A liquid moving in pipe from a low static or low pressure head source has kinetic energy which can easily be converted to a high pressure head for a low volume flow. A swing check valve simply slams shut. The kinetic energy of the incompressible is converted to a pressure head.
This low tech idea was used to lift water for irrigation two hundred years ago.
Does anyone know the conversion efficiency?
Bret Cahill
Miserable, but practically free. I seem to recall that it takes about 10 gallons spilled at the bottom to pump 1 gallon at the top.
I wasn't so much interested in how much liquid was "lost" as how much mechanical energy was lost as heat.
If 10 gallons dropped one foot at the bottom to raise one gallon 10 feet, then the free energy conversion efficiency would be 100%.
My first guess is even a well designed ram pump would have a miserable energy conversion efficiency as well, but I'ld like to know if it's less than 15%, 40 - 60%, etc.
Bret Cahill
What I have is this scenario:
Ram Pump mechanism (air chamber, poppet and check valve) located 10' below the water supply with a 50' long supply line will pump to the source 50' above the pump. Length of the supply pipe is an important part of the performance.
Also, I googled ("ram pump" +efficiency) this technical paper with lots of equations that ought to make you happy:
And this efficiency equation as well:
Pump Performance - Some information suggests that typical ram pumps discharge approximately 7 gallons of water through the waste valve for every gallon pressurized and pumped. The percentage of the drive water delivered actually varies based on the ram construction, vertical fall to pump, and elevation to the water outlet. The percentage of the drive water delivered varies from approximately 22% when the vertical fall to the pump is 1/2 (50%) of the elevation to the water outlet down to 2% when the vertical fall is 0.04 (4%) of the elevation to the water outlet. Rife Hydraulic Engine Manufacturing Company literature
Q is the available drive flow in gallons per minute, F is the fall in feet from the water source to the ram, E is the elevation from the ram to the water outlet, and D is the flow rate of the delivery water in gallons per minute. 0.6 is an efficiency factor and will differ somewhat between various ram pumps. For instance, if 12 gallons per minute is available to operate a ram pump, then pump is placed 6 feet below the water source, and the water will be pumped up an elevation of 20 feet, the amount of water that may be pumped with an appropriately-sized ram pump is
0.6 x 12 gpm x 6 ft / 20 ft = 2.16 gpmThe same pump with the same drive flow will provide less flow if the water is to be pumped up a higher elevation. For instance, using the data in the previous example but increasing the elevation lift to 40 feet:
0.6 x 12 gpm x 6 ft / 40 ft = 1.08 gpm
That's better than a lot of centrifugal pumps going from low entropy electrical power to higher head.
Bret Cahill
On average, hydraulic ram pumps show an eficiency of 0.6 but 0.85 is not unheard of.
Brian Whatcott
Maybe the losses come from the compressibility of the water or the elasticity of the pipe.
It hard to say at first glance if efficiency will change with head. In any event several working at 60% can always be combined in series for higher pressure system working at 60%.
Bret Cahill
Dear BretCahill:
More likely n * 0.6, or much less efficiency.
David A. Smith
If you pipe the left over water down to each pump below for recycling it should be 0.6 overall.
If you are just throwing the left over water over the side from each higher station the efficiency will drop really fast > something ^ n.
Bret Cahill
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