Question

Consider a heat pump that operates on the reversed Carnot cycle with \(\mathrm{R}-134 \mathrm{a}\) as the working fluid executed under the saturation dome between the pressure limits of 140 and 800 kPa. R-134a changes from saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycle is (a) \(28 \mathrm{kJ} / \mathrm{kg}\) (b) \(34 \mathrm{kJ} / \mathrm{kg}\) \((c) 49 \mathrm{kJ} / \mathrm{kg}\) \((d) 144 \mathrm{kJ} / \mathrm{kg}\) \((e) 275 \mathrm{kJ} / \mathrm{kg}\)

Short answer

Answer: The net work input for the cycle is approximately \(34 \, \frac{\text{kJ}}{\text{kg}}\).

Step-by-step solution

Find the temperatures at the pressure limits.

Let's denote the lower pressure as \(P_1\) and the higher pressure as \(P_2\). To find the corresponding temperatures, we need the saturation temperature corresponding to the given pressures. You can use the R-134a properties table, or you can use an online calculator or software (such as REFPROP) to find the saturation temperatures. In this case: \(P_1 = 140 \, \text{kPa} \Rightarrow T_1 = -15.6 \, ^{\circ}\text{C}\) (saturated vapor) \(P_2 = 800 \, \text{kPa} \Rightarrow T_2 = 39.4 \, ^{\circ}\text{C}\) (saturated liquid)

Calculate the Carnot heat pump coefficient of performance (COP).

Since we are dealing with a reversed Carnot cycle, the coefficient of performance of the heat pump is given by: \(\text{COP}_{\text{HP}} = \frac{T_2}{T_2 - T_1}\) First, convert temperatures to Kelvin: \(T_1 = -15.6 \, ^{\circ}\text{C} + 273.15 = 257.55 \, \text{K}\) \(T_2 = 39.4 \, ^{\circ}\text{C} + 273.15 = 312.55 \, \text{K}\) Now calculate the COP: \(\text{COP}_{\text{HP}} = \frac{312.55}{312.55 - 257.55} = \frac{312.55}{55} = 5.683\)

Calculate the heat rejected in the cycle.

The heat rejected in the cycle can be calculated using the enthalpy difference between the saturated vapor and saturated liquid states at pressure \(P_2\), i.e., \(Q_{\text{out}} = h_4 - h_3\). \(h_3\) and \(h_4\) can be found from the R-134a properties table as \(h_3 = h_{\text{g}} \text{ at } P_2 = 293.34 \, \frac{\text{kJ}}{\text{kg}}\) and \(h_4 = h_{\text{f}} \text{ at } P_2 = 85.53 \, \frac{\text{kJ}}{\text{kg}}\). So, \(Q_{\text{out}} = 293.34 - 85.53 = 207.81 \, \frac{\text{kJ}}{\text{kg}}\)

Calculate the work input for the cycle.

Since we now have the COP and \(Q_{\text{out}}\), we can calculate the work input per unit mass as follows: \(W_{\text{in}} = \frac{Q_{\text{out}}}{\text{COP}_{\text{HP}}} = \frac{207.81}{5.683} = 36.56 \, \frac{\text{kJ}}{\text{kg}}\) Comparing the calculated value with the options provided, we find that it is closest to option (b): Net work input for the cycle is approximately \(34 \, \frac{\text{kJ}}{\text{kg}}\).