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If I had a 10 kW hydro electric generator (including turbine drive), what formula would I use to calculate how much water per second would need to be pumped into the turbine at 700 kpa to generate the 10 kW of electric power that the generator is rated to produce (assuming the turbine was 92% efficient & the motor was 90% efficient)?

This is not a homework problem. I'm just curious how this is calculated.

I noticed a power formula at the following url, however, given that "head" in the equation doesn't seem to make sense in a pump driven system, I'm wondering if anyone can help me further. It seems that pump pressure & flow rate should be in the calculation, however, when I leave out the "head" height, the answer seems high. Would a 1 m\$^3\$/s flow rate through a 92% efficient H\$_2\$O turbine produce 9,021.52 watts?

See Water turbine - power on Wikipedia.

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  • \$\begingroup\$ 1m\$^3\$ per what? Second, minute, hour? \$\endgroup\$
    – Transistor
    Commented May 1, 2016 at 9:38
  • \$\begingroup\$ "Head" gives you the pump pressure. It's not a distance it's a height. If it's a pump driven system. throw the waterworks away and drive the generator from whatever drives the pump... \$\endgroup\$
    – user16324
    Commented May 1, 2016 at 10:28
  • \$\begingroup\$ @transistor - above txt updated. \$\endgroup\$
    – DIYser
    Commented May 2, 2016 at 11:50
  • \$\begingroup\$ @Brian Drummond - "distance" changed to "height" above. \$\endgroup\$
    – DIYser
    Commented May 2, 2016 at 11:51

3 Answers 3

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The basic formula for power is just like P = V.I except it is: -

Power = pressure (pascals) x flow rate (cubic metres per second)

To calculate flow rate needed to deliver 10 kW, divide power by pressure i.e. flow rate is 10/700 = 0.0143 cubic metres per second.

I'll leave you to work out what it is given the motor and turbine efficiences.

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  • \$\begingroup\$ TY, Andy! . . . \$\endgroup\$
    – DIYser
    Commented May 2, 2016 at 4:52
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700 kPa is the pressure of a reservoir of water placed at an height of 70 metres over the turbine that is free flow at the output. You can place it in the formula as 70 metres of head. Power is generated by the loss of energy of water when it comes down from 70 metres or when loses 700 kPa when driven by a pump. The formula you quoted gives W but 10 kW is generally meant as 10 kWh to get quantity of water required in an hour.

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  • \$\begingroup\$ W/s and kw/h are meaningless. \$\endgroup\$
    – user16324
    Commented May 1, 2016 at 10:29
  • \$\begingroup\$ 1 J/S = 1 W/S. 1 W/S * 3600 S/h = 1 Wh. 1 * 1000 Wh = 1 kWh. 1 kWh * 3600 S/h = 3,600,000 W/S. 10 kW/S * 3600 S/h = 36,000,000 W/S. 36,000,000 W/S / 3,600,000 W/S (<--which equals 1 kWh) = 10 kWh. Therefore 10 kW/S of power = 10 kWh of energy after 1 hr of power generation. \$\endgroup\$
    – DIYser
    Commented May 5, 2016 at 3:51
  • \$\begingroup\$ @DIYser : No. 1 J/S = 1W not 1W/S. Therefore 10kW of power = 10kWh after an hour. \$\endgroup\$
    – user16324
    Commented May 5, 2016 at 9:24
  • \$\begingroup\$ @Brian Drummond - According to the dictionary, 1 Watt-second is "a unit of energy equal to the energy of one watt acting for one second; the equivalent of one joule." That seems to indicate a W/S = a J/S. Am I missing something? ref: dictionary.com/browse/watt-second?s=t This definition for Joule also seems to indicate the same thing: dictionary.com/browse/joule?s=t \$\endgroup\$
    – DIYser
    Commented May 6, 2016 at 0:40
  • \$\begingroup\$ Yes. 1 Watt-Second = 1 Joule. So 1 Joule/Second = 1 Watt-Second/Second = 1 Watt. Not 1 Watt/Second. Dimensional analysis is very basic and quite important. \$\endgroup\$
    – user16324
    Commented May 6, 2016 at 9:10
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The specific question

Would a 1 m3/s flow rate through a 92% efficient H2O turbine produce 9,021.52 watts?

is not answerable as it contains no term for pressure (or, equivalently, head). There will be one value of pressure for which the answer is yes.

1m^3 of water (1000 litres) weighs 1000kg or a ton.

So the question becomes, at what value of head (or pressure) does a ton of water , used at 92% efficiency, deliver 9kJ?

At 100% efficiency you would have 9/0.92 = 9.8kj. So from potential energy Ep=mgh we can see you haven't eliminated the "head" term from the linked equation, you have just set it to 1 metre.

And the pressure of a 1 metre high column of water is force/area, or mg/area = 1000kg * 9.81 / 1 m^2 = 9.8 kPa.

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  • \$\begingroup\$ Your answer checks out (9,021.52W / .92 eff / 9.8 kpa = ~1 m\$^3\$/s flow rate), however, why would you assume the value of h=1m and the area must be 1 m\$^2\$? (I would not have instinctively known to assume h=1m or use 1 m\$^2\$). \$\endgroup\$
    – DIYser
    Commented May 2, 2016 at 22:22
  • \$\begingroup\$ Pressure of a column of water is independent of area so you'd get the same answer for any area (because the weight of water would change proportionally. As for the height of the column, you set that yourself, I just discovered it from your question. \$\endgroup\$
    – user16324
    Commented May 2, 2016 at 22:46
  • \$\begingroup\$ That makes sense--1 m^3 is definitely 1m in height :). Thank you! \$\endgroup\$
    – DIYser
    Commented May 3, 2016 at 1:16
  • \$\begingroup\$ Not always, but 1 m^3 occupying 1m^2 area definitely is. If you distributed it over 10x10cm it would be 100x the height, so 100x the pressure, so (potentially) 100x the power at the same flow rate. \$\endgroup\$
    – user16324
    Commented May 3, 2016 at 10:03
  • \$\begingroup\$ How do you reconcile your last statement with your earlier statement "Pressure of a column of water is independent of area so you'd get the same answer for any area (because the weight of water would change proportionally." The statements seem to contradict each other. If the mass of water & the flow rate do not change--regardless of the column dimensions-then the power level would seem to remain the same, wouldn't it? \$\endgroup\$
    – DIYser
    Commented May 3, 2016 at 19:02

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