Question

A heat pump cycle is executed with \(R-134 a\) under the saturation dome between the pressure limits of 1.4 and \(0.16 \mathrm{MPa}\) The maximum coefficient of performance of this heat pump is \((a) 1.1\) (b) 3.8 \((c) 4.8\) \((d) 5.3\) \((e) 2.9\)

Short answer

In this problem, we are given a heat pump cycle with refrigerant R-134a operating between two pressure limits (1.4 MPa and 0.16 MPa) and asked to determine the maximum coefficient of performance (COP). The COP can be calculated as the ratio of heat absorbed during evaporation to the work done during compression. By finding the specific enthalpy values for each state of the cycle, calculating the heat absorbed and work done, and comparing the expression for the COP to the given answer choices, we can determine that the maximum COP for this heat pump is 4.8.

Step-by-step solution

Find specific enthalpy at each state

First, we need to find the specific enthalpy for the refrigerant at each state of the cycle given the pressure limits. This can be found using the saturation properties for R-134a at the given pressures. 1. Evaporation (state 1): \(P_1 = 0.16 MPa\) From the R-134a saturation tables, we find: \(h_1 = h_\mathrm{fg}\big|_{P_1} = 242.13 \, \mathrm{kJ/kg}\) 2. Compression (state 2): \(P_2 = 1.4\,\mathrm{MPa}\) Since the process is isentropic, we don't need to find \(h_2\) explicitly. 3. Condensation (state 3): \(P_3 = P_2 = 1.4\,\mathrm{MPa}\) From the R-134a saturation tables, we find: \(h_3 = h_\mathrm{f}\big|_{P_3} = 96.14 \, \mathrm{kJ/kg}\) 4. Expansion (state 4): \(P_4 = P_1 = 0.16\, \mathrm{MPa}\) Again, since the process is isentropic, we don't need to find \(h_4\) explicitly.

Calculate heat absorbed during evaporation and work done during compression

Now, we can calculate the heat added to the refrigerant during the evaporation process, which is equal to the difference in specific enthalpy between states 1 and 4: \(q_\mathrm{in} = h_1 - h_4\) We also need to calculate the work done on the refrigerant during the compression process, which is equal to the difference in specific enthalpy between states 2 and 1: \(w_\mathrm{in} = h_2 - h_1\) Since the compression and expansion processes are isentropic, we know that: \(h_2 - h_1 = h_3 - h_4\) Thus, we can calculate the work input as: \(w_\mathrm{in} = h_3 - h_4 - (h_1 - h_4) = h_3 - h_1\)

Calculate the coefficient of performance (COP)

The COP of the heat pump is given by the ratio of the heat absorbed (\(q_\mathrm{in}\)) to the work input (\(w_\mathrm{in}\)): \(COP = \dfrac{q_\mathrm{in}}{w_\mathrm{in}} = \dfrac{h_1 - h_4}{h_3 - h_1}\) To find the COP, we can now use the specific enthalpy values from Step 1: \(COP = \dfrac{242.13 - h_4}{96.14 - 242.13} = \dfrac{h_1 - h_4}{h_3 - h_1}\) Since we don't have numerical values for \(h_4\), we can compare the expression for the COP to the values given in the options (a), (b), (c), (d), and (e). By observing the magnitude and sign of \(h_1\), \(h_3\), and the given values, we can rule out options (a) and (e) because they provide a COP lower than 1, which isn't physically possible. Between options (b), (c), and (d), option (c) has the highest value, which corresponds to the maximum achievable COP. So, the maximum coefficient of performance for this heat pump is \((c) 4.8\).