Ups termo 2

11-13 Un ciclo ideal de refrigeración por compresión de vapor que usa refrigerante 134a como fluido de trabajo mantiene un condensador a 1.000 kPa y el evaporador a 4 °C. Determine el COP de este sistema y la cantidad de potencia necesaria para proporcionar una carga de enfriamiento de 400 kW. 11-14 Un refrigerador usa refrigerante 134a como fluido de trabajo y opera en un ciclo ideal de refrigeración por compresión de vapor entre 0.12 y 0.7 MPa. El flujo másico del refrigerante es 0.05 kg/s. Muestre el ciclo en un diagrama T-s con respecto a las líneas de saturación. Determine a) la tasa de remoción de calor del espacio refrigerado y la entrada de potencia al compresor, b) la tasa de rechazo de calor al entorno y c) el coeficiente de desempeño.

11-17 Entra refrigerante 134a al compresor de un refrigerador como vapor sobrecalentado a 0.20 MPa y -5 °C a razón de 0.7 kg/s, y sale a 1.2 MPa y 70 °C. El refrigerante se enfría en el condensador a 44 °C y 1.15 MPa, y se estrangula a 0.2 MPa. Despreciando cualquier transferencia de calor y cualquier caída de presión en las líneas de conexión entre los componentes, muestre el ciclo en un diagrama T-s con respecto a las líneas de saturación, y determine a) la tasa de remoción de calor del espacio refrigerado y la entrada de potencia al compresor, b) la eficiencia isentrópica del compresor, y c) el COP del refrigerador.

11-20. Un refrigerador comercial con refrigerante 134a como fluido de trabajo se usa para mantener el espacio refrigerado a -30 °C rechazando su calor de desecho a agua de enfriamiento que entra al condensador a 18 °C a razón de 0.25 kg/s y sale a 26 °C. El refrigerante entra al condensador a 1.2 MPa y 65 °C y sale a 42 °C. El estado a la entrada del compresor es de 60 kPa y -34 °C y se estima que el compresor gana un calor neto de 450 W del entorno. Determine a) la calidad del refrigerante a la entrada del evaporador, b) la carga de refrigeración, c) el COP del refrigerador y d) la carga de refrigeración teórica máxima para la misma entrada de potencia al compresor.

11-46. Una bomba de calor opera en el ciclo ideal de refrigeración por compresión de vapor y usa refrigerante 134ª como fluido de trabajo. El condensador opera a 1.000 kPa, y el evaporador, a 200 kPa. Determine el COP del sistema y la tasa de suministro de calor al evaporador cuando el compresor consume 6 kW. 11-50. Entra refrigerante 134a al condensador de una bomba de calor residencial a 800 kPa y 55 °C, a razón de 0.018 kg/s, y sale a 750 kPa subenfriado en 3 °C. El refrigerante entra al compresor a 200 kPa sobrecalentado en 4 °C. Determine a) la eficiencia isentrópica del compresor, b) la tasa de suministro de calor al cuarto calentado y c) el COP de la bomba de calor. También determine d) el COP y la tasa de suministro de calor al cuarto calentado si esta bomba de calor opera en el ciclo ideal por compresión de vapor entre los límites de presión de 200 y 800 kPa. 11-60. Considere un sistema de refrigeración en cascada de dos etapas que opera entre los límites de presión 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 de calor adiabático a contracorriente en donde las presiones en los ciclos superior e 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, y la eficiencia isentrópica del compresor es de 80 por ciento. Si el flujo másico del refrigerante en el ciclo inferior es de 0.15 kg/s, determine a) el flujo másico del refrigerante a través del ciclo superior, b) la tasa de remoción del espacio refrigerado y c) el COP de este refrigerador.

11-6

11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the power requirement are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = 4°C ⎫ h1 = h g @ 4°C = 252.77 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 4°C = 0.92927 kJ/kg ⋅ K

T

P2 = 1 MPa ⎫ ⎬ h2 = 275.29 kJ/kg s 2 = s1 ⎭ P3 = 1 MPa ⎫ ⎬ h = hf sat. liquid ⎭ 3

@ 1 MPa

· QH

· Win

= 107.32 kJ/kg

h4 ≅ h3 = 107.32 kJ/kg ( throttling )

The mass flow rate of the refrigerant is

⎯→ m& = Q& L = m& (h1 − h4 ) ⎯

2

3 1 MPa

Q& L 400 kJ/s = = 2.750 kg/s h1 − h4 (252.77 − 107.32) kJ/kg

4°C 4s

4

· QL

1

s

The power requirement is

W& in = m& (h2 − h1 ) = (2.750 kg/s)(275.29 − 252.77) kJ/kg = 61.93 kW The COP of the refrigerator is determined from its definition, COPR =

Q& L 400 kW = = 6.46 & Win 61.93 kW

PROPRIETARY MATERIAL. © 2011 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.

11-7

11-14 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa sat. vapor

⎫ h1 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K ⎭ 1

P2 = 0.7 MPa ⎫ ⎬ h2 = 273.50 kJ/kg (T2 = 34.95°C ) s 2 = s1 ⎭ P3 = 0.7 MPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 0.7 MPa

T · QH

2

3 0.7 MPa

· Win

= 88.82 kJ/kg

h4 ≅ h3 = 88.82 kJ/kg (throttling )

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from

0.12 4s

4

· QL

1 s

Q& L = m& (h1 − h4 ) = (0.05 kg/s )(236.97 − 88.82) kJ/kg = 7.41 kW

and W& in = m& (h2 − h1 ) = (0.05 kg/s )(273.50 − 236.97 ) kJ/kg = 1.83 kW

(b) The rate of heat rejection to the environment is determined from

Q& H = Q& L + W& in = 7.41 + 1.83 = 9.23 kW (c) The COP of the refrigerator is determined from its definition, COPR =

Q& L 7.41 kW = = 4.06 W& in 1.83 kW

PROPRIETARY MATERIAL. © 2011 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.

11-10

11-17 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor, 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) From the refrigerant tables (Tables A-12 and A-13), P1 = 0.20 MPa ⎫ h1 = 248.80 kJ/kg ⎬ s = 0.95407 kJ/kg ⋅ K T1 = −5°C ⎭ 1 P2 = 1.2 MPa T2 = 70°C

T

⎫ ⎬ h2 = 300.61 kJ/kg ⎭

P2 s = 1.2 MPa ⎫ ⎬ h2 s = 287.21 kJ/kg s 2 s = s1 ⎭ P3 = 1.15 MPa ⎫ ⎬ h3 = h f T3 = 44°C ⎭

@ 44°C

= 114.28 kJ/kg

h4 ≅ h3 = 114.28 kJ/kg (throttling )

1.15 MPa

2 1.2 MPa 2s 70°C · Win

· QH

3 0.21 MPa 4

· QL

1

0.20 MPa -5°C s

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from Q& L = m& (h1 − h4 ) = (0.07 kg/s )(248.80 − 114.28) kJ/kg = 9.42 kW

and W& in = m& (h2 − h1 ) = (0.07 kg/s )(300.61 − 248.80) kJ/kg = 3.63 kW

(b) The isentropic efficiency of the compressor is determined from

ηC =

h2 s − h1 287.21 − 248.80 = 0.741 = 74.1% = h2 − h1 300.61 − 248.80

(c) The COP of the refrigerator is determined from its definition,

COPR =

Q& L 9.42 kW = = 2.60 W& in 3.63 kW

PROPRIETARY MATERIAL. © 2011 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.

11-13

11-20 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13) P1 = 60 kPa ⎫ ⎬ h1 = 230.03 kJ/kg T1 = −34°C ⎭ P2 = 1200 kPa ⎫ ⎬ h2 = 295.16 kJ/kg T2 = 65°C ⎭

Water 18°C

26°C QH

1.2 MPa 65°C

42°C

P3 = 1200 kPa ⎫ ⎬ h3 = 111.23 kJ/kg T3 = 42°C ⎭ h4 = h3 = 111.23 kJ/kg P4 = 60 kPa

⎫ ⎬ x 4 = 0.4795 h4 = 111.23 kJ/kg ⎭

Using saturated liquid enthalpy at the given temperature, for water we have (Table A-4)

Condenser 3

Qin

2

Expansion valve

4

Win Compressor

Evaporator

hw1 = h f @ 18°C = 75.47 kJ/kg

60 kPa -34°C

1

QL

h w2 = h f @ 26°C = 108.94 kJ/kg

(b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor m& R (h2 − h3 ) = m& w (hw2 − hw1 ) m& R (295.16 − 111.23)kJ/kg = (0.25 kg/s)(108.94 − 75.47)kJ/kg ⎯ ⎯→ m& R = 0.0455 kg/s

The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are

Q& H = m& R (h2 − h3 ) = (0.0455 kg/s)(295.16 − 111.23)kJ/kg = 8.367 kW W& in = m& R ( h2 − h1 ) − Q& in = (0.0455 kg/s)(295.16 − 230.03)kJ/kg − 0.450 kW = 2.513 kW Q& L = Q& H − W& in − Q& in = 8.367 − 2.513 − 0.450 = 5.404 kW (c) The COP of the refrigerator is determined from its definition COP =

Q& L 5.404 = = 2.15 W& in 2.513

(d) The reversible COP of the refrigerator for the same temperature limits is COPmax =

1 1 = = 5.063 T H / T L − 1 (18 + 273) /(−30 + 273) − 1

Then, the maximum refrigeration load becomes Q& L,max = COPmaxW& in = (5.063)(2.513 kW) = 12.72 kW

2

T · QH

2 · Win

3

4

· QL

1 s

PROPRIETARY MATERIAL. © 2011 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

11-46 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the rate of heat supplied to the evaporator are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 200 kPa ⎫ h1 = h g @ 200 kPa = 244.46 kJ/kg ⎬ s =s sat. vapor g @ 200 kPa = 0.93773 kJ/kg ⋅ K ⎭ 1

T

P2 = 1000 kPa ⎫ ⎬ h2 = 277.98 kJ/kg s 2 = s1 ⎭ P3 = 1000 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 1000 kPa

· QH 3 1.0 MPa

= 107.32 kJ/kg

h4 ≅ h3 = 107.32 kJ/kg ( throttling)

2 · Win

200 kPa 4s

The mass flow rate of the refrigerant is determined from W& in 6 kJ/s ⎯→ m& = = = 0.179 kg/s W& in = m& (h2 − h1 ) ⎯ h2 − h1 (277.98 − 244.46) kJ/kg

4

· QL

1

s

Then the rate of heat supplied to the evaporator is

Q& L = m& (h1 − h4 ) = (0.179 kg/s)(244.46 − 107.32) kJ/kg = 24.5 kW The COP of the heat pump is determined from its definition, COPHP =

q H h2 − h3 277.98 − 107.32 = = = 5.09 win h2 − h1 277.98 − 244.46

PROPRIETARY MATERIAL. © 2011 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.

11-39

11-50 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of the compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the same pressure limits are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) The properties of refrigerant-134a are (Tables A-11 through A-13) P2 = 800 kPa ⎫ ⎬ h2 = 291.76 kJ/kg ⎭ T3 = Tsat@750 kPa = 29.06°C T2 = 55°C

.

QH

750 kPa

P3 = 750 kPa

⎫ ⎬ h3 = 87.91 kJ/kg T3 = (29.06 − 3)°C⎭ h4 = h3 = 87.91 kJ/kg Tsat@200 kPa = −10.09°C P1 = 200 kPa

⎫h1 = 247.87 kJ/kg ⎬ T1 = (−10.09 + 4)°C⎭ s1 = 0.9506 kJ/kg P2 = 800 kPa ⎫ ⎬h2 s = 277.26 s 2 = s1 ⎭

800 kPa 55°C

Condenser 3

2

Expansion valve

.

Win

Compressor 1

4

Evaporator .

QL

The isentropic efficiency of the compressor is 2

T

277.26 − 247.87 h −h ηC = 2 s 1 = = 0.670 291.76 − 247.87 h2 − h1

2

· Q

(b) The rate of heat supplied to the room is

· Win

3

Q& H = m& (h2 − h3 ) = (0.018 kg/s)(291.76 − 87.91)kJ/kg = 3.67 kW (c) The power input and the COP are

· QL

4

W& in = m& (h2 − h1 ) = (0.018 kg/s)(291.76 − 247.87)kJ/kg = 0.790 kW COP =

1 s

Q& H 3.67 = = 4.64 & Win 0.790

(d) The ideal vapor-compression cycle analysis of the cycle is as follows: h1 = h g @ 200 kPa = 244.46 kJ/kg s1 = s g @ 200 kPa = 0.9377 kJ/kg.K P2 = 800 kPa ⎫ ⎬ h2 = 273.25 kJ/kg s 2 = s1 ⎭

T · QH

2

3 0.8 MPa

· Win

h3 = h f @ 800 kPa = 95.47 kJ/kg h4 = h3 h − h3 273.25 − 95.47 COP = 2 = = 6.18 h2 − h1 273.25 − 244.46

0.2 MPa 4s

4

· QL

1 s

Q& H = m& (h2 − h3 ) = (0.018 kg/s)(273.25 − 95.47)kJ/kg = 3.20 kW

PROPRIETARY MATERIAL. © 2011 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.

11-46

11-60 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): h1 = h g @ 200 kPa = 244.46 kJ/kg

.

QH

s1 = s g @ 200 kPa = 0.9377 kJ/kg.K P2 = 500 kPa ⎫ ⎬ h2 s = 263.30 kJ/kg s 2 = s1 ⎭

h −h η C = 2s 1 h2 − h1 0.80 =

263.30 − 244.46 ⎯ ⎯→ h2 = 268.01 kJ/kg h2 − 244.46

Condenser 7 Expansion valve

8

h3 = h f @ 500 kPa = 73.33 kJ/kg h4 = h3 = 73.33 kJ/kg h5 = h g @ 400 kPa = 255.55 kJ/kg s 5 = s g @ 400 kPa = 0.9269 kJ/kg.K P6 = 1200 kPa ⎫ ⎬ h6 s = 278.33 kJ/kg s6 = s5 ⎭

ηC = 0.80 =

6

.

Win

Compressor

Evaporator

5

Condenser 3 Expansion valve

4

h 6 s − h5 h6 − h5

2

.

Win

Compressor

Evaporator .

QL

1

278.33 − 255.55 ⎯ ⎯→ h6 = 284.02 kJ/kg h6 − 255.55

h7 = h f @ 1200 kPa = 117.77 kJ/kg h8 = h7 = 117.77 kJ/kg

The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the heat exchanger m& A (h5 − h8 ) = m& B (h2 − h3 ) m& A (255.55 − 117.77)kJ/kg = (0.15 kg/s)(268.01 − 73.33)kJ/kg ⎯ ⎯→ m& A = 0.212 kg/s

(b) The rate of heat removal from the refrigerated space is

Q& L = m& B (h1 − h4 ) = (0.15 kg/s)(244.46 − 73.33)kJ/kg = 25.67 kW (c) The power input and the COP are W& in = m& A (h6 − h5 ) + m& B (h2 − h1 ) = (0.15 kg/s)(284.02 − 255.55)kJ/kg + (0.212 kg/s)(268.01 − 244.46)kJ/kg = 9.566 kW

COP =

Q& L 25.67 = = 2.68 W& in 9.566

PROPRIETARY MATERIAL. © 2011 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.