Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at

Question

Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and-10°C at a rate of 0.112 kg/s, and it leaves at 0.7 MPa and 50°C. The refrigerant is cooled in the condenser to 24°C and 0.65 MPa, and it is throttled to 0.15 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, kW (Round to one decimal place) kW (Round to two decimal places) Win (b) the isentropic efficiency of the compressor, % (Round to one decimal place) and (c) the COP of the refrigerator. COPR (Round to two decimal places)

Explanation / Answer

All data taken from refrigerant tables

At state 1

Pressure P1 = 0.140 MPa

Temperature T1 = - 10.0°C

Enthalpy H1 = 246.360 kJ/kg

Entropy S1 = 0.972360 kJ/kg-K

At state 2

Pressure P2 = 0.70 MPa

Temperature T2 = 50.0°C

Enthalpy H2 = 288.530 kJ/kg

At state 2s

P2s = 0.70 MPa

S2s =S1 = 0.972360 kJ/kg-K

H2s = 281.160 kJ/kg

At state 3

P3 = 0.650 MPa

T3 = 24.0°C

H3 = Hf = 84.980 kJ/kg

Throttling is an isenthalpic process

H4 = H3 = 84.980 kJ/kg

Part a

Rate of heat removal

Q = m (H1 - H4) = 0.120 kg/s x (246.360 - 84.980)

= 19.4 kW

Power input to compressor

W =m (H2 - H1) = 0.120 kg/s x (288.530 - 246.360)

= 5.06 kW

Part b

Isentropic efficiency = (H2s - H1) / (H2 - H1)

= (281.160 - 246.360) / (288.530 - 246.360)

= 0.8252

= 82.5%

Part c

COP = Q/W

= 19.4/5.06

= 3.83

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