Question 5.15: In a vapor-compression refrigeration cycle, consider refrige...
In a vapor-compression refrigeration cycle, consider refrigerant R134a at a pressure of 0.8 MPa and in the saturated liquid state flowing through an expansion valve, as shown in Figure 5.17. The refrigerant is then throttled down to a pressure of 0.12 MPa. (a) Write the mass, energy, entropy, and exergy balance equations, (b) determine the temperature of the refrigerant leaving the throttle valve, and (c) calculate the specific entropy generation and specific exergy destruction during this process.
a) Write the mass, energy, entropy, and exergy balance equations.
MBE : \dot{m}_{1}=\dot{m}_{2}=\dot{m}EBE : \dot{m}_{1} h_{1}=\dot{m}_{2} h_{2} \rightarrow h_{1}=h_{2}
which makes the process isenthalpic as discussed earlier.
EnBE : \dot{m}_{1} S_{1}+\dot{S}_{g e n}=\dot{m}_{2} S_{2}ExBE : \dot{m}_{1} e x_{1}=\dot{m}_{2} e x_{2}+\dot{E} x_{d}
b) Calculate the temperature of the refrigerant leaving the throttling valve.
By using a properties software or properties tables (such as Appendix B-3b) the enthalpy at the inlet of the throttling valve is found as follows:
\left.\begin{array}{c}P_{1}=0.8 MPa \\ x_{1}=0 \text { (saturated liquid) }\end{array}\right\} h_{1}=95.5 kJ / kgThen by using the energy balance equation, h_2 is calculated as follows:
\dot{m}_{1} h_{1}=\dot{m}_{2} h_{2} \rightarrow \text { by using MBE } \rightarrow h_{1}=h_{2}=95.5 kJ / kg\left.\begin{array}{l}P_{2}=0.12 MPa \\ h_{2}=95.5 \frac{ kJ }{ kg }\end{array}\right\} T _{2}=- 2 2 . 3 2 { }^{\circ} C
c) Calculate the specific entropy generation and exergy destruction as follows:
\left.\begin{array}{c} P_{1}=0.8 MPa \\ x_{1}=0 \text { (saturated liquid) }\end{array}\right\} s_{1}=0.354 \frac{ kJ }{ kgK }\left.\begin{array}{rl} P_{2} & =0.12 MPa \\ h_{2} & =95.5 \frac{ kJ }{ kg } \end{array}\right\} s_{2}=0.3838 \frac{ kJ }{ kgK }
\dot{m}_{1} s_{1}+\dot{S}_{\text {gen }}=\dot{m}_{2} s_{2}
\dot{S}_{g e n}=\dot{m}_{2} S_{2}-\dot{m}_{1} S_{1}
s _{\text {gen }}=s_{2}-s_{1}=0.3838-0.354= 0 . 0 2 9 8 \frac{ k J }{ k g K }
e x _{ d }=T_{o} S_{\text {gen }}=298 K \times 0.0298 \frac{ kJ }{ kgK }= 8 . 8 8 \frac{ k J }{ k g }