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Calculate the actual annual energy output of the power plant.

Physics

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  1. Calculate the actual annual energy output of the power plant.

    The total capacity is 915 MW915 \text{ MW}, and the load factor is 72.5%72.5\%. There are 8760 hours in a year (365 days×24 hours/day365 \text{ days} \times 24 \text{ hours/day}).

    First, calculate the actual average power output:

    Pactual=Pcapacity×Load FactorP_{actual} = P_{capacity} \times \text{Load Factor} Pactual=915 MW×0.725=663.375 MWP_{actual} = 915 \text{ MW} \times 0.725 = 663.375 \text{ MW}

    Next, calculate the annual energy output:

    Eout=Pactual×Hours in a yearE_{out} = P_{actual} \times \text{Hours in a year} Eout=663.375 MW×8760 hours/yearE_{out} = 663.375 \text{ MW} \times 8760 \text{ hours/year} Eout=5811390 MWh/yearE_{out} = 5811390 \text{ MWh/year}

    Convert MWh to MJ (since 1 MWh=3600 MJ1 \text{ MWh} = 3600 \text{ MJ}):

    Eout=5811390 MWh/year×3600 MJ/MWhE_{out} = 5811390 \text{ MWh/year} \times 3600 \text{ MJ/MWh} Eout=2.0921004×1010 MJ/yearE_{out} = 2.0921004 \times 10^{10} \text{ MJ/year}
  2. Calculate the annual heat input required from the coal.

    The plant efficiency (η\eta) is 40%40\%.

    η=EoutQin\eta = \frac{E_{out}}{Q_{in}} Qin=EoutηQ_{in} = \frac{E_{out}}{\eta} Qin=2.0921004×1010 MJ/year0.40Q_{in} = \frac{2.0921004 \times 10^{10} \text{ MJ/year}}{0.40} Qin=5.230251×1010 MJ/yearQ_{in} = 5.230251 \times 10^{10} \text{ MJ/year}
  3. Calculate

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