Known An ideal vapor-compression refrigeration cycle operates with Refrigerant 134a. The states of the refrigerant entering the compressor and leaving the condenser are specified, and the mass flow rate is given.
Find Determine the compressor power, in kW, the refrigeration capacity, in tons, coefficient of performance, and the coefficient of performance of a Carnot vapor refrigeration cycle operating between warm and cold regions at the specified temperatures.
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 accompanying sketch.
2. Except for the expansion through the 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 leaves the condenser.
Analysis Let us begin by fixing each of the principal states located on the accompanying schematic and T–s diagrams. At the inlet to the compressor, the refrigerant is a saturated vapor at 0°C, so from Table A-10, h_{1}=247.23 kJ / kg \text { and } s_{1}=0.9190 kJ / kg \cdot K.
The pressure at state 2s is the saturation pressure corresponding 1 to 26^{\circ} C \text {, or } p_{2}=6.853 bar. State 2s is fixed by p_{2} and the fact that the specific entropy is constant for the adiabatic, internally reversible compression process. The refrigerant at state 2s is a superheated vapor with h_{2 s }=264.7 kJ/kg. State 3 is saturated liquid at 26°C, so h_{3}=85.75 kJ/kg. The expansion through the valve is a throttling process (assumption 2), so h_{4}=h_{3}.
a. The compressor work 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 )(264.7-247.23) kJ / kg \left|\frac{1 kW }{1 kJ / s }\right|
= 1.4 kW
b. The refrigeration capacity is the heat transfer rate to the refrigerant passing through the evaporator. This is given by
\dot{Q}_{ in }=\dot{m}\left(h_{1}-h_{4}\right)
=(0.08 kg / s )|60 s / \min |(247.23-85.75) kJ / kg \left|\frac{1 \text { ton }}{211 kJ / \min }\right|
= 3.67 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{247.23-85.75}{264.7-247.23}=9.24
d. For a Carnot vapor refrigeration cycle operating at T_{ H }=299 K and T_{ C }=273 K, the coefficient of performance determined from Eq. 10.1 is
\beta_{\max }=\frac{T_{ C }}{T_{ H }-T_{ C }} (10.1)
2 \beta_{\max }=\frac{T_{ C }}{T_{ H }-T_{ C }}=10.5
1 The value for h_{2 s } can be obtained by double interpolation in Table A-12 or by using Interactive Thermodynamics: IT.
2 As expected, the ideal vapor-compression cycle has a lower coefficient of performance than a Carnot cycle operating between the temperatures of the warm and cold regions. The smaller value can be attributed to the effects of the external irreversibility associated with desuperheating the refrigerant in the condenser (Process 2s–a on the T–s diagram) and the internal irreversibility of the throttling process.
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 refrigeration capacity and coefficient of performance.
• compare with the corresponding Carnot refrigeration cycle.
Quick Quiz
Keeping all other given data the same, determine the mass flow rate of refrigerant, in kg/s, for a 10-ton refrigeration capacity. Ans. 0.218 kg/s.
