Electrical Power required to drive a DIY electromagnetic air compressor

AI Thread Summary
The discussion focuses on calculating the power requirements for a DIY electromagnetic air compressor, specifically the force needed to move a permanent magnet piston through a cylinder and the electrical power necessary to generate that force. The user aims for a single-stage compressor design but is considering multiple stages to achieve a target pressure of 8-10 bar and a flow rate of over 300 liters per hour. Various contributors provide insights on the efficiency of compressors, suggesting that 1 scfm at 150 psi requires approximately 0.218 horsepower, translating to around 750 watts for the user's goals. The conversation also highlights the complexity of achieving efficient operation with electromagnets and the lack of commercially available tubular electromagnetic compressors. Ultimately, the feasibility of the project hinges on effective calculations and design considerations for the electromagnet system.
Marc bacon
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I'm trying to calculate the battery requirements for an experimental home built air compressor. I understand that electromagnetic air compressors operate using attraction and repulsion forces created by electromagnets using the basic magnetic principle that like poles of two magnets repel and opposite poles attract. I 'm trying to calculate two things. Firstly the force required to repel a permanent magnet, which acts as a piston, through a cylinder with diameter 2cm and length 30cm. Secondly how much electrical power in watts it would take to create sufficient magnetic force to repel the piston at one end and attract at the other end.. I believe the SI unit for magnetic force is N (Newton meters). Is there a formula I could use to determine these two measures? .
 
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Marc bacon said:
Firstly the force required to repel a permanent magnet, which acts as a piston, through a cylinder with diameter 2cm and length 30cm
Hi Marc bacon.
smiley_sign_welcome.gif


But you're wanting the magnet to do a lot more than just move itself---you want it to compress air in the process, and that's what will consume energy!

If it weren't compressing air, then it takes almost no energy to move a magnet slowly along a horizontal frictionless surface.

Is this going to be single-stage compression? What pressure are you aiming for? How many litres/hour of compressed air are you planning for?

I can't help you, but I'm sure these figures will be needed in any estimate.
 
Hi there,

Thanks for your reply and welcome :-) yes my initial idea was a single stage compression. But that looks unlikely, multiple stages seem more realistic. I'm aiming for 8- 10 bar. If the project is feasible with 24 or 48 volt lithium batteries from one cylinder I'm hoping for more than 300 litres an hour. I'm counting on one compression stroke taking much less than one second.
 
This might be cheating, but reciprocating compressors are pretty common devices, so you can google for one that meets your requirements and see what the power input is.
 
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upload_2016-8-20_19-33-39.png

P - Piston (permanent magnet)
N - North
S - South

Hi Russ,
Thanks for your feedback. I've googled and googled but would you believe there are currently no commercially available tubular electromagnetic air compressors, unless I'm missing something. There are many planar ones used for clutches, fish tanks and aquariums etc. These operate over very short distances. I'm trying to replicate a hand pump to pump up bicycle tyres, hence the cylinder 2cm diameter by 30cm length. By reciprocating piston do you mean one that uses a crankshaft? I'm trying to avoid that. The diagram above shows what I'm trying to achieve. Its a sort of free piston that gets pushed from and pulled to alternating poles of two electro magnets. The piston is a permanent magnet and the electromagnets change polarity from north to south and back to north and so pushes and pulls the free piston through the cylinder which I'm hoping will be powerful enough to compress air. The cylinder has a input valve for sucking in air and an output valve for compressing air into a tank. Any thoughts on how I could calculate these two variables? (force and watts)
 
(1) Google ' Solenoid driven air compressors '

(2) To get long working strokes you need at least two solenoids working push pull .

(3) A better arrangement is to use several solenoid sections packet switched .
 
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I opened my Womack fluid power designers manual and there is a chart for what you are asking.
1 scfm of air in a single stage compressor @ 150psig will require .218 hp
Efficiency of all compressors are considered to be 85%
I think you can look at almost all air motors and compressors and find this to be pretty accurate, hope it helps a little.
 
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Marc bacon said:
upload_2016-8-20_19-33-39-png.104906.png

P - Piston (permanent magnet)
N - North
S - South

The piston is a permanent magnet and the electromagnets change polarity from north to south and back to north and so pushes and pulls the free piston through the cylinder which I'm hoping will be powerful enough to compress air. The cylinder has a input valve for sucking in air and an output valve for compressing air into a tank.
Don't forget that electromagnets and permanent magnets alike all come with a North and a South. So if the piston has one end N and its other end S, then the polarities at the cylinder extremes should work together: both simultaneously N or simultaneously S.
 
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RonL said:
I opened my Womack fluid power designers manual and there is a chart for what you are asking.
1 scfm of air in a single stage compressor @ 150psig will require .218 hp
Efficiency of all compressors are considered to be 85%
I think you can look at almost all air motors and compressors and find this to be pretty accurate, hope it helps a little.

Hi RonL , thanks for your feedback and yes its a great help. Are you sure that this type of compressor will work as efficiently as an ICE or electric motor? If so: then as 1 scfm requires .218HP then 5 scfm will require around 1.1HP @ 150psi, after taking into account 15% losses I will need around 750 watts which will be feasible to run on a 48 volt 15 amp/h battery. If the compressor will run at 1,500 rpm I can get around 140 litres a minute and just over 2 litres per second if my calculations are correct.
 
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NascentOxygen said:
Don't forget that electromagnets and permanent magnets alike all come with a North and a South. So if the piston has one end N and its other end S, then the polarities at the cylinder extremes should work together: both simultaneously N or simultaneously S.

upload_2016-8-21_14-57-25.png


Hello, Yes the piston will have both a north and south pole as shown above. Will this affect the efficiency? Is this why this type of compressor isn't commercially available?
 
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RonL said:
I opened my Womack fluid power designers manual and there is a chart for what you are asking.
1 scfm of air in a single stage compressor @ 150psig will require .218 hp
Efficiency of all compressors are considered to be 85%
I think you can look at almost all air motors and compressors and find this to be pretty accurate, hope it helps a little.
Engineering toolbox has a calculator that gives you the theoretical minimum: 0.210 hp. That's close enough to what you pulled off the table.
http://www.engineeringtoolbox.com/horsepower-compressed-air-d_1363.html
Marc bacon said:
I've googled and googled but would you believe there are currently no commercially available tubular electromagnetic air compressors, unless I'm missing something.
Marc bacon said:
Hi RonL , thanks for your feedback and yes its a great help. Are you sure that this type of compressor will work as efficiently as an ICE or electric motor?
These calculations were from the standpoint of the air. The air doesn't know anything other than that it got compressed by a piston, so the methods are functionally identical from the air's point of view. For a DIY project efficiency is always problematic, but I don't see a reason why a linear motor should be less efficient than a normal rotating one if it is constructed well.
 
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upload_2016-8-21_14-57-25-png.104924.png


That's still not right. [emoji86]

The cylinder's end poles need to be the same.
 

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