Question

Air enters the evaporator section of a window air conditioner at \(100 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\) with a volume flow rate of \(6 \mathrm{m}^{3} / \mathrm{min} .\) The refrigerant- \(134 \mathrm{a}\) at \(120 \mathrm{kPa}\) with a quality of 0.3 enters the evaporator at a rate of \(2 \mathrm{kg} / \mathrm{min}\) and leaves as saturated vapor at the same pressure. Determine the exit temperature of the air and the rate of entropy generation for this process, assuming ( \(a\) ) the outer surfaces of the air conditioner are insulated and ( \(b\) ) heat is transferred to the evaporator of the air conditioner from the surrounding medium at \(32^{\circ} \mathrm{C}\) at a rate of \(30 \mathrm{kJ} / \mathrm{min}\).

Short answer

Answer: The exit temperature of the air can be found using the formula \(T_{outlet} = T_{inlet} + \frac{\dot{Q}_{evap}}{\dot{m}_{air} \cdot C_{p_{air}}}\). After calculating the values, the exit temperature will be determined. 2) What is the rate of entropy generation for the evaporation process in case (a), where the outer surfaces of the air conditioner are insulated? Answer: The rate of entropy generation for case (a) is given by the formula \(\dot{S}_{gen(a)} = \dot{m}_{ref} \cdot (s_{out} - s_{in})\). After calculating the values, the rate of entropy generation for case (a) will be determined. 3) What is the rate of entropy generation for the evaporation process in case (b), where heat is transferred to the evaporator from the surrounding medium at 32°C at a rate of 30 kJ/min? Answer: The rate of entropy generation for case (b) is given by the formula \(\dot{S}_{gen(b)} = \dot{m}_{ref} \cdot (s_{out} - s_{in}) + \frac{\dot{Q}_{sur}}{T_{sur}}\). After calculating the values, the rate of entropy generation for case (b) will be determined.

Step-by-step solution

Calculate the heat transfer rate in the evaporator

We have the following information: - Volume flow rate of air, \(\dot{V}_{air} = 6 \thinspace m^{3}/min\) - Temperature of air entering the evaporator, \(T_{inlet} = 27 ^{\circ}C\) - Pressure of air entering the evaporator, \(P_{inlet} = 100 \thinspace kPa\) - Mass flow rate of refrigerant, \(\dot{m}_{ref} = 2 \thinspace kg/min\) - Refrigerant pressure in the evaporator, \(P_{evap} = 120 \thinspace kPa\) - Refrigerant quality in the evaporator, \(x_{evap} = 0.3\) From the refrigerant tables, find the specific enthalpies and entropies at the given conditions: \(h_{f_{in}}\), \(h_{fg_{in}}\), \(s_{f_{in}}\), and \(s_{fg_{in}}\) at \(P_{evap} = 120 \thinspace kPa\). Next, calculate the specific enthalpy and entropy of refrigerant at the inlet of the evaporator: \(h_{in} = h_{f_{in}} + x_{evap} \cdot h_{fg_{in}}\) \(s_{in} = s_{f_{in}} + x_{evap} \cdot s_{fg_{in}}\) Since the refrigerant leaves the evaporator as a saturated vapor, its specific enthalpies and entropies will be: \(h_{out} = h_{g_{out}}\) and \(s_{out} = s_{g_{out}}\) at \(P_{evap} = 120 \thinspace kPa\) Using the conservation of energy equation, we can find the heat transfer rate in the evaporator: \(\dot{Q}_{evap} = \dot{m}_{ref} \cdot (h_{out} - h_{in})\)

Find the exit temperature of air

Using the energy conservation equation for air (assume constant specific heat capacity \(C_{p_{air}}\)): \(\dot{Q}_{evap} = \dot{m}_{air} \cdot C_{p_{air}} \cdot (T_{outlet} - T_{inlet})\) The mass flow rate of air can be found using the ideal gas law: \(\dot{m}_{air} = \frac{\dot{V}_{air} \cdot P_{inlet}}{R_{air} \cdot T_{inlet}}\) Substitute the values and solve for \(T_{outlet}\): \(T_{outlet} = T_{inlet} + \frac{\dot{Q}_{evap}}{\dot{m}_{air} \cdot C_{p_{air}}}\)

Calculate the rate of entropy generation

For case (a), the process within the air conditioner is adiabatic, so the rate of entropy generation is given by: \(\dot{S}_{gen(a)} = \dot{m}_{ref} \cdot (s_{out} - s_{in})\) For case (b), there is heat transfer to the evaporator from the surrounding medium at a rate of \(30 \thinspace kJ/min\). To calculate the rate of entropy generation, we need to include the entropy change of the surrounding medium. \(\dot{S}_{gen(b)} = \dot{m}_{ref} \cdot (s_{out} - s_{in}) + \frac{\dot{Q}_{sur}}{T_{sur}}\) Where \(\dot{Q}_{sur} = 30 \thinspace kJ/min\) and \(T_{sur} = 32^{\circ}C = 305.15 \thinspace K\) Calculate the values of \(\dot{S}_{gen(a)}\) and \(\dot{S}_{gen(b)}\). The exit temperature of air and the rate of entropy generation under both given conditions have been determined.