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18 July, 20:17

A hydraulic turbine-generator unit placed at the bottom of a 75-m-high dam accepts water at a rate of 1020 L/s and produces 630 kWof electricity. Determine A) the overall efficiency of the turbine-generator unit and B) the turbine efficiency if the generator efficiency is 96 percent, and C) the power losses due to inefficiencies in the turbine and generator.

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  1. 18 July, 20:26
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    A. Overall efficiency = 83.95%

    B. 87.45%

    C. 120.465kw

    Explanation:

    H=75m

    Flow rate = 120L/s=1.02m^2/s

    Po=Output power = 630kw

    As power input Pi = Total power available

    =gQH/1000

    Where

    g=9.81

    Q=120L/s=1.02m^2/s

    H=75m

    = (9.81m/s^2 * 1.02m^2/s*75m) / 1000

    750.465kw

    A. Overall efficiency of generator turbine unit (Po/Pi)

    = Output power / power input

    = 630/750.465 = 0.8395

    Overall efficiency of the unit = 0.8395 * 100

    =83.95%

    B. Turbine efficiency if generator efficiency is 96%

    no * nt = 0.96

    nT = 87.45%

    C. Power losses = Power input - Power output

    Pi - Po

    =750.465 - 630

    =120.465kw

    Power losses due to inefficiencies in turbine of the generator is 120.465kw
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