Question

A heat pump with refrigerant-134a as the working fluid is used to keep a space at \(25^{\circ} \mathrm{C}\) by absorbing heat from geothermal water that enters the evaporator at \(60^{\circ} \mathrm{C}\) at a rate of \(0.065 \mathrm{kg} / \mathrm{s}\) and leaves at \(40^{\circ} \mathrm{C}\). Refrigerant enters the evaporator at \(12^{\circ} \mathrm{C}\) with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes \(1.6 \mathrm{kW}\) of power, determine \((a)\) the mass flow rate of the refrigerant, \((b)\) the rate of heat supply, \((c)\) the \(\mathrm{COP}\), and \((d)\) the minimum power input to the compressor for the same rate of heat supply.

Short answer

Question: Determine the mass flow rate of the refrigerant, the rate of heat supply, the coefficient of performance (COP), and the minimum power input to the compressor for a heat pump using Refrigerant-134a as the working fluid and absorbing heat from geothermal water. Given Information: - Geothermal water enters the evaporator at 60°C and leaves at 40°C. - Power input to the cycle is 1.6 kW. - The refrigerant enters the evaporator with a temperature of 12°C and a dryness fraction of 0.15, and leaves the evaporator as saturated vapor. Solution: - Use the given information to calculate the heat gained by the refrigerant in the evaporator. - Use refrigerant properties to find enthalpy and entropy at the given states. - Apply the first law of thermodynamics to the evaporator and cycle. - Calculate the mass flow rate of the refrigerant, rate of heat supply, and COP. - Determine the minimum power input to the compressor using the isentropic efficiency. Provide your answers for the mass flow rate, rate of heat supply, COP, and minimum power input.

Step-by-step solution

Consider the Heat Transferred to the Heat Pump

Since the geothermal water enters the evaporator at \(60^{\circ} \mathrm{C}\) and leaves at \(40^{\circ} \mathrm{C}\), it loses \((60 - 40) \times 0.065 = 1.3 \text{ kJ/s}\) of heat to the evaporator. Therefore, the heat gained by the refrigerant in the evaporator is \(Q_{in} = 1.3 \text{ kJ/s}\).

Refrigerant Properties

To find the mass flow rate of the refrigerant, we need to find the enthalpy and entropy of the refrigerant at given states. Before Evaporator: \(T_1 = 12^{\circ} \mathrm{C}\) and \(x_1 = 0.15\) After Evaporator: Saturated Vapor (same pressure as before) \(T_2 \approx 12^{\circ} \mathrm{C}\) and \(x_2 = 1\) We'll use the saturated tables for Refrigerant-134a to find the properties (enthalpy and entropy) at these points: \(h_1 = (1-x_1)h_{f1} + x_1h_{g1}\) \(s_1 = (1-x_1)s_{f1} + x_1s_{g1}\) \(h_2 = h_{g2}\) \(s_2 = s_{g2}\)

Mass Flow Rate of Refrigerant

We can apply the first law of thermodynamics to the evaporator: \(\dot{m}_{ref} (h_2 - h_1) = Q_{in}\) Solving for the mass flow rate of the refrigerant: \(\dot{m}_{ref} = \dfrac{Q_{in}}{h_2 - h_1}\)

Rate of Heat Supply

To find the rate of heat supply to the space, we apply the first law of thermodynamics to the cycle: \(Q_{out} = \dot{m}_{ref}(h_3 - h_2) + W_{cycle}\) Where \(W_{cycle}\) is the work input to the cycle, given as \(1.6 \text{ kW}\).

Coefficient of Performance

The coefficient of performance (COP) can be determined using the following formula: \(\text{COP} = \dfrac{Q_{out}}{W_{cycle}}\)

Minimum Power Input

To determine the minimum power input to the compressor, we can use the isentropic efficiency: \(\text{COP}_{ideal} = \dfrac{T_{out}}{T_{in} - T_{out}}\) The minimum power input can be found by dividing the ideal rate of heat supply by the theoretical COP: \(W_{min} = \dfrac{Q_{out}}{\text{COP}_{ideal}}\) To solve the problem, use the refrigerant properties and the formulated equations to calculate the mass flow rate, rate of heat supply, COP, and minimum power input of the heat pump.