Air enters the evaporator section of a window air conditioner at 100 kPa and 27°C with a volume flow rate of 6 m3/min. The refrigerant-134a at 120 kPa with a quality of 0.3 enters the evaporator at a rate of 2 kg/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°C at a rate of 30 kJ/min.
FIGURE P7–180
a)
The exit temperature of air and the entropy generated during the process.
Answer to Problem 180RP
The exit temperature of air is
The entropy generated during the process is
Explanation of Solution
Write the expression to calculate the initial enthalpy of the refrigerant.
Here, initial enthalpy is
Write the expression to calculate the initial entropy of the refrigerant.
Here, initial entropy is
Write the expression to calculate the mass flow rate of air.
Here, mass flow rate of air is
Write the expression for the mass balance of the system.
Here, mass flow rate into the control system is
Write the expression for the energy balance equation for closed system.
Here, rate of energy transfer into the control volume is
Write the expression for the rate of entropy balance for the system.
Here, rate of entropy in the system is
Conclusion:
From Table A-12, “Saturated refrigerant-134a-Pressure table”, Obtain the following properties at saturated pressure of 120 kPa
Saturated liquid enthalpy,
Evaporated enthalpy,
Saturated vapor enthalpy,
Saturated vapor entropy,
Saturated liquid entropy,
Evaporated entropy,
Substitute
Substitute
Refrigerant –134a enters and leaves at the same pressure. Hence,
From Table A-1E, “the molar mass, gas constant and critical–point properties table”, select the gas constant of air at room temperature as
Substitute
Substitute
Here, mass flow rate of refrigerant is
Substitute
Here, mass flow rate of air is
Substitute
From the Table A-2, “Ideal-gas specific heats of various common gases”, select the value of the specific heat at constant pressure value of air as
Substitute 27 C for
Hence, the exit temperature of air is
For the steady flow system, change of entropy in the system is zero.
Substitute
Substitute
Hence, the entropy generated during the process is
b)
The exit temperature of air and the entropy generated during the process.
Answer to Problem 180RP
The exit temperature of air is
The entropy generated during the process is
Explanation of Solution
Write the expression for the energy balance equation for closed system.
Here, rate of energy transfer into the control volume is
Write the expression for the rate of entropy balance for the system.
Here, rate of entropy in the system is
Conclusion:
Substitute
Here, rate of heat gain from the surrounding is
Substitute 27 C for
Hence, the exit temperature of air is
For the steady flow system, change of entropy in the system is zero.
Substitute
Substitute
Hence, the entropy generated during the process is
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Chapter 7 Solutions
Thermodynamics: An Engineering Approach
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