Suppose the dam is 80% efficient at converting the water’s potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 45.0 MWMW of electricity? This is a typical value for a small hydroelectric dam.


  1. Answer:

    \dot m = 163877.114\,\frac{kg}{s}


    Let assume that the water dam has a fall of 35 meters. The required instantaneous potential energy is:

    \dot E_{g} = \frac{\dot E_{el}}{\mu_{el}}

    \dot E_{g}= \frac{45\times 10^{6}\,W}{0.8}

    \dot E_{g} = 56.25\cdot 10^{6}\,W

    The instantaneous potential energy has the following model:

    \dot E_{g} = \dot m \cdot g \cdot \Delta h

    The mass flow rate needed to produced the required energy production:

    \dot m = \frac{56.25\times 10^{6}\,W}{(9.807\,\frac{m}{s^{2}} )\cdot (35\,m)}

    \dot m = 163877.114\,\frac{kg}{s}

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