Question 6.2: A back-pressure turbine cogeneration plant (refer to Figures......

1. From the steam Mollier h–s diagram (Appendix 11)
Live steam enthalpy at p_{1} = 180 bars and t_{1} = 550°C is h_{1} = 3420 kJ/kg
Exhaust steam enthalpy at isentropic expansion from p_{1} to p_{2} is h_{2s} = 2430 kJ/kg
2. Actual exhaust steam enthalpy at irreversible expansion

h_{2}\,=\,h_{1}-{\bf h}_{it}(h_{1}-h_{2s})\,=\,3420-0.9(3420-2430)\,=\,2529 kJ/kg

3. Boiler feedwater enthalpy at p_{2} = 2 bar: h_\mathrm{fw} = 504.7 kJ/kg (from water/steam table, Appendix 7)
4. Turbine output power

P_{t}\,=\,m(h_{1}-h_{2})\,=\,280\times(3420-2529)\,=\,249,480\,\mathrm{kW}

5. Rate of heat addition per unit time

Q_{\mathrm{in}}\,=\,m(h_{1}-h_{\mathrm{fw}})=\,280\times(3420-504.7)=816,284\,{\mathrm{kJ}}/{\mathrm{s}}

6. Plant thermal efficiency

\mathbf{h}_{\mathrm{th}}\,=\,P_{\mathrm{t}}/Q_{\mathrm{in}}\,=\,249,480/816,284\,=\,0.306

7. Plant electrical efficiency

\mathbf{h}_{\mathrm{el}}\,=\,\mathbf{h}_{\mathrm{th}}\mathbf{h}_{\mathrm{g}}\,=\,0.306\times0.98\,=\,0.3

8. Plant electric power output

P_{\mathrm{el}}\,=\,P_{t}\ \mathbf{h}_{\mathrm{g}}\,=\,249,480\times0.98\,=\,244,490\,\mathrm{kW}

9. Plant useful heat output rate

Q_{u}=m(h_{2}-h_\mathrm{fw})(1-q_\mathrm{loss})=280(2529-504.7)(1-0.15)

= 470,447 kJ/s

10. Heat-to-power ratio

Q_{u}/P_{\mathrm{el}}\,=\,470,447/244,490\,=\,1.92

11. Plant energy utilization factor

\mathrm{EUF}=(P_{\mathrm{el}}+Q_{\mathrm{u}})/Q_{\mathrm{in}}=(244,490+470,447)/816,284=0.876