cfm compared to psi chart?

[Piper]

I'm wondering if there exists a chart which compares CFM and PSI? I understand the differences with each of these units but here's where I'm coming from. I have a compressor that states "4.2 CFM at 120 PSI, or 6.4 CFM at 90 PSI" (these are just examples). So, the new compressor I'm considering buying states 19 CFM at 175 PSI. Well, most of my tools run at 90 PSI so I'm not too sure how the 175 helps me determine the "useful" CFM.

I've done a bit of searching but come up with very little. So, does anyone have any quazi scientific info on these 2 measurements? I've been told that if you're running a tool at roughly 1/2 the rated PSI quoted (so in my case 175/2 = 87.5) that the CFM would basically double or be at 38 CFM. This somehow seems kinda logical initially but I don't see that it's likely. FWIW, this is a 2 stage 80 gal compressor with a 5 hp motor, 220v.

Piper

 

[russlaferrera]

I don't know of any charts. MY understanding is about design and size (cubic inches). A two stage designed compressor will develop higher pressure. The size of the piston provides the CFM.

The requirements or specs given tell you the output at a pressure rating. Most air tools operate @ 90 psi. Withe a 175 psi reserve you are allowed more reserve because at the tank bleeds down to the cut in pressure their is no interruption in working, as the working pressure is always over 90 psi.providing the compressor is large enough.

I hoped I answered your question?

 

[W-Cummins]

The only chart of CFM VS PSI that would be created would be for a specific model of compressor pump. I'm sure that the compressor manufactures have these charts of their pumps performance. I have not seen them used in the marketing of the compressors though. Most machines are listed at their free air displacement, 90 psi and max psi, for the pumps I have seen the specs on.

I've been told that if you're running a tool at roughly 1/2 the rated PSI quoted (so in my case 175/2 = 87.5) that the CFM would basically double or be at 38 CFM. This somehow seems kinda logical initially but I don't see that it's likely

It doesn't work that way at all. The CFM at the different pressures is directly related to the individual pumps efficiency.

 

[timgr]

There's a simple gas relationship that says P1 V1 = P2 V2. It just means that you can trade off pressure for volume, and vice-versa. With no other considerations, you're right that the volume doubles at half the pressure.

I think the numbers don't add up just because the compressor has a harder time producing a higher pressure in a given time. Note that 4.2 * 120 = 504, and 6.4 * 90 = 576, so the amount of extra work done by the compressor (conversion loss) goes up as the pressure goes up. Not sure why this is - maybe the compression cycle is less efficient as the pressure goes up. But anyway, yeah the volume will double if you let the pressure drop by 1/2.

 

[W-Cummins]

There's a simple gas relationship that says P1 V1 = P2 V2. It just means that you can trade off pressure for volume, and vice-versa. With no other considerations, you're right that the volume doubles at half the pressure.

NO NO NO NO NO NO NO NO NO.......

UNITS!!! CFM = Cubic Feet per MINUTE

is the thing we are interested in here

I think the numbers don't add up just because the compressor has a harder time producing a higher pressure in a given time. Note that 4.2 * 120 = 504, and 6.4 * 90 = 576, so the amount of extra work done by the compressor (conversion loss) goes up as the pressure goes up. Not sure why this is - maybe the compression cycle is less efficient as the pressure goes up. But anyway, yeah the volume will double if you let the pressure drop by 1/2.

The numbers don't' add up because it's not a calculated value, it would be a measured value. To make a chart like that the manufacture would RUN the pump and record the observed values and plot them.... The equation P1V1=P2V2 is used on a system that those values (pressure and volume) are the only ones changing. Start changing the stuff that happens in the compressor you move into a set up that would be VERY difficult to calculate.

William...

 

[timgr]

William, if you have minutes on both sides of the equation, it should not matter. As a rough estimate, I believe it is correct. I understand what you are saying, but the question here is what the compressor can deliver, not what the pump can deliver.

Do you agree that the compressor can deliver 19 CFM at 175 psi? I presume that this is at 100% duty cycle - essentially a steady-state. Then a regulated output, set to 90 psi, is only limited by the P*V limit of the source. The regulator lets the 175 psi air expand into the delivery space until the set pressure is reached, ie 90 psi.

I'm not saying that it's an exact correspondence, but I think it's a good estimate. Even the the 504 and 576 are within about 15% of a P*V relationship.

 

[logical]

William, if you have minutes on both sides of the equation, it should not matter. As a rough estimate, I believe it is correct. I understand what you are saying, but the question here is what the compressor can deliver, not what the pump can deliver.

Do you agree that the compressor can deliver 19 CFM at 175 psi? I presume that this is at 100% duty cycle - essentially a steady-state. Then a regulated output, set to 90 psi, is only limited by the P*V limit of the source. The regulator lets the 175 psi air expand into the delivery space until the set pressure is reached, ie 90 psi.

I'm not saying that it's an exact correspondence, but I think it's a good estimate. Even the the 504 and 576 are within about 15% of a P*V relationship.

William is right. CFM is basically the amount of production it can deliver at a given level.

 

[timgr]

William is right. CFM is basically the amount of production it can deliver at a given level.

Sorry, I think you miss the point. I'm not disputing the output of the compressor. If you take any volume of gas and reduce its pressure by one-half, you will double the volume (ignoring the change in temperature).

The regulator releases gas into the lines at the supply pressure until the receiving volume (the line) reaches a certain pressure. It's a simple feedback system. Does not matter if it is flowing or not - in fact, the regulator will only work when gas flows.

If the downstream pressure is halved, the volume will double. The rate of production does not change, only the pressure changes. If you count the molecules, the amount of gas (number of molecules) does not change.

 

[roofster]

I agree with William. On the few compressoers I have seen that list a CFM at 90 or 100PSI it is no where near double the CFM at 175 psi. P1V1=P2V2 is the ideal gas law. It works more like this. A 10 gallon tank at 100 psi will be 50 psi in a 20 gallon tank.

I saw somewhere, maybe Eaton compressors, that displaced CFM is bore times stroke times pump rpm. Seems to me it would be more Pi/4 x bore squared x stroke (areaxlength) x rpm. Which is fine until you have a mulit cylinder 2 stage with different bores and strokes. Then you Best start measuring. Then consider pump efficiency and loses to get an actual CFM. Finally consider it takes more work to get to 175 psi, but yet you still only have a 5 hp engine, so the pump up time increases but not per the ideal gas law.

 

[Piper]

So this is more involved as I thought it was going to be. I guess I should talk to the manufacturer. I think generally that we all agree that at 90 psi the cfm will be just as high (and likely higher) than at 175 psi. I just kind of find it strange that they'd indicate 175 which is essentially a pressure that isn't used. Why not rate it at 90? I dunno. I'll phone the manufacturer on Tuesday and see what they say.

I just want to buy the compressor once and have enough air for my sandblaster, air tools etc that I don't ever say, "gee, I should have got the ..." Buy once, cry once.

Piper

 

[cw_racefan]

The relationship between CFM and PSI has nothing to do with ideal gas equations or the pressure volume relationship. It is completely compressor dependent, thus as William said, the only table would be for a given compressor. Flow rate is different than volume. A more better/more efficient compressor can produce a higher flow rate at a given pressure than a less efficient compressor. To the original posters question, you need to look at the requirements of the tools you want to use, they will all have a CFM @ PSI rating. Compare those requirements to the specs of the compressor you're looking at, and make sure it will supply the flow rate for the highest CFM tool you want to use. BTW, as an example, one impact wrench may require 4 cfm @90psi, where another may require 8cfm @90psi, so your requirements are tool dependent as well.

Craig

 

[Piper]

so to finish off with what I found from the manufacturer. He told me that they always rate 2 stage compressors at 175 psi and that at 90 psi I could expect an additional 10% CFM. So, 18.7 CFM at 90 psi.

Thanks for all your thoughts!

Piper

 

[Lyaec350]

If you have two tanks and just dump air from one to the other, the P1V1=P2V2 relationship holds true, but in a time dependent system (a compressor pump w/ flow) the PV relationship won't hold up.

Think about it this way, you're pumping up an empty, closed, tank from 0 to 100psi. At first the compressor will find it very easy (little work) to push a unit (cubic foot of air) into the tank. Since the pump can only do a fixed amount of work in a given time, as the pressure increases and the amount of work to push in each unit of air increases, the time to pump that unit of air in will increase as well, resulting in a decrease in the cubic feet per minute pumped in to the tank. Basically it is load dependent and it is not a linear relationship.

It is based on the "load" that the compressor sees at the other end of the system.

 

[roofster]

Piper,

As you shop and compare you will notice that some single stage compresors will have higher cfm ratings at 90 psi than more expensive 2 stage units. I don't know for sure, but the claimed advantages of a 2 stage (175 psi) compressor are cooler air due to less work and therefore dryer air, maybe less noise and more storage. I read that there is more air in a 60 gallon tank at 175 psi than in an 80 gal at 90 psi. I'm looking for a unit to do sandblasting and have decided on a 2 stage one.

 

[timgr]

Sorry to bring this up again guys, but none of the explanations previously given were very satisfying. So, I think it might be helpful to thrash it a little more.

First, let me say that my reasoning about this seems to have been completely wrong. I think my understanding of the physics is right, but my (and I think all the other posters) interpretation of the compressor specs is wrong.

I did a little research, and some arithmetic, and here's what I think is going on.

First, the CFM specs for the compressor always are in terms of free air at the intake. So, 19 CFM means an actual flow of air at 185 psi going into the tank is 19 CFM * 15 psi / (185 + 15 psi) = 1.4 CFM. I guess this way of specifying the flow is as good as any, since when all the work is taken from the air (ie at the exhaust of the tool), it's back at atmospheric pressure.

The specified pressure is the 'set point' of the compressor. When the manufacturer says that the CFM increases by 10% at 90psi, that means that the amount of free air that the compressor can pump increases to 19 + 1.9 = 21 CFM. (I think that 18.7 is 17 PSI + 10%? Maybe the OP meant 17, not 19). This makes sense, because when the set point is lowered, the pump is working against less back pressure, so the volume pumped goes up a little.

However, with the set point of the compressor at 90 psi instead of 175 psi, the reserve of volume in the tank is cut by about 1/2: the same volume will only hold half as much air at half the pressure. You would hardly ever want to do this, since the amount of work (P*V=W) that the compressor can do on a full tank of air is also halved. The only time this would make sense would be at near 100% duty cycle, and where you need more than 19 CFM and less than 21 CFM - ie not often.

So, when you set the regulator to 90 psi because the tool requires it, that only affects the rate at which air is released, and it's not really related to the pressure that the compressor is spec'd to. The higher pressure only means that you have more reserve in the tank, since (I presume) the tools are spec'd in atmospheric air at the outlet, just like the compressor is spec'd in atmospheric air at the inlet.

 

[W-Cummins]

Sorry to bring this up again guys, but none of the explanations previously given were very satisfying. So, I think it might be helpful to thrash it a little more.

Well if you must

First, let me say that my reasoning about this seems to have been completely wrong. I think my understanding of the physics is right, but my (and I think all the other posters) interpretation of the compressor specs is wrong.

Well one out of three isn't too bad.... Your physics was wrong in thinking that you can use a static equation, P1V1=P2V2 in a dynamic system. You can't make it a dynamic equation by just adding time to it.

I did a little research, and some arithmetic, and here's what I think is going on.

First, the CFM specs for the compressor always are in terms of free air at the intake. So, 19 CFM means an actual flow of air at 185 psi going into the tank is 19 CFM * 15 psi / (185 + 15 psi) = 1.4 CFM. I guess this way of specifying the flow is as good as any, since when all the work is taken from the air (ie at the exhaust of the tool), it's back at atmospheric pressure.

Well your still at it, You CAN NOT add time to the static Ideal gas eq. So all you get above is CF NOT CFM!!!

Here is the correct eq you used
To find cubic feet of either compressed air or free air

cu.ft. comp. air X (psig + 14.7) = cu.ft. free air X 14.7

• 
  [*]cu.ft. comp. air [ Cubic Feet of Compressed Air ]
  [*]cu.ft. free air [ Cubic Feet of Free Air ]

The specified pressure is the 'set point' of the compressor. When the manufacturer says that the CFM increases by 10% at 90psi, that means that the amount of free air that the compressor can pump increases to 19 + 1.9 = 21 CFM. (I think that 18.7 is 17 PSI + 10%? Maybe the OP meant 17, not 19). This makes sense, because when the set point is lowered, the pump is working against less back pressure, so the volume pumped goes up a little.

Ok here again the train departs from the track....

The manufactures I have seen don't say that the CFM increases by 10% They say that the free air spec is the HIGHEST CFM the pump can put out. Then if they list a 90 psi rating it's lower than the free air spec. and then the Max rated pressure has an even lower CFM. Exactally what you would expect to see as the pump works harder to compress the air to the higher pressure, it can't match the free air CFM rating IE. a 100% pumping effecency at say 175psi. So you see ratings like 20cfm free air, 18cfm @ 90 psi and 17. cfm @ 175 psi..... on a 2 stage 5 hp compressor....

since the amount of work (P*V=W)

Dam I will say at least your consistent ..... P*V NOT = work

William....

 

[logical]

Toss those formulas and wrap yer mind around this:

You pump the tank up to 150 psi, then you start bleeding off air at 150 psi with the pump running. Now keep increasing the rate that you bleed off (use) the 150 psi air. At some point the pump can't keep up and your tank (air supply) drops below 150 psi. The bleed rate where this happens is your max CFM at 150 psi. So if this happens when you get up to bleeding off 60 cubic feet per minute...then your compressor is capable of 60 CFM @ 150 psi.

Do the test again at 75 psi and it will be a different CFM. The relatinship might be 2:1 but if it is, it is just a happy coincidence. The relationship curve is unique to every compressor...pump size, pump configuration, pump design, pump efficiency, motor HP, motor performance under load, etc. The only place p1v1 = p2v2 is when you are filling your bicycle tire with a portable harbor freight 5 gallon tank and the air stops flowing.

 

[timgr]

Hey William - I hope you don't take this the wrong way. I appreciate your input on this... and I could be wrong, but here's my response.

Well one out of three isn't too bad.... Your physics was wrong in thinking that you can use a static equation, P1V1=P2V2 in a dynamic system. You can't make it a dynamic equation by just adding time to it.

In this case, sure you can. It's just a rate. There's no fancy dynamic dependence. It has no functional dependence other than "t", a linear term. See below.

cu.ft. comp. air X (psig + 14.7) = cu.ft. free air X 14.7

• 
  [*]cu.ft. comp. air [ Cubic Feet of Compressed Air ]
  [*]cu.ft. free air [ Cubic Feet of Free Air ]

Here you've restated what I wrote, in a different form: CFM1 * P1 = CFM2 * P2. If "t" is a unit of time, CFM1 = V1/t. CFM2 = V2/t. P1 V1/t = P2 V2/t. P1 V1 = P2 V2. The 14.7 is an offset due to the atmospheric pressure, always present - free air pressure is already 14.7 psig. You've contradicted your claim in the first paragraph.

In my reply, I rounded 14.7 to 15 - something I'm inclined to do because of my training/work.

Ok here again the train departs from the track....

The manufactures I have seen don't say that the CFM increases by 10% They say that the free air spec is the HIGHEST CFM the pump can put out. Then if they list a 90 psi rating it's lower than the free air spec. and then the Max rated pressure has an even lower CFM. Exactally what you would expect to see as the pump works harder to compress the air to the higher pressure, it can't match the free air CFM rating IE. a 100% pumping effecency at say 175psi. So you see ratings like 20cfm free air, 18cfm @ 90 psi and 17. cfm @ 175 psi..... on a 2 stage 5 hp compressor....

The OP wrote that the manufacturer's rep said that the CFM rating of the compressor goes up by about 10% when the setpoint pressure is reduced from 175 psig to 90 psig. Not my number - just trying to understand the rep's claim.

I'm not sure what you are disputing here. My reading says that the CFM rating is free air at the input, ie CFM of atmospheric air. You have a problem with that? My reading also says that the pressure rating is not the regulated pressure at the output, but the setpoint pressure of the compressor. Do you have a problem with that?

Dam I will say at least your consistent ..... P*V NOT = work

Sure it is. Work = force * displacement. Let's define some new variables.
F = force
A = area
L = length aka displacement
Now, P=F/A, V = L*A.
P*V = F*L*A/A = F*L = W, qed.

You can also do this by dimensional analysis.
Work is joules, in units of N-M (newton-meters).
Pressure is N/M^2. Volume is M^3.
N/M^2 * M^3 = N-M.

 

[timgr]

Toss those formulas and wrap yer mind around this:

You pump the tank up to 150 psi, then you start bleeding off air at 150 psi with the pump running. Now keep increasing the rate that you bleed off (use) the 150 psi air. At some point the pump can't keep up and your tank (air supply) drops below 150 psi. The bleed rate where this happens is your max CFM at 150 psi. So if this happens when you get up to bleeding off 60 cubic feet per minute...then your compressor is capable of 60 CFM @ 150 psi.

Do the test again at 75 psi and it will be a different CFM. The relatinship might be 2:1 but if it is, it is just a happy coincidence. The relationship curve is unique to every compressor...pump size, pump configuration, pump design, pump efficiency, motor HP, motor performance under load, etc. The only place p1v1 = p2v2 is when you are filling your bicycle tire with a portable harbor freight 5 gallon tank and the air stops flowing.

The question then is, where do you measure the CFM? Air changes volume at different pressures and temperatures. As I understand it, volume is measured at the input. It does not really matter if you measure it at the input or the output as long as the P and T are the same. Mass (the actual amount of air) is always conserved.

 

[logical]

.......... Mass (the actual amount of air) is always conserved.

No, mass is released and then if the compressor has the balls to keep up, it is replenished. Again, you need to think of the ability of the pump to keep up with a rate of use....flow. (a given CFM rate of 150 psi air is not the same volume as that at a different psi).

My 12v cigarette lighter pump will make 150 psi but not very fast.

 

[logical]

Reminds me of:
Elaine: Come on, come on kitty...[slice disappears] ooh...how about that; it worked! Wow, can I borrow that thing for a while?
Kramer: Oh no, I don't think so.
Elaine: Why not?
Kramer: Well, you're not checked at on it.
Elaine: What do I have to know?
Kramer: Well, where the meat goes?
Elaine: Right there.
Kramer: Where do you turn it on?
Elaine: Right there.
Kramer: But where does the meat go?

 

[logical]

Displacement of air is interesting if we were talking about a fan but means next to nothing when talking about compressors. Again...focus on the pump's ability to compress air to a given psi at a given rate....how fast can it make pressurized air.

 

[timgr]

Displacement of air is interesting if we were talking about a fan but means next to nothing when talking about compressors. Again...focus on the pump's ability to compress air to a given psi at a given rate....how fast can it make pressurized air.

Man, you're the one who's going round and round.

All I'm claiming is that you can equate potential and kinetic energy. P*V is potential energy. Work is kinetic energy. Displacement is a convenient unit to demonstrate that. If you read all the earlier posts you'll understand why it came up.

 

[logical]

I'm wondering if there exists a chart which compares CFM and PSI?

Yes, but it has more than one line on it and they aren't straight.

Agreed?

 

[W-Cummins]

Hey William - I hope you don't take this the wrong way. I appreciate your input on this... and I could be wrong, but here's my response.

No problem here I too have made mistakes in this post see below!

In this case, sure you can. It's just a rate. There's no fancy dynamic dependence. It has no functional dependence other than "t", a linear term. See below.



Here you've restated what I wrote, in a different form: CFM1 * P1 = CFM2 * P2. If "t" is a unit of time, CFM1 = V1/t. CFM2 = V2/t. P1 V1/t = P2 V2/t. P1 V1 = P2 V2. The 14.7 is an offset due to the atmospheric pressure, always present - free air pressure is already 14.7 psig. You've contradicted your claim in the first paragraph.

In my reply, I rounded 14.7 to 15 - something I'm inclined to do because of my training/work.

I didn't restate your equation I posted the CORRECT equation as it's used. There is no CFM in the equation as you added to yours.

The eq that your trying to adapt (p1v1=p2v2) to this story, is a SPECIAL case of the ideal gas law that is RESTRICTED to Static systems. The systems must be at equilibrium for it to work. In a dynamic pumping system you will have many other factors to deal with. Just for Fun why not go back to the general gas law eq. PV=nRT and deal with the delta T of 320deg F added to the gas as it's compressed from STP to 175 psi like my compressor does.

The Timgr eq as we can call it, as your its inventor has yet to be proven by its creator. We know that it fails to explain the pumping seen by a real compressor. Because the CFM produced @ 180 psi compared to the value @ 90 psi is not 50%

The OP wrote that the manufacturer's rep said that the CFM rating of the compressor goes up by about 10% when the setpoint pressure is reduced from 175 psig to 90 psig. Not my number - just trying to understand the rep's claim.

Yep my mistake here I missed the original poster's added post about the manufacture and his 10% difference

I'm not sure what you are disputing here. My reading says that the CFM rating is free air at the input, ie CFM of atmospheric air. You have a problem with that?

ME either see above

My reading also says that the pressure rating is not the regulated pressure at the output, but the set point pressure of the compressor. Do you have a problem with that?

This one remains rather fuzzy what is the set point pressure?

Sure it is. Work = force * displacement. Let's define some new variables.
F = force
A = area
L = length aka displacement
Now, P=F/A, V = L*A.
P*V = F*L*A/A = F*L = W, qed.

No you said that P*V=W Once again you have a STATIC system there! NO work done! There is no change in either pressure or volume ( no displacement) so no work done on the system.

I see that you have corrected your self below and called it correctly as POTENTIAL energy

P*V is potential energy.

William....

 

[timgr]

I didn't restate your equation I posted the CORRECT equation as it's used. There is no CFM in the equation as you added to yours.

I don't see how it's different from what I posted, but we can disagree.

The eq that your trying to adapt (p1v1=p2v2) to this story, is a SPECIAL case of the ideal gas law that is RESTRICTED to Static systems. The systems must be at equilibrium for it to work. In a dynamic pumping system you will have many other factors to deal with. Just for Fun why not go back to the general gas law eq. PV=nRT and deal with the delta T of 320deg F added to the gas as it's compressed from STP to 175 psi like my compressor does.

The Timgr eq as we can call it, as your its inventor has yet to be proven by its creator. We know that it fails to explain the pumping seen by a real compressor. Because the CFM produced @ 180 psi compared to the value @ 90 psi is not 50%

Your response bothers me because I feel you are making this WAY more complicated than it needs to be. Any physical problem can be drilled down to a point where the complexity becomes overwhelming - that's not profound. Making appropriate simplifications can be hard though.

Maybe we're simply getting off track. I thought we were trying to get some understanding of what the specs mean, not of what goes on inside the compressor. In terms of the specs, everything is spec'd in terms of volume at STP (mass) and the time dependence is not complicated. It's just a rate.

ME either see above

I presume you're pointing to the STP comment above, and you are agreeing with me, or I'm agreeing with you, or some such.

This one remains rather fuzzy what is the set point pressure?

It's the pressure that the compressor maintains in the tank, +/- the operational range; ie the setting of the pressurestat.

No you said that P*V=W Once again you have a STATIC system there! NO work done! There is no change in either pressure or volume ( no displacement) so no work done on the system.

I see that you have corrected your self below and called it correctly as POTENTIAL energy

Sorry, I fail to see the distinction. Energy is energy. The unit of work is energy. The equation is not invalidated because you can make the conditions more complicated. It is indeed true that the amount of work you can do with the air is equal to the stored work, which is the pressure times the volume. I can't change that. PE and KE are something of a distinction without a difference, though useful for a lay explanation (sorry if that sounds haughty).

We could instead do this in terms of power - we'll have to pick a unit, say K for kilowatts... then PV/t = K ... but that has a time dependence, so I'm in jeopardy of stirring up a whole new pot of ugly complexity. (nothing snarky intended - I just wanted to use that icon!)

Let's do it anyway, just for fun. We know that conservation of mass says that we can't release any more gas than we can pump into the tank at STP. This is a suitable use for P1V1/t=P2V2/t by your rules, since it's a static situation. So P2 = 185 psig and V2 = 15/(185+15) * 19 = 1.4 CF. We can deliver power as 1.4 CFM of air at 185 psig. The easiest way to proceed here is to convert this to Pascals and Kg to get watts.
185 psi * 6.895 kPa/psi = 1276 kPa
1.4 CF / 35.5 CF/M^3 = 0.04 M^3
1276 kPa * 0.04 M^3 / min / 60 sec/min = 0.85 kW of power = K.
Now, 1 kW = 1.34 HP, so the compressor can deliver 1.14 HP as air to the tool at a constant rate.

This also gives us an efficiency rating. If the motor really produces 5 HP, then we can deliver 1.34/5*100% = 27% of the motor power to the tool. This is reassuring, since I would have guessed that the efficiency would be something less than 50%, but still a significant fraction of the motor power.

Cheers - Tim

 

[W-Cummins]

No you said that P*V=W Once again you have a STATIC system there! NO work done! There is no change in either pressure or volume ( no displacement) so no work done on the system.

I see that you have corrected your self below and called it correctly as POTENTIAL energy

Sorry, I fail to see the distinction. Energy is energy. The unit of work is energy. The equation is not invalidated because you can make the conditions more complicated. It is indeed true that the amount of work you can do with the air is equal to the stored work, which is the pressure times the volume. I can't change that..

Well lets just leave it at you don't get it. I'm not here to teach you basic physics. So I'm moving on.

William....

 

[Certified]

i am not sure if this will help, i found this on a forum, basicaly you take your listed cfm @ 175 X 1.81 = cfm @ 90

exemple:
14 CFM@175PSI X1.81 = 25.34CFM@90PSI

i hope it helps

http://www.physicsforums.com/archive/index.php/t-155966.html

 

[fomocoforrester]

This thread illustrates one of my pet peeves.

Why, oh why! do people insist on calling a compressor a pump?

A compressor increases the pressure of a compressible fluid - ie a gas.

A pump increases the pressure of an incompressible fluid - ie a liquid.

Grrrrrrrr!!!

 

[Megadodo]

While the SI unit for work and energy are the same, the "joule". Work is the transfer of energy. If energy is present and not being transferred it is potential energy. The difference is subtle but must be fully understood to make rational statements about practical applications.

 

[er3456df]

Timgr- A thought experiment for ya- If you increase the pressure in your tank, do you get MORE, or LESS airflow through the hose?

Now look at P1V1=P2V2...... Increasing P1 causes V1 to reduce. How can this be? The reason is that your formula only works when mass is constant. It's a valid formula, but it doesn't apply to this discussion.

 

[EOC_Jason]

When I was doing research on a new air compressor it was interesting to see how each manufacturer listed specs... CFM @ 90PSI, 100PSI, 175PSI, MAX PSI, etc... Some were free air cfm, displaced cfm, and actual cfm, oh it was just irritating... Lots of digging and emails finally yielded results I needed... But anyhow...

To answer the OP's question

For 5HP - 10 HP units, 2 stage models... Typically there was only maybe a couple CFM gained at 100 PSI vs 175 PSI. Yes, a measly 2 CFM AT BEST... Some were only maybe around HALF A CFM...

My two cents... leave it at 175 PSI and stick a regulator on the end for what ever pressure you need.

 

[Devildogs]

It seems to me nobody is interested in the answer to the original question, which by the way, is sledgehamer simple. You match the required tool psi and cfm reuirement to the compressor max output rating. You guys sound like a bunch of mad process engineers.

 

[fomocoforrester]

Or - Rotating Equipment Engineers....

 

[Thruxton]

Certified is on the right track. There are really three values needed to relate cfm (volume) to psi (pressure), and the one missing is hp (power). There are also some corrections to apply for a "real world" calculation, but the basic calculation involves those three.

 

[darkk]

Well......a 220 volt 5hp 2 stage 80 gallon compressor will surely produce enough CFM to run your most air hungry tools at 90psi as long as you're not trying to run 2 or 3 at one time. If it'srated at 19CFM @ 175 psi, that is the max it will ever produce, it will produce slightly less at a lower psi. But I can assure you that the compressor in question will produce no where near 38 CFM at any time in it's useful life. All the figuring in the world isn't going to help you here. Compressor companies always state the highest CFM they can. Just look at thespecs of any compressor you can find on the internet. Some will give you figures of a mid range psi/cfm and high range to compare.

 

[fomocoforrester]

Certified is on the right track. There are really three values needed to relate cfm (volume) to psi (pressure), and the one missing is hp (power). There are also some corrections to apply for a "real world" calculation, but the basic calculation involves those three.

Exactly - some people are not balancing the dimensions - length - force - time - on both sides of the equations.

 

[DekeT]

There's a simple gas relationship that says P1 V1 = P2 V2. It just means that you can trade off pressure for volume, and vice-versa. With no other considerations, you're right that the volume doubles at half the pressure.

I think the numbers don't add up just because the compressor has a harder time producing a higher pressure in a given time. Note that 4.2 * 120 = 504, and 6.4 * 90 = 576, so the amount of extra work done by the compressor (conversion loss) goes up as the pressure goes up. Not sure why this is - maybe the compression cycle is less efficient as the pressure goes up. But anyway, yeah the volume will double if you let the pressure drop by 1/2.

I have tried to follow all your math in the rest of this thread and I think I get it. To address the specific question of the OP it appears you are saying that he will get double flow of cfm if he runs his tools at 85 psi. I did not notice anything else from your following threads that says otherwise. yeah? nay?

 

[SGKent]

Sorry, I think you miss the point. I'm not disputing the output of the compressor. If you take any volume of gas and reduce its pressure by one-half, you will double the volume (ignoring the change in temperature).

The regulator releases gas into the lines at the supply pressure until the receiving volume (the line) reaches a certain pressure. It's a simple feedback system. Does not matter if it is flowing or not - in fact, the regulator will only work when gas flows.

If the downstream pressure is halved, the volume will double. The rate of production does not change, only the pressure changes. If you count the molecules, the amount of gas (number of molecules) does not change.

in a static world. One has manifolds, reeds, tubing, hoses etc that all introduce variables. Take something as simple as laminar flow - like a stream moves faster at the outside of a turn and slower at the inside. Those things are dynamic and changing just like a river changes with flow.

Generally what the OP wants is the most CFM he can get at the pressure he will be using it at with his decision balanced between price and service history. If he wants something with a longer warranty he may have to trade off CFM - and say get a cast iron sleeved compressor vs an aluminum sleeved one. One can almost never have enough CFM cause limiting it will limit the tools he can use. Don't do like I did and buy one that worked well for what I had at the time but later discover it won't support the new tools so now I have two compressors tied together at the tanks so I can run high volume. Buy the right compressor once.

 

[Bob C]

Pressure and Volume are indirectly proportional. As Pressure increases, Volume decreases and vise versa. If a compressor states ACFM(Actual CFM) 10.6 at 90 and 12.4 at 40. It's real simple. It means it will deliver 12.4 ACFM at 40 psi continuous and 10.6 ACFM at 90 psi continuous. If you have any doubt whatsoever, put a flow meter on the line and measure it. These raw equasions being put forth here are lacking as they do not account for your local Atmospheric pressure and humidity factors to start with. Those equasions are designed for laboratory use. Some of you guys have garages so clean they could be labs, but that's another story.