TERMO
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![Page 1: TERMO](https://reader033.fdocuments.ec/reader033/viewer/2022042821/563db83e550346aa9a91e459/html5/thumbnails/1.jpg)
Pauta Tarea nº2 (Grupal 2 personas)OBS.I. Responda de forma clara y precisa y enuncie todos los postulados utilizados según corresponda. OBS.2. No se recibirán tareas atrasadas
Pregunta nºI Una planta termoeléctrica de vapor de agua opera en el ciclo Rankine con recalentamiento. El vapor entra a la turbina de alta presión a 12.5 MPa y 550ºC, a razón de 7.7 kg/s y sale a 2 MPa. El vapor luego se recalienta a 450ºC antes de expandirse en la turbina de baja presión. Las eficiencias isentrópicas de la turbina y la bomba son 85% y 90%, respectivamente. El vapor sale del condensador como líquido saturado. Si el contenido de humedad del vapor a la salida de la turbina no debe exceder el 5%, determine:
a) La presión del condensador, en kPab) La producción neta de Potencia, en MWc) La eficiencia térmica
Obs. Para enumerar las corrientes, especifique la corriente de salida del condensador, con el número (1), salida de la bomba (2) y así correlativamente, según el esquema visto en la clase.
Solución
!
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Así, el resultado de la interpolación reiterada, nos permite determinar que:
Prof. José Luis Salazar N.
![Page 2: TERMO](https://reader033.fdocuments.ec/reader033/viewer/2022042821/563db83e550346aa9a91e459/html5/thumbnails/2.jpg)
!
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Con ésto:
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Assumptions!1!I=).>A!57)-.=28C!+58>2=2589!)J29="!2!K28)=2+!.8>!75=)8=2.3!)8)-CA!+(.8C)9!.-)!8)C32C2?3)"!
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Assumptions!1!I=).>A!57)-.=28C!+58>2=2589!)J29="!2!K28)=2+!.8>!75=)8=2.3!)8)-CA!+(.8C)9!.-)!8)C32C2?3)"!
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Prof. José Luis Salazar N.
![Page 3: TERMO](https://reader033.fdocuments.ec/reader033/viewer/2022042821/563db83e550346aa9a91e459/html5/thumbnails/3.jpg)
Pregunta nº2 Una planta eléctrica de turbina de gas opera en in ciclo simple Brayton entre los límites de presión de 100 y 1.2 MPa. EL fluido de trabajo es aire, que entra al compresor a 30ºC y una razón de 150 m3/min y sale de la turbina a 500ºC. Usando calores específicos variables para el aire y suponiendo una eficiencia isentrópica de compresión de 82% y una eficiencia isentrópica de turbina de 88%, determine:
a) La producción de potencia, en kWb) La relación de Retrotrabajo, definida por la fracción de trabajo de la turbina que se emplea para
accionar el compresor.Solución
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-75
9-105 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio,
and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are
negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis (a) For this problem, we use the properties
from EES software. Remember that for an ideal gas,
enthalpy is a function of temperature only whereas
entropy is functions of both temperature and pressure.
Process 1-2: Compression
KkJ/kg 7159.5kPa 100
C30
kJ/kg 60.303C30
11
1
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212
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))
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hhh
hh s*
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ss hh
h
hh
hh
43
3
43
43T
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)
)"'('
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We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with
the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The
solution by hand would require a trial-error approach.
h_3=enthalpy(Air, T=T_3)
s_3=entropy(Air, T=T_3, P=P_2)
h_4s=enthalpy(Air, P=P_1, s=s_3)
The mass flow rate is determined from
+ ,+ ,kg/s 875.2
K 27330K/kgmkPa 0.287
)s/m 0kPa)(150/6 100(3
3
1
11 "-!!
""RT
Pm
!!!
The net power output is
kW 1100)kJ/kg60.30324kg/s)(686. 875.2()( 12inC, ")")" hhmW !!
kW 1759)kJ/kg62.792.7kg/s)(1404 875.2()( 43outT, ")")" hhmW !!
kW 659")")" 11001759inC,outT,net WWW !!!
(b) The back work ratio is
0.625"""kW 1759
kW 1100
outT,
inC,
bwW
Wr
!
!
(c) The rate of heat input and the thermal efficiency are
kW 2065)kJ/kg24.686.7kg/s)(1404 875.2()( 23in ")")" hhmQ !!
0.319"""kW 2065
kW 659
in
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Q
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23
4
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100 kPa
30°C
500°C
1.2 MPa
De aquí, podemos determinar que:
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-75
9-105 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio,
and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are
negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis (a) For this problem, we use the properties
from EES software. Remember that for an ideal gas,
enthalpy is a function of temperature only whereas
entropy is functions of both temperature and pressure.
Process 1-2: Compression
KkJ/kg 7159.5kPa 100
C30
kJ/kg 60.303C30
11
1
11
!"#$%
"
&"
"'('&"
sP
T
hT
kJ/kg 37.617kJ/kg.K 7159.5
kPa 12002
12
2 "#$%
""
"sh
ss
P
kJ/kg 24.68660.303
60.30337.61782.0 2
212
12C "'('
))
"'(')
)" h
hhh
hh s*
Process 3-4: Expansion
kJ/kg 62.792C500 44 "'('&" hT
ss hh
h
hh
hh
43
3
43
43T
62.79288.0
)
)"'('
)
)"*
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with
the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The
solution by hand would require a trial-error approach.
h_3=enthalpy(Air, T=T_3)
s_3=entropy(Air, T=T_3, P=P_2)
h_4s=enthalpy(Air, P=P_1, s=s_3)
The mass flow rate is determined from
+ ,+ ,kg/s 875.2
K 27330K/kgmkPa 0.287
)s/m 0kPa)(150/6 100(3
3
1
11 "-!!
""RT
Pm
!!!
The net power output is
kW 1100)kJ/kg60.30324kg/s)(686. 875.2()( 12inC, ")")" hhmW !!
kW 1759)kJ/kg62.792.7kg/s)(1404 875.2()( 43outT, ")")" hhmW !!
kW 659")")" 11001759inC,outT,net WWW !!!
(b) The back work ratio is
0.625"""kW 1759
kW 1100
outT,
inC,
bwW
Wr
!
!
(c) The rate of heat input and the thermal efficiency are
kW 2065)kJ/kg24.686.7kg/s)(1404 875.2()( 23in ")")" hhmQ !!
0.319"""kW 2065
kW 659
in
net
Q
Wth !
!
*
1
Combustion
chamber
Turbine
23
4
Compress.
100 kPa
30°C
500°C
1.2 MPa
El flujo másico será:
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-75
9-105 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio,
and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are
negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis (a) For this problem, we use the properties
from EES software. Remember that for an ideal gas,
enthalpy is a function of temperature only whereas
entropy is functions of both temperature and pressure.
Process 1-2: Compression
KkJ/kg 7159.5kPa 100
C30
kJ/kg 60.303C30
11
1
11
!"#$%
"
&"
"'('&"
sP
T
hT
kJ/kg 37.617kJ/kg.K 7159.5
kPa 12002
12
2 "#$%
""
"sh
ss
P
kJ/kg 24.68660.303
60.30337.61782.0 2
212
12C "'('
))
"'(')
)" h
hhh
hh s*
Process 3-4: Expansion
kJ/kg 62.792C500 44 "'('&" hT
ss hh
h
hh
hh
43
3
43
43T
62.79288.0
)
)"'('
)
)"*
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with
the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The
solution by hand would require a trial-error approach.
h_3=enthalpy(Air, T=T_3)
s_3=entropy(Air, T=T_3, P=P_2)
h_4s=enthalpy(Air, P=P_1, s=s_3)
The mass flow rate is determined from
+ ,+ ,kg/s 875.2
K 27330K/kgmkPa 0.287
)s/m 0kPa)(150/6 100(3
3
1
11 "-!!
""RT
Pm
!!!
The net power output is
kW 1100)kJ/kg60.30324kg/s)(686. 875.2()( 12inC, ")")" hhmW !!
kW 1759)kJ/kg62.792.7kg/s)(1404 875.2()( 43outT, ")")" hhmW !!
kW 659")")" 11001759inC,outT,net WWW !!!
(b) The back work ratio is
0.625"""kW 1759
kW 1100
outT,
inC,
bwW
Wr
!
!
(c) The rate of heat input and the thermal efficiency are
kW 2065)kJ/kg24.686.7kg/s)(1404 875.2()( 23in ")")" hhmQ !!
0.319"""kW 2065
kW 659
in
net
Q
Wth !
!
*
1
Combustion
chamber
Turbine
23
4
Compress.
100 kPa
30°C
500°C
1.2 MPa
Y la potencia neta:
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-75
9-105 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio,
and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are
negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis (a) For this problem, we use the properties
from EES software. Remember that for an ideal gas,
enthalpy is a function of temperature only whereas
entropy is functions of both temperature and pressure.
Process 1-2: Compression
KkJ/kg 7159.5kPa 100
C30
kJ/kg 60.303C30
11
1
11
!"#$%
"
&"
"'('&"
sP
T
hT
kJ/kg 37.617kJ/kg.K 7159.5
kPa 12002
12
2 "#$%
""
"sh
ss
P
kJ/kg 24.68660.303
60.30337.61782.0 2
212
12C "'('
))
"'(')
)" h
hhh
hh s*
Process 3-4: Expansion
kJ/kg 62.792C500 44 "'('&" hT
ss hh
h
hh
hh
43
3
43
43T
62.79288.0
)
)"'('
)
)"*
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with
the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The
solution by hand would require a trial-error approach.
h_3=enthalpy(Air, T=T_3)
s_3=entropy(Air, T=T_3, P=P_2)
h_4s=enthalpy(Air, P=P_1, s=s_3)
The mass flow rate is determined from
+ ,+ ,kg/s 875.2
K 27330K/kgmkPa 0.287
)s/m 0kPa)(150/6 100(3
3
1
11 "-!!
""RT
Pm
!!!
The net power output is
kW 1100)kJ/kg60.30324kg/s)(686. 875.2()( 12inC, ")")" hhmW !!
kW 1759)kJ/kg62.792.7kg/s)(1404 875.2()( 43outT, ")")" hhmW !!
kW 659")")" 11001759inC,outT,net WWW !!!
(b) The back work ratio is
0.625"""kW 1759
kW 1100
outT,
inC,
bwW
Wr
!
!
(c) The rate of heat input and the thermal efficiency are
kW 2065)kJ/kg24.686.7kg/s)(1404 875.2()( 23in ")")" hhmQ !!
0.319"""kW 2065
kW 659
in
net
Q
Wth !
!
*
1
Combustion
chamber
Turbine
23
4
Compress.
100 kPa
30°C
500°C
1.2 MPa
La relación de retrotrabajo será:
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-75
9-105 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio,
and the thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are
negligible. 3 Air is an ideal gas with variable specific heats.
Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).
Analysis (a) For this problem, we use the properties
from EES software. Remember that for an ideal gas,
enthalpy is a function of temperature only whereas
entropy is functions of both temperature and pressure.
Process 1-2: Compression
KkJ/kg 7159.5kPa 100
C30
kJ/kg 60.303C30
11
1
11
!"#$%
"
&"
"'('&"
sP
T
hT
kJ/kg 37.617kJ/kg.K 7159.5
kPa 12002
12
2 "#$%
""
"sh
ss
P
kJ/kg 24.68660.303
60.30337.61782.0 2
212
12C "'('
))
"'(')
)" h
hhh
hh s*
Process 3-4: Expansion
kJ/kg 62.792C500 44 "'('&" hT
ss hh
h
hh
hh
43
3
43
43T
62.79288.0
)
)"'('
)
)"*
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with
the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The
solution by hand would require a trial-error approach.
h_3=enthalpy(Air, T=T_3)
s_3=entropy(Air, T=T_3, P=P_2)
h_4s=enthalpy(Air, P=P_1, s=s_3)
The mass flow rate is determined from
+ ,+ ,kg/s 875.2
K 27330K/kgmkPa 0.287
)s/m 0kPa)(150/6 100(3
3
1
11 "-!!
""RT
Pm
!!!
The net power output is
kW 1100)kJ/kg60.30324kg/s)(686. 875.2()( 12inC, ")")" hhmW !!
kW 1759)kJ/kg62.792.7kg/s)(1404 875.2()( 43outT, ")")" hhmW !!
kW 659")")" 11001759inC,outT,net WWW !!!
(b) The back work ratio is
0.625"""kW 1759
kW 1100
outT,
inC,
bwW
Wr
!
!
(c) The rate of heat input and the thermal efficiency are
kW 2065)kJ/kg24.686.7kg/s)(1404 875.2()( 23in ")")" hhmQ !!
0.319"""kW 2065
kW 659
in
net
Q
Wth !
!
*
1
Combustion
chamber
Turbine
23
4
Compress.
100 kPa
30°C
500°C
1.2 MPa
Pregunta nº3Un sistema de refrigeración en cascada (ver apuntes de clase) de dos etapas, opera entre los límites de 1.2 MPa y 200 kPa con refrigerante 134a como fluido de trabajo. El rechazo de calor del ciclo inferior al superior tiene lugar en un intercambiador adiabático a contracorriente en donde las presiones en los ciclos superior en inferior son 0.4 y 0.5 MPa, respectivamente. En ambos ciclos el refrigerante es un líquido saturado a la salida del condensador y un vapor saturado a la entrada del compresor. La eficiencia isentrópica del compresión es del 80%. Si el flujo másico del refrigerante en el ciclo inferior es 0.15 kg/s, determine:
a) El flujo másico del ciclo superior, en kg/sb) La tasa de remoción de calor del espacio refrigerado, en kW
Prof. José Luis Salazar N.
![Page 4: TERMO](https://reader033.fdocuments.ec/reader033/viewer/2022042821/563db83e550346aa9a91e459/html5/thumbnails/4.jpg)
c) El coeficiente de operación (COP) de este refrigerador
Solución
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-35
11-49 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through
the upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):
kJ/kg.K 9377.0
kJ/kg 46.244
kPa 200 @1
kPa 200 @1
!!
!!
g
g
ss
hh
kJ/kg 30.263kPa 500
212
2 !"#$
!
!sh
ss
P
kJ/kg 01.26846.244
46.24430.26380.0 2
2
12
12
!%&%''
!
'
'!
hh
hh
hh sC(
kJ/kg 33.73
kJ/kg 33.73
34
kPa 500 @3
!!
!!
hh
hh f
kJ/kg.K 9269.0
kJ/kg 55.255
kPa 400 @5
kPa 004 @5
!!
!!
g
g
ss
hh
kJ/kg 33.278kPa 1200
656
6 !"#$
!
!sh
ss
P
kJ/kg 02.28455.255
55.25533.27880.0 6
6
56
56
!%&%''
!
'
'!
hh
hh
hh sC(
kJ/kg 77.117
kJ/kg 77.117
78
kPa 1200 @7
!!
!!
hh
hh f
The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the
heat exchanger
kg/s 0.212!%&%'!'
'!'
AA
BA
mm
hhmhhm
!!
!!
kJ/kg)33.7301kg/s)(268. 15.0(kJ/kg)77.117.55255(
)()( 3285
(b) The rate of heat removal from the refrigerated space is
kW 25.67!'!'! kJ/kg)33.7346kg/s)(244. 15.0()( 41 hhmQ BL!!
(c) The power input and the COP are
kW 566.9kJ/kg)46.24401kg/s)(268. 212.0(kJ/kg)55.25502kg/s)(284. 15.0(
)()( 1256in
!')'!
')'! hhmhhmW BA!!!
2.68!!!566.9
67.25COP
in
L
W
Q
!
!
.
5
6 7
8
QH
Condenser
Evaporator
Compressor
Expansion
valve Win
1
2 3
4
Condenser
Evaporator
Compressor
Expansion
valve
QL
Win
.
.
.
El flujo másico de refrigerante se determina con un balance de energía en el Intercambiador de calor (Evaporador-Condensador) según:
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-35
11-49 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through
the upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):
kJ/kg.K 9377.0
kJ/kg 46.244
kPa 200 @1
kPa 200 @1
!!
!!
g
g
ss
hh
kJ/kg 30.263kPa 500
212
2 !"#$
!
!sh
ss
P
kJ/kg 01.26846.244
46.24430.26380.0 2
2
12
12
!%&%''
!
'
'!
hh
hh
hh sC(
kJ/kg 33.73
kJ/kg 33.73
34
kPa 500 @3
!!
!!
hh
hh f
kJ/kg.K 9269.0
kJ/kg 55.255
kPa 400 @5
kPa 004 @5
!!
!!
g
g
ss
hh
kJ/kg 33.278kPa 1200
656
6 !"#$
!
!sh
ss
P
kJ/kg 02.28455.255
55.25533.27880.0 6
6
56
56
!%&%''
!
'
'!
hh
hh
hh sC(
kJ/kg 77.117
kJ/kg 77.117
78
kPa 1200 @7
!!
!!
hh
hh f
The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the
heat exchanger
kg/s 0.212!%&%'!'
'!'
AA
BA
mm
hhmhhm
!!
!!
kJ/kg)33.7301kg/s)(268. 15.0(kJ/kg)77.117.55255(
)()( 3285
(b) The rate of heat removal from the refrigerated space is
kW 25.67!'!'! kJ/kg)33.7346kg/s)(244. 15.0()( 41 hhmQ BL!!
(c) The power input and the COP are
kW 566.9kJ/kg)46.24401kg/s)(268. 212.0(kJ/kg)55.25502kg/s)(284. 15.0(
)()( 1256in
!')'!
')'! hhmhhmW BA!!!
2.68!!!566.9
67.25COP
in
L
W
Q
!
!
.
5
6 7
8
QH
Condenser
Evaporator
Compressor
Expansion
valve Win
1
2 3
4
Condenser
Evaporator
Compressor
Expansion
valve
QL
Win
.
.
.
La tasa de remoción de calor del espacio refrigerado será:
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-35
11-49 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through
the upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):
kJ/kg.K 9377.0
kJ/kg 46.244
kPa 200 @1
kPa 200 @1
!!
!!
g
g
ss
hh
kJ/kg 30.263kPa 500
212
2 !"#$
!
!sh
ss
P
kJ/kg 01.26846.244
46.24430.26380.0 2
2
12
12
!%&%''
!
'
'!
hh
hh
hh sC(
kJ/kg 33.73
kJ/kg 33.73
34
kPa 500 @3
!!
!!
hh
hh f
kJ/kg.K 9269.0
kJ/kg 55.255
kPa 400 @5
kPa 004 @5
!!
!!
g
g
ss
hh
kJ/kg 33.278kPa 1200
656
6 !"#$
!
!sh
ss
P
kJ/kg 02.28455.255
55.25533.27880.0 6
6
56
56
!%&%''
!
'
'!
hh
hh
hh sC(
kJ/kg 77.117
kJ/kg 77.117
78
kPa 1200 @7
!!
!!
hh
hh f
The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the
heat exchanger
kg/s 0.212!%&%'!'
'!'
AA
BA
mm
hhmhhm
!!
!!
kJ/kg)33.7301kg/s)(268. 15.0(kJ/kg)77.117.55255(
)()( 3285
(b) The rate of heat removal from the refrigerated space is
kW 25.67!'!'! kJ/kg)33.7346kg/s)(244. 15.0()( 41 hhmQ BL!!
(c) The power input and the COP are
kW 566.9kJ/kg)46.24401kg/s)(268. 212.0(kJ/kg)55.25502kg/s)(284. 15.0(
)()( 1256in
!')'!
')'! hhmhhmW BA!!!
2.68!!!566.9
67.25COP
in
L
W
Q
!
!
.
5
6 7
8
QH
Condenser
Evaporator
Compressor
Expansion
valve Win
1
2 3
4
Condenser
Evaporator
Compressor
Expansion
valve
QL
Win
.
.
.
Y el COP
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-35
11-49 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through
the upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):
kJ/kg.K 9377.0
kJ/kg 46.244
kPa 200 @1
kPa 200 @1
!!
!!
g
g
ss
hh
kJ/kg 30.263kPa 500
212
2 !"#$
!
!sh
ss
P
kJ/kg 01.26846.244
46.24430.26380.0 2
2
12
12
!%&%''
!
'
'!
hh
hh
hh sC(
kJ/kg 33.73
kJ/kg 33.73
34
kPa 500 @3
!!
!!
hh
hh f
kJ/kg.K 9269.0
kJ/kg 55.255
kPa 400 @5
kPa 004 @5
!!
!!
g
g
ss
hh
kJ/kg 33.278kPa 1200
656
6 !"#$
!
!sh
ss
P
kJ/kg 02.28455.255
55.25533.27880.0 6
6
56
56
!%&%''
!
'
'!
hh
hh
hh sC(
kJ/kg 77.117
kJ/kg 77.117
78
kPa 1200 @7
!!
!!
hh
hh f
The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the
heat exchanger
kg/s 0.212!%&%'!'
'!'
AA
BA
mm
hhmhhm
!!
!!
kJ/kg)33.7301kg/s)(268. 15.0(kJ/kg)77.117.55255(
)()( 3285
(b) The rate of heat removal from the refrigerated space is
kW 25.67!'!'! kJ/kg)33.7346kg/s)(244. 15.0()( 41 hhmQ BL!!
(c) The power input and the COP are
kW 566.9kJ/kg)46.24401kg/s)(268. 212.0(kJ/kg)55.25502kg/s)(284. 15.0(
)()( 1256in
!')'!
')'! hhmhhmW BA!!!
2.68!!!566.9
67.25COP
in
L
W
Q
!
!
.
5
6 7
8
QH
Condenser
Evaporator
Compressor
Expansion
valve Win
1
2 3
4
Condenser
Evaporator
Compressor
Expansion
valve
QL
Win
.
.
.
Prof. José Luis Salazar N.