Question

An actual refrigerator operates on the vaporcompression refrigeration cycle with refrigerant-22 as the working fluid. The refrigerant evaporates at \(-15^{\circ} \mathrm{C}\) and condenses at \(40^{\circ} \mathrm{C}\). The isentropic efficiency of the compressor is 83 percent. The refrigerant is superheated by \(5^{\circ} \mathrm{C}\) at the compressor inlet and subcooled by \(5^{\circ} \mathrm{C}\) at the exit of the condenser. Determine (a) the heat removed from the cooled space and the work input, in \(\mathrm{kJ} / \mathrm{kg}\) and the COP of the cycle. Determine ( \(b\) ) the same parameters if the cycle operated on the ideal vapor-compression refrigeration cycle between the same evaporating and condensing temperatures. The properties of \(R-22\) in the case of actual operation are: \(h_{1}=402.49 \mathrm{kJ} / \mathrm{kg}, h_{2}=454.00 \mathrm{kJ} / \mathrm{kg}, h_{3}=243.19 \mathrm{kJ} / \mathrm{kg}\) The properties of \(R-22\) in the case of ideal operation are: \(h_{1}=399.04 \mathrm{kJ} / \mathrm{kg}, h_{2}=440.71 \mathrm{kJ} / \mathrm{kg}, h_{3}=249.80 \mathrm{kJ} / \mathrm{kg}\) Note: state 1: compressor inlet, state 2: compressor exit, state 3: condenser exit, state 4: evaporator inlet.

Short answer

Answer: The isentropic efficiency of the compressor affects the performance of the vapor-compression refrigeration cycle by making the COP lower for the actual cycle compared to the ideal cycle. In the example provided, the COP was 3.09 for the actual cycle and 3.58 for the ideal cycle.

Step-by-step solution

Calculate the work done by the compressor

The work done by the compressor (\(W_c\)) is the difference between the enthalpy at the exit and entrance of the compressor. \(W_c = h_2 - h_1 = 454.00 - 402.49 = 51.51 \, \mathrm{kJ/kg}\)

Determine the heat rejected in the condenser

The heat rejected in the condenser is the difference between the enthalpy at the entrance and exit of the condenser. \(Q_{rejected} = h_2 - h_3 = 454.00 - 243.19 = 210.81 \, \mathrm{kJ/kg}\)

Calculate the heat removed from the cooled space

The heat removed from the cooled space (\(Q_{removed}\)) can be calculated using the energy balance in a steady-flow process. \(Q_{removed} = Q_{rejected} - W_c = 210.81 - 51.51 = 159.30 \, \mathrm{kJ/kg}\)

Determine the COP for the actual cycle

The COP for the actual cycle can be calculated using the formula: \(\text{COP} = \frac{Q_{removed}}{W_c} = \frac{159.30}{51.51} = 3.09\) ##Part (b)##

Calculate the work done by the compressor for an ideal cycle

Using the given properties of R-22 for the ideal case, the work done by the compressor for an ideal cycle is: \(W_{c,ideal} = h_2 - h_1 = 440.71 - 399.04 = 41.67 \, \mathrm{kJ/kg}\)

Determine the heat rejected in the condenser for an ideal cycle

The heat rejected in the condenser for an ideal cycle is: \(Q_{rejected,ideal} = h_2 - h_3 = 440.71 - 249.80 = 190.91 \, \mathrm{kJ/kg}\)

Calculate the heat removed from the cooled space for an ideal cycle

The heat removed from the cooled space for an ideal cycle is: \(Q_{removed,ideal} = Q_{rejected,ideal} - W_{c,ideal} = 190.91 - 41.67 = 149.24 \, \mathrm{kJ/kg}\)

Determine the COP for the ideal cycle

The COP for the ideal cycle can be calculated using the formula: \(\text{COP}_{ideal} = \frac{Q_{removed,ideal}}{W_{c,ideal}} = \frac{149.24}{41.67} = 3.58\) The results show that the performance of an actual vapor-compression cycle is affected by the isentropic efficiency of the compressor; the COP is lower for the actual cycle (3.09) compared to the ideal cycle (3.58).