Question

Determine the COP of a refrigerator that removes heat from the food compartment at a rate of \(5040 \mathrm{kJ} / \mathrm{h}\) for each \(\mathrm{kW}\) of power it consumes. Also, determine the rate of heat rejection to the outside air.

Short answer

Question: Calculate the coefficient of performance (COP) and the rate of heat rejection of a refrigerator that cools the food compartment by removing heat at a rate of 5040 kJ/h and consumes 1 kW of power. Answer: The coefficient of performance (COP) of the refrigerator is 1.4, and the rate of heat rejection to the outside air is 8640 kJ/h.

Step-by-step solution

Calculate COP of the refrigerator

To calculate the COP, we use the formula COP = \(Q_L/W\). Given, the rate of heat removal from the food compartment, \(Q_L = 5040\,\mathrm{kJ/h}\). And the power consumed by the refrigerator, W = 1 kW. First, we need to convert the power to the same unit as the heat rate, which is kJ/h. Since 1 kW = 1000 J/s, we can convert it to kJ/h by multiplying by 3600 (number of seconds in an hour): \(W = 1\,\mathrm{kW} \times 1000\,\mathrm{J/s} \times \frac{1\,\mathrm{kJ}}{1000\,\mathrm{J}} \times \frac{3600\,\mathrm{s}}{1\,\mathrm{h}} = 3600\,\mathrm{kJ/h}\) Now, we can compute the COP: \(\mathrm{COP} = \frac{Q_L}{W} = \frac{5040\,\mathrm{kJ/h}}{3600\,\mathrm{kJ/h}}\)

Evaluate the COP

Using the values from Step 1, compute the COP: \(\mathrm{COP} = \frac{5040\,\mathrm{kJ/h}}{3600\,\mathrm{kJ/h}} = 1.4\) This means that the refrigerator's efficiency is such that it can remove 1.4 units of heat from the food compartment for every unit of power that it consumes.

Calculate the rate of heat rejection to the outside air

To calculate the rate of heat rejection, \(Q_H\), we use the energy conservation principle: \(Q_H = Q_L + W\). We obtained the values for \(Q_L\) and \(W\) in Step 1, so we can substitute those values into the formula and compute the rate of heat rejection: \(Q_H = Q_L + W = 5040\,\mathrm{kJ/h} + 3600\,\mathrm{kJ/h}\)

Evaluate the rate of heat rejection

Using the values from Step 3, compute the rate of heat rejection: \(Q_H = 5040\,\mathrm{kJ/h} + 3600\,\mathrm{kJ/h} = 8640\,\mathrm{kJ/h}\) This means that the refrigerator rejects 8640 kJ/h of heat to the outside air. In conclusion, the COP of the refrigerator is 1.4, and the rate of heat rejection to the outside air is 8640 kJ/h.