Question 5.12: The hot exhaust gases leaving an internal combustion engine ...
The hot exhaust gases leaving an internal combustion engine at 400 °C, 150 kPa, and at a rate of 0.8 kg/s are used to produce saturated steam at 200 °C in an insulated heat exchanger, as shown in Figure 5.14. Water enters the heat exchanger at the ambient temperature of 20 °C and the exhaust gases leave the heat exchanger at 350 °C. (a) Write the mass, energy, entropy, and exergy balance equations, (b) determine the rate of steam production, and (c) calculate the entropy generation rate and exergy destruction rate in the heat exchanger.
a) Write the mass, energy, entropy, and exergy balance equations for the heat exchanger.
\text { MBE }: \dot{m}_{1}=\dot{m}_{2}=\dot{m}_{\text {gas }} \text { and } \dot{m}_{3}=\dot{m}_{4}=\dot{m}_{\text {water }}EBE : \dot{m}_{1} h_{1}+\dot{m}_{3} h_{3}=\dot{m}_{2} h_{2}+\dot{m}_{4} h_{4}
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}
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}
Here, we denote the inlet and exit states of exhaust gases by (1) and (2) and that of the water by (3) and (4).
b) The properties of water are obtained from the steam tables (such as Appendix B-1a) or EES database to be:
T_{3}=20{ }^{o} C, \text { liquid } \rightarrow h_{3}=83.91 \frac{ kJ }{ kg }, s_{3}=0.29649 \frac{ kJ }{ kg K}
T_{4}=200{ }^{\circ} C , \text { saturated vapor } \rightarrow h_{4}=2792.0 \frac{ kJ }{ kg }, s_{4}=6.4302 \frac{ kJ }{ kgK }
An energy balance on the heat exchanger gives the rate of steam production as follows:
\dot{m}_{1} h_{1}+\dot{m}_{3} h_{3}=\dot{m}_{2} h_{2}+\dot{m}_{4} h_{4}\dot{m}_{a} c_{p}\left(T_{1}-T_{2}\right)=\dot{m}_{w}\left(h_{4}-h_{3}\right)
0.8 \frac{ kg }{ s } \times 1.063 \frac{ kJ }{ kgK }(400-350)=\dot{m}_{w}(2792.0-83.91)
\dot{m}_{w}=0.01570 kg / s
c) The entropy generation rate can be calculated by using the EnBE:
\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}The entropy change of the system when rearranged is equal to \dot{S}_{\text {gen }} as follows:
\dot{S}_{\text {gen }}=\dot{m}_{2} S_{2}+\dot{m}_{4} S_{4}-\dot{m}_{1} S_{1}-\dot{m}_{3} S_{3}The question provided enough state point properties in order to obtain all the entropy values for the state points:
\dot{S}_{\text {gen }}= 0.01570 \frac{kg}{s}\left( 6.13 \frac{kJ}{kgK}\right) + 0.8 \frac{kg}{s}\left( -0.08206 \frac{kJ}{kgK}\right)\dot{S}_{\text {gen }}= 0.031 \ kW/K
The specific exergy changes of air and water streams as they flow in the heat exchanger are:
\Delta e x_{a}=c_{p}\left(T_{2}-T_{1}\right)-T_{o}\left(s_{2}-s_{1}\right)=1.063 kJ / kg K(-50 K )-293 K (-0.08206 kJ / kg K)
=-29.106 kJ / kg
\Delta e x_{w}=\left(h_{4}-h_{3}\right)-T_{o}\left(s_{4}-s_{3}\right)
=(2792.0-83.91) kJ / kg -293 K (6.4302-0.29649) kJ / kg K=910.913 kJ / kg
The exergy destruction rate is determined from the ExBE as given below:
\left(\dot{m}_{a} e x_{1}+\dot{m}_{w} e x_{3}\right)-\left(\dot{m}_{a} e x_{2}+\dot{m}_{w} e x_{4}\right)-\dot{E} x_{d}=0Rearranging and substituting, one obtains:
\dot{E} x_{d}=\dot{m}_{a} \Delta e x_{a}+\dot{m}_{w} \Delta e x_{w}=\left(0.8 \frac{ kg }{ s }\right) \times\left(-29.106 \frac{ kJ }{ kg }\right)+\left(0.01570 \frac{ kg }{ s }\right) \times\left(910.913 \frac{ kJ }{ kg }\right)\dot{E} x _{d}=8.98 k W