Question 6.8: A fuel cell-based CHP plant produces power and hot water for......
A fuel cell-based CHP plant produces power and hot water for a small residential block. The plant operates as follows:
• Electric capacity of the CHP plant P_{el} = 5 kW
• CHP operating hours per day \tau = 9 h
• Heat-to-power ratio of the CHP plant HPR = 0.84
• Overall efficiency of the CHP plant EUF = 0.91
• Cold and hot water temperatures t_{cw} = 24°C and t_{hw} = 60°C
• Isobaric specific heat of the water c_{p} = 4.19 kJ/kgK
• Fuel cell uses natural gas with a heating value LHV of 49 MJ/kg
Calculate (i) the rate of useful heat production, (ii) the daily electricity and use-ful heat production, (iii) the amount of hot water produced per day, and (iv) the fuel consumption rate per day.
1. Rate of useful heat production per unit time
Q_{u}=\mathrm{H P R}\,P_{\mathrm{el}}=0.84\times5=4.2~\mathrm{kW}2. Daily electricity production
E_{\mathrm{el,}\mathrm{d}}=P_{\mathrm{el}}\ \tau=5\times9=45\mathrm{~kWh}3. Daily useful heat production
Q_{u,\mathrm{d}}=Q_{u}\;\tau=4.2\times9=37.8~\mathrm{kWh}4. Amount of hot water produced per day
m_{w}=Q_{u,d}/[{ c}_{ p}~(t_{hw}-t_{cw})]=37.8\times3600/[4.19\times(60-24)]=902.15~\mathrm{kg/day}5. Rate of fuel consumption per day
m_\mathrm{f}=(E_{\mathrm{el,d} }+Q_{\mathrm{u,d} })/(\mathrm{EUF}\times\mathrm{LHV})=(45+37.8)\times3600/(0.91\times49,000)=6.685\mathrm{~kg/day}