Question

Air enters the evaporator section of a window air conditioner at 14.7 psia and \(90^{\circ} \mathrm{F}\) with a volume flow rate of \(200 \mathrm{ft}^{3} / \mathrm{min}\). Refrigerant- \(134 \mathrm{a}\) at 20 psia with a quality of 30 percent enters the evaporator at a rate of \(4 \mathrm{lbm} / \mathrm{min}\) and leaves as saturated vapor at the same pressure. Determine (a) the exit temperature of the air and ( \(b\) ) the rate of heat transfer from the air.

Short answer

Answer: (a) The exit temperature of the air is 34.15°F. (b) The rate of heat transfer from the air is 191.76 Btu/min.

Step-by-step solution

Calculate the enthalpy of the refrigerant entering the evaporator

Since we know the refrigerant-134a enters the evaporator section at 20 psia and has a quality of 30 percent, we can use the tables for refrigerant-134a to find the specific enthalpy of the refrigerant entering the evaporator. The enthalpy at the inlet is given by the equation \(h_{in} = h_f + x_{in}(h_g - h_f)\), where \(h_f\) and \(h_g\) are enthalpy values for saturated liquid and vapor states, respectively, and \(x_{in} = 0.3\) (the quality). Using the tables for refrigerant-134a at 20 psia, we find: \(h_f = 47.23 \mathrm{Btu/lbm}\) \(h_g = 115.43 \mathrm{Btu/lbm}\) Now, calculate the inlet enthalpy: \(h_{in} = 47.23 \mathrm{Btu/lbm} + 0.3(115.43 \mathrm{Btu/lbm} - 47.23 \mathrm{Btu/lbm}) = 67.49 \mathrm{Btu/lbm}\)

Calculate the enthalpy of the refrigerant leaving the evaporator

Since the refrigerant leaves the evaporator as saturated vapor, we can find the exit enthalpy from the tables for refrigerant-134a at 20 psia. From the tables, \(h_{out} = h_g = 115.43 \mathrm{Btu/lbm}\)

Determine the rate of heat transfer in the evaporator

We can now determine the rate of heat transfer in the evaporator using the equation for the conservation of energy, which states that the heat transfer rate, \(Q_{evap}\), equals the product of the mass flow rate of the refrigerant, \(\dot{m}\), and the change in enthalpy of the refrigerant, \(\Delta h\): \(Q_{evap} = \dot{m}(\Delta h) = \dot{m}(h_{out} - h_{in})\) Using the given mass flow rate of the refrigerant, \(4 \mathrm{lbm/min}\), \(Q_{evap} = 4 \mathrm{lbm/min}(115.43 \mathrm{Btu/lbm} - 67.49 \mathrm{Btu/lbm}) = 191.76 \mathrm{Btu/min}\)

Calculate the specific volume of the air

We can use the ideal gas law to find the specific volume of the air entering the evaporator: \(v_{air} = \frac{R_{air}T_{air}}{P_{air}}\), where \(R_{air}\) is the specific gas constant for air, \(T_{air}\) is the temperature of the air, and \(P_{air}\) is the pressure of the air. Using the given temperature and pressure of the air, \(90^{\circ} \mathrm{F}\) and 14.7 psia, and the specific gas constant for air, \(R_{air} = 53.35 \mathrm{ft\cdot lbf/lbm\cdot R}\), we can calculate the specific volume of the air: \(v_{air} = \frac{53.35 \mathrm{ft\cdot lbf/lbm\cdot R}(90+459.67\,\mathrm{R})}{(14.7\,\mathrm{psia}) (144\,\mathrm{lb_f/ft^2\cdot psi})} = 13.51 \mathrm{ft^3/lbm}\)

Calculate the mass flow rate of the air

Now, we can calculate the mass flow rate of the air using the equation \(\dot{m}_{air} = \frac{\dot{V}_{air}}{v_{air}}\), where \(\dot{V}_{air}\) is the volumetric flow rate of the air. Using the given volumetric flow rate, \(200 \mathrm{ft^3/min}\), \(\dot{m}_{air} = \frac{200 \mathrm{ft^3/min}}{13.51 \mathrm{ft^3/lbm}} = 14.81 \mathrm{lbm/min}\)

Calculate the exit temperature of the air

We can find the exit temperature of the air from the air-specific heat capacity equation \(Q_{evap} = \dot{m}_{air}C_{p_{air}}(T_{out,air} - T_{in,air})\), where \(C_{p_{air}}\) is the specific heat capacity of air at constant pressure. Using the specific heat capacity for air, \(C_{p_{air}} = 0.24 \mathrm{Btu/lbm\cdot R}\), \(T_{out,air} = \frac{Q_{evap}}{\dot{m}_{air}C_{p_{air}}} + T_{in,air} = \frac{191.76 \mathrm{Btu/min}}{14.81 \mathrm{lbm/min} \times 0.24 \mathrm{Btu/lbm\cdot R}} + 90^{\circ} \mathrm{F} = 34.15^{\circ} \mathrm{F}\) (a) The exit temperature of the air is \(34.15^{\circ} \mathrm{F}\). (b) The rate of heat transfer from the air is \(191.76\) Btu/min.