and ( f )
the second-law efficiency of the plant. Also determine (g) the second-law (exergetic)
efficiencies of the compressor, the turbine, and the regenerator, and (h) the rate of the
exergy flow with the combustion gases at the regenerator exit.
Consideraciones:
T
871C
3
Qin
400C
Qin
5
4
30C
Qout
h=533.98 kJ/kg
P=100 kPa
P= 700 kPa
Pr=1.386
Pr=10.37
s1=1.70203kJ/kg s2=2.27967kJ/kg
Edo. 3
T= 871C
1140K
h=1207.57
kJ/kg
P= 700 kPa
Pr=193.1
Edo. 4
= hs=705.438
kJ/kg
h=780.758
kJ/kg
P= 700 kPa
Pr=27.7
kJ
kJ
h2 s - h1 523.90157 kg - 300.19 kg
hC =
=
= 0.956 x 100 % = 95.6 %
h2 a - h1
533.98 kJ
- 300.19 kJ
kg
kg
Pr2 Pr1
Pr1 P2 H1.386L H700 kPaL
=
Pr2 =
=
= 9.702
P2
P1
P1
100 kPa
Pr4 Pr3
P P
H193.9L H100 kPaL
=
Pr4 = r3 4 =
= 27.7
P4
P3
P3
700 kPa
Edo. 5
T=400C=670K
h=681.14 kJ/kg
P=100 kPa
Pr=24.46
kJ y
kJ
kJ z
kJ
i
i
y
z
h4 = h3- hT Hh3 - h4 sL = j
j1207.57
z - H0.85L j
j1207.57
- 705.43
z = 780.758
kg {
kg
kg {
kg
k
k
J681.14 kJ - 533.98 kJ N
h5 - h2 a
kg
kg
=
=
= 0.59
h4 a - h2 a J780.758 kJ - 533.98 kJ N
kg
kg
kJ
j1207.57 kJ - 780.788 kJ y
z-i
j533.98 kJ - 300.19 kJ y
z
wNET = Hh3 - h4L - Hh2 - h1L = i
j
z
j
z = 193.022
kg
kg { k
kg
kg {
kg
k
kJ
kJ
kJ
qin = Hh3 - h5L = 1207.57
- 681.14
= 526.43
kg
kg
kg
kJ
wNET 193.022 kg
hTH =
=
= 0.37 x 100 % = 37 %
qin
526.43 kJ
hT =
h3 - h4
h3 - h4 s
kg
9155 A four-cylinder, four-stroke spark-ignition engine operates on the ideal Otto cycle
with a compression ratio of 11 and a total displacement volume of 1.8 liter. The air is at
90 kPa and 50C at the beginning of the compression process. The heat input is 1.5 kJ
per cycle per cylinder. Accounting for the variation of specific heats of air with
temperature, determine (a) the maximum temperature and pressure that occur during
the cycle, (b) the net work per cycle per cyclinder and the thermal efficiency of the
cycle, (c) the mean effective pressure, and (d) the power output for an engine speed of
3000 rpm.
9156 A gas-turbine plant operates on the regenerative Brayton cycle with two stages of
reheating and two-stages of intercooling between the pressure limits of 100 and 1200
kPa. The working fluid is air. The air enters the first and the second stages of the
compressor at 300 K and 350 K, respectively, and the first and the second stages of the
turbine at 1400 K and 1300 K, respectively. Assuming both the compressor and the
turbine have an isentropic efficiency of 80 percent and the regenerator has an
effectiveness of 75 percent and using variable specific heats, determine (a) the back
work ratio and the net work output, (b) the thermal efficiency, and (c) the second-law
efficiency of the cycle. Also determine (d) the exergies at the exits of the combustion
chamber (state 6) and the regenerator (state 10) (See Figure 943 in the text).
1015 A steam power plant operates on a simple ideal Rankine cycle between the
pressure limits of 3 MPa and 50 kPa. The temperature of the steam at the turbine inlet is
300C, and the mass flow rate of steam through the cycle is 35 kg/s. Show the cycle on
a T-s diagram with respect to saturation lines, and determine (a) the thermal efficiency
of the cycle and (b) the net power output of the power plant.
Consideraciones:
Qin
2
1
Edo. 1
h1=340.54 kJ/kg
v1=0.001030 m3/kg
P1=50kPa
4
Qout
Edo. 2
h2=343.5285kJ/kg
Edo. 3
Edo.4
h3=2994.3kJ/kg
h4=2277.3776 kJ/kg
T3=300C
P3=3 MPa
P4= 50kPa
s3=65412 kJ/kgK
s4=s3
3
m
lkJ
kJ
i
y
z
j0.001030
z H3000 kPa - 50 kPaL J
wpin = v1 HP2 - P1L = j
N = 3.0485
3
kg {
1 kPa.m
kg
k
kJ
kJ
kJ
+ 3.0385
= 343.5285
kg
kg
kg
6.5912 kJ - 1.0912 kJ
s4 - sf
kg K
kg K
x4 =
=
= 0.8382
kJ
sfg
6.5019
h2 = h1 + wpin = 340.49
kJ
kJ y
kJ
i
z
h4 = hf + x4 hfg = 340.54
+ H0.8382L j
j2304.7
z = 2277.3776
kg
kg {
kg
k
kJ
kJ
kJ
qin = Hh3 - h2L = 2994.3
- 343.5285
= 2650.7715
kg
kg
kg
kJ
kJ
kJ
qout = Hh4 - h1L = 2277.3776
- 340.54
= 1936.8376
kg
kg
kg
kJ
kJ
kJ
wNET = qin - qout = 2650.7715
- 1936.8376
= 713.9339
kg
kg
kg
kg i
kJ
kJ
y
j713.9339
z = 24987.6865
w NET = m wNET = J35
Nj
z
= 24.98 MW
s k
kg {
s
kg K
1046 A steam power plant operates on an ideal regenerative Rankine cycle with two
open feedwater heaters. Steam enters the turbine at 10 MPa and 600C and exhausts to
the condenser at 5 kPa. Steam is extracted from the turbine at 0.6 and 0.2 MPa. Water
leaves both feedwater heaters as a saturated liquid. The mass flow rate of steam
through the boiler is 22 kg/s. Show the cycle on a T-s diagram, and determine (a) the
net power output of the power plant and (b) the thermal efficiency of the cycle.
7
10MPa
6
0.6MPa
4 5
0.2MPa
2 3
5kPa
Edo.1
h1=kJ/kg
P1=5kPA
v1=m3/kg
Edo.6
h6=kJ/kg
Edo. 2
h2=kJ/kg
Edo. 3
h3=kJ/kg
P3=0.2MPA
v3=m3/kg
Edo.8
P8=0.6MPa
s8=s7
x8=0.9627
h8=kJ/kg
Edo.7
P7=10MPa
h7=kJ/kg
T7=600C
s7=kJ/kgK
j0.001005
wpin = v1 HP2 - P1L = j
k
h2 = h1 + wpin = 137.75
i
m3
kg
8
1-y
9
1-y-z
10
Edo.4
h4=kJ/kg
Edo.9
P9=0.6MPa
s9=s7
h9=kJ/kg
x9=
Edo.5
h5=kJ/kg
P5=0.6MPA
v5=m3/kg
Edo.10
P10=5kPa
s10=s7
h10=kJ/kg
x10=
lkJ
kJ
y
z
z H200 kPa - 5 kPaL J
N = 0.195975
1 kPa.m3
kg
{
kJ
kJ
kJ
+ 0.195975
= 137.9459
kg
kg
kg
j0, 001061
wpin = v3 HP4 - P3L = j
k
h4 = h3 + wpin = 04.71
i
kJ
kJ
kJ
+ 0.4244
= 505.1344
kg
kg
kg
j0.001101
wpin = v5 HP6 - P5L = j
k
m3 y
lkJ
kJ
z
z H600 kPa - 200 kPaL J
N = 0.4244
kg {
1 kPa.m3
kg
m3 y
lkJ
kJ
z
z H10000 kPa - 600 kPaL J
N = 10.3494
kg {
1 kPa.m3
kg
kJ
kJ
kJ
+ 10.3494
= 680.7294
kg
kg
kg
kJ
kJ
6.9045
- 1.9308
s -s
kg K
kg K
x9 = 9 f =
= 0.9699
sfg
4.8285 kJ
h6 = h5 + wpin = 670.38
kg K
h9 = hf + x9 hfg = 670.38
kJ
kJ y
kJ
i
z
+ H0.9699L j
j2085.8
z = 2693.4569
kg
kg {
kg
k
kJ
s10 - sf 6.9045 kg K - 0.4762
x10 =
=
kJ
sfg
7.9176 kg
K
kJ
kg K
= 0.8119
kJ
kJ y
kJ
i
z
+ H0.8119L j
j2423.0
z = 2104.9838
kg
kg {
kg
k
m8
m8 h8 + h2 m2 = m5 h5 yh8 + H1 - yL h4 = h5 y =
m5
Ein = Eout
504.71 kJ - 137.9459 kJ
h3 - h2
kg
kg
y=
=
= 0.14469
Hh9 - h2L
J2693.4569 kJ - 137.9459 kJ N
kg
kg
mi hi = me he
m9 h9 + h2 m2 = m3 h3
zh9 + H1- y - zL h2 = H1 - yL h3
504.71 kJ - 137.9459 kJ
h3 - h2
kg
kg
z=
H1 - yL =
H1 - 0.14469L = 0.12375
h9 - h2
J2693.4569 kJ - 137.9459 kJ N
kg
kg
kJ
kJ
kJ
i
y
z
qin = Hh7 - h6L = j
j3625.8
- 680.2794
z = 2945.0706
kg
kg {
kg
k
j
qout = H1 - y - zL Hh10 - h1L = H1 - 0.14469 - .12375L i
j2104.9838
k
m4
m5
kJ
kJ y
z = 1439.144 kJ
- 137.75
z
kg
kg {
kg
kJ
kJ
kJ
wNET = qin - qout = 2945.0706
- 1439.144
= 1505.9265
kg
kg
kg
kg i
kJ
y
j1505.9265
z = 33130.38507 kW
W NET = m wNET = J22
Nj
z
s k
kg {
1439.144 kJ
qout
kg
hTH = 1 = 1= 0.5113 X100 % = 51.13 %
qin
2945.0706 kJ
kg
1050 A steam power plant operates on an ideal reheatregenerative Rankine cycle and
has a net power output of 80 MW. Steam enters the high-pressure turbine at 10 MPa and
550C and leaves at 0.8 MPa. Some steam is extracted at this pressure to heat the
closed feedwater heater. The rest of the steam is reheated to 500C and is expanded in
the low-pressure turbine to the condenser pressure of 10 kPa. Show the cycle on a T-s
diagram with respect to saturation lines, and determine (a) the mass flow rate of steam
through the boiler and (b) the thermal efficiency of the cycle. Assume that the
feedwater leaves the heater at the condensation temperature of the extracted steam
and that the extracted steam leaves the heater as a saturated liquid and is pumped to
the line carrying the feedwater.
Consideraciones:
5
10
4
9
2 3
10MPa
0.8MPa
10kPa
7
6
1-y
8
s
Edo.1
h1=191.81 kJ/kg
Edo. 2
h2=201.89
kJ/kg
Edo. 3
h3=720.87 kJ/kg
P1=10kPA
v1=0.001010m3/kg
P3=0.8MPA
v3=0.001115m3/kg
Edo.6
P6=0.8MPa
h6=2811.9 kJ/kg
Edo.7
P7=0.8MPa
h7=3481.3
kJ/kg
T7=500C
s7=7.8692
kJ/kgK
i
j0.001010
wpin = v1 HP2 - P1L = j
k
h2 = h1 + wpin = 191.81
i
Edo.4
h4=731.128
kJ/kg
h4=h9=h10
Edo.5
h5=3375.1
kJ/kg
P5=10MPA
s5=6.5995
kJ/kgK=s6
Edo.8
P8=10kPa
s8=s7
x8=0.9627
h8=2494.727 kJ/kg
m3 y
lkJ
kJ
z
z H10000 kPa - 10 kPaL J
N = 10.0899
kg {
1 kPa.m3
kg
kJ
kJ
kJ
+ 10.0899
= 201.8999
kg
kg
kg
m3 y
lkJ
kJ
z
z H10000 kPa - 800 kPaL J
N = 10.258
3
kg {
1 kPa.m
kg
k
kJ
kJ
kJ
h4 = h3 + wpin = 720.87
+ 10.258
= 731.128
kg
kg
kg
kJ - 0.6492 kJ
7.8692
s -s
kg K
kg K
x8 = 8 f =
= 0.9627
j0.001115
wpin = v3 HP4 - P3L = j
kJ
7.4996 kg
K
kJ
kJ y
i
z = 2494.727 kJ
h8 = hf + x8 hfg = 191.81
+ H0.9627L j
j2392.1
z
kg
kg
kg
k
{
sfg
Ein = Eout
h9 - h2
y=
=
Hh6 - h3L - Hh9 - h2L
J2811.9
731.128
kJ
kg
- 201.8999 kJ
kg
m3
m4
= 0.20197
- 720.87
- 201.8999 kJ N
kg
kJ
kJ y
kJ
kJ y
kJ
i
i
j
z
j
z
qin = Hh5 - h4L + H1 - yL Hh7 - h6L = j3375.1
- 731.128
z + H1 - 0.201L j3481.3
- 2811.9
z = 3178.173
kg
kg {
kg
kg {
kg
k
k
kJ
kJ
kJ
i
y
j2494.727
z
qout = H1 - yL Hh8 - h1L = H1 - 0.201L j
- 198.81
z = 1837.796
kg
kg {
kg
k
kJ
kg
kJ N + J731.128 kJ
kg
kg
y=
kJ
kJ
kJ
- 1837.796
= 1340.377
kg
kg
kg
kg
= 59.6847
s
1837.796
3178.173
kJ
kg
kJ
kg
1060 Determine the exergy destruction associated with the regenerative cycle described
6MPa
4
2
0.4MPa
1-y
20kPa
1
Edo.1
h1=251.42 kJ/kg
Edo. 2
h2=251.8064
kJ/kg
P1=20kPA
v1=0.001017m3/kg
Edo. 3
h3=604.66 kJ/kg
Edo.4
h4=610.7304
kJ/kg
P3=0.4MPa
v3=0.001080m3/kg
Edo.6
P6=0.4MPa
h6=2665.67 kJ/kg
Edo.5
h5=3302.9
kJ/kg
P5=6MPa
s5=6.7219
kJ/kgK=s6=s7
T=450C
Edo.7
P7=20kPa
h7=2213.971
kJ/kg
x7=0.8324
m3 z
lkJ
kJ
i
y
j0.001017
z H400 kPa - 20 kPaL J
wpin = v1 HP2 - P1L = j
N = 0.38646
kg {
1 kPa.m3
kg
k
kJ
kJ
kJ
h2 = h1 + wpin = 251.42
+ 0.38646
= 251.80646
kg
kg
kg
3
m
lkJ
kJ
i
y
z
j0.001080
z H6000 kPa - 400 kPaL J
wpin = v3 HP4 - P3L = j
N = 6.0704
3
kg {
1 kPa.m
kg
k
kJ
kJ
kJ
+ 6.0704
= 610.7304
kg
kg
kg
6.7219 kJ - 1.7765 kJ
s -s
kg K
kg K
x6 = 6 f =
= 0.966
kJ
h4 = h3 + wpin = 604.66
sfg
5.1191
kJ
kJ y
kJ
i
z
h6 = hf + x6 hfg = 604.66
+ H0.966L j
j2133.4
z = 2665.67
kg
kg {
kg
k
kg K
kJ
s7 - sf 6.7219 kg K - 0.8320
x7 =
=
sfg
7.0752 kJ
kJ
kg K
= 0.8324
kJ
kJ y
kJ
i
z
h7 = hf + x7 hfg = 251.42
+ H0.8324L j
j2357.5
z = 2213.971
kg
kg {
kg
k
kg K
Ein = Eout
m6 h6 + h2 m2 = m3 h3
yh6 + H1 - yL h2 = h3
m6
m3
604.66 kJ - 251.8064 kJ
h3 - h2
kg
kg
y=
=
= 0.14617
Hh6 - h2L
J2665.67 kJ - 251.8064 kJ N
kg
kg
kJ
kJ y
kJ
i
j3302.9
z
qin = Hh5 - h4L = j
- 610.7304
z = 2692.1696
kg
kg {
kg
k
kJ
kJ y
kJ
i
j2213.971
z
qout = H1 - yL Hh7 - h1L = H1 - 0.14617L j
- 251.42
z = 1675.669
kg
kg {
kg
k
kJ
kJ
kJ
wNET = qin - qout = 2692.1696
- 1675.669
= 1016.5006
kg
kg
kg
1675.669 kJ
qout
kg
hTH = 1 = 1= 0.3775 X100 % = 37.75 %
qin
2692.1696 kJ
kg
kJ
2692.1696 kJ z
i 1675.669 kg
y
j
q
q
kJ
kg z
i
OUT
IN
y
j
z
z
iCICLO = T0 j
j
z = H290 KL j
= 1155.1828
j
z
j
z
j
z
TH {
290 K
1500 K
kg
k TL
k
{
1065C Consider a cogeneration plant for which the utilization factor is 0.5. Can the
exergy destruction associated with this plant be zero? If yes, under what conditions?
S, si el ciclo no involucra irreversibilidades tales como: desaceleracin, friccin, y
transferencia de calor a travs de una diferencia finita de temperatura.
1093 Consider a steam power plant that operates on a regenerative Rankine cycle and
has a net power output of 150 MW. Steam enters the turbine at 10 MPa and 500C and
the condenser at 10 kPa. The isentropic efficiency of the turbine is 80 percent, and that
of the pumps is 95 percent. Steam is extracted from the turbine at 0.5 MPa to heat the
feedwater in an open feedwater heater. Water leaves the feedwater heater as a
saturated liquid. Show the cycle on a T-s diagram, and determine (a) the mass flow rate
of steam through the boiler and (b) the thermal efficiency of the cycle. Also, determine
the exergy destruction associated with the regeneration process. Assume a source
temperature of 1300 K and a sink temperature of 303 K.
10MPa
4
2
0.5MPa
6s
1-y
10kPa
7s
7
s
Edo.1
h1=191.81 kJ/kg
Edo. 2
h2=192.3304
kJ/kg
P1=10kPA
s2=1.8604kJ/kgK
v1=0.001010m3/kg P2=10kPa
Edo.6
Edo.7
P6=0.4MPa
P7s=10kPa
h6=2798.288 kJ/kg h7=2346.81
kJ/kg
x6=0.9554
x7=0.7934
Edo. 3
h3=640.09 kJ/kg
P3=0.5MPa
v3=0.001093m3/kg
s3=0.6492kJ/kgK
Edo.4
h4=651.02
kJ/kg
Edo.5
h5=3375.1
kJ/kg
P5=10MPa
s5=6.5995
kJ/kgK=s6=s7s
T=500C
h2 = h1 + wpin = 191.81
i
kJ
kJ
kJ
+ 0.5209
= 192.3304
kg
kg
kg
lkJ
kJ
y
z
z H500 kPa - 10 kPaL J
N = 0.5209
3
1 kPa.m
kg
k
{
v3 HP4 - P3L j
m3 z
lkJ
kJ
i
y H10000 kPa - 500 kPaL
z
wpin =
= j0.001093
J
N = 10.93
3
hth
kg {
.95
1 kPa.m
kg
k
j0.001010
wpin = v1 HP2 - P1L = j
h4 = h3 + wpin = 640.09
m3
kg
kJ
kJ
kJ
+ 10.93
= 651.02
kg
kg
kg
kJ
kJ
s6 - sf 6.5995 kg K - 1.8604 kg K
x6 =
=
= 0.9554
kJ
sfg
4.9603 kg
K
kJ
kJ y
kJ
i
z
h6 = hf + x6 hfg = 3375.1
+ H0.9554L j
j2108.0
z = 2798.2884
kg
kg {
kg
k
kJ - 0.6492 kJ
66.5995
s -s
kg K
kg K
x7 s = 7 s f =
= 0.7934
kJ
sfg
7.4996 kg K
h5 - h7
kJ
kJ
kJ y
kJ
i
ht =
h7 = h5 - ht Hh5 - h7 sL = 3375.1
- H0.8L j
j3375.1
- 2089.739
5z
z = 2346.8116
h5 - h7 s
kg
kg
kg {
kg
k
Ein = Eout
m6 h6 + h2 m2 = m3 h3
640.09
h -h
y= 3 2 =
Hh6 - h2L
J2654.0856
kJ
kg
kJ
kg
yh6 + H1 - yL h2 = h3
m6
m3
- 192.3304
- 192.3304 kJ N
= 0.1818
kJ
kJ y
kJ
j
z
qin = Hh5 - h4L = i
j3375.1
- 651.02
z = 2724.08
kg
kg {
kg
k
kJ
kJ y
kJ
i
z
qout = H1 - yL Hh7 - h1L = H1 - 0.1818L j
j2346.8116
- 191.81
z = 1763.036
kg
kg {
kg
k
wNET = qin - qout = 2724.08
kg
kJ
kJ
kJ
- 1763.036
= 961.0437
kg
kg
kg
1763.036 kJ
qout
kg
hTH = 1 = 1= 0.3528 X100 % = 35.28 %
kJ
qin
2724.08 kg
150000 kJ
W NET
kg
s
m=
=
= 156.08
wNET 961.0437 kJ
s
kg
i
j
iregen = T0 j
j
j me se - mi si +
y
qalr z
z
z = T0 Hs3 - ys6 - H1 - yL s2L
TL z
{
=
k=
kJ
kJ y
kJ y
kJ
i
i
z - H1 - 0.1818L i
j
zy
z
iregen = H303 KL j
j1.8604
- H0.1818L j
j6.9308
z
j0.6492
z
z = 120.8048
kgK
kgK {
kgK {{
kg
k
k
k
1095 Consider an ideal reheatregenerative Rankine cycle with one open feedwater
heater. The boiler pressure is 10 MPa, the condenser pressure is 15 kPa, the reheater
pressure is 1 MPa, and the feedwater pressure is 0.6 MPa. Steam enters both the highand low-pressure turbines at 500C. Show the cycle on a T-s diagram with respect to
saturation lines, and determine (a) the fraction of steam extracted for regeneration and
(b) the thermal efficiency of the cycle.
1MPa
10MPa
5 7
6 8
0.6MPa
15kPa
9
s
Edo.1
h1=225.94 kJ/kg
Edo. 2
h2=226.53319
kJ/kg
P1=15kPA
v1=0.001014m3/k
g
Edo.6
P6=1MPa
h6=2783.8124
kJ/kg
Edo. 3
h3=670.38 kJ/kg
Edo.4
h4=680.7294
kJ/kg
P3=0.6MPa
v3=0.001101m3/k
g
Edo.7
P7=1MPa
h7=3479.1
kJ/kg
s7=7.7642m3/k
g
i
Edo. 8
P8=0.6MPa
h7=3310.1569
kJ/kg
s8=s7=s9
Edo. 9
P7s=15kPa
h7=2518.783
8 kJ/kg
x9=0.966506
Edo.5
h5=3375.1
kJ/kg
P5=10MPa
s5=6.5995
kJ/kgK=s6=s7
s
T=500C
lkJ
kJ
y
z
z H600 kPa - 15 kPaL J
N = 0.59319
3
1 kPa.m
kg
k
{
kJ
kJ
kJ
h2 = h1 + wpin = 225.94
+ 0.59319
= 226.53319
kg
kg
kg
3
v3 HP4 - P3L i
m
lkJ
kJ
y
z
j0.001101
z H10000 kPa - 600 kPaL J
wpin =
=j
N = 10.3494
3
hth
kg {
1 kPa.m
kg
k
j0.001014
wpin = v1 HP2 - P1L = j
m3
kg
kJ
kJ
kJ
+ 10.3494
= 680.7294
kg
kg
kg
kJ - 0.7549 kJ
7.7642
s9 - sf
kg K
kg K
x9 =
=
= 0.966506
kJ
sfg
7.2522 kg
K
h4 = h3 + wpin = 670.38
h9 = hf + x9 hfg = 225.94
kJ
kJ y
i
z = 2518.7838 kJ
+ H0.9665064L j
j2372.3
z
kg
kg {
kg
k
Ein = Eout
yh8 + H1 - yL h2 = h3
m8 h8 + h2 m2 = m3 h3
m8
m3
670.38 kJ - 226.53319 kJ
h3 - h2
kg
kg
y=
=
= 0.143936
kJ
Hh8 - h2L
J3310.1569
- 226.53319 kJ N
kg
kg
j
qin = Hh5 - h4L + Hh7 - h6L = i
j3375.1
kJ
kJ y
kJ
kJ y
kJ
i
z
z
- 680.7299
z+j
j3479.1
- 2783.8124
z = 3389.6577
kg
kg { k
kg
kg {
kg
j
qout = H1 - yL Hh9 - h1L = H1 - 0.143936L i
j2518.7838
q
hTH = 1 - out = 1 qin
1962.81428
3389.6577
kJ
kg
kJ
kg
kJ
kJ y
kJ
z
- 225.94
z = 1962.81428
kg
kg {
kg
1112 A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal
vapor-compression refrigeration cycle between 0.12 and 0.7 MPa. The mass flow rate of
the refrigerant is 0.05 kg/s. Show the cycle on a T-s diagram with respect to saturation
lines. Determine (a) the rate of heat removal from the refrigerated space and the power
input to the compressor, (b) the rate of heat rejection to the environment, and (c) the
coefficient of performance.
T
.
QH
2
3
0.7MPa
.
Win
0.12MPa
4s
.1
QL
s
Edo.1
h1=236.97 kJ/kg
Edo. 2
h2=273.5377 kJ/kg
Edo. 3
h3=88.82 kJ/kg
P1=0.12MPA
s1=0.94779 kJ/kgK
P2=0.7MPA
s2=s1
P3=0.7MPa
kg j
kJ
kJ y
i
z
N j236.97
- 88.82
z = 7.4075 kW
s k
kg
kg {
Edo.4
h4h3
kg i
kJ
kJ y
z
Nj
j273.5377
- 236.97
z = 1.828 kW
s k
kg
kg {
QL
7.4075 kW
=
= 4.0522
1.828 kW
Win
1132 A heat pump using refrigerant-134a heats a house by using underground water at
8C as the heat source. The house is losing heat at a rate of 60,000 kJ/h. The
refrigerant enters the compressor at 280 kPa and 0C, and it leaves at 1 MPa and 60C.
The refrigerant exits the condenser at 30C. Determine (a) the power input to the heat
pump, (b) the rate of heat absorption from the water, and (c) the increase in electric
power input if an electric resistance heater is used instead of a heat pump.
T
casa
.
QH
60C
30C
2
3
1MPa
.
Win
0C
0.28MPa
4s
.1
QL
Edo.1
h1=250.83 kJ/kg
Edo. 2
h2=293.38 kJ/kg
Edo. 3
h3=93.58 kJ/kg
P1=280kPA
T1=0C
P2=1MPA
T2=60C
P3=1MPa
T3=30C
60000
kJ
Edo.4
h4h3
Q
kg
3600 S
mR = H =
= 0.0834
qH 293.38 kJ - 93.58 kJ
s
kg
kg
kg i
kJ
kJ y
z
Win = m Hh2 - h1L = J0.0834
Nj
j293.38
- 250.83
z = 3.54938 kW
s k
kg
kg {
kg i
kJ
kJ y
z
QL = m Hh1 - h4L = J0.0834
Nj
j250.83
- 93.58
z = 13.1172 kW
s k
kg
kg {
60000 kJ
= 16.6666 kW
3600 S
Wincrem = We - Win = 16.6666 k- 3.54938 kW = 13.11728kW
We = QH =
pressure limits of 0.8 and 0.14 MPa. The working fluid is refrigerant-134a. The
refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber
operating at 0.4 MPa. Part of the refrigerant evaporates during this flashing process, and
this vapor is mixed with the refrigerant leaving the low-pressure compressor. The
mixture is then compressed to the condenser pressure by the high-pressure compressor.
The liquid in the flash chamber is throttled to the evaporator pressure, and it cools the
refrigerated space as it vaporizes in the evaporator. Assuming the refrigerant leaves the
evaporator as saturated vapor and both compressors are isentropic, determine (a) the
fraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) the
amount of heat removed from the refrigerated space and the compressor work per unit
mass of refrigerant flowing through the condenser, and (c) the coefficient of
performance.
4
5
0.8MPa
2
7
0.4MPa
6
0.14MPa
qL
s
h1=239.16 kJ/kg
h2=256.58 kJ/kg
h7=63.94 kJ/kgh8
h9=256.4105 kJ/kg
kJ
h6 - hf 95.47 kg - 63.94
x6 =
=
hfg
191.62 kJ
kg
me he = mi hi
Ein = Eout
kJ
kg
h5h695.47 kJ/kg
= 0.16454
i
k
h9 = x6 h3 + H1 - x6L h2 = H0.16454L j
j255.55
=
h3=255.55 kJ/kg
kJ y
kJ y
kJ
i
z
z
z + H1 - 0.16454L j
j256.58
z = 256.4105
kg {
kg {
kg
k
j
qL = H1 - x6L Hh1 - h8L = H1 - 0.1645L i
j239.16
k
kJ
kJ y
kJ
z
- 63.94
z = 146.389
kg
kg {
kg
j
win = H1 - 0.1645L i
j256.58
k
146.389
q
COPR = L =
win 37.4617
kJ
kg
kJ
kg
kJ
kJ y
kJ
kJ y
kJ
i
z
z
- 239.16
z +j
j279.3185
- 256.4105
z = 37.4617
kg
kg { k
kg
kg {
kg
= 3.9076