The Actual Vapor-Compression Refrigeration Cycle Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa and is throttled to 0.15 MPa.

Question

The Actual Vapor-Compression Refrigeration Cycle Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa and 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, (b) the isentropic efficiency of the compressor, and (c) the coefficient of performance of the refrigerator.

Accepted Answer

A refrigerator operating on a vapor-compression cycle is considered. The rate of refrigeration, the power input, the compressor efficiency, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis The T-s diagram of the refrigeration cycle is shown in Fig. 11–8. We note that the refrigerant leaves the condenser as a compressed liquid and enters the compressor as superheated vapor. The enthalpies of the refrigerant at various states are determined from the refrigerant tables to be

[latex]\begin{aligned}&\left.\begin{array}{l}P_{1}=0.14 MPa \T_{1}=-10^{\circ} C\end{array} ight\} \quad h_{1}=246.36 kJ / kg \&\left.\begin{array}{l}P_{2}=0.8 MPa \T_{2}=50^{\circ} C\end{array} ight\} \quad h_{2}=286.69 kJ / kg \&\left.\begin{array}{l}P_{3}=0.72 MPa \T_{3}=26^{\circ} C\end{array} ight\} \quad h_{3} \cong h_{f @ 26^{\circ} C }=87.83 kJ / kg \&h_{4} \cong h_{3}( ext { throttling }) \quad \longrightarrow \quad h_{4}=87.83 kJ / kg\end{aligned}[/latex]

(a) The rate of heat removal from the refrigerated space and the power input to the compressor are determined from their definitions:

[latex]\dot{Q}_{L}=\dot{m}\left(h_{1}-h_{4} ight)=(0.05 kg / s )[(246.36-87.83) kJ / kg ]=7.93 kW[/latex]

and

[latex]\dot{W}_{ ext {in }}=\dot{m}\left(h_{2}-h_{1} ight)=(0.05 kg / s )[(286.69-246.36) kJ / kg ]= 2 . 0 2 k W[/latex]

(b) The isentropic efficiency of the compressor is determined from

[latex]\eta_{C} \cong \frac{h_{2 s}-h_{1}}{h_{2}-h_{1}}[/latex]

where the enthalpy at state 2s ([latex]P_{2 s}=0.8 MPa ext { and } s_{2 s}=s_{1}=0.9724[/latex]kJ/kg · K) is 284.21 kJ/kg. Thus,

[latex]\eta_{C}=\frac{284.21-246.36}{286.69-246.36}=0.939 ext { or } 93.9 \%[/latex]

(c) The coefficient of performance of the refrigerator is

[latex]COP _{ R }=\frac{\dot{Q}_{L}}{\dot{W}_{ ext {in }}}=\frac{7.93 kW }{2.02 kW }= 3 . 9 3[/latex]

Discussion This problem is identical to the one worked out in Example 11–1, except that the refrigerant is slightly superheated at the compressor
inlet and subcooled at the condenser exit. Also, the compressor is not isentropic. As a result, the heat removal rate from the refrigerated space increases (by 10.4 percent), but the power input to the compressor increases even more (by 11.6 percent). Consequently, the COP of the refrigerator decreases from 3.97 to 3.93.

Question 11.2: The Actual Vapor-Compression Refrigeration Cycle Refrigerant...

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