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Question

A refrigerator uses R134a as its refrigerant and operates on an ideal vapour compression refrigeration cycle between 0.14MPa and 0.8MPa. If the mass flow rate of the refrigerant is 0.05kg/s, the rate of heat rejection to the environment is \(kW\).
Given data :
At P=0.14MPa,h=236.04 kJ/kg,s=0.9322 kJ/kgK
At P=0.8 MPa,h=272.05 kJ/kg(superheated vapour)
At P=0.8 MPa,h=93.42 kJ/kg(saturated liquid)

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Solution

Given data :
m=0.05 kg/s
At pe=0.14 MPa
h1=236.04 kJ/kg
s1=0.9322 kJ/kg
At pc=0.8 MPa
h3=h4=hf=93.42 kJ/kg
h2=272.05 kJ/kg

Rate of heat rejection to the environment,

Q23=m[h2h3]=0.05[272.0593.42]

=8.93 kW

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