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Pump water weight upward pressure needed

by tempjock2000
Tags: pump water weight
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tempjock2000
#1
Mar2-12, 02:37 AM
P: 6
I'm designing a bucket pump for a water well I'll be attaching a pumpjack to. I'm just a regular shop guy so I was kind of stumped on a question of how much upward pressure will be required to move the water up the drop pipe. If the pump cylinder is the same diameter of the drop pipe I figure I'll need to move the entire weight of the water in the pipe BUT I plan to have the cylinder actually only about half the diameter of the drop pipe. My question is if I have a 2 inch pump cylinder (2 feet long) tapering out/upward to a 4 inch drop pipe will I need to ad the weight of the water in the cylinder to the entire weight of the water in the 4 inch drop pipe? I'm not a physics guy so I wasn't sure if there is some law here I don't know about to figure this one out. Would the entire weight of the 4 inch drop pipe water no longer be needed to move it upward since the pump cylinder was more narrow? Thanks for you help in advance.

Tom
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Averagesupernova
#2
Mar2-12, 05:15 PM
P: 2,506
The weight working on the piston is proportional to the area of the piston. It does not matter how much larger or smaller the pipe leaving the piston chamber is.
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Have a look at this: http://www.engineeringtoolbox.com/pu...ure-d_663.html

p = 0.434 h SG (1)

where

p = pressure (psi)

h = head (ft)

SG = specific gravity

Soooo, if you have a 100 foot well with a 2 inch piston for an area of 3.1415 square inches, plug the numbers in. Specific gravity of water is 1, so it is 43.4 PSI. A piston with an area of 3.1415 square inches gets you 136.3411 LBS that the piston has to lift. Of course this is only the water, you need to lift the pump rod too.
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BTW, why are you calling this a bucket pump?
tempjock2000
#3
Mar2-12, 05:57 PM
P: 6
Quote Quote by Averagesupernova View Post
The weight working on the piston is proportional to the area of the piston. It does not matter how much larger or smaller the pipe leaving the piston chamber is.

So the weight of the 4 inch wide column of water (100 ft. tall) in the drop pipe isn't creating downward pressure on the piston? The piston pushing the 2 inch cylinder of water into the 4 inch drop pipe above only has to move the weight of the 2 inch portion? There is no check valve between the 2 inch cylinder and 4 inch pipe. I've attached (I think) a crude drawing of what I'm planning.

Thanks again,

Tom
Attached Thumbnails
pump.JPG  

tempjock2000
#4
Mar2-12, 06:07 PM
P: 6
Pump water weight upward pressure needed

also, I was was using the term piston pump but after doing some research it looks like the term would most likely be "bucket pump" since the plunger is actually pulling the water up the pipe. I do shop work on our farm. More of a tinkerer with this stuff than a designer/manufacturer. Again, I appreciate all your help. I now regret never taking physics in school, LOL.
Averagesupernova
#5
Mar2-12, 08:57 PM
P: 2,506
Of course the water in the 4 inch pipe is creating downward pressure. But only on the area of the piston. You could have something the size of Lake Erie or as small as a straw for a drop pipe and as long as the height of the water doesn't change the pressure on the piston will be the same. Usually in a piston-type deep well pump the pump cylinder down in the ground is larger than the drop pipe. Typically a 2 inch cylinder in the ground and a 1.25 inch pipe. The cylinder in the ground is only a lift pump. For a pressure system there will be another cylinder on top that actually pumps to system pressure. Upwards stroke lifts the water, the downward stroke pushes the water out.
tempjock2000
#6
Mar2-12, 09:21 PM
P: 6
This is a prototype for a much larger ag pump I want to build. The 4 inch drop pipe is actually the well casing. The 2 inch pump cylinder will be tapered out to the sides of the casing. A rubber gasket will seal the sides with a Tyvek liner to the top of the casing to seal off the perforations above the pump cylinder. We have a 12" unused well on the property I want to rehab to bring more land into production. I want to experiment with an 8 inch by 8 ft. pump cylinder on that one using the 12" casing instead of 200 ft. of drop pipe (100 ft. to water) I'd have to buy if we use a turbine pump. I hope this makes sense. It's kind of crazy but I just think it has to be cheaper to pump jack the water up than the 100 hp electric motor to turbine pump it. Part of the reason why I posted on here was to get a second, more educated opinion of the concept instead of potentially looking crazy to my friends, LOL.
Averagesupernova
#7
Mar2-12, 09:44 PM
P: 2,506
I guess I don't see why you don't just go with a conventional well pipe. A 4 inch casing won't get the pump to move more water than a standard well pipe made to go down inside a 4 inch casing with the same size pistion. How will you install and pull a pump installed this way? Not sure how much water you need, but this type of pump is not real fast compared to a submersible turbine.
tempjock2000
#8
Mar2-12, 11:22 PM
P: 6
Due to earthquake damage (we're in the So Cal Desert) the casing has become bowed just enough to prevent using a turbine. I just think it's a shame to destroy a 50 year old well so I'm being cheap and trying to salvage it. A new well and turbine pump could cost upwards of 250k which we don't have. The bottom part of the pump will be suspended using brackets running down the side of the casing. The actual piston will not be attached and easily removed by drawing out the sucker rod. I want to see what gpm we get with the prototype. I'm hoping for 40 strokes per minute basically moving the volume from the pump cylinder out the top of the well (8 inch X 8 ft of water≈ 20 gallons hopefully on big well). I need the whole 12" of casing width on the big well to compensate for the bow so there is a straight line to the piston. Even if the bow gets worse I'm thinking rollers along the side of the sucker rod would keep it working. We probably wouldn't be able to finance a new well for a couple more years so I'm using my spare time to experiment with the old one. The fun is really going to start when I start building the "nodding donkey" to move this thing.
jim hardy
#9
Mar3-12, 09:26 AM
Sci Advisor
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jim hardy's Avatar
P: 3,632
The piston pushing the 2 inch cylinder of water into the 4 inch drop pipe above only has to move the weight of the 2 inch portion?
I think you have the idea correct , even if the words are imprecise.

Looking at your picture in post #3

FORCE on piston is pressure X its area.
This is exact same principle as the hydraulic jack you use to lift a truck.
Lifting the piston does lift the whole column of water
but as water moves from skinny section to fat section it spreads out
so one inch of upward travel at piston gives less than one inch of upward travel in fat section of pipe, where there's more cubic inches per foot of pipe.

That's "mechanical advantage".

think of it this way
consider the water in wide section of pipe as a piston instead of water
and FORCE = PRESSURE X AREA
use small FORCE on small-area piston to apply larger FORCE to large-area piston above.

Force on your small piston in pounds is:
its area(in square feet) X height of water(in feet) X 62.4 pounds

only trouble i see is roughness of the old pipe - can you get a seal betwen pipe wall and piston?
tempjock2000
#10
Mar3-12, 04:25 PM
P: 6
You guys are great. These were all "maybes" in my plans but I didn't have the engineer/physics background to realize it.

The pump cylinder will be new pipe. The seal will have to be where the tapered portion intersects with the old well casing. The gasket will be like a 2 wide rubber belt pushed up against the old well casing by the weight of the water and Tyvek sleeve. I added a few more things to the diagram to hopefully explain it.

Thanks Again
Attached Thumbnails
pump2.JPG  


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