Question

A heat pump using refrigerant-134a heats a house by using underground water at \(8^{\circ} \mathrm{C}\) as the heat source. The house is losing heat at a rate of \(60,000 \mathrm{kJ} / \mathrm{h}\). The refrigerant enters the compressor at \(280 \mathrm{kPa}\) and \(0^{\circ} \mathrm{C}\), and it leaves at \(1 \mathrm{MPa}\) and \(60^{\circ} \mathrm{C}\). The refrigerant exits the condenser at \(30^{\circ} \mathrm{C}\). Determine \((a)\) the power input to the heat pump, (b) the rate of heat absorption from the water, and (c) the increase in electric power input if an electric resistance heater is used instead of a heat pump.

Short answer

Based on the given information and calculations, the power input to the heat pump is 5.7 kW, the rate of heat absorption from the water is 16.67 kJ/s, and the increase in electric power input if an electric heater is used instead of a heat pump is 10.97 kW.

Step-by-step solution

Find enthalpy values of refrigerant

First, we need to determine the enthalpy values for the refrigerant at different points in the cycle. Using refrigerant-134a property tables, we can find the enthalpy values for the given temperatures and pressures. From the table, the refrigerant has the following properties: - State 1 (Compressor inlet): \(h_1 = 238.02\,\text{kJ/kg}\) (given \(T_1 = 0^{\circ}\text{C}\) and \(P_1 = 280\,\text{kPa}\)) - State 2 (Compressor outlet): \(h_2 = 289.35\,\text{kJ/kg}\) (given \(T_2 = 60^{\circ}\text{C}\) and \(P_2 = 1\,\text{MPa}\)) - State 3 (Condenser outlet): \(h_3 = 88.18\,\text{kJ/kg}\) (given \(T_3 = 30^{\circ}\text{C}\) and \(P_3 = 1\,\text{MPa}\), assuming isobaric process)

Calculate the work done on the refrigerant and heat extracted from the water

The work done on the refrigerant by the compressor can be calculated by the difference in enthalpy between states 2 and 1: \(W_\text{in} = h_2 - h_1 = 289.35\,\text{kJ/kg} - 238.02\,\text{kJ/kg} = 51.33\,\text{kJ/kg}\) The heat extracted from the water by the heat pump can be calculated by the difference in enthalpy between states 1 and 3: \(Q_\text{out} = h_1 - h_3 = 238.02\,\text{kJ/kg} - 88.18\,\text{kJ/kg} = 149.84\,\text{kJ/kg}\)

Calculate the power input to the heat pump

To calculate the power input to the heat pump, we'll use the energy transfer equation: \(\text{COP}_\text{HP} = \frac{Q_\text{out}}{W_\text{in}}\) The heat pump's coefficient of performance (COP) is the ratio of heat output to the work input. Since we're given the rate of heat loss in the house, we can use it to find the mass flow rate of the refrigerant: \(Q_\text{loss} = 60,000\,\text{kJ/h} = 16.67\,\text{kJ/s}\) \(\dot{m} = \frac{Q_\text{loss}}{Q_\text{out}} = \frac{16.67\,\text{kJ/s}}{149.84\,\text{kJ/kg}} = 0.1111\,\text{kg/s}\) Now we can calculate the power input to the heat pump, \(P_\text{in}\): \(P_\text{in} = W_\text{in} \times \dot{m} = 51.33\,\text{kJ/kg} \times 0.1111\,\text{kg/s} = 5.70\,\text{kW}\) The power input to the heat pump is \(5.7\,\text{kW}\).

Calculate the increase in electric power input if an electric heater is used

If an electric resistance heater is used instead of a heat pump, the COP is 1, since all the input power is converted to heat: \(\text{COP}_\text{EH} = \frac{Q_\text{loss}}{P_\text{in}} = 1\) Therefore, the power input for an electric heater is equal to the rate of heat loss: \(P_\text{in}^\text{EH} = Q_\text{loss} = 16.67\,\text{kW}\) The increase in electric power input is: \(\Delta P_\text{in} = P_\text{in}^\text{EH} - P_\text{in} = 16.67\,\text{kW} - 5.70\,\text{kW} = 10.97\,\text{kW}\) The increase in electric power input if an electric heater is used instead of the heat pump is \(10.97\,\text{kW}\). In summary, (a) the power input to the heat pump is \(5.7\,\text{kW}\), (b) the rate of heat absorption from the water is \(16.67\,\text{kJ/s}\), and (c) the increase in electric power input if an electric heater is used instead of a heat pump is \(10.97\,\text{kW}\).