Known A modified vapor-compression refrigeration cycle operates with Refrigerant 134a as the working fluid. The evaporator temperature and condenser pressure are specified, and the mass flow rate is given.
Find Example 10.1Determine the compressor power, in kW, the refrigeration capacity, in tons, and the coefficient of performance. Compare results with those of .
Schematic and Given Data:
Engineering Model
1. Each component of the cycle is analyzed as a control volume at steady state. The control volumes are indicated by dashed lines on the sketch accompanying Example 10.1.
2. Except for the process through the expansion valve, which is a throttling process, all processes of the refrigerant are internally reversible.
3. The compressor and expansion valve operate adiabatically.
4. Kinetic and potential energy effects are negligible.
5. Saturated vapor enters the compressor, and saturated liquid exits the condenser.
Analysis Let us begin by fixing each of the principal states located on the accompanying T–s diagram. Starting at the inlet to the compressor, the refrigerant is a saturated vapor at −10°C, so from Table A-10, h_{1}=241.35 kJ / kg \text { and } s_{1}=0.9253 kJ / kg \cdot K.
The superheated vapor at state 2s is fixed by p_{2}=9 bar and the fact that the specific entropy is constant for the adiabatic, internally reversible compression process. Interpolating in Table A-12 gives h_{2 s }=272.39 kJ/kg.
State 3 is a saturated liquid at 9 bar, so h_{3}=99.56 kJ/kg. The expansion through the valve is a throttling process; thus, h_{4}=h_{3}.
a. The compressor power input is
\dot{W}_{ c }=\dot{m}\left(h_{2 s }-h_{1}\right)
where \dot{m} is the mass flow rate of refrigerant. Inserting values
\dot{W}_{ c }=(0.08 kg / s )(272.39-241.35) kJ / kg \left|\frac{1 kW }{1 kJ / s }\right|
= 2.48 kW
b. The refrigeration capacity is
\dot{Q}_{\text {in }}=\dot{m}\left(h_{1}-h_{4}\right)
=(0.08 kg / s )|60 s / \min |(241.35-99.56) kJ / kg \left|\frac{1 \text { ton }}{211 kJ / \min }\right|
= 3.23 ton
c. The coefficient of performance β is
\beta=\frac{\dot{Q}_{\text {in }}}{\dot{W}_{ c }}=\frac{h_{1}-h_{4}}{h_{2 s }-h_{1}}=\frac{241.35-99.56}{272.39-241.35}=4.57
Comparing the results of the present example with those of Example 10.1, we see that the power input required by the compressor is greater in the present case. Furthermore, the refrigeration capacity and coefficient of performance are smaller in this example than in Example 10.1. This illustrates the considerable influence on performance of irreversible heat transfer between the refrigerant and the cold and warm regions.
Skills Developed
Ability to…
• sketch the T–s diagram of the ideal vapor-compression refrigeration cycle.
• fix each of the principal states and retrieve necessary property data.
• calculate compressor power, refrigeration capacity, and coefficient of performance.
Quick Quiz
Determine the rate of heat transfer from the refrigerant passing through the condenser to the surroundings, in kW. Ans. 13.83 kW.
