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In a hydroelectric dam, water falls 26.0 m and then spins a turbine to generate electricity. What is ∆U of 1.0 kg of water? 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 54.0 MW of electricity? This is a typical value for a small hydroelectric dam.

  • Physics -

    potentialenergy= mgh=1kg*9.8N/kg*26m

    massflow*g*h/time * .80=54*10^6
    Flowrate=54^10^6/gh * 1/.80

  • Physics -

    If what you are calling ∆U includes the gravitational potential energy*, then it is -(density*g*(delta H)
    = -(10^3 kg/m^3)*26 m*9.81 m/s^2
    The product will be in Joules.

    For P = 54*10^6 Watts, you need a mass flow rate such that
    P = (mass flow rate)*(∆U)

    *Usually, in thermodynamics, U represents internal energy only, which has nothing to do with elevation.

  • Physics -

    I forgot to include the 80% efficiency factor, but BobPursley did it right

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