Question 5.7: A balloon with a 12 kg of refrigerant R134a in the saturated...
A balloon with a 12 kg of refrigerant R134a in the saturated vapor state at a pressure of 240 kPa, as shown in Figure 5.9, is expanded in a process, during which a total of 300 kJ of heat is transferred to the refrigerant. Take the source temperature of heat as 300 °C.
(a) Write the mass, energy, entropy, and exergy balance equations for this closed system and
(b) calculate the final state temperature, entropy generation, and exergy destruction. Take T_o = 25 °C.
a) Write the mass, energy, entropy, and exergy balance equations.
The change in kinetic energy of the system can be neglected since the system velocity does not change (not accelerating nor decelerating) throughout the process. The change in the potential energy of the system can be neglected as well since the system elevation does not change throughout the process.
MBE : m_{1}=m_{2}= constantEBE : Q_{i n}+m_{1} u_{1}=m_{2} u_{2}+W_{b}
\text { EnBE : } Q_{i n} / T_{s}+m_{1} S_{1}+S_{g e n}=m_{2} S_{2}
ExBE : E x_{Q_{i n}}+m_{1} e x_{1}=m_{2} e x_{2}+W_{b}+E x_{d}
The boundary work is done on a constant pressure process, then:
W_{b}=P \times\left(V_{2}-V_{1}\right)=P V_{2}-P V_{1}And by using the definition of the enthalpy we can rearrange the energy equation and substitute the enthalpy values, as follows:
h=u+PV
Q_{\text {in }}+m_{1} u_{1}+P V_{1}=m_{2} u_{2}+P V_{2}Q_{i n}=m_{2} u_{2}-m_{1} u_{1}+\left(P V_{2}-P V_{1}\right)
Q_{\text {in }}=\left(m_{2} u_{2}+P V_{2}\right)-\left(m_{1} u_{1}+P V_{1}\right)
Q_{\text {in }}=m_{2} h_{2}-m_{1} h_{1}
Q_{\text {in }}=m \times\left(h_{2}-h_{1}\right)
b) Calculate the final state temperature, entropy generation and exergy destruction. Given the properties of R134a at the initial state, which are a saturated vapor (x = 1) and a pressure of 240 kPa, then:
h_{1}=h_{g} @ 240 kPa =247.32 \frac{ kJ }{ kg }By substituting the values of terms in the energy balance equation:
Q_{i n}=m \times\left(h_{2}-h_{1}\right)300 kJ =12 kg \times\left(h_{2}-247.32 \frac{ kJ }{ kg }\right)
h_{2}=272.32 kJ / kg
As mentioned in the problem statement, the expansion of the balloon can be assumed at a constant pressure:
P_{1}=P_{2}=240 kPa\left\{\begin{array}{c} P_{2}=240 kPa \\ h_{2}=272.32 \frac{ kJ }{ kg } \end{array}\right\} T _{2}= 2 3 . 4 { }^{\circ} C
In order to find the entropy generated by the system, one can use the entropy balance equation along with the R134a properties from the property tables (such as Appendix B-3c) or EES. Take s_{1}=0.9346 \frac{ kJ }{ kgK } \text { and } s_{2}=1.023 \frac{ kJ }{ kgK } \text { accordingly. }
m_{1} S_{1}+Q_{i n} / T_{s}+S_{g e n}=m_{2} S_{2}
S_{g e n}=m_{2} S_{2}-m_{1} S_{1}-Q_{i n} / T_{s}
S_{\text {gen }}=12 kg \times(1.023-0.9346)\left(\frac{ kJ }{ kgK }\right)+\frac{300 kJ }{(300+273) K }
S_{g e n}=1.58 \frac{k J}{K}
One can then simply plug in the values in the exergy destruction relation as follows:
E x_{d}=S_{g e n} T_{0}E x_{d}=1.58 \frac{k J}{K} \times 298.15 K
E x_{d}=471.08 k J