How can I increase the water pressure in a low head hydroelectric system?

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To increase water pressure in a low head hydroelectric system, utilizing a ram pump could be a potential solution, although its effectiveness in generating sufficient force for turbine operation is uncertain. The estimated power output from the stream is approximately 0.07 HP or 0.05 kWh, which is insufficient for household needs without significant losses. The current flow rate of 60-80 gallons per minute may be underestimated, but even with higher flow, the low head limits practical energy generation. A prototype turbine demonstrated varying RPMs depending on water source, indicating that pressure and flow adjustments could enhance performance. Exploring methods to maximize water pressure from the stream remains crucial for improving energy output.
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Hi Everyone. I'm new here and glad to find a community with ideas and creativity.

I have a stream behind my house that I feel could be utilized for hydro electrical power. I am no engineer but I've been reading up on it's potential for electrical energy. Trouble is that this stream has very little head in my area (only like 3 - 5 ft.) for me to utilize in order to get the kinetic energy i need to move a turbine (I've been looking into impulse systems) So... I've been looking up ways to fix that. I saw that a Ram pump could lift water to create artificial head but I don't know if I could lift enough water to get a solid jet of water to turn a turbine. Anyone have any suggestions on where I should start to either create enough force, or should I go in a different direction? The river flows at probably 60 - 80 gallons a minute.
 
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Based on some numbers I've run, it does not look as though you have enough flow (and head) to make it a very useful project.

I estimated the potential (ideal) power at approximately 0.07 Hp or 0.05 kWh. Here is how I made this estimate:
Power = mass flow * head
The mass flow is the 60 - 80 gal/min, converting to lbs/sec it is 60 gal/min * 8 lbs/gal = 8 lbs/sec
The head = 5 ft.
Power = 8 lbs/sec * 5 ft = 40 lbs-ft/sec
or in HP = 40 lbs-ft/sec / (550 lbs-ft/sec/HP) = 0.07 HP
or in kWh = 0.07 HP * 0.746 kWh/HP = 0.05 kWh
That is about 1- 60W light bulb's worth of power, and that is before any losses are subtracted.

What may be useful, if you have an emergency or night time use of a small amount of power, would be using the hydro power to charge a battery bank that would supply the emergency or intermittent power.

You may also want to double check that flow rate. 60 - 80 gal/min sounds pretty low for even a small stream.
 
DickL-

Thank you for the reply. You are right about the flow, it is probably more than that but even so It doesn't sound as if I can just throw a turbine in the stream and make enough power to operate my house.

I made a prototype turbine with 6 flat paddles and sat it in the water and timed the RPMs a few days ago. This thing was very primitive (we are talking paddles made of license plates and an axle of pvc all stuffed into a pickle bucket with the bottom removed.) It was getting 75 RPM. I put my garden hose to it and got the RPMs much higher, like 400 - 500. Wish I could figure out a way to create more water pressure like that out of my garden hose using just the water flowing and not much head distance...
 
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