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# Run a hydraulic ram pump on no water drop? Answered

So I was wondering if you could run a hydraulic ram pump by digging down 3ft beside your well then pump water up about 50 feet at about a 40 degree incline. This way there would need to be no change in elevation between your ram pump and your well. Thanks!!

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## 13 Replies

Ultimately all that really matters (not including molecular friction between the pipe and water) is the difference in height, how many feet does the water have to fall, and how high is it pumped. Incline and distance just make the calculation more difficult. (no one likes trigonometry lol)

Anyway, the way one of these pumps work is basicly it uses the weight of water falling down and out of the system (waste valve) to push a percentage of that water up hill a lot further. If you are familiar with switch mode power supplies, you can think of a RAM pump as a boost converter, where there is a medium pressure source (a reservoir) and a ground connection (waste valve), and the fall can be thought of as a inductor in electronics terms. When the flow is suddenly stopped, the inertia of the water will keep it from stopping immediately, since that water cannot immediately stop when the waste valve is stopped, pressure skyrockets to insanely high levels where it is forced to squeeze through a check valve and since there is no place for the water to go except into another opening up to a point of much higher pressure, usually up a big hill far higher than the original source of water.

So to answer your question, if you do not have adequate fall, not only do you have limited initial pressure on the RAM pump, (difference in height between the water reservoir and waste valve) you also do not have enough intera being produced when the check valve shuts after letting water flow for awhile. That means it will not work, and if it does, it will not work very well. The energy to pump water uphill comes from the fact that some water is allowed to escape and the energy of that falling water is 'extracted' and powers the pump. Hope I cleared everything up!

camping crazy (author)2014-12-01

Thank you that did clear everything up!! I definitely wont bother digging two holes now:)

-max- (author)2014-12-01

Geez the instructable commenting system is so buggy. It posted the same message like 10 times. I deleted all but one of those. Sorry if you got like 1000 emails of the same comment!

camping crazy (author)2014-12-01

Not a problem. I hooked instructables up to a junk email I never use.

-max- (author)2014-12-01

Geez the instructable commenting system is so buggy. It posted the same message like 10 times. I deleted all but one of those. Sorry if you got like 1000 emails of the same comment!

-max- (author)2014-12-01

Geez the instructable commenting system is so buggy. It posted the same message like 10 times. I deleted all but one of those. Sorry if you got like 1000 emails of the same comment!

petercd (author)2014-11-29

A simple search in google yields a wiki: It takes in water at one "hydraulic head" (pressure) and flow rate, and outputs water at a higher hydraulic head and lower flow rate.

This basically means that you need a head for the device to work, if you input zero your output will be zero.

camping crazy (author)2014-11-30

So a straight hole down beside the well would not be enough head??

petercd (author)2014-11-30

where's the water going to at the bottom of the hole?

as soon as the waste water covers the outlet the pump will stop working.

camping crazy (author)2014-11-30

I was going to put the pump on a rack one or two feet above the bottom of the hole the sides of the hole would be made of a steel drum with the top and bottom cut out. With one or two feet of clearance and ample drainage it shouldn't be a problem.

petercd (author)2014-11-30

ample drainage ...lol

I dont think you quite understand how much water flows through these devices, check some videos on the Tube. Ive never yet heard of soil draining at the rate you need without a nearby mineshaft.

Jack A Lopez (author)2014-11-30

Lifting water requires work, specifically W= M*g*h, where M is the mass of the water, h is the height I want to lift it, and g is the acceleration of gravity.

Or you can think about it in terms of volume, then

W= V*P where P = (M/V)*g*h = d*g*h

where d=M/V is the density of water.

P has units of pressure. Notice that P and h only differ by a factor of d*g=constant. For that reason, P, or h, is essentially a measure of how much work you have do per unit volume of water.

Water is heavy stuff, and it takes work to lift it. That work has to come from somewhere.

The usual application for a hydraulic ram is for drawing water out of a river, or stream, and then throwing it up the side of a hill. The work comes from a large volume of water, V1, which drops a short height, P1. The short drop occurs in a pipe in the river, or right next to it.

Then that work is used to throw a much smaller volume of water, V2, up the side of a hill with height P2.

So without losses, W = V1*P1 = V2*P2, or (P2/P1) = (V1/V2),

and that's basically the game being played here. You can throw a small volume of water V2 up the side of a hill P2, at the expense of a large volume of water V1, which flows back in the river over a drop P1.

I don't think this trick will work with a lake, or the bottom of a well, which is kind of the same thing, an underground lake, a huge amount of water all at the same height or depth. It won't work because there is no P1 and nowhere to place to dump the V1, no water available to do work.

camping crazy (author)2014-11-30

Okay I understand it now, I thought maybe I could cheat but I guess not. Ill probably just buy a cheap pitcher pump and walk down with a bucket to get water. Not a big deal. Thanks for the help!!