Question 5.16: Propane enters a heat exchanger, as shown in Figure 5.18, at...
Propane enters a heat exchanger, as shown in Figure 5.18, at 4 °C at a mass flow rate of 32.6 kg/s to be heated with exhaust gases for later use at a constant pressure of 1 MPa. The exhaust gases flow at a rate of 2.484 × 10^5 kg/h and a temperature of 400 °C. Treat both propane and exhaust gases as ideal gases. (a) Write the mass, energy, entropy, and exergy balance equations, (b) calculate the exit temperatures of propane and exhaust air, (c) calculate the heat rate transferred to the propane, and (d) determine the entropy generation rate and exergy destruction rate.
a) Write mass, energy, entropy, and exergy balance equations.
\text { MBE }: \dot{m}_{1}=\dot{m}_{2}=\dot{m}_{p}, \dot{m}_{3}=\dot{m}_{4}=\dot{m}_{a}\text { EBE : } \dot{m}_{1} h_{1}+\dot{m}_{3} h_{3}=\dot{m}_{2} h_{2}+\dot{m}_{4} h_{4}
\text { EnBE : } \dot{m}_{1} s_{1}+\dot{m}_{3} s_{3}+\dot{S}_{\text {gen }}=\dot{m}_{2} s_{2}+\dot{m}_{4} s_{4}
\text { ExBE : } \dot{m}_{1} e x_{1}+\dot{m}_{3} e x_{3}=\dot{m}_{2} e x_{2}+\dot{m}_{4} e x_{4}+\dot{E} x_{d}
b) Calculate the exit temperatures of the propane and exhaust air.
From property tables or from property database software one can find the saturation temperature of propane.
T_{2}=T_{\text {sat } @ P=1 M P a}=26.4^{\circ} CAs for the exhaust gases one can find the temperature using the energy balance equations as follows:
\dot{m}_{2} h_{2}-\dot{m}_{1} h_{1}=\dot{m}_{3} h_{3}-\dot{m}_{4} h_{4}\dot{m}_{p} c_{p}\left(T_{2}-T_{1}\right)=\dot{m}_{a} c_{p}\left(T_{3}-T_{4}\right)
By using a properties software or properties tables (such as Appendix H-1) the specific heat of propane is found to be, c_{p}=2.632 \frac{ kJ }{ kgK }
32.6 \frac{ kg }{ s } \times 2.632 \frac{ kJ }{ kgK } \times(299.4-277) K =2.484 \times 10^{5} \frac{ kg }{ h } \times \frac{1 h }{3600 s } \times 1.005 \frac{ kJ }{ kgK }\left(400+273-T_{4}\right) K
T_{4}=372.24^{\circ} C
c) Calculate the heat rate transferred to the propane.
\dot{m}_{3} h_{3}=\dot{m}_{4} h_{4}+\dot{Q}_{\text {out }} \rightarrow \dot{m}_{3} h_{3}-\dot{m}_{4} h_{4}=\dot{Q}_{\text {out }} \rightarrow \dot{Q}_{\text {out }}=\dot{m}_{a} c_{p}\left(T_{3}-T_{4}\right)\dot{Q}_{\text {out }}=2.484 \times 10^{5} \frac{ kg }{ h } \times \frac{1 h }{60 min } \times \frac{1 min }{60 s } \times 1.005 \frac{ kJ }{ kgK }(400-372.24) K
\dot{Q}_{\text {out }}=1922 kW =1.92 MW
d) Calculate the entropy generation rate and exergy destruction rate.
\dot{m}_{1} S_{1}+\dot{m}_{3} S_{3}+\dot{S}_{g e n}=\dot{m}_{2} S_{2}+\dot{m}_{4} S_{4}By using a properties based software (such as EES) or properties tables,
S_{1}=S_{f @ T=10^{\circ} C }=0.151 kJ / kgK \text { and } S_{2}=S_{g @ 1 M P a}=6.586 kJ / kgK\dot{S}_{g e n}=\dot{m}_{a} \times\left(s_{4}-s_{3}\right)+\dot{m}_{p}\left(s_{2}-s_{1}\right)
\dot{S}_{\text {gen }}=\dot{m}_{a} \times\left(c_{p} \ln \left(\frac{T_{4}}{T_{3}}\right)-R \ln \left(\frac{P_{4}}{P_{3}}\right)\right)+\dot{m}_{p}\left(s_{2}-s_{1}\right)
\dot{S}_{\text {gen }}=2.484 \times 10^{5} \frac{ kg }{ h } \times \frac{1 h }{3600 s } \times\left(1.005 \frac{ kJ }{ kgK } \times \ln \left(\frac{372.24+273}{400+273}\right)\right)+32.6 \frac{ kg }{ s } \times\left(2.632 \frac{ kJ }{ kgK } \times \ln \left(\frac{26.4+273}{4+273}\right)\right) \frac{ kJ }{ kgK }
\dot{S}_{g e n}=3.751 kW / K
Furthermore, the exergy destruction rate for this heat exchanger is calculated as follows:
\dot{ E } x _{ d }=T_{o} \dot{S}_{\text {gen }}=298 K \times 3.751 kW / K = 1 1 1 7 .8 ~ k W