Known A vapor-compression refrigeration cycle has an isentropic compressor efficiency of 80%.
Find Determine the compressor power, in kW, the refrigeration capacity, in tons, the coefficient of performance, and the rates of exergy destruction within the compressor and expansion valve, in kW.
Schematic and Given Data:
Engineering Model
1. Each component of the cycle is analyzed as a control volume at steady state.
2. There are no pressure drops through the evaporator and condenser.
3. The compressor operates adiabatically with an isentropic efficiency of 80%. The expansion through the valve is a throttling process.
4. Kinetic and potential energy effects are negligible.
5. Saturated vapor at −10°C enters the compressor, and liquid at 30°C leaves the condenser.
6. The environment temperature for calculating exergy is T_{0}=299 K \left(26^{\circ} C \right).
Analysis Example 10.2Let us begin by fixing the principal states. State 1 is the same as in , so h_{1}=241.35 kJ / kg \text { and } s_{1}=0.9253 kJ / kg \cdot K.
Owing to the presence of irreversibilities during the adiabatic compression process, there is an increase in specific entropy from compressor inlet to exit. The state at the compressor exit, state 2, can be fixed using the isentropic compressor efficiency
\eta_{ c }=\frac{\left(\dot{W}_{ c } / \dot{m}\right)_{ s }}{\dot{W}_{ c } / \dot{m}}=\frac{\left(h_{2 s }-h_{1}\right)}{\left(h_{2}-h_{1}\right)}
where h_{2 s } is the specific enthalpy at state 2s, as indicated on the accompanying T–s diagram. From the solution to Example 10.2, h_{2 s }=272.39 kJ/kg. Solving for h_{2} and inserting known values
h_{2}=\frac{h_{2 s }-h_{1}}{\eta_{ c }}+h_{1}=\frac{(272.39-241.35)}{(0.80)}+241.35=280.15 kJ / kg
State 2 is fixed by the value of specific enthalpy h_{2} and the pressure, p_{2}=9 bar. Interpolating in Table A-12, the specific entropy is s_{2}=0.9497 kJ / kg \cdot K.
The state at the condenser exit, state 3, is in the liquid region. The specific enthalpy is approximated using Eq. 3.14, together with saturated liquid data at 30°C, as follows: h_{3} \approx h_{ f }=91.49 kJ/kg.
h(T, p) \approx h_{ f }(T) (3.14)
Similarly, with Eq. 6.5, s_{3} \approx s_{ f }=0.3396 kJ / kg \cdot K.
The expansion through the valve is a throttling process; thus, h_{4}=h_{3}. The quality and specific entropy at state 4 are, respectively,
x_{4}=\frac{h_{4}-h_{ f 4}}{h_{ g 4}-h_{ f 4}}=\frac{91.49-36.97}{204.39}=0.2667
and
\begin{aligned}s_{4} &=s_{ f 4}+x_{4}\left(s_{ g 4}-s_{ f 4}\right) \\&=0.1486+(0.2667)(0.9253-0.1486)=0.3557 kJ / kg \cdot K\end{aligned}
a. The compressor power is
\dot{W}_{ c }=\dot{m}\left(h_{2}-h_{1}\right)
=(0.08 kg / s )(280.15-241.35) kJ / kg \left|\frac{1 kW }{1 kJ / s }\right|=3.1 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 |(241.35-91.49) kJ / kg \left|\frac{1 ton }{211 kJ / \min }\right|
= 3.41 ton
c. The coefficient of performance is
1 \beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{(241.35-91.49)}{(280.15-241.35)}=3.86
d. The rates of exergy destruction in the compressor and expansion valve can be found by reducing the exergy rate balance or using the relationship \dot{E} _{ d }=T_{0} \dot{\sigma}_{ cv } \text {, where } \dot{\sigma}_{ cv } is the rate of entropy production from an entropy rate balance. With either approach, the rates of exergy destruction for the compressor and valve are, respectively,
\left(\dot{ E }_{ d }\right)_{ c }=\dot{m} T_{0}\left(s_{2}-s_{1}\right) \quad \text { and } \quad\left(\dot{ E }_{ d }\right)_{\text {valve }}=\dot{m} T_{0}\left(s_{4}-s_{3}\right)
Substituting values
2 \left(\dot{ E }_{ d }\right)_{ c }=\left(0.08 \frac{ kg }{ s }\right)(299 K )(0.9497-0.9253) \frac{ kJ }{ kg \cdot K }\left|\frac{1 kW }{1 kJ / s }\right|
= 0.58 kW
and
\left(\dot{ E }_{ d }\right)_{\text {valve }}=(0.08)(299)(0.3557-0.3396)=0.39 kW
1 While the refrigeration capacity is greater than in Example 10.2, irreversibilities in the compressor result in an increase in compressor power compared to isentropic compression. The overall effect is a lower coefficient of performance than in Example 10.2.
2 The exergy destruction rates calculated in part (d) measure the effect of irreversibilities as the refrigerant flows through the compressor and valve. The percentages of the power input (exergy input) to the compressor destroyed in the compressor and valve are 18.7 and 12.6%, respectively.
Skills Developed
Ability to…
• sketch the T–s diagram of the vapor-compression refrigeration cycle with irreversibilities in the compressor and subcooled liquid exiting the condenser.
• fix each of the principal states and retrieve necessary property data.
• calculate compressor power, refrigeration capacity, and coefficient of performance.
• calculate exergy destruction in the compressor and expansion valve.
Quick Quiz
What would be the coefficient of performance if the isentropic compressor efficiency were 100%? Ans. 4.83.
