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,

Question

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.

Accepted Answer

a) Write the mass, energy, entropy, and exergy balance equations.

[latex]MBE : \dot{m}_{1}=\dot{m}_{2}=\dot{m}[/latex]

[latex]EBE : \dot{m}_{1} h_{1}=\dot{m}_{2} h_{2} ightarrow h_{1}=h_{2}[/latex]

which makes the process isenthalpic as discussed earlier.

[latex]EnBE : \dot{m}_{1} S_{1}+\dot{S}_{g e n}=\dot{m}_{2} S_{2}[/latex]

[latex]ExBE : \dot{m}_{1} e x_{1}=\dot{m}_{2} e x_{2}+\dot{E} x_{d}[/latex]

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:

[latex]\left.\begin{array}{c}P_{1}=0.8 MPa \ x_{1}=0 ext { (saturated liquid) }\end{array} ight\} h_{1}=95.5 kJ / kg[/latex]

Then by using the energy balance equation, [latex]h_2[/latex] is calculated as follows:

[latex]\dot{m}_{1} h_{1}=\dot{m}_{2} h_{2} ightarrow ext { by using MBE } ightarrow h_{1}=h_{2}=95.5 kJ / kg[/latex]

[latex]\left.\begin{array}{l}P_{2}=0.12 MPa \ h_{2}=95.5 \frac{ kJ }{ kg }\end{array} ight\} T _{2}=- 2 2 . 3 2 { }^{\circ} C[/latex]

c) Calculate the specific entropy generation and exergy destruction as follows:

[latex]\left.\begin{array}{c} P_{1}=0.8 MPa \ x_{1}=0 ext { (saturated liquid) }\end{array} ight\} s_{1}=0.354 \frac{ kJ }{ kgK }[/latex]

[latex]\left.\begin{array}{rl} P_{2} & =0.12 MPa \ h_{2} & =95.5 \frac{ kJ }{ kg } \end{array} ight\} s_{2}=0.3838 \frac{ kJ }{ kgK }[/latex]

[latex]\dot{m}_{1} s_{1}+\dot{S}_{ ext {gen }}=\dot{m}_{2} s_{2}[/latex]

[latex]\dot{S}_{g e n}=\dot{m}_{2} S_{2}-\dot{m}_{1} S_{1}[/latex]

[latex]s _{ ext {gen }}=s_{2}-s_{1}=0.3838-0.354= 0 . 0 2 9 8 \frac{ k J }{ k g K }[/latex]

[latex]e x _{ d }=T_{o} S_{ ext {gen }}=298 K imes 0.0298 \frac{ kJ }{ kgK }= 8 . 8 8 \frac{ k J }{ k g }[/latex]

Question 5.15: In a vapor-compression refrigeration cycle, consider refrige...

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