What waterpressure will turn waterwheel to produce 2kW of power with 1.6litres a sec?

  • #1

Main Question or Discussion Point

I require 2000W of Electricity from a waterwheel (which can be any size). I only have 1.6 litres-per-second MAX, but any pressure.

What pressure of water will I need to turn that waterwheel so it produces 2kW with 1.6 litres of water per second?

If possible, please take efficiency into account, specify the size of wheel you have chosen and explain how you’ve done it.

Thank you
 

Answers and Replies

  • #2
alxm
Science Advisor
1,842
9


If you can have any pressure it doesn't matter what kind of waterwheel you use. You can get any amount of energy out of it.
 
  • #3


I mean, there is no maximum. But obviously I would like to keep it as low as possible and still produce the 2kW with the 1.6 litres p/s.... otherwise I am wasting the excess unnecessary energy....
 
  • #4
alxm
Science Advisor
1,842
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Pressure = Energy / Volume = Power / Flow = 2000/1.6E-3 = 1.25 MPa pressure differential.
So there's your theoretical minimum. Assuming water is incompressible.
 
  • #5
russ_watters
Mentor
19,306
5,334


Those are odd constraints. You aren't confusing static fill pressure with the dP across the turine, are you?
 
  • #6


I don't believe so, no.

For russ_ watters : what do you calculate the minimum pressure required to be if I needed 2000W and only had 2.5litres per second of water?
 
  • #7
5,601
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1.6/2.5 the first answer...

efficiency is another matter and depends how you define it....simple way is to assume a steady state rotation at fixed revolutions....
the water wheel has frictional losses and any generator has electrical losses...
including heat....likely you can find typical generator or alternator losses online...how much a typical water wheel loses in efficiency might be difficult to find....Sounds like you are delivering power locally but if it's distributed over long distances I believe half the power might be lost in transmission lines....virtually none is lost in wiring if power is delivered all local and is not stepped up and down via transformers with their own losses...
 
  • #8
688
0


Pressure = Energy / Volume = Power / Flow = 2000/1.6E-3 = 1.25 MPa pressure differential.
So there's your theoretical minimum. Assuming water is incompressible.
This formular is ok providing the coeficient is 100 %. Good for first approximation.
 
  • #9
Q_Goest
Science Advisor
Homework Helper
Gold Member
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Hi matthew'. So tell us about the use for this thing. How are you getting the water pressure? Why 1.6 liters per second?

The first law of thermo reduces to Hin - Hout = work for an expander/turbine. Throw efficiency into the mix...

If I assume an outlet pressure of 0 psig and an isentropic efficiency of 80% (fairly nominal for typical turbines) then I get an inlet of 227 psig needed to produce the 2 kW. I had to do a few conversions to my spreadsheet and this is only a 2 minute analysis, so there's a chance I screwed something up but I think it's ok.
 

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