Question 12.6: An advanced power plant with a CFB boiler is designed to ope......
An advanced power plant with a CFB boiler is designed to operate under the following conditions:
• Overall (electrical) efficiency of the power plant \eta_{el} is 0.42
• Fuel consumption rate of the plant is m_{f} = 20 kg/s
• Useful steam enthalpy drop in the steam turbine is \Delta h_{s}=1470\,\mathrm {kJ/kg}
• Fuel for CFB boiler is petcoke with LHV of 30 MJ/kg
• CFB boiler efficiency \eta_{b} is 0.92
• Electric generator efficiency \eta_{g} is 0.98
• Power plant capacity factor CF is 0.85
Calculate (i) the fuel energy input rate, (ii) the plant electric power output, (iii) the plant thermal efficiency, (iv) the power output of the ST, (v) the boiler steam rate, and (vi) the plant annual power generation.
1. Fuel energy input rate
Q_\mathrm{f}=m_\mathrm{f}\,\mathrm{LHV}=20\times30=600\,\mathrm{MJ}/ \mathrm{s}2. Plant electric power output
P_{\mathrm{el}}=\eta_{\mathrm{el}}\ Q_{\mathrm{f}}=0.42\times600=252\ \mathrm{MW}3. Thermal efficiency of the power plant cycle
\eta_{\mathrm{th}}=\eta_{\mathrm{el}}/(\eta_{\mathrm{g}}\;\eta_{\mathrm{b}})=0.42/(0.98\times0.92)=0.4664. Power output of the ST
P_{ST}=Q_{\mathrm{f}}\;\eta_{\mathrm{b}}\;\eta_{\mathrm{ th}}=600\times0.92\times0.466=277.1\ \mathrm{MW}5. Boiler steam rate
m_{s}=Q_{f}\;\eta_{b}/\Delta h_{s}=600,000\;\mathrm{kJ}/s\times0.92/1470\;\mathrm{kJ/kg} =375.5\;\mathrm{kg}/\mathrm{s}=1352~\mathrm{t/h}6. Plant annual power generation
E_{y r}=8760\ \mathrm{CF}\,P_{\mathrm{el}}=8760\times0.85\times252=1,876,392\ \mathrm{MWh/A}\approx187.64\ \mathrm{GW~h/A}Example 12.7 presents the performance calculation for a CFB cogeneration plant.