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Pumping water downhill to increase psi on micro hydro 2

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watermanjl

Agricultural
Aug 16, 2012
3
I am trying to come up with a solution to increase pressure and power on my micro hydro turbine. I am siphoning from a pond using 6" pvc. The penstock runs about 300 feet with 80 to 90 feet of head. I currently have 40 psi at the bottom generating about 1000 watts(I think). Is it possible to use a pump to add pressure to the line given the power consumed by the pump is low or solar? Can I regulate how much pressure the pump puts out?
 
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Firstly you need to establish what the flow requirement is for the turbine at various pressures, once this is known you can then look at what you might need in terms of a pump to increase the flow and head at the turbine.
Yes you can regulate the pump output - but at the end of the day, installing a pump, motor, cables, flow regulation etc may not be economical. A second turbine may be a better option.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
 
may not be economical? Its aganist the laws of thermodynamics! You WILL use more power in you pump than you will generate. So unless its a pump that runs on a powersource thats free (and cant be utilised at the point of delivery for your microturbine) it will not be a good business. IMO if you can rig up solar power at the pump - you can also do it at the turbine...

One exception could be if you really really NEED more power at your turbine - and cant get it any other way (but it will have a negative payback).

Im beginning to wonder - is this trolling (engineer style)?

Best regards

Morten
 
Forget that idea! IT IS EXTREMELY BAD.

Any extra power applied in that manner, other than direct hydraulic power (obtained by getting either more water head or water flow on/to the turbine), comes at a very, very, very large cost.

If you add say a 500 W pump, due to pump efficiency of, say 0.54 for a low power pump, the pump will only give an output of 270 Watts.

That 270 Watts New output power will be applied to the turbine. Due to turbine efficiency, say 0.70, only 270*.70 = 189 Watts will be recovered as output from the turbine.

True you got 189 New Watts output, but it cost you 500 Watts input.
The Efficiency of that transaction was 189/500 = 37.8 %
The efficiencies combine by the multiple of each power source.
0.700 * 0.540 = 0.378
It would be much better just doing this by adding more water flow (or more head) directly to the turbine, without the pump. Adding the same 500 W in direct hydropower on the turbine would get you 500 * 0.7 = 350 W instead of only the 189.

To make that attractive, you would have to buy and install the pump and its power source at a cost cheaper than adding the direct hydropower in either flow, or head, or a combination of both. And you would still have to pay for running that pump power source generator, a gasoline generator, or a solar generator, or grid power.

Let's say you go solar for that. You need 500 Watts of power supplied to the pump. For normal Si solar panels, efficiency say 17%, you have to install 500/0.17 = 2941 watts, say 3000 watts. Now with 3000 installed and only the same 189 W output from the pump, that's 189/3000 = 0.063 = 6.3 % efficiency on that little addition. And that probably cost you 3000 x $4/watt installed cost = 12,000. That's a bunch of money for only 189 watts total output.

If you just used the 500 Watts coming off the solar panels, you would get 500 watts, not just the measely 189 watts from the turbine.

Assuming no batteries and inverters for the solar panels. Add that for night work and/or an AC pump. No electric line losses counted either.

Furthermore, turbines are built for a specific water flow and head. If you change either, the turbine efficiency will be less than 0.70

Either go full hydro, full solar, some independent combination of both, but NEVER, NEVER, NEVER feed one to another and expect to gain much. The penalty of hooking them together is Eff 1 x Eff 2 x Eff 3 and so on.

True you can do it, but as you can see, you play a fools game.


"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
 

Your linesize is plenty large enough for a 300' run. If you have a properly sized Pelton wheel turbine and nozzle you should be able to get at least another 500-1000 watts out of your system. The link gives you calcs to come up with an accurate figure for your actual situation, but the table in the link gives you an idea of what you can do- the power available from a 4" penstock 500 ft long with 100 ft of head is about 4300 watts in the water jet or about 1900 watts extracted by the Pelton wheel.
 
"If" He just needs a turbine with 10 x the capacity and 10 x the flow. Why not 100 x, or 1000 x? Being that the pipe is the cheapest thing in this system, and he's not getting anywhere near what the power delivery capacity of that pipe is now, his 6" pipe doesn't seem to be the problem. Is it not reasonable to assume that his turbine flow capacity is the limiting factor and it's already maxed out at about 0.2 cfs flow right now.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
 
I agree with BigInch. If more power is required, installation of a larger pipe will provide the necessary hydraulic power, assuming the water supply is present.

It is not going to be feasible to add any pump considering the pump cost, the pumping efficiency is probably around 50-60%, and you only have a 1000 Watts system.
 
You can estimate System Output like this,

0.7 x water power
0.7 x 0.6 = 42% of pump power rating
0.7 x 0.6 x 0.9 = 38% of motor power rating

Grid Powered Pump (include wire loss of power to motor, eff 0.95) I forgot wire losses in my calcs above.
0.7 x 0.6 x 0.9 x 0.95 = 36% of power to motor

Gasoline powered pump (38% gasoline to generator, 90% generator, include generator to pump wire loss)
0.7 x 0.6 x 0.9 x 0.95 x 0.38 x 0.90 = 12% (of gasoline energy added)

Solar Powered Pump (solar panels 17% eff, wire loss panel to pump 95% eff)
0.7 x 0.6 x 0.9 x 0.95 x 0.17 = 6% of solar power panel capacity

To find final delivered power after wire losses from turbine (subtract another 5%)

Each power source in the chain reduces overall efficiency and forces the successive cost/W of the added power source ever higher and higher.

(Assuming rough average efficiencies for low power, low capacity equipment and normal wire sizing criteria.)

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
 
You are getting 40 psig under a static head, however, what pressure would you have when water is flowing thru the impulse turbine assuming that what you are installing. I would rather go with a reaction turbine such as the Ossberger type in your case and increase my pipe size for a greater volume of water with such turbine.This turbines are less complicated to maintain for domestic use.
 
biginch: the pump is a total non-starter for obvious energy balance reasons. I thought that was established in the 1st couple of posts.

I wouldn't assume that 300' of 6" penstock is the least expensive thing in his system. I have difficulty imagining that a 2000 watt Pelton wheel costs as much as what it took him to install this penstock. I know a guy who makes these wheels as a hobby business, and wheels that size can even be made of plastic.

At 6", he can get a lot more flow and hence a lot more power out of that penstock by buying a properly sized turbine and nozzles, and coupling them with an alternator which spins at the correct speed to match his flowrate.

 
That's what I said, the turbine is the limiting factor. Why not buy 1 x 10 GW machine? Money and maybe because he still needs it to run at minimum flow.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
 
What's this microturbine supposed to be doing? Why not apply the solar power directly to whatever the microturbine is doing? Wouldn't that be more efficient? The power loss in converting electrical to mechanical to mechanical to something else probably is rather large.

TTFN
faq731-376
7ofakss
 
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