Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 6.11, Problem 149RP

A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 60°C at a rate of 0.065 kg/s and leaves at 40°C. Refrigerant enters the evaporator at 12°C with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes 1.6 kW of power, determine (a) the mass flow rate of the refrigerant, (b) the rate of heat supply, (c) the COP, and (d) the minimum power input to the compressor for the same rate of heat supply.

FIGURE P6–152

Chapter 6.11, Problem 149RP, A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25C by absorbing

(a)

Expert Solution
Check Mark
To determine

The mass flow rate of the refrigerant.

Answer to Problem 149RP

The mass flow rate of the refrigerant is 0.0338kg/s_.

Explanation of Solution

Determine the rate of heat absorbed from the water.

Q˙H=m˙w(h2h1)w (I)

Here, the mass flow rate of the water is m˙w, the enthalpy of saturated liquid that is entering the inlet of the condenser is hw,1 and the enthalpy of saturated liquid which is leaving the condenser is hw,2.

Determine the mass flow rate of a refrigerant.

m˙R=Q˙Hh1h2 (II)

Conclusion:

From the Table A-11, “Saturated refrigerant R-134a”, obtain the value of saturated pressure of the refrigerant at the inlet temperature of 12°C as below.

P1=443.3kPa.

Here, the pressure of refrigerant is constant in evaporation.

P1=P2.

From the Table A-11, “Saturated refrigerant R-134a” to obtain the value of specific enthalpy of the refrigerant at the outlet pressure of 443.3kPa and the dryness faction of 1 as,

h2=257.33kJ/kg.

From the Table A-11, “Saturated refrigerant R-134a” to obtain the value of specific enthalpy of saturated liquid and specific enthalpy change upon vaporization of the refrigerant at the inlet temperature of 12°C as,

hf,1=68.18kJ/kghfg,1=189.09kJ/kg

Calculate the specific enthalpy of refrigerant at evaporator inlet.

h1=hf,1+x1×hfg,1 (III)

Here, the specific enthalpy of saturated liquid is hf,1, the dryness fraction is x1, and the specific enthalpy change upon vaporization is hfg,1.

Substitute 68.18kJ/kg for hf,1, 189.09kJ/kg for hfg,1, and 0.15 for x1 in Equation (III).

h1=(68.18kJ/kg)+(0.15)×(189.09kJ/kg)=(68.18kJ/kg)+(28.3635kJ/kg)=96.5435kJ/kg

From the Table A-4, “Saturated water-temperature” to obtain the value of specific enthalpy of saturated liquid of water at the inlet temperature of 60°C as,

hw,1=251.18kJ/kg

From the Table A-4, “Saturated water-temperature” to obtain the value of specific enthalpy of saturated liquid of water at the outlet temperature of 40°C as,

hw,2=167.53kJ/kg

Substitute 0.065kg/s for m˙w, 251.18kJ/kg for hw,1, and 167.53kJ/kg for hw,2 in Equation (I).

Q˙L=(0.065kg/s)(251.18kJ/kg167.53kJ/kg)=(0.065kg/s)(83.65kJ/kg)=5.437kJ/s=5.437kJ/s×(1kW1kJ/s)

      =5.437kW

Substitute 5.437kW for Q˙L, 96.54kJ/kg for h1, and 257.33kJ/kg  for h2 in Equation (II).

m˙R=5.437kW(257.3396.54)kJ/kg=5.437kW×(1kg/s1kW)160.79kJ/kg=0.0338kg/s

Thus, the mass flow rate of the refrigerant is 0.0338kg/s_.

(b)

Expert Solution
Check Mark
To determine

The heating load of the heat pump.

Answer to Problem 149RP

The heating load of the heat pump is 7.04kW_.

Explanation of Solution

Determine the heating load of the heat pump.

Q˙H=Q˙L+W˙in (IV)

Here, the power input consumed by compressor is W˙in.

Conclusion:

Substitute 5.437kW for Q˙L and 1.6kW for W˙in in Equation (IV).

Q˙L=(5.437kW)+(1.6kW)=7.037kW7.04kW

Thus, the heating load of the heat pump is 7.04kW_.

(c)

Expert Solution
Check Mark
To determine

The COP of a heat pump operating between the same temperature limits.

Answer to Problem 149RP

The COP of a heat pump operating between the same temperature limits is 4.40_.

Explanation of Solution

Determine the coefficient of performance of the heat pump.

COPHP=Q˙HW˙in (V)

Conclusion:

Substitute 7.04kW for Q˙H and 1.6kW for W˙in in Equation (IV).

COPHP=7.04kW1.6kW=4.40

Thus, the COP of a heat pump operating between the same temperature limits is 4.40_.

(d)

Expert Solution
Check Mark
To determine

The minimum power input to the compressor.

Answer to Problem 149RP

The minimum power input to the compressor is 0.740kW_.

Explanation of Solution

Determine the maximum coefficient of performance of the heat pump operating between the same temperature limits.

COPmax=1TLTH1 (VI)

Here, the temperature of higher temperature body is TH and the temperature of lower temperature body is TL.

Determine the minimum power input to the condenser for the same heat pump load.

W˙in,min=Q˙HCOPmax (VII)

Conclusion:

Substitute 60°C for TH and 25°C for TL in Equation (VI).

COPmax=11(25°C)(60°C)=11(25°C+273)(60°C+273)=11(298K333K)=9.51

Substitute 7.04kW for Q˙H and 9.51 for COPmax in Equation (VII).

W˙in,min=7.04kW9.51=0.740kW

Thus, the minimum power input to the compressor is 0.740kW_.

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Students have asked these similar questions
A heat pump uses R134a as refrigerant.  The refrigerant flows into the condenser at 900 kPa and 60°C and flows out of it (the condenser)as saturated liquid at the same pressure. The pressure in the evaporator is 70 kPa. The mass flow of the refrigerant is 0.025 kg/s and the compressor power is 1.4 kWa) Calculate the heat pump's power factor, COPHP.b) Calculate how much heat is absorbed/taken up from the air outside.c) Calculate the vapor quality of the refrigerant into the evaporator.
A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25 C by absorbing heat from geothermal water that enters the evaporator at 60 C at a rate of 0.065 kg/s and leaves at 40 C. Refrigerant enters the evaporator at 12 C with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes 1.6 kW of power, determine; (a) the mass flow rate of the refrigerant, (b) the rate of heat supply, (c) the COP, and (d) the minimum power input to the compressor for the same rate of heat supply.
A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 60°C at a rate of 0.065 kg/s and leaves at 40°C. Refrigerant enters the evaporator at 12°C with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes 1.6 kW of power, determine the mass flow rate of the refrigerant.

Chapter 6 Solutions

Thermodynamics: An Engineering Approach

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