Question

A heat pump is used to maintain a house at a constant temperature of \(23^{\circ} \mathrm{C}\). The house is losing heat to the outside air through the walls and the windows at a rate of \(85,000 \mathrm{kJ} / \mathrm{h}\) while the energy generated within the house from people, lights, and appliances amounts to \(4000 \mathrm{kJ} / \mathrm{h}\). For a COP of \(3.2,\) determine the required power input to the heat pump.

Short answer

Answer: The required power input to the heat pump is 7.03125 kW.

Step-by-step solution

Find the net heat loss from the house

The house loses heat to the outside air and gains heat from the people, lights, and appliances. The net heat loss from the house can be calculated as: Net heat loss = Heat loss to outside air - Heat gained from sources = 85,000 kJ/h - 4,000 kJ/h = 81,000 kJ/h

Find the heat supplied by the heat pump

In order to maintain a constant temperature, the heat supplied by the heat pump must be equal to the net heat loss from the house. Therefore, the heat supplied by the heat pump is: Heat supplied by the heat pump = Net heat loss from the house = 81,000 kJ/h

Convert the rate of heat supplied to watts

Since we need to find the power input to the heat pump, we need to convert the rate of heat supplied from kJ/h to watts. To do this, use the following conversion: 1 kJ/h = (1/3600) kW So, the rate of heat supplied in kW is: Heat supplied by the heat pump (in kW) = 81,000 kJ/h * (1/3600) kW/(kJ/h) = 22.5 kW

Find the required power input using COP

The COP (Coefficient of Performance) of the heat pump is given as 3.2. The formula for COP is: COP = (Heat supplied by the heat pump) / (Power input to the heat pump) Now, solving for the power input to the heat pump, we get: Power input to the heat pump = (Heat supplied by the heat pump) / (COP) = 22.5 kW / 3.2 = 7.03125 kW So, the required power input to the heat pump is 7.03125 kW.