Question

For a refrigeration cycle, the ratio of actual COP to the COP of a reversible refrigerator operating between the same temperature limits is 0.8. The condenser and evaporator temperatures are \(51^\circ C\) and \(-30^\circ C\), respectively. If the cooling capacity of the plant is 2.4 kW, then the power input to the refrigerator is:

• A. 1.00 kW
• B. 1.33 kW
• C. 1.25 kW
• D. 2.08 kW

Hint:

To find the power input to a refrigerator, use the formula \( \text{Power input} = \frac{\text{Cooling capacity}}{\text{COP}} \).

Answer 1

The Correct Option is A

We are given the ratio of the actual COP to the ideal COP of a reversible refrigerator, which is 0.8. The cooling capacity of the plant is 2.4 kW. The COP of a refrigeration cycle is given by: \[ \text{COP}_{\text{ideal}} = \frac{T_{\text{evaporator}}}{T_{\text{condenser}} - T_{\text{evaporator}}}. \] Converting the temperatures to Kelvin: \[ T_{\text{evaporator}} = -30^\circ C + 273.15 = 243.15 \, \text{K}, T_{\text{condenser}} = 51^\circ C + 273.15 = 324.15 \, \text{K}. \] The ideal COP is: \[ \text{COP}_{\text{ideal}} = \frac{243.15}{324.15 - 243.15} = \frac{243.15}{81} = 3.00. \] The actual COP is: \[ \text{COP}_{\text{actual}} = 0.8 \times \text{COP}_{\text{ideal}} = 0.8 \times 3.00 = 2.4. \] The cooling capacity \(Q_{\text{cooling}}\) is given as 2.4 kW, so the power input is: \[ \text{Power input} = \frac{Q_{\text{cooling}}}{\text{COP}_{\text{actual}}} = \frac{2.4}{2.4} = 1.33 \, \text{kW}. \] Final Answer: \text{(A) 1.33 kW}