Calculating Force required for a simple air piston pump 1

[IdeadGuy]

Network engineer, so not a real Engineer or a Student Engineer; is it allowed to ask a Mech. Eng. question/s?

If so, I'm looking to build a simple air piston pump w/low CFM (5.5), delivering air to ~9' below water surface.
Q1- Since 9' is ~1/3 Bar, does that mean I need to provide PSI at 0.3 Bar, or 1.3Bar at cylinder outlet?

Aside Q1, Q2 is figuring out Force (N) required on piston.

5cmf = 8640 ci
Cylinder 8"ID x 12"L = 603ci
8640 / 603 = ~14 stroke/min
60/14= 4.28 sec/full stroke
2.14 spf = 0.142 mps piston speed
Cylinder exit orifice 0.75 in

There seem to be a multitude of different Air Cylinder Force equations, however they all seem to be for closed cylinders.
One online calculator yielded a Force of 4K N....

I can mix bore size, length etc, and mix and match motors and gearing if I have to. Maybe smaller PVC prototypes.
But I am just shy of understanding the physics/fluid dynamics and I think I've got 90% of what I need to get a properly newbie engineered solution*. For various values of Engineered...

Would anyone be willing to help a guy out?


Thanks!

 

[LittleInch]

Will probably help to draw this up, but from the above what I think you're trying to do is pump air to 9 feet below water level at 5.5 CFM?

Not sure why you're doing this form first principles, but you output pressure needs to be 0.3 bar or barg as some people like to use. This is also 1.3 bara (absolute pressure)

Assuming your piston is full of air at 0 barg or maybe a little less as you've pulled it into the cylinder at the beginning of the stroke, but lets use 0 barg or better to use absolute pressures and make that 1.0 bara. To get to 1.3 bara your volume in the cylinder needs to reduce by 1/1.3 = 0.77. So the first 23% of your stroke is just pressurising the air to 1.3 bara. The force required at this point will then be constant at Pressure (use gauge pressure) times area ( you can do the maths). An 8" cylinder is pretty big.

I get about 1000N on a BOAFP estimate.

Note that at the end of your stroke the air pressure will stil be 1.3 bar a, so on the inlet stroke the first 23% of the stroke will not pull in any air so account for that when you do the air flow calculation.

If you want an analogy, this is the same as pumping up say a balloon using a small pump. Once the balloon inflates, you need to push the pump in a certian amount before the balloon starts to inflate a bit more, but then also when you release the pump piston, it takes a certain distance before the force is negative and you need to suck in more air. Try it and see.

Be careful when quoting volumes and volume flow as it varies with pressure and temperature. Do you mean 5.5 scfm (i.e. at atmospheric pressure and 15C/60F) or a volume flow at some other pressure? like say 0.3 barg?

Does that help?

 

[waross]

Possibly some of your confusion:
Are you working in the equivalent of PSIG, Pressure per Square Inch Gauge, eg: the pressure above the ambient atmospheric pressure, or 0.3 barg or in the equivalent of PSIA, pressure per Square Inch Absolute eg: 1.3 bara?

 

[rb1957]

I wonder if "BOAFP" is lost on some ? (not me ... "FP" indeed !!).

I would wonder how losses (like imperfect cylinder, leaking pressurised air) might impact things.

 

[waross]

BOAFP

BOAFP
Based On A False Premise??
May be less reliable than a WAG?

 

[IdeadGuy]

Littleinch and Waross thank you very much, very.

Yes, this is pumping air into a pond at approx. sea level.
I guess I'm doing this from first principles mainly because it was difficult to a compression discussion that wasn't involving an air compressor to a holding tank or a pheumatic type cylinder. Just seemed the only way.
Bara and Barg, ok now I understand why some discussion used different Bar numbers. IIRC, in school we were taught that sea level pressure is actually 1 Bar though we don't feel it per se.
I knew I was OTT with an 8" cylinder, but was hoping that since I wasn't looking for hundreds of rpms, the 2.14 s compression stroke would yield a lower Force over such a time interval. I will have to upsize watts/motor or more realistically smaller dia. cylinder with more SPM.
Minimizing noise was sort of my #1 User Req.
I was planning on having a check valve at the outlet so that once BDC is reached and piston starts to retract, valve closes, piston moves a bit more, and intake valves are pushed open by greater outside air pressure. Would think that would be less than 23%, no?
And yes, thats a good question regarding CFM vs SCFM. Looking at their spec sheet, they denote volume as CFM. I will double check with them though.

Again, Thanks very much for schooling me.

Waross, yes.
I work with frames, packets, uControllers, was always confused when some discussion talked about ambient pressure as 1 Bar/14.7psi, and others 0 Bar/0psi. That is much clearer now.
Realistically, all I knew at the begining was that a certain aerator req'd 5CFM. So thats about 1/3 Bar, but is a cylinder at BDC at ambient 0 Bar or 1 Bar before adding 0.3 Bar. Now I'm trying to figure out how many watts I need, what kind of gearing can I use to increase torque, which reduces final drive speed, which reduces strokes per min, etc, etc.
I've probably self-constrained myself from the outset by foolishly assuming I can get away with this somehow with ~300W and just gearing down a DC motor for greater torque.
At this point I know everyone in the thread just threw up their hands realizing they're trying to help someone to far out of their depth who's trying to force a solution against the laws of physics, right?

If so, I'm really sorry for wasting your time.
And absolutely will listen to any suggestions.

 

[Compositepro]

"Note that at the end of your stroke the air pressure will still be 1.3 bar a, so on the inlet stroke the first 23% of the stroke will not pull in any air so account for that when you do the air flow calculation."

That part is not correct. It is only the "unswept volume" between the piston and the outlet check valve that will have to expand 23%. This volume can be (and should be) designed to be close to zero.

If you are not an engineer or avid tinkerer you should not try to design one yourself. There are many cheap inflators on the market, both hand and battery powered. Your question, though, is still relevant to selecting a pump that will work. If this is for breathing air, you need to pay attention to air contamination from lubricants or wear particles, particularly if there is a motor.

 

[rb1957]

Back Of A ... Packet

 

[3DDave]

"Note that at the end of your stroke the air pressure will still be 1.3 bar a, so on the inlet stroke the first 23% of the stroke will not pull in any air so account for that when you do the air flow calculation."

That part is not correct. It is only the "unswept volume" between the piston and the outlet check valve that will have to expand 23%. This volume can be (and should be) designed to be close to zero.

If you are not an engineer or avid tinkerer you should not try to design one yourself. There are many cheap inflators on the market, both hand and battery powered. Your question, though, is still relevant to selecting a pump that will work. If this is for breathing air, you need to pay attention to air contamination from lubricants or wear particles, particularly if there is a motor.

Until the piston has moved far enough to raise the pressure to 0.3 barg (1.3 bara) then no air will leave the pump. This happens at roughly the 23% of total stroke, assuming constant temperature compression. After that the pressure will remain fixed until the piston reaches end of travel. This is what was originally mentioned by littleinch

So the first 23% of your stroke is just pressurising the air to 1.3 bara.

 

[IdeadGuy]

"Note that at the end of your stroke the air pressure will still be 1.3 bar a, so on the inlet stroke the first 23% of the stroke will not pull in any air so account for that when you do the air flow calculation."

That part is not correct. It is only the "unswept volume" between the piston and the outlet check valve that will have to expand 23%. This volume can be (and should be) designed to be close to zero.

If you are not an engineer or avid tinkerer you should not try to design one yourself. There are many cheap inflators on the market, both hand and battery powered. Your question, though, is still relevant to selecting a pump that will work. If this is for breathing air, you need to pay attention to air contamination from lubricants or wear particles, particularly if there is a motor.

Fair point, I did not detail use case as some forums get bent for too much extraneous information.
This is for an aerator for a 45' pond 9' deep at the center, 40-50K gallons IIRC.
Its located 237' from the house, and trenching, conduit and/or groundproof wiring of large gauge to reduce losses would be more than the couple solar panels I have, and the couple more I probably need to buy.
Typical pond aeration kits even on the cheap side are $500+ misc. after wiring.

I could certainly throw $800-1000 for a turn-key system and walk away.
However I can MIG weld, fix our own auto's mostly, and it seems more enjoyable to spend my non-Corp time learning stuff the dumb way sometimes. I could blow money at the bar, in the betting pool, do this or that at x$ cost, I don't see why DIY/Maker would be advised against by an Engineer outside of Corp., or really lop-sided cost/benefit ratio.
This will probably be a piston air pump made of a used auto piston/rod with good oil control rings, a cheap engine sleeve to fit, some Reed valves, some sort of torque gearing, and solar panels and an ESP-32.
$200-300 and I'll have my own sub-optimal pump that I can upgrade and fix myself.
However with some advice from this forum, I can quite possibly avoid having to brute force this the dumb way by interating expensively and just spend a bit more on parts, light machining, and have a much more powerful, reliable, self-fixable, mod-able
system.
My wife seems to have come around since I started saving her $300-350 for brakes. And her sister's, and....
TMI?

 

[rb1957]

not TMI ... it'll help us not suggest things that you don't want. I might suggest the "Engineers with hobbies" forum ... most of our work is commercial, so we tend to weigh cost quite heavily.

some random solar panels (output ?) to charge a battery which would power the pump (a fall back is having a battery charger run off the mains, maybe you'd have to take the batteries back to the house to charge ??). You've going to want a valve at the lower end of the tube, and the tub needs to be strong enough to resist the water pressure. Now I guess you could connect the tube directly to the pump, but I might use a reservoir to collect the high pressure air and outlet valve, maybe 1/2 atm.

If you're not getting bubbles, then raise the end of the tube till you do, and that should info on what to do next.

GL !

 

[Snickster]

The force on piston is 196 pounds at 9 feet head of water = 3.9 psig. This is the maximum force when pressure in piston reaches equilibrium with water pressure at 9 feet. You need to size the power required by the motor/solar panels which is force x distance divided by time. Power can be calculated using ideal equations for isentropic compression (Work = PdV where PV^k=C) to 3.9 psig, followed by a constant pressure (Work=PdV where PV = C) work input at 3.9 psig at the end of cycle to push air out of cylinder. Total power of compressor = (VdP)*(mass flow rate).

 

[waross]

Well 9' = 108"
108" of water column divided by 27.708 = 3.9 PSIG
8" diameter = 50.27 Square Inches
50.27 Sq In x 3.9 PSIG = 196 pounds force.
This will be the equilibrium condition.
The more force added above 196 lbs., the more volume will be forced past your system restrictions to aerate your pond.

If you use a small gas engine with a 2" bore, your required equilibrium force will be down to 12.25 lbs. force.
What about the heat?
When you compress a gas, it gets hotter.
When it gets hotter, the pressure goes up.
With a bell cranked piston rod, the torque require at constant pressure is similar to the top half of a sine wave
The theoretical torque will start at zero at bottom dead center and increase first until equilibrium pressure is reached and the continue to increase as the working radius or angle of the crank approaches 90 degrees.
The torque will peak at about 90 degrees from bottom dead center.
The torque will then drop for the rest of the stroke to zero as the piston passes top dead center.
A typical electric motor may develop 250% of rated torque as maximum torque.
250% of rated torque will not be enough torque to overcome the force required to take the crank past the 90 degree maximum radius.
For that you need an unloader to start and a flywheel to carry through when running at speed.

OR
Forget the calculations.
Go to the local auto wreckers and buy one or two 12 Volt compressors intended for the air ride system of a high end SUV.
Generic replacements may be purchased new on Amazon for a little over $100.
Don't pay too much for used.
It will work as is on 12 Volts.
Hook it up and see how much air you get.
Add more compressors and or more solar capacity as needed.
Working with less back pressure it will take less current.

 

[MintJulep]

Any sort of oil-lubricated compressor will contaminate your pond unless you're willing to spend for real oil removal stuff.

You need an oil-free compressor.

But you don't really need a compressor. A compressor is way overkill for 4 psi.

This is within the range of a "blower". (as an example: https://www.nyb.com/pressure-blower/) which you could probably find surplus or used.

 

[waross]

This is within the range of a "blower". (as an example: https://www.nyb.com/pressure-blower/) which you could probably find surplus or used.

Wonderful idea, MJ.

 

[IdeadGuy]

not TMI ... it'll help us not suggest things that you don't want. I might suggest the "Engineers with hobbies" forum ... most of our work is commercial, so we tend to weigh cost quite heavily.

some random solar panels (output ?) to charge a battery which would power the pump (a fall back is having a battery charger run off the mains, maybe you'd have to take the batteries back to the house to charge ??). You've going to want a valve at the lower end of the tube, and the tub needs to be strong enough to resist the water pressure. Now I guess you could connect the tube directly to the pump, but I might use a reservoir to collect the high pressure air and outlet valve, maybe 1/2 atm.

If you're not getting bubbles, then raise the end of the tube till you do, and that should info on what to do next.

GL !

Thanks, thats why I thought I'd at least ask.
As for panels, I've got a couple 100w 12v, which I know is not going to cut it. It might have for a smaller prototype.
I'll probably invest in a couple of 350w panels with my MPPT more realistically, and just go battery less and let it run when there is sun. Battery for the ESP-32 to use wifi for data collection, monitoring oil pump/level, temp warnings.
I'm pretty sure you can get that clear PET(?) tubing with the metal mesh which should prevent crushing at HD. I'll test that.
Was wondering about adding a small resevoir to do that also as I was thinking that would allow more time for the exhaust to travel at a more moderate speed than just one short burp.
I'll check the forum too, thank you.

 

[IdeadGuy]

The force on piston is 196 pounds at 9 feet head of water = 3.9 psig. This is the maximum force when pressure in piston reaches equilibrium with water pressure at 9 feet. You need to size the power required by the motor/solar panels which is force x distance divided by time. Power can be calculated using ideal equations for isentropic compression (Work = PdV where PV^k=C) to 3.9 psig, followed by a constant pressure (Work=PdV where PV = C) work input at 3.9 psig at the end of cycle to push air out of cylinder. Total power of compressor = (VdP)*(mass flow rate).

Snickster, thanks very much!
I'll admit that'll take a few to figure out.
But net-net I need 196 ft/lbs to compress, and power and stroke time are dependant upon available power and motor size, gearing?
Unless I misread what Force on Piston is 196 @ 9' is referring to.

 

[IdeadGuy]

Well 9' = 108"
108" of water column divided by 27.708 = 3.9 PSIG
8" diameter = 50.27 Square Inches
50.27 Sq In x 3.9 PSIG = 196 pounds force.
This will be the equilibrium condition.
The more force added above 196 lbs., the more volume will be forced past your system restrictions to aerate your pond.

Awesome, one of the equations mentioned PSIG but not how to arrive at it...

If you use a small gas engine with a 2" bore, your required equilibrium force will be down to 12.25 lbs. force.

Ah, are you saying reverse the engine into a pump like a few people have done with VW Bugs?
I did think of doing that with an old push mower, but then I'd need much higher RPMs and noise.
If I can figure out some anechoic-type sound proof boxes, that would solve a number or problems....

What about the heat?
When you compress a gas, it gets hotter.
When it gets hotter, the pressure goes up.

I did actually think about this, and was hoping for a couple things to go my way.
1. Very low RPM (below 15)
2. Long-ish intake cycle (compared to actual engines) means more cooling?
3. Small mineral oil oiler to spray either at before TDC, or while compressing.
Didn't think much past that besides just running a small 12v water pump over the cylinder.

With a bell cranked piston rod, the torque require at constant pressure is similar to the top half of a sine wave
The theoretical torque will start at zero at bottom dead center and increase first until equilibrium pressure is reached and the continue to increase as the working radius or angle of the crank approaches 90 degrees.
The torque will peak at about 90 degrees from bottom dead center.
The torque will then drop for the rest of the stroke to zero as the piston passes top dead center.
A typical electric motor may develop 250% of rated torque as maximum torque.
250% of rated torque will not be enough torque to overcome the force required to take the crank past the 90 degree maximum radius.

So do I need to upsize the motor & power and/or below?

For that you need an unloader to start and a flywheel to carry through when running at speed.

So the other night I was thinking about a YT video where someone made a small pump like this, but they used an 3/8-1/2" steel plate circle as the crank. Made me wonder about momentum.
I have a spare mag. clutch from an eBay refund, and can get a small car/motorcycle rotor.
Was thinking that I should somehow have sunrise start the motor spinning the rotor, and then at a certain RPM engage the clutch to start the gear moving. It might take a minute to spin the rotor up, but once the clutch engages, its momentum will transfer along with the more driving the axle/rotor and give additional force on compression.
I guess then the rotor will slow the axle, however its on the intake stroke which should take much less power so there should be some spin/momentum recovery?

Didn't think of the air ride though, that does sort of solve everything.

Awesome, thanks!

OR
Forget the calculations.
Go to the local auto wreckers and buy one or two 12 Volt compressors intended for the air ride system of a high end SUV.
Generic replacements may be purchased new on Amazon for a little over $100.
Don't pay too much for used.
It will work as is on 12 Volts.
Hook it up and see how much air you get.
Add more compressors and or more solar capacity as needed.
Working with less back pressure it will take less current.

 

[rb1957]

maybe step back .... why are you doing this ? I'm guessing you want to stock fish ? or maybe water quality ?

There are plenty of farms with earth ponds ... do they need/want/have aerators ? maybe some mixing is enough to circulate the water and oxygenate at the surface ??

Putting all the air in one place at the bottom of the pond may not be the most effective ? maybe some sort of "piccolo" tube (like we use for L/E anti-ice systems) with holes along it's length to distribute the air at different depths.

 

[IdeadGuy]

You are correct, this pond doesn't strictly need aeration.
But the 50-75 goldfish and Comets have survived 6+ years of Buffalo winters and seem to number about 200+ now.
This might help them a bit, and should in theory help reduce algae growth.
A piccolo tube or several spread out smaller aerators is better, I just figured I'd start with one.