Looking for a super high-efficiency air pump

• Syed F. Karim

Syed F. Karim

I am looking for a manufacturer who will be able to custom design an unusually efficient air pump. The most efficient ones on the market right now (or at least the ones I have come across) require about 60 Joules to move just one liter of air. I need something that can do the same amount of work, but only use 2 Joules. Any suggestions? Is this even possible?[?]

You'll have to be more specific about what you are looking for: volume/flowrate, pressure. The calculations to figure out if it is possible are relativly straightforward: calculating the energy to pressurize air is the same as calculating energy in a spring.

I haven't ever checked, but I'd imagine commercially available pumps are probably pretty efficient. Its unlikely that you'll find something 30x more efficient since that would imply most pumps are only about 3% efficient. I'd expect 75%.

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specifics of the air pump situation

This is what I am trying to do. I would like to extinguish a flame by replacing the oxygen that the flame consumes with an inert gas, in this case helium. Imagine a bucket turned over and placed on a stand. Now place a flame--a candle will be fine--in side of the bucket. The way that I intend to extinguish the flame is by pumping helium into the bottom of the bucket; the helium will rise and push the oxygen inside of the bucket out. Eventually a point will be reached where the flame is no longer submersed in oxygen, but in helium. Unfortunately, though, the helium is not being kept under pressure, and so can not simply be released into the bucket; it must be moved.
There is very little pressure for the pump to overcome, nothing more than normal atmospheric pressure to move the air out of the way. And as far as pressurizing the air goes, I don't believe that this is at all like a spring because the air inside the bucket is not being compressed, it is only being pushed out of they way and does have an avenue for escape; definitely not a situation where is being trapped in a container with nowhere to run.
So as far as specifications go, I have very few. The pressure that the pump must overcome is only the pressure imposed by air, and the flow rate required for this operation is absolutley unimportant--I don't care if it takes a whole week to do, I just want to do it with as little energy as physically possible.
To give you an example of what the market has to offer, here are the specs of a fairly typical DC diaphragm air pump:
1 Liter/min (or 1 Liter/60 seconds)
.2 Amps
3 Volts
(.2 Amps) * (3 Volts)=.6 Watts
time needed to move 1 Liter is 60 seconds
(.6 Watts) * (60 seconds)= 36 Wattseconds or 36 Joules

Is there anyway to decrease the energy consumption at all? Remember, most of these pumps are working at a certain pressure and they also have a flow rates which are of no importance to me. Can I bring that number down to 4 Joules? Incidently, as far as just moving air goes, there are quite a few fans (like the ones inside of computers) that are MUCH more effiecient thatn these pumps. They can actually meet my energy parameters, the only problem is that I don't know whether or not the fans can generate enough pressure to push the helium out the end of a hose and keep the air from entering it.

I hope I haven't given you too much information and put you to sleep.[zz)] . And thank you.

Originally posted by Syed F. Karim
Unfortunately, though, the helium is not being kept under pressure, and so can not simply be released into the bucket; it must be moved.
How is it being stored if not under pressure?

In any case, moving it is just a matter of finding a fan of the right specs, not a compressor.

What you are trying to do is similar to how halon is used. Halon is stored pressurized in tanks and relased very quickly into the space. Halon is good because it doesn't need to completely replace the air in the room to be effective - only 10% concentration or so.

Also, helium won't push the air up, it will rise above the air in the same way oil will rise to the top when poured in a container of water. Generally you want to fill the room from the bottom up.

Originally posted by Syed F. Karim
...Unfortunately, though, the helium is not being kept under pressure, and so can not simply be released into the bucket; it must be moved.
There is very little pressure for the pump to overcome, nothing more than normal atmospheric pressure to move the air out of the way.

Not quite. You have to overcome the duct losses between the storage location and the overturned bucket, the friction losses of the bucket imparted to the air being moved, through whatever openings this occurs, and compression of the air as it enters its new location (could be minimal depending on the volume being moved and the volume of the new container (or atmosphere).

Originally posted by Syed F. Karim
...I don't know whether or not the fans can generate enough pressure to push the helium out the end of a hose and keep the air from entering it.

Check out the friction losses on that hose.

If the volume is not too high on the space you are trying to protect, you could store your helium in an inverted piston simply holding against the fill pressure of the helium that when released would press the helium from the one space to the other as the piston drops. This would only require the kinetic energy stored in the piston arrangement.

Originally posted by russ_watters
How is it being stored if not under pressure?

Well maybe I explained things incorrectly. Instead of being compressed, the helium will be stored--in let's say a box--at normal atmospheric pressure. So if the the volume of the box is 5 liters, that's all the helium I am going to have to work with.

So you say a fan is all I need? Great. But how do I go about calculating the exact speciifications of the fan; not the electrical specs, the pressure specs (to make sure that air does not somehow make its way into the container through the hose). I'm not looking for precise calcualtions, just a strategy on how to attack this problem.

And lastly, if the bucket is turned over, won't the helium--which will be rising and accumulating at the top--eventually displace the air in the bucket. Thanks again.

*Must clarify* So if the the volume of the box is 5 liters, that's all the helium I am going to have to work with.

What I meant is that I will only have 5 liters of helium at 1 atmosphere. Am I making any sense?

Originally posted by Artman
If the volume is not too high on the space you are trying to protect, you could store your helium in an inverted piston simply holding against the fill pressure of the helium that when released would press the helium from the one space to the other as the piston drops. This would only require the kinetic energy stored in the piston arrangement.

If it is only a relatively small volume of helium that I am trying to move (let's say 3 liters), do you think the piston arrangement would cost me less energy. I mean, I am only going to be able to work woth about 10 Joules max. Won't a piston be fairly heavy, and therefore require more work (to actually get it into the inverted position)? Maybe all the added friction(s) of the hose-method really will lead to significant energy losses; I don't know, that's why I'm asking.

Originally posted by Syed F. Karim
Originally posted by russ_watters
How is it being stored if not under pressure?

Well maybe I explained things incorrectly. Instead of being compressed, the helium will be stored--in let's say a box--at normal atmospheric pressure. So if the the volume of the box is 5 liters, that's all the helium I am going to have to work with.

So you say a fan is all I need? Great. But how do I go about calculating the exact speciifications of the fan; not the electrical specs, the pressure specs (to make sure that air does not somehow make its way into the container through the hose). I'm not looking for precise calcualtions, just a strategy on how to attack this problem.

And lastly, if the bucket is turned over, won't the helium--which will be rising and accumulating at the top--eventually displace the air in the bucket. Thanks again.

You have to overcome the friction loss of the wall of the hose, exit opening loss at the end of the hose, reducer loss if the hose is smaller than the fan... You know, if you replace the helium in the box with the air from the bucket, you could actually push and pull the air from the bucket.

I like the piston idea - and no, the mass of the piston can be as small as you want.

You can't just blow air out of a box unless there is something to replace it. A bag would work for example. But in either case, it would be far more efficient to have it pushed out of the bag or box with a piston since a piston doesn't have any of the energy loss problems associated with a fan. You get near perfect energy transfer. You still have losses in the hose though of course, but a piston can easily overcome them.

You could even hang the piston vertically and require no input energy at all. The system would be set up with gravitational potential energy and just opening a valve would allow all the helium to flow from its container.

With your example, the helium would fill the bucket from the top down, pushing the air down out the bottom of the bucket.

quote russ_watters
I like the piston idea - and no, the mass of the piston can be as small as you want.

Just off the top of your head, would it be possible to drive a piston--let's say vertically to get it to a position where it will have potential energy--with just 5-10 Joules of energy? I'm guessing that it is more than enough, but why not ask the experts, I always say

Also, could bellows be classified under the piston-idea? Or are they more of a pump?

A bellows works pretty much the same as the piston, yeah. And you can give it whatever work you want almost down to zero. Using Bernouli's equation, velocity is a function of force (pressure) in the bellows/piston.

So its really up to you: the variables are pressure (force) and nozzle opening diameter. From these and Bernouli's you get velocity and flowrate.

If the hose isn't too long and is significantly larger in diameter than the nozzle opening, losses are negligible.

Russ, you'd probably know the answer to this better than I would...

Couldn't you hook up a pulse-width modulator to an 'off-the-shelf' pump? I think that would reduce the power output and drawn. I don't know if it would damage the pump, though.

Originally posted by enigma

Couldn't you hook up a pulse-width modulator to an 'off-the-shelf' pump? I think that would reduce the power output and drawn. I don't know if it would damage the pump, though. [/B]

Hello Mr. Enigma,
What is a pulse-width modulator and would it really bring down the power consumption (a lot)? And I'm not all that concerned with damaging the pump, it's only got to do its job a few times or so (move the helium, I mean).

Is helium considered a "greenhouse gas", cause that is why "Halon" Is NOT used (widely) anymore. Watched one on televison, a guy doing a Halon based fire extiguisher company, put him out of business they did, cause of the Halon being a greenhouse gas. (Be careful)

The bellows arrangement could be coupled with a lever arm and counterweight to make the operating power requirements very low, especially if the friction losses of the pipe are reduced by use of a large enough diameter pipe. The bellows could be filled with the lever in equillibrium, the pipe valved shut, and then the lever could be slid into unbalanced position pressing down on the bellows awaiting release. I should think this could be done with minimal power if bearings are employed on the arm's lateral motion.

When the valve is released, gravity would do the work of overcoming the pipe losses.

What is a pulse-width modulator

It works like a dimmer switch. It clips off a portion of the AC wave, effectively trimming the power to whatever it's feeding. IIRC: The AC wave charges an RC circuit, and once the voltage on the capacitor hits a set voltage, it trips a transistor (SCR, iirc) which cuts off the power until the wave hits the negative trigger voltage. I'll have to look back in my electronics notes to make sure that's entirely correct... the concept is right though.

and would it really bring down the power consumption (a lot)? And I'm not all that concerned with damaging the pump, it's only got to do its job a few times or so (move the helium, I mean).

That's what I'm not sure. Aside from labs (running simple fans), I haven't worked with them. I don't know if a pump will even work if given less power than it is designed for.

The other ideas may work better. I'll leave it to the PE's here to weigh in with authority.

Originally posted by Mr. Robin Parsons
Is helium considered a "greenhouse gas", cause that is why "Halon" Is NOT used (widely) anymore.
Halon is a greenhouse gas, but that's not the reason it isn't being used - its a CFC and destroys the ozone layer.

Helium is manufactured by separating it from the air. So when you use it it goes right back where it came from in the first place.
Aside from labs (running simple fans), I haven't worked with them. I don't know if a pump will even work if given less power than it is designed for.
I used a device called a "Power Planner" once that was supposed to reduce power usage by a compressor that wasn't fully loaded by sensing voltage (via power factor I think) and cutting the input voltage wave accordingly. It didn't work, but the compressor was close enough to fully loaded that there probably wasn't much extra loss in efficiency.

In any case, I like the piston/bellows idea because there isn't any input power - just a control circuit.

Syed F. Karim, may I ask what the purpose of all this is?

Originally posted by russ_watters

Syed F. Karim, may I ask what the purpose of all this is?

Yes, sir, you may. Although I did pose a hypothetical situation at the beginning of this thread, the situation itself is not really what I was interested in. I am an inventor, and being an individual of limited means and education, I have no choice but to pose parallel problems as threads in the hopes that some kind soul may be able to offer some useful insight. Naturally the most direct method of solving some of my problems would be to simply retain a consultant, but unfortunately \$100/hour is just not real easy to come up with.

As far as this particular puzzle goes, I am leaning towards a D.C. motor-driven piston to move the required volume of gas. The intricate bellows-idea sure did sound interesting, but complexity increases the number of patent attorney-hours significantly. And they run about twice as much as engineering consultants.

I sincerely wish I could be more open with some of my ideas, but it's just that the patent office has so many damn rules that I have no choice but to follow.

The piston/bellows could be as simple as two pipes with one slightly smaller in diameter such that it fits into the other one. Cap the end of the top one, gasket the other one and plug the end that goes into the larger one with just an opening to your gas tube. Put a heavy enough weight on the upper tube to overcome the gasket friction and the tube friction, put a normally closed solenoid valve on the gas tube to be opened when the power is applied and control the valve with a heat detector in the bucket.

Fire starts, heat detector energises the valve, valve opens, weight presses the pipe down on the other pipe pushing the gas out through the tube into the overturned bucket and the fire goes out.

Originally posted by russ_watters
(SNIP) Halon is a greenhouse gas, but that's not the reason it isn't being used - its a CFC and destroys the ozone layer. (SNoP)
Ooooops! yup! your right! so sorry...