Question 5.17: Consider a two connected steam turbines with the state prope...
Consider a two connected steam turbines with the state properties as shown in the Figure 5.19. Assume that the changes in kinetic and potential energies are negligible. (a) Write mass, energy, entropy, and exergy balance equations for each turbine separately, (b) calculate the work rates produced by both turbines, (c) find the temperature of the water leaving turbine 2, and (d) calculate the total entropy generation rate and total exergy destruction rate.
a) Write mass, energy and entropy balance equations for each turbine separately.
MBE : \dot{m}_{1}=\dot{m}_{2}=\dot{m}_{3}=\dot{m}_{w}\text { EBE (turbine 1) }: \dot{m}_{w} h_{1}=\dot{m}_{w} h_{2}+\dot{W}_{\text {out }, 1}+\dot{Q}_{\text {out }, 1}
\text { EBE (turbine 2) }: \dot{m}_{w} h_{2}=\dot{m}_{w} h_{3}+\dot{W}_{\text {out }, 2}+\dot{Q}_{\text {out }, 2}
\operatorname{EBE}(\text { overall }): \dot{m}_{w} h_{1}=\dot{m}_{w} h_{3}+\dot{W}_{\text {out }, t}+\dot{Q}_{\text {out }, 1}+\dot{Q}_{\text {out }, 2}
\text { EnBE (turbine 1): } \dot{m}_{w} S_{1}+\dot{S}_{\text {gen }, 1}=\dot{m}_{w} S_{2}+\dot{Q}_{\text {out }, 1} / T_{a v, 1}
\text { EnBE (turbine 2) : } \dot{m}_{w^{S}}+\dot{S}_{\text {gen, } 2}=\dot{m}_{w^{S} 3}+\dot{Q}_{\text {out }, 2} / T_{a v, 2}
\text { EnBE (overall) }: \dot{m}_{w} S_{1}+\dot{S}_{g e n}=\dot{m}_{w} S_{3}+\dot{Q}_{o u t, 1} / T_{a v, 1}+\dot{Q}_{o u t, 2} / T_{a v, 2}
\operatorname{ExBE}(\text { turbine } 1): \dot{m}_{w} e x_{1}=\dot{m}_{w} e x_{2}+\dot{W}_{\text {out }, 1}+\dot{E} x_{\dot{Q}_{\text {out }, 1}}+\dot{E} x_{d}
\text { ExBE (turbine 2): } \dot{m}_{w} e x_{2}=\dot{m}_{w} e x_{3}+\dot{W}_{\text {out }, 2}+\dot{E} x_{\dot{Q}_{\text {out }, 2}}+\dot{E} x_{d}
\operatorname{ExBE}(\text { overall }): \dot{m}_{w} e x_{1}=\dot{m}_{w} e x_{3}+\dot{W}_{\text {out }, t}+\dot{E} x_{\dot{Q}_{\text {ou }, 1}}+\dot{E} x_{\dot{Q}_{\text {out }, 2}}+\dot{E} x_{d}
where t stands for total and w for water.
b) Calculate the work rates produced by both turbines.
From a properties based software (such as EES) or properties tables (such as Appendix B-1c), one obtains:
\left.\begin{array}{c} P_{1}=8000 kPa \\ T_{1}=600{ }^{\circ} C \end{array}\right\} \begin{array}{c}h_{1}=3642.4 kJ / kg \\ s_{1}=7.0221 kJ / kg K \end{array}\left.\begin{array}{c} P_{2}=300 kPa \\ x_{2}=1 \end{array}\right\} h_{2}=2724.9 kJ / kg
\dot{m}_{w} h_{1}=\dot{m}_{w} h_{2}+\dot{W}_{\text {out }, 1}+\dot{Q}_{\text {out }, 1}
\left(13 \frac{ kg }{ s } \times 3642.4 \frac{ kJ }{ kg }\right)=\left(13 \frac{ kg }{ s } \times 2724.9 \frac{ kJ }{ kg }\right)+\left(800 \frac{ kJ }{ s }\right)+\dot{W}_{ out , 1}
\dot{W}_{\text {out }, 1}=11128 k W
h_{3}=h_{f}+x\left(h_{f g}\right)
h_{3}=191.81+0.85 \times(2392.1)=2225.095 kJ / kg
\dot{m}_{w} h_{2}=\dot{m}_{w} h_{3}+\dot{W}_{\text {out }, 2}+\dot{Q}_{\text {out }, 2}
\left(13 \frac{ kg }{ s } \times 2724.9 \frac{ kJ }{ kg }\right)=\left(13 \frac{ kg }{ s } \times 2225.095 \frac{ kJ }{ kg }\right)+\left(200 \frac{ kJ }{ s }\right)+\dot{W}_{ out , 2}
\dot{W}_{\text {out }, 2}=6297 kW
The total turbine work is calculated as follows:
\dot{W}_{\text {out }, t}=\dot{W}_{\text {out }, 1}+\dot{W}_{\text {out }, 2}=17425 k Wc) Find the temperature of the water leaving turbine 2.
By using a properties based software or properties tables (such as Appendix B-1b):
\left.\begin{array}{c} P_{3}=10 kPa \\ x_{3}=0.85 \end{array}\right\} T _{3}= 4 5 . 8 2 { }^{\circ} Cd) Calculate the total entropy generation rate and the total exergy destruction rate.
s_{3}=s_{f}+x\left(s_{f g}\right)s_{3}=0.6492+0.85 \times 8.1488=7.57568 kJ / kg
\dot{m}_{w} S_{1}+\dot{S}_{g e n}=\dot{m}_{w} S_{3}+\dot{Q}_{o u t, 1} / T_{a v g, 1}+\dot{Q}_{o u t, 2} / T_{a v g, 2}
\dot{S}_{g e n}=13 \frac{ kg }{ s } \times(7.57568-7.0221) \frac{ kJ }{ kgK }+\frac{800 kW }{(873+406.6) / 2}+\frac{200 kW }{(318.8+406.6) / 2}
\dot{S}_{\text {gen }}=8.998 k W / K
The total exergy destruction rate then becomes
\dot{E} x _{ d }=T_{o} \dot{S}_{\text {gen }}=298 K \times 8.998 kW / K=2681.4 k W