How to use the hydroelectric power equation

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Discussion Overview

The discussion centers around the hydroelectric power equation and its application to a specific setup involving a reservoir and flow rate calculations. Participants explore the implications of flow rate on power generation, reservoir drainage time, and the physical dimensions required for the penstock tunnel.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks assistance in deriving the flow rate variable for a hydroelectric setup involving a reservoir of 8000 cubic meters and a head of 70 meters.
  • Another participant suggests looking up Bernoulli's equation for background understanding and refers to a specific page for sample calculations.
  • Some participants express concern that a flow rate of 50 cubic meters per second is very high and propose that achieving a reasonable flow velocity would require a penstock diameter of 6.5 meters.
  • There is a calculation presented that indicates the reservoir would empty in approximately 160 seconds if the flow rate is maintained at 50 cubic meters per second, assuming no inflow.
  • Clarifications are made regarding the dimensions of the reservoir and its height relative to the turbine.

Areas of Agreement / Disagreement

Participants generally agree on the calculation that the reservoir would empty in about 160 seconds at a flow rate of 50 cubic meters per second. However, there is a discussion about the feasibility of achieving such a high flow rate and the appropriate penstock diameter, indicating some uncertainty and differing views on practical implementation.

Contextual Notes

Participants have not resolved the implications of flow rate on the design of the hydroelectric system, and there are assumptions regarding the absence of inflow that may affect the calculations.

gloo
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I would like to get some help with the power equation that derives the power generated in Watts

P= density (water) * head (meters) * flow rate (cubic meter per second) * g * efficiency

I understand how to plug in everything except for the flow rate variable. I would like to get help in deriving 2 things for the following set up: If I have a reservoir that is 20m by 20m by 20m (8000 cubic meters of water) and it is 70 meters high.

1. How big would my pen stock tunnel diameter half to be achieve a 50 cubic meter per second drop? Or is that a function of the head size and the diameter?

2. If I achieved the 50 cubic meter per second...would the reservoir empty out in about 160 seconds? (8000 cubic meter / 50 seconds??)
 
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Is this something you are building yourself? If so, 50 cubic meters per second is a very high flow rate. To get a reasonable flow velocity (say 5 ft/sec or 1.5 m/sec) you'd want a pipe 6.5 meters diameter. Yes, it would drain an 8000 m3 reservoir in 160 seconds.

when you say 20 x 20 x 20 meters, 70 meters high, do you mean the 8000 m3 tank is on a 70 meter hill, above your turbine?
 
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gmax137 said:
Is this something you are building yourself? If so, 50 cubic meters per second is a very high flow rate. To get a reasonable flow velocity (say 5 ft/sec or 1.5 m/sec) you'd want a pipe 6.5 meters diameter. Yes, it would drain an 8000 m3 reservoir in 160 seconds.

when you say 20 x 20 x 20 meters, 70 meters high, do you mean the 8000 m3 tank is on a 70 meter hill, above your turbine?

Yes those assumptions are correct. I was just working out some theoretical costs of building a hydro electric dam and calculating payback periods etc.
 
gloo said:
2. If I achieved the 50 cubic meter per second...would the reservoir empty out in about 160 seconds? (8000 cubic meter / 50 seconds??)

Yes. If there was no inflow.

50 cubic meters per second is pretty fast. What's the flow rate into the reservoir?
 

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