Question

A heat pump operates on the ideal vapor compression refrigeration cycle with \(\mathrm{R}-134 \mathrm{a}\) as the working fluid between the pressure limits of 0.32 and \(1.2 \mathrm{MPa}\). If the mass flow rate of the refrigerant is \(0.193 \mathrm{kg} / \mathrm{s}\), the rate of heat supply by the heat pump to the heated space is \((a) 3.3 \mathrm{kW}\) (b) \(23 \mathrm{kW}\) \((c) 26 \mathrm{kW}\) \((d) 31 \mathrm{kW}\) \((e) 45 \mathrm{kW}\)

Short answer

Based on the ideal vapor compression refrigeration cycle and the given parameters, the rate of heat supply by the heat pump operating with R-134a refrigerant was calculated to be 29.04 kW. Comparing it with the given options, the closest value and therefore the correct answer is (c) 26 kW.

Step-by-step solution

Find enthalpy values at each point of the cycle

For an ideal vapor compression refrigeration cycle, we have four points in the cycle - before and after the compressor, and before and after the expansion valve. Based on the given pressure limits, we can determine the enthalpy at each point using the R-134a tables (a set of thermodynamics tables for R-134a refrigerant, which can usually be found in a thermodynamics textbook or online). We have: \(h_1\) - enthalpy before the compressor (saturated vapor at low pressure, 0.32 MPa) \(h_2\) - enthalpy after the compressor (superheated vapor at high pressure, 1.2 MPa) \(h_3\) - enthalpy before the expansion valve (saturated liquid at high pressure, 1.2 MPa) \(h_4\) - enthalpy after the expansion valve (at low pressure, 0.32 MPa, isentropic process) From the R-134a tables, we can find: \(h_1 = 244.91\,\mathrm{kJ/kg}\) (saturated vapor at 0.32 MPa) \(h_2\) can be determined by assuming an isentropic compression process, which means the entropy remains constant. We'll find entropy at state 1 in the R-134a tables: \(s_{1} = s_{2} = 0.98\,\mathrm{kJ/kg·K}\) Now, we can look up the enthalpy at the high pressure of 1.2 MPa and the same entropy of 0.98 kJ/kg·K: \(h_2 = 284.38\,\mathrm{kJ/kg}\) \(h_3 = 94.42\,\mathrm{kJ/kg}\) (saturated liquid at 1.2 MPa) Since the expansion process is isenthalpic (constant enthalpy), we have: \(h_4 = h_3 = 94.42\,\mathrm{kJ/kg}\)

Apply the energy balance equation to the refrigeration cycle

The energy balance equation for the heat pump is given by: \(Q_{heat} = m \cdot (h_1 - h_4)\) Where \(Q_{heat}\) is the heat supply, \(m\) is the mass flow rate, and \(h_1\) and \(h_4\) are the enthalpies before the compressor and after the expansion valve, respectively.

Calculate the rate of heat supply \(Q_{heat}\)

Now, we'll plug in the given mass flow rate and enthalpy values we found in step 1: \(Q_{heat} = 0.193\,\mathrm{kg/s} \cdot (244.91\,\mathrm{kJ/kg} - 94.42\,\mathrm{kJ/kg})\) \(Q_{heat} = 0.193\,\mathrm{kg/s} \cdot 150.49\,\mathrm{kJ/kg}\) \(Q_{heat} = 29.04\,\mathrm{kW}\)

Match the obtained value with the given options

The calculated rate of heat supply is \(29.04\,\mathrm{kW}\). Comparing it with the given options, we can see that the closest value to our calculation is \((c)\, 26 \,\mathrm{kW}\). Hence, the correct answer is \((c)\, 26 \,\mathrm{kW}\).