Author Topic: A hydroelectric system getting built  (Read 186 times)

Offline plasma

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A hydroelectric system getting built
« on: January 11, 2020, 11:26:31 PM »
Hi just thought I'll post a hydroelectric system I'm building, its taking awhile, but the relative I'm building it for isn't in a rush.
Below is the blades with the skirt yet to be finished.

Offline klugesmith

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Re: A hydroelectric system getting built
« Reply #1 on: January 12, 2020, 01:53:35 AM »
Nice.
Can you tell us the intended flow rate and pressure head?

Offline plasma

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Re: A hydroelectric system getting built
« Reply #2 on: January 12, 2020, 02:03:03 AM »
Its a river, using 1meter head, the flow rate is guessmaite of 0.1m/sec ,but that could be miles off ;), its about 1-2m deep.
1000kg/m3 * 0.1m3/sec * 1m = 100watt

The https://www.powerspout.com/collections/complete-smart-drive-pmas/products/complete-42pole-pma-1-2kw Genny.
« Last Edit: January 12, 2020, 02:09:10 AM by plasma »

Offline klugesmith

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Re: A hydroelectric system getting built
« Reply #3 on: January 12, 2020, 04:02:42 AM »
I can't make sense of your arithmetic, because kilogram meter per second is dimensionally different from power.  Acceleration of gravity (g) is missing.  But the bottom line seems not too crazy.

Power = pressure x flow.
Water head of 1 meter is the same as pressure of about 10000 N/m^2 = 10 kPa.
Flow for 100 watts before losses: 0.01 m^3/s = 10 liters/s.

That's the flow in a cross-sectional area of 0.1 square meter at velocity of 0.1 m/s.

But velocity of free flowing stream isn't relevant, is it?  Will your weldment go at the end of a penstock, or flume and penstock, connecting places where the river surface elevation is different by a meter?
« Last Edit: January 12, 2020, 04:16:49 AM by klugesmith »

Offline plasma

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Re: A hydroelectric system getting built
« Reply #4 on: January 12, 2020, 05:01:29 AM »
Its a free flowing stream, I justed used 1m but the gradient on there property might be that over the whole thing.
The foumla used was power = density * flow * gravity * head ,as there's no head I removed gravity and head??

Offline Uspring

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Re: A hydroelectric system getting built
« Reply #5 on: January 12, 2020, 01:35:20 PM »
Quote
The foumla used was power = density * flow * gravity * head ,as there's no head I removed gravity and head??

The formula is fine. In your calculation

Quote
1000kg/m3 * 0.1m3/sec * 1m = 100watt

you left out "gravity" or more precisely the gravitational acceleration, which is about 10m/s². So you get:

power = 1000kg/m³ * 0.1m³/s * 10m/s² * 1m = 1000 kg m²/s³ = 1000 W.


Offline klugesmith

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Re: A hydroelectric system getting built
« Reply #6 on: January 12, 2020, 06:23:30 PM »
Yup. But Plasma gave numbers without stating the problem, so we have to guess.

The thousand-watt answer is for a flow of 0.1m³/s dropping vertically by 1 meter. 
How do either of those factors relate to the river and the pictured hydroelectric part?

It will be nice to see more progress on the project.  Congratulations for doing things instead of talking about other people's things.

I have a buddy whose electric bill is offset by power from a little creek in hilly terrain.  A penstock made from plastic pipe brings water from far upstream, where the creek level is (IIRC) more than 20 meters higher.  Enough that a Pelton wheel is the appropriate kind of turbine; some are listed in Plasma's product link.

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Re: A hydroelectric system getting built
« Reply #6 on: January 12, 2020, 06:23:30 PM »

 


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