forces

The air pressure at the compressor outlet should be so that at the bottom its pressure will be at least =X.

Lets say what are you trying to get done?

Air is a lot less dense than water, so a column of air 20,000 feet long at STP has a force at the bottom of 14.7 PSIG. A column of water about 32 feet long has that same pressure, and of mercury about 30 inches.

Thus you need 500/32 x 14.7 = 230 PSI to equalizee it against the water pressure 500 feet down

This is a good approximation. It sounds like you are designing an air lift pump. If so, you should know that to pump a 500 ft depth you should inject the air at about 350 ft. The reasons are too complex to get into unless you seriously need the help.

.434lbs/ft for water X 350 ft = 152 psi to run the pump. The compressor will need to make around 200 to clear the air line of water and have enough reserve capacity to run the pump.

You will need to inject about 3-5% of the flow of water you desire to maintain best lifting efficiency but, if you do not need a lot of head you can pump less cfm.

The maximum static head you could acheive for the spec you give would be about 100+ ft. Ballpark, no promises. Very little data on water wells this deep and flow is VERY dependent on the static reservoir level of the ground water. Efficiency falls of very rapidly as the water table falls.

Best wishes,

Mr. Gee

"an air lift pump" thank you for giving it a name. I think you got the idea and thank you for not getting complex and technical. As a general question, the idea is the important, not the numbers. For that the watertable level is a big deal. I guess I should have said 500(or 100) ft below ground water level.

A twist on this question, is more simple this way?

A ball filled with air(or some thing light) pressed down into water. the force needed to push it into water = the force it will come back right? (not counting losses). so air pumped down will take less (lets you energy) to do that than energy recovered as it comes up, (not counting losses)

thanks

Sorry to those who got too specific on me, and thanks.

I filed patents on the process you are describing 3 years ago. They were recently denied as someone in the oil and gas industry had precedence in the art on which I originally filed.

The answer to the question you next asked is, yes, there is a net positive energy available in such a system as the lift pump side of the system is driven by gravity which is a renewable energy source.

The engineering of the entire system is extremely complex and the realization of a working system will be very expensive.

There are aspects of this system that I have designed but not filed patents on that would still create some ownership interest in the means for me, but the overall business aspects of the venture are rather discouraging.

Best wishes,

Mr. Gee

A couple of things came to mind when I read your question, i.e.,--- The equation that defines force is F=MA with M=mass and A= acceleration------ Force and pressure are commonly (incorrectly) used to express (define) the same value(s).Pressure is expressed in lbs/sq inch -grams/sq meter, etc--- i.e., mass or weight/given area

All of the above is taught in engineering physics.

The height (head) of the water will determine the amount of force required to move it upward (disregarding the distance +- from sea level). Same as can be determined by the pressure exerted by a water tank, i.e., approximately 34 feet =1 atmosphere or 14.7 lbs/sq.inch ----- Years ago,, this was taught in the 9th grade.

THE POINT IS-- THAT THE DIAMETER HAS NO EFFECT ON THE PRESSURE AT THE BOTTOM OF THE HOLE

dear, in 5" dia of pipe to 500feet down, there should be a class to be mentioned. if not then, consider 500feet=152.43meter.&5"=125nb(bore dia).then there will X pressure i.e 15kg/cm2=217psi.there is pump know as centrifugal pump to deliver water to that rate. if there is a air then there will let off that air through air piston.

therefore, the condition of that system will different. if there is air then there also will be vapour formation then the ratio of NET POSITIVE SUCTION HEAD WILL OCOURS.

&

ASK QUE WHICH ARE APPLICABLE?

Well say there is a hole and you want pump air down that 5 inch hole to a point 500 ft deep. ok

Where is the water comming from ? The hole ?

Is the hole in a water pipe ?

Is this a well and there is a leak in the casing ?

Our Guest really needs to provide a little more information. An explanation of what he is really after would help others provide the answer Guest is looking for. My $.02.

Dear Guset,

This is a simple question : the answer is - you have to overcome the pressure of 500' of water column in the first place.A meniscule excess over that will allow air to be pumped at the level you disired.In terms of pressure ( not force) it will be 216.45 psi + o.1 psi( let's say) or around 217 psi ( pounds per sqaure inch)

C S Cowlagi