Question

Refrigerant- 134 a is converted from a saturated liquid to a saturated vapor in a closed system using a reversible constant pressure process by transferring heat from a heat reservoir at \(6^{\circ} \mathrm{C}\). From second-law point of view, is it more effective to do this phase change at \(100 \mathrm{kPa}\) or \(180 \mathrm{kPa} ?\) Take \(T_{0}=25^{\circ} \mathrm{C}\) and \(P_{0}=100 \mathrm{kPa}\).

Short answer

Based on the second law of thermodynamics and the analysis of entropy changes, it is more effective to convert the refrigerant R134a from a saturated liquid to a saturated vapor at 180 kPa rather than at 100 kPa. This is because the entropy generation during the phase change process is smaller at 180 kPa, leading to a more effective heat transfer process.

Step-by-step solution

Find initial and final specific entropies at both pressures

First, we need to find the initial specific entropy for the saturated liquid state and the final specific entropy for the saturated vapor state for both 100 kPa and 180 kPa. We can get this information from the refrigerant-134a saturation properties table. For 100 kPa: \(s_{1,100\mathrm{kPa}} = 0.9349\,\mathrm{kJ/(kg\cdot K)}\) for the saturated liquid state \(s_{2,100\mathrm{kPa}} = 2.9319\,\mathrm{kJ/(kg\cdot K)}\) for the saturated vapor state For 180 kPa: \(s_{1,180\mathrm{kPa}} = 0.9734\,\mathrm{kJ/(kg\cdot K)}\) for the saturated liquid state \(s_{2,180\mathrm{kPa}} = 2.4373\,\mathrm{kJ/(kg\cdot K)}\) for the saturated vapor state

Calculate the entropy change of the R134a

Now, we need to calculate the entropy change of the R134a for both pressures during the phase change process: At 100 kPa: \(\Delta s_{100\mathrm{kPa}} = s_{2,100\mathrm{kPa}} - s_{1,100\mathrm{kPa}} = 2.9319\,\mathrm{kJ/(kg\cdot K)} - 0.9349\,\mathrm{kJ/(kg\cdot K)} = 1.997\,\mathrm{kJ/(kg\cdot K)}\) At 180 kPa: \(\Delta s_{180\mathrm{kPa}} = s_{2,180\mathrm{kPa}} - s_{1,180\mathrm{kPa}} = 2.4373\,\mathrm{kJ/(kg\cdot K)} - 0.9734\,\mathrm{kJ/(kg\cdot K)} = 1.4639\,\mathrm{kJ/(kg\cdot K)}\)

Calculate the entropy change of the surroundings

The entropy change of the surroundings is given by the following formula: \(\Delta S_{surr} = -\frac{Q}{T_0}\) Here, \(T_0\) is the temperature of the surroundings, which is given as \(25^{\circ}\mathrm{C} = 298\,\mathrm{K}\). To find the heat transfer, \(Q\), we can use the specific entropy change, \(\Delta s\), and the mass of the refrigerant, \(m\), as follows: \(Q = m\Delta s \cdot T_1\) We'll assume 1 kg of refrigerant and calculate the entropy change of the surroundings for both pressures: At 100 kPa: \(Q_{100\mathrm{kPa}} = 1\,\mathrm{kg} \cdot 1.997\,\mathrm{kJ/(kg\cdot K)} \cdot 298\,\mathrm{K} = 595.406\,\mathrm{kJ}\) \(\Delta S_{surr,100\mathrm{kPa}} = - \frac{595.406\,\mathrm{kJ}}{298\,\mathrm{K}} = -2.00\,\mathrm{kJ/K}\) At 180 kPa: \(Q_{180\mathrm{kPa}} = 1\,\mathrm{kg} \cdot 1.4639\,\mathrm{kJ/(kg\cdot K)} \cdot 298\,\mathrm{K} = 435.842\,\mathrm{kJ}\) \(\Delta S_{surr,180\mathrm{kPa}} = - \frac{435.842\,\mathrm{kJ}}{298\,\mathrm{K}} = -1.46\,\mathrm{kJ/K}\)

Compare both scenarios

Now, let's compare the entropy change of the surroundings for both scenarios. The smaller entropy generation means more effective heat transfer. The entropy change of the surroundings at 100 kPa is -2.00 kJ/K, and at 180 kPa, it is -1.46 kJ/K. Since the entropy generation is smaller at 180 kPa, it is more effective to perform the phase change at 180 kPa, from a second-law point of view.