Question 5.10: Consider an adiabatic steam turbine, as shown in Figure 5.12...
Consider an adiabatic steam turbine, as shown in Figure 5.12, which receives 10 kg/s of superheated steam at a temperature of 500 °C and a pressure of 10 000 kPa. The steam turbine converts part of the energy in the steam to the shaft work output. The steam exits the steam turbine as a saturated vapor at a pressure of 10 kPa and a mixture quality of 0.95. Take the dead-state temperature of steam to be saturated liquid water at 25 °C. (a) Write the mass, energy, entropy, and exergy balance equations, (b) find the turbine work rate, and (c) calculate the entropy generation rate and exergy destruction rate.
a) Write the mass, energy, entropy, and exergy balance equations.
MBE : \dot{m}_{1}=\dot{m}_{2}=\dot{m}EBE : \dot{m}_{1} h_{1}=\dot{m}_{2} h_{2}+\dot{W}_{o u t}
\text { EnBE : } \dot{m}_{1} s_{1}+\dot{S}_{\text {gen }}=\dot{m}_{2} s_{2}
ExBE : \dot{m}_{1} e x_{1}=\dot{m}_{2} e x_{2}+\dot{W}_{o u t}+\dot{E} x_{d}
b) Find the turbine work rate.
In order to find the work rate produced by the steam turbine we use the EBE to find the work production rate.
Using P_1 and T_1, h_1 is obtained as 3374 kJ/kg from the tables (such as Appendix B-1b) or EES software. In addition, h_2 is obtained from the following equation:
h_{2}=h_{f}+x h_{f g}=191.81 \frac{k J}{k g}+0.95 \times 2392.1 \frac{k J}{k g}=2464 \frac{k J}{k g}Using the EBE,wecalculate the specific power produced by the steamturbine as follows:
\dot{m}_{1} h_{1}=\dot{m}_{2} h_{2}+\dot{W}_{\text {out }}From the MBE, \dot{m}_{1}=\dot{m}_{2}=\dot{m} and by dividing by the mass flow rate the EBE becomes: h_{1}=h_{2}+w_{\text {out }} with the given data of temperatures and pressures, the enthalpy values are obtained and substituted to calculate the specific work:
3374=2464+w_{\text {out }}w_{\text {out }}=909.8 kJ / kg
and the turbine work rate becomes:
\dot{W}_{\text {out }}=\dot{m} \times w_{\text {out }}=10 \frac{ kg }{ s } \times 909.8 \frac{ kJ }{ kg }= 9 . 0 9 8 M Wc) Calculate the entropy generation rate and exergy destruction rate.
Using P_1 and T_1, s_1 is obtained as 6.5995 kJ/kgK from the tables (such as Appendix B-1b) or EES software.
s_{2}=s_{f}+x s_{f g}=0.6492 \frac{k J}{k g K}+0.95 \times 7.4996 \frac{k J}{k g K}=7.7738 \frac{k J}{k g K}Using the EnBE, one may extract the following:
\dot{S}_{\text {gen }}=\dot{m}\left(s_{2}-s_{1}\right)=10(7.7738-6.5995)= 1 1 . 7 4 3 k W / KThen, using the ExBE the exergy destroyed through the expansion process is calculated as follows:
\dot{m}_{1} e x_{1}=\dot{m}_{2} e x_{2}+\dot{W}_{o u t}+\dot{E} x_{d}From the MBE, \dot{m}_{1}=\dot{m}_{2}=\dot{m} and by dividing by the mass flow rate, the ExBE becomes:
e x_{1}=e x_{2}+w_{\text {out }}+e x_{d}e x_{d}=1412-151.1-909.8=350.7 kJ / kg
\dot{E} x_{d}=\dot{m} \times e x_{d}=10 \frac{ kg }{ s } \times 350.7 \frac{ kJ }{ kg }= 3 . 5 1 M W