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15 March, 03:48

1000.0 gpm (gallons-per-minute) of water is released from the reservoir down the hill in order to generate electricity during the day (peak hours) for a 12-hour period each day. The drop of the water is 50.0 m and has negligible friction. Calculate how much energy (in kWh/day) can be generated on a daily basis, if the turbine is 85% efficient. Assume the density of the water is 1000.0 kg/m3.

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  1. 15 March, 05:08
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    E = 378.7 kWh / day

    Explanation:

    Formula to calculate the energy generated (E):

    E=P*t Formula (1)

    P = Power in kilowatts (kW)

    t = period the time to generate (h)

    The formula for calculating hydraulic power of the turbine is as follows:

    P = Q * p*η Formula (2)

    Where:

    P = Power in kilowatts (W)

    Q: Water flow (m³/s)

    p : Water pressure in Pascals (Pa)

    η : efficient of the turbine (%/100)

    The formula to calculate the Water pressure is:

    p = ρ*g*H Formula (3)

    p: pressure in Pa (N/m²)

    ρ = density of the fluid (kg / m³)

    g: acceleration due to gravity (m/s²)

    H : The drop of the fluid (m)

    Data

    Q = 1000 gpm

    1 gpm = 7,5768 * 10⁻⁵ m³/s

    Q = 1000 * 7,5768 * 10⁻⁵ m³/s

    Q = 7,5768 * 10⁻² m³/s

    H = 50 m The drop of the water

    η = 85% = 85/100 = 0.85

    ρ = 1000.0 kg/m³ : density of the water

    g = 9.8 m/s²

    t = 12-hour

    Calculation of the Water pressure

    We replace data in the formula (3)

    p = (1000 kg/m³) * (9.8 m/s²) (50 m)

    p = (1000 kg/m³) * (9.8 m/s²) (50 m)

    p = 490000 (kg*m/s²) / m² = 490000 (N/m²) = 490*10³ Pa

    Calculation of the Power

    We replace data in the the formula (2):

    P = Q * p*η

    P = (7,5768 * 10⁻² m³/s) * (490*10³N/m²) * (0.85)

    P = 31557.37 (N*m) / s = 31557.37 W

    1 W = 1 kW / 1000

    P = (31557.37 / 1000) kW

    P = 31.56 kW

    Calculation of the Energy

    We replace data in the the formula (1):

    E=P*t

    E=31.56 kW * 12 h

    E = 378.7 kWh / day
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