A refrigerator operating on standard vapour compression cycle has a co-efficiency performance of 6.5 and is driven by a 50 kW compressor. The enthalpies of saturated liquid and saturated vapour refrigerant at the operating condensing temperature of 35°C are 62.55 kJ/kg and 201.45

Question

A refrigerator operating on standard vapour compression cycle has a co-efficiency performance of 6.5 and is driven by a 50 kW compressor. The enthalpies of saturated liquid and saturated vapour refrigerant at the operating condensing temperature of 35°C are 62.55 kJ/kg and 201.45 kJ/kg respectively. The saturated refrigerant vapour leaving evaporator has an enthalpy of 187.53 kJ/kg. Find the refrigerant temperature at compressor discharge. The [latex] c_{p} [/latex] of refrigerant vapour may be taken to be 0.6155 kJ/kg°C. (UPTU)

Accepted Answer

Given : C.O.P. = 6.5 ; W = 50 kW, [latex] h_{3} ^{′ }[/latex]= 201.45 kJ/kg,
[latex] h_{f_4} = h_{1} = 69.55 kJ/kg ; h_{2} [/latex]= 187.53 kJ/kg
[latex] c_{p} [/latex] = 0.6155 kJ/kg K

Temperature, [latex] t_{3} [/latex]:
Refrigerating capacity = 50 × C.O.P.
= 50 × 6.5 = 325 kW

Heat extracted per kg of refrigerant
= 187.53 – 69.55 = 117.98 kJ/kg
Refrigerant flow rate = [latex] \frac{325}{117.98} [/latex] = 2.755 kg/s

Compressor power = 50 kW
∴ Heat input per kg = [latex] \frac{50}{2.755} [/latex] = 18.15 kJ/kg

Enthalpy of vapour after compression
= [latex] h_{2} [/latex] + 18.15 = 187.53 + 18.15
= 205.68 kJ/kg

Superheat = 205.68 – [latex] h_{3} ^{′ }[/latex] = 205.68 – 201.45
= 4.23 kJ/kg
But 4.23 = 1 × [latex] c_{p} (t_{3} – t_{3} ^{′ }) = 1 × 0.6155 × (t_{3} [/latex] – 35)

∴ [latex] t_{3} = \frac{4.23}{0.6155} [/latex] + 35 = 41.87 °C.
Note. The compressor rating of 50 kW is assumed to be the enthalpy of compression, in the absence of any data on the efficiency of compressor.

Question 12.23: A refrigerator operating on standard vapour compression cycl...

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