A combined heat and power (CHP) plant fueled by a fuel oil with heating value of 40 MJ/kg generates 90 kW electricity and 180 kJ/s of useful heat. The overall efficiency (called energy utilization factor, EUF) of the CHP plant is 0.9.
The two separate plants produce power with an efficiency \eta_{el} of 0.4 and useful heat with an efficiency \eta_{h} of 0.85.
Calculate the fuel consumption rate of the CHP plant and the fuel savings compared to the separate power and heat production.
1. Fuel consumption rate of CHP plant (P_{el} = 90 kW, Q_{u} = 180 kJ/s)
m_{\mathrm{CHP}}=(P_{\mathrm{el}}+Q_{\mathrm{u}})/(\mathrm{EUF~{*}~H V})=(90+180)/(0.9\ {*}\ 42,000)=7.14\ {*}\ 10^{-3} \mathrm{~kg/s}2. Fuel consumption rate of two separate plants (\eta_{el} = 0.4 and \eta_{h} = 0.85)
m_{\mathrm{sep}}=(P_{\mathrm{el}}/\mathrm{h}_{\mathrm{el}}+Q_{\mathrm{u}}/\mathrm{h}_{\mathrm{h}})/\mathrm{HV}=(90/0.4+180/0.85)/42,000)=10.4\ {*}\ \,10^{-3}\,\mathrm{kg/s}3. Fuel savings of CHP plant
\Delta m_{\mathrm{sav}}=m_{\mathrm{sep}}-\,m_{\mathrm{CHP}}=(10.4-7.14)\ {*}\ 10^{-3}=3.26\ {\mathrm{*}}\ 10^{-3}\ {\mathrm{kg/s}}CHP requires 31.4% less fuel.