Water Tank Overflow Air Piston concept/question

[ScottyP99]

I am trying to determine the minimum piston weight/force that would be required to push the air from the air tank into the top of the supply tank, filling up the supply tank and purging the discarded water thru the outlet supply pipe. We can assume no friction loss and sufficient air in the air tank, with the discharge going to atmosphere.

Any help is GREATLY APPRECIATED !

 

[berkeman]

Welcome to PF.

What is the application? What are your thoughts so far for how to do the calculations?

 

[ScottyP99]

Thank you !

Been many years since I have used most of my engineering learnings and I am pretty rusty.

Its part of a larger concept but the basis is to use gravity fed piston (with some special help) that will purge the tank without use of electricity.

 

[berkeman]

Do you just want to purge the bottom tank, or do you want to transfer the water to the upper/overflow tank?

 

[ScottyP99]

Sorry yes it needs to flow up the discharge pipe to a tank at a higher level.

 

[ScottyP99]

I have considered using the direct piston weight to push the water up the discharge pipe, but was unsure the feasibility, and the air tank concept seemed would require less piston weight.

 

[berkeman]

A quick ballpark calculation (not counting friction, etc.) is to calculate the change in the gravitational potential energy (GPE) when the water is in the upper tank versus the lower tank. Then calculate the work that needs to be done by the piston driving the water out (piston moves vertically by the height of the water in the lower tank), which will give you the downward driving force needed.

Alternately, calculate the change in GPE from the initial state of water in the lower tank and weight poised above that water, to the final state with the water in the upper tank and the weight lowered by the height of the water.

 

[ScottyP99]

Thank you very much !

Do you feel the weight of the piston would need to be greater/same/less for the direct plunge concept compared to using it to inject air into the lower tank ?

 

[Chestermiller]

Thank you very much !

Do you feel the weight of the piston would need to be greater/same/less for the direct plunge concept compared to using it to inject air into the lower tank ?

Where is the air initially?

 

[ScottyP99]

It will be in a tank or bladder that will be compressed by the piston.

 

[Chestermiller]

It will be in a tank or bladder that will be compressed by the piston.

please add this to the schematic

 

[ScottyP99]

That is what the Air Tank is. Should I say something different ?

 

[berkeman]

That is what the Air Tank is. Should I say something different ?

I only see a water tank and an empty overflow tank. Where is this Air Tank bladder?

 

[ScottyP99]

Sorry I uploaded the wrong PDF !

 

[berkeman]

So in your new diagram, the Air Tank is at atmospheric pressure. When you release the weighted piston, that pressurizes the air and starts pumping the water out of the lower tank. You want to compare that to the case where there is no air in the lower tank to start?

 

[ScottyP99]

I want to understand the force needed to pump the air from the air tank into the supply tank, which in turn pushes the water from the supply tank thru the supply pipe to the discharge to atmosphere.

 

[ScottyP99]

And the weight of the piston, the air holes between the air tank and supply tank, as well as supply pipe to discharge are customizable to find an optimal solution.

 

[berkeman]

Why do you think that it will be more efficient to have the atmospheric pressure air space initially between the weight and the water in the bottom tank? Compressing that air would seem to take extra work, IMO... Can you do the calculation I suggested early on with the delta GPE numbers?

 

[russ_watters]

I want to understand the force needed to pump the air from the air tank into the supply tank, which in turn pushes the water from the supply tank thru the supply pipe to the discharge to atmosphere.

Pressure is density times height for a column of water. Force is pressure times area. That should enable you to calculate the force needed to be applied to the piston to lift the water up the discharge pipe.

The air bladder doesn't do anything as far as I can tell.

And the weight of the piston, the air holes between the air tank and supply tank, as well as supply pipe to discharge are customizable to find an optimal solution.

Is there a flow rate requirement? You haven't said anything about it. Otherwise, what does "optimal" mean?

 

[ScottyP99]

Berkeman, yes I will do that. To do this, would I need to also include the energy changes from the falling piston and air compression for a total system calc ? And I don't think we are compressing water, we are compressing air to inject into the atmospheric water tank. Unless I am missing something which is very possible.

Russ, there is now flow rate required. And optimal wasnt the right word. I guess the minimum weight required to force the purge of the water thru the discharge line. To follow your line though, if the piston weight were 10000 lbs, and the air hole between the air tank and supply tank were 2", what would the PSI of the air going into the supply tank be ?

 

[berkeman]

When you compress air, it heats up. Is that heat put to good use in your scenario?

$$PV = nRT$$
https://en.wikipedia.org/wiki/Ideal_gas_law

 

[russ_watters]

Russ, there is now flow rate required. And optimal wasnt the right word. I guess the minimum weight required to force the purge of the water thru the discharge line. To follow your line though, if the piston weight were 10000 lbs, and the air hole between the air tank and supply tank were 2", what would the PSI of the air going into the supply tank be ?

We teach here, we don't just give out answers. Pressure is force divided by area. You have the force, what's the area of the piston? And then what's the pressure?

 

[ScottyP99]

Thanks Russ, I have done this calc out many ways but wasnt sure it applied the same way for the air tank, etc.

The piston area is 1808 sq in
If the piston force is 10000 lbs
The piston pressure would 5.53 psi

So using the same logic, the pressure at the air hole would also be 5.53 psi, correct ? Or would the force of the piston apply equally thruout the air tank, and apply 10000lb of force at the 2" air hole ?

So if ambient psi is 14.7 psi, would the piston not push any air into the tank if the actual psi at air hole is 5.53 ?

 

[ScottyP99]

And if I haven't already said it to everyone, THANK YOU VERY MUCH !

 

[russ_watters]

Thanks Russ, I have done this calc out many ways but wasnt sure it applied the same way for the air tank, etc.

The piston area is 1808 sq in
If the piston force is 10000 lbs
The piston pressure would 5.53 psi

So using the same logic, the pressure at the air hole would also be 5.53 psi, correct ? Or would the force of the piston apply equally thruout the air tank, and apply 10000lb of force at the 2" air hole ?

In a fluid the applied pressure is independent of the area (different for the solid piston). So yes, the pressure of the air flowing through the hole is still 5.53 psi.

Now calculate the pressure required to drive the water up the pipe. I think you have a problem...

So if ambient psi is 14.7 psi, would the piston not push any air into the tank if the actual psi at air hole is 5.53 ?

No, we're talking gauge pressure here. Pressure above ambient. (air is pushing down on the piston too)

 

[ScottyP99]

OK, so the pressure we would need to overcome would be the pressure caused by the weight of the water column, which would be the weight of the water above the water level.

Pipe area is 12.56 sq in
Weight of water is ~ 8lb/cu ft
Volume of 50ft pipe rise is ~ 4.36 cu ft
Weight of 1 gal water ~ 8lbs
Weight of water column is ~ 279 lbs
Pressure at tower is 279/12.56 ~ 22.21 psi

 

[ScottyP99]

Sorry the 2nd line should say volume is 8 gal/cu ft

 

[russ_watters]

OK, so the pressure we would need to overcome would be the pressure caused by the weight of the water column, which would be the weight of the water above the water level.

Pipe area is 12.56 sq in
Weight of water is ~ 8lb/cu ft
Volume of 50ft pipe rise is ~ 4.36 cu ft
Weight of 1 gal water ~ 8lbs
Weight of water column is ~ 279 lbs
Pressure at tower is 279/12.56 ~ 22.21 psi

You seem to be switching back and forth between your scenarios. The height in your more recent one was 15 ft (still a problem). In any case, if you know the weight density of water is 62.4 lb/cu ft you can do the calculation easier: 62.4*50/144= 21.7 psi

 

[ScottyP99]

Yes sorry I had initially had put 50ft, but was adjusting calcs on my xl and and uploaded 15ft but referred to 50ft.

So at 50 ft, with a piston area of 1808 sq in we would need a force of 21.7 * 1808 = 37968 correct ?

 

[russ_watters]

So at 50 ft, with a piston area of 1808 sq in we would need a force of 21.7 * 1808 = 37968 correct ?

Correct.

Any chance you could make the tank taller and skinnier...?

 

[ScottyP99]

Hi Russ sorry for the delay. It can, but not by any major scale, so we are stuck.

But, could you confirm my calcs below IF we were to pump air from a compressor into the supply tank to force the water up the pipe:

Assume we need min 20 psi min
Tank volume is 279 cu ft
Air is incompressible

Compressor specs (values taken from commercial comp online)
PSI	40
CFM @ 40psi	15.4
Volts	230
AMPS	25
Time to fill Supply Tank (mins)	6
Power (watts/hour)	5750
KWh used	0.61

So it would take the above compressor 6 minutes at 40 psi to fill the supply tank and purge the water up thru the outlet line ?

I know there will be issues with some water not being able to purged, but the overall volume should be using these values correct ?l

 

[ScottyP99]

And I get the min 20 PSI from the weight of the water in the column/the column area (279 lb/12.2 sq in), and used a 40 psi compressor value to account for compression of air etc