Calculating KWH Generated by a Water Turbine?

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I'm attempting to calculate the amount of power (in KWH) produced by a water turbine based on x litres of water falling x metres vertically through a tube of x cm diameter

So far I've managed the following:
H = height (distance fallen) (m)
Time = Secs
R = Radius of the pipe (m)

Velocity (m/s) = (2 x H x G) ^ 0.5
Volume / Sec = Velocity x Time x Pi x R^2
mass (kg) / Sec = (Volume / sec) x 1000
Energy = (Mass x Velocity^2)/2 = wattage generated

My questions is... how do i convert this wattage generated in KWH generated?


Once I have this am i correct in thinking I can do:
(KWH generated / 3600) * (time taken for x litres of water to fall through the pipe).
To calculate the KWH generated by the x litres of water falling?

Any help / guidance anyone could give me would be greatly appreciated. - As you can (maybe) tell I'm by no means a physics graduate but i do have a basic understanding.

Any help anyone could give me would be greatly appreciated.

Many Thanks

P.s For the purposes of this I'm happy to work with a turbine having 100% efficiency.
 
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Power isn't measured in kWh; that's a unit of energy.

Energy (in Watt-hours) = power (in Watts) * time (in hours).

So, if your turbine produces 100kW (power) constantly for 3 hours (time), you will have produced 300kWh of energy.

The short way of finding out the amount of energy available in your example would just be to use:

Gravitational potential energy = (mass of water)*(vertical distance travelled)*g

Ok?
 
thats a good point, i never thought of working it out with PE, that would have made it much simpler.

Thanks for your comments they are extremely useful.

Did u happen to look at my wattage calculation? I'm fairly confident it's correct but would be useful if someone could confirm it.

Thanks again for your help.
 
Well your 'wattage calculation' is wrong, because (mass*velocity^2)/2 equals kinetic energy, not power.

Power is the rate of consumption/production of energy:

Power = energy / time


Also, your volumetric flow rate calculation is incorrect. Volumetric flow rate equals flow velocity multiplied by the cross-sectional area of the pipe. I don't know why you've multiplied this by time. Then, mass flow rate is just volume flow rate multiplied by density; in this instance it's water so 1 litre per second is one kilogram per second.
 
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Do a search for "hydro power calculator". You'll find many online and several will go into the details...frictional losses, head pressure, etc...some theoretical and some real-world.

There's loooooots of info out there online and sometimes the hardest part of finding it is the phrasing of the search term.

Toymkr