GATE ME 2016 Set 3 — Question 57
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Thermodynamics → Refrigeration & Air Conditioning → Vapour Compression Cycle
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Question
In the vapour compression cycle shown in the figure, the evaporating and condensing temperatures are 260 K and 310 K, respectively. The compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). The specific heat of the liquid refrigerant is 4.8 kJ/kg-K and may be treated as constant. The enthalpy of evaporation for the refrigerant at 310 K is 1054 kJ/kg.
The difference between the enthalpies at state points 1 and 0 (in kJ/kg) is ____________
The difference between the enthalpies at state points 1 and 0 (in kJ/kg) is ____________
Correct answer
1095 to 1130
Solution
From the T-s diagram, state 0 is saturated liquid at K and state 1 is a liquid-vapour mixture at the same temperature. Process 1-2 is isentropic compression to dry saturated vapour at K.
Since process 1-2 is isentropic, .
We need to find the enthalpy difference .
Using the relation for entropy in the mixture region:
The entropy difference can be expanded as:
kJ/kg-K
kJ/kg-K
Total entropy difference: kJ/kg-K
Finally, kJ/kg.
Since process 1-2 is isentropic, .
We need to find the enthalpy difference .
Using the relation for entropy in the mixture region:
The entropy difference can be expanded as:
kJ/kg-K
kJ/kg-K
Total entropy difference: kJ/kg-K
Finally, kJ/kg.
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