What Water Pressure is Needed to Generate 2kW with 1.6 Liters per Second?

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Discussion Overview

The discussion revolves around determining the water pressure required to generate 2kW of electricity using a waterwheel with a maximum flow rate of 1.6 liters per second. Participants explore theoretical calculations, efficiency considerations, and the implications of different waterwheel designs.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant requests the pressure needed to produce 2kW with 1.6 liters per second, emphasizing the importance of efficiency and wheel size.
  • Another participant suggests that if any pressure can be used, the type of waterwheel becomes irrelevant for energy generation.
  • A different viewpoint expresses a desire to minimize pressure while still achieving the required power output to avoid wasting energy.
  • A calculation is presented, indicating that a theoretical minimum pressure differential of 1.25 MPa is necessary, assuming incompressibility of water.
  • Concerns are raised about confusing static fill pressure with the pressure differential across the turbine.
  • A participant inquires about the minimum pressure required if the flow rate were increased to 2.5 liters per second while maintaining the 2kW output.
  • Another participant discusses the complexities of efficiency, noting frictional losses in the waterwheel and electrical losses in generators, and highlights potential power loss in transmission lines.
  • Reiteration of the theoretical minimum pressure calculation is provided, with a note that this assumes 100% efficiency for a first approximation.
  • One participant introduces a thermodynamic perspective, suggesting that an inlet pressure of 227 psig would be needed with an isentropic efficiency of 80% to achieve the desired power output.

Areas of Agreement / Disagreement

Participants express differing views on the implications of pressure requirements, efficiency, and the assumptions involved in calculations. There is no consensus on the optimal approach or the exact pressure needed, as various factors and interpretations are considered.

Contextual Notes

Limitations include assumptions about water incompressibility, the definition of efficiency, and the potential for losses in different components of the system. The discussion does not resolve these complexities.

matthewgill
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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
 
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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.
 


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...
 


Pressure = Energy / Volume = Power / Flow = 2000/1.6E-3 = 1.25 MPa pressure differential.
So there's your theoretical minimum. Assuming water is incompressible.
 


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


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?
 


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...
 


alxm said:
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 formula is ok providing the coeficient is 100 %. Good for first approximation.
 


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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