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How much maximum electricity we can produce from a water channel with flow of 10 cfs and flow head of 100m ?

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How much maximum electricity we can produce from a water channel with flow of 10 cfs and flow head of 100m ?

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  1. You left out a term.  what is the size and shape of the channel.  We need that to calculate velocity head.


  2. E=energy

    P = power = E/t

    t = time

    m=mass of water

    V = volume of water

    rho = density

    h=height

    h= 100m * ft/0.3048m = 328ft

    Total energy E = mgh

    P = E/t = mgh/t

    P=(rho*V)gh/t

    P = (rho*g)*(V/t)*h

    P=62.4lb/ft3 * 10 ft3/s * 328 ft

    P = 205,000 ft*lb/s * HP/(550 ft*lb/s)

    P = 372HP * 0.746kW/HP = 277.7kW

    This assumes all of the energy is converted. For this, the channel must be a closed pipe of large diameter to make all the head available (or water going into a lake behind a dam). So, from this number, you have to subtract any losses. These include at a minimum, head loss from having an open channel, friction loss in the pipe/channel, turbine efficiency and generator conversion efficiency.

    Edit: The intermediate numbers reported are rounded, but the calculations are exact except the 999.552 density assumes pure fresh water at a temp near 20°C. Other poster is close, only off by a percent or so because he rounds everything. The losses in an actual system will dwarf this error.

  3. 1cfs = 0.028m³/s

    10cfs x 0.028 = 0.28m³/s. Density of water = 1,000kg/m³.

    Water head = 100m.

    P.E. = 0.28m³/s x 1,000kg/m³ x 9.81m/s² x 100m.

    = 274,680 J/s

    1J/s = 1 Watt.

    Assuming 100% efficiency 274,680J/s = 274.7kW. of Power.

    ('Power' is the rate of Transfer of Energy, measured in Watts or J/s (Note the 'Assuming Maximum Efficiency'...PE = KE).

    *...The thumbs down is totally unwarranted. This answer is more correct than any other.

    The asker says 'Maximum Electricity Produced'...i.e. No losses, i.e. 100% efficiency which is what I have given him.

    You're not very confident about your own answers are you ?.

    I've worked with Power Transfer most of my working life.

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