dumb question on compressor displacement cfm

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900 strokes per minutes (which Grant referred to as 900 rpm in another post) is 900 compressions and 900 intakes per minute. In Grant's original calculation (which was something like (4*3.5*pi/12^3)*900*2 CFM), the factor of 2 is for 2 cylinders.
Each compression in one cylinder pumps 4*3.5*pi = 43.98 cubic inches of air, or about .0255 cubic feet. Hence 4*3.5*pi/1728*900 CFM or 22.9 CFM per cylinder, or about 45.8 CFM for two cylinders.
Reply to
James Waldby
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Remember that the compressor may not "pump" anything at all depending on the input pressure (atmospheric on a single stage compressor) and the output pressure. If the outlet pressure requirement / back pressure is sufficiently high, and particularly if there is a moderate amount of unswept dead space, there will be no output at all from the compressor and no input either as the compressor will simply keep compressing and decompressing the same volume of air. This makes more sense when you consider multi stage compressors and gas boosters.
Reply to
Pete C.
Stroke is the distance a piston travels up or down every 1/2 revolution. Each cyl. contributes 1 "volume" of air each stroke or 1/2 revolution.
Bob Swinney
It seems to me, Robert, that more correctly each piston sucks air in on 1/2 a revolution of the crank and blows air out on the other 1/2. But the piston travels the full stroke length each time, and of course has the same bore. And there are two cylinders.
((pi * (bore/2)^2 * stroke) / 1728) * rpm TIMES NUMBER OF CYLINDERS
i.e. total cyl vol. in cubic inches, converted to cubic feet, times rpm = cfm of the free air displacement variety
I wish I could understand how you figure a cylinder only pumps half its volume per crankshaft revolution. If that were true it would make mathemtical sense.
Grant
Robert Sw> Probably the statement, "Each piston compresses on each stroke" means that:
Reply to
Robert Swinney
Ah, but there is something even you may not know yet, Mr. Kinch. One horsepower from a gas engine doesn't equal one horsepower from an electric motor (assuming this is not something sold at Home Depot, in which case the horsepower rating is largely fictional). The good people at Quincy a long time ago told me that the electric equivalent of an 18 hp gas engine is a 7.5hp electric motor.
My compressor's specs are now known:
770 rpm -> 21.4 CFM @ 100psi 900 rpm -> 25.4 CFM @ 100psi
So to get 25.4 CFM you would indeed need most of a 7.5hp electric motor.
Why is a gas engine horsepower different from an electric motor horsepower?
Oh, boy.
Grant
Reply to
Grant Erwin
Yeah, that's something it took me some time to understand. I assumed (wrongly) that if it was rated as X CFM @ 90 PSI, they were measuring the volume at 90 PSI. But that's just not true. It's as you say, the volume of air sucked in they are measuring. The compressor is rated at different PSI values only because the CFM drops as the back pressure of the tank rises.
I'll answer the original question if no one else made it clear in the thread (I have another 20 posts to read first).
Reply to
Curt Welch
Lots of ways to describe the difference but one is that the gasoline engine generally has a throttle and is therefore economically not designed to deliver its maximum horsepower constantly. They generally have a fairly constant torque over a wide RPM range so the horsepower is pretty much proportional to the speed. Onan and probably others have built nearly identical engines rated at 1800 and at 3600 RPM, with the higher speed giving about twice the horsepower. The electric motor can deliver its rated horsepower (or slightly more) constantly but can not deliver a lot more even for a short time. The gasoline motor is commonly rated for the maximum power it can deliver, but it can not deliver that amount constantly and have a satisfactory service life. There are industrial internal combustion engines which are rated for continuous horsepower.
A useful rating system would give both peak and continous horsepower and motors are sometimes so rated (and over-rated). Actually, the horsepower to drive a compressor at a certain speed and pressure is exactly the same regardless of the type of motor. But a 5HP electric motor can deliver that 5HP constantly with good life but a 5HP gasoline motor would be running "wide-open" and would not last long.
Lots of other ways to look at it; electric motors can withstand short term torque overload without stalling better than a gasoline engine, for example. Just quite different speed, torque, and other characteristics for the two types of motors.
Don Young
Reply to
Don Young
My engine (a Wisconsin THD) is an industrial gas engine designed to run at one speed (at a time). It's horsepower varies somewhat with RPM, from say 14.1 hp @ 2200 rpm up to 18.0 at 3200 rpm. It is fitted with a variable speed control but it isn't like a gas pedal - it only varies the speed over a small range. This engine never idles.
I stand by my earlier statement. A hp from a gas engine - ANY gas engine - is less than half the hp from an electric motor.
GWE
Reply to
Grant Erwin
You create some of the more fun questions here Grant!
A few people seemed to have touched on the answer, but no one seems to have made it clear. Let me try.
The problem lies with the dead space at the top of the cylinder. You can't calculate the CFM unless you also have an good measure of the effective volume of that dead space (and I have no clue what is typical).
The problem Grant, is what Pete says here. Not all the air in the stroke actually gets pushed out into the tank. The worst case is that none of it gets pushed out. But even with the normal case, not all of the air which was in the stroke volume, will get pushed out.
The calculations that Pete posted yesterday I believe were invalid because they seemed to be based on the assumption that CFM was a measure of compressed volume (which is a good guess - but just not true).
Lets do an example with simple numbers. Lets say we have a stroke volume of 90 cubic inches, and 10 cubic inches of dead space (not trying to be realistic here). This means that the 100 ci of the cylinder gets compressed down to 10 ci at TDC. So the pressure will go from 15 PSI, to 150 PSI in this compressor. That means that the max it can do is 150 PSI. At 150 PSI, no air will go into the tank per the exmaple Pete gave.
At 140 PSI of tank pressure, the air won't start to flow until the cylinder pressure reaches 140 PSI. 140/15 is 9.3e so that's a 1 to 9.3 compression ratio we need. But because the piston must compress both the stroke volume AND the dead space volume, we need more than 1:9.3e of stroke compression. We need enough piston movement to create a 1:9.3e compression on that 110 ci of air before air will start to flow to the tank. The cylinder volume needs to be reduced from 110 to 110/9.33 or 11.79 ci before air flows. From that point, the stroke is only 1.79 ci away from TDC. So as it pumps the air out, the pressure effectively stays at 140 PSI, but only 1.79 ci of air gets pumped out. The rest stays in the dead space and doesn't go into the tank.
1.79 ci of air at 140 PSI, is the same as 1.79 * 9.33 = 16.7 ci of air at 15 PSI.
So where the stroke of this example was 100 ci per stroke, the air moved per stroke was only 16.7 ci when the pressure reached 140 PSI. So the efficiency of the compressor drops as the pressure rises. And it's all because of the dead space at the top of the cylinder. The larger the dead space, the faster the CFM numbers drop as tank pressure goes up.
Now I assume for air compressors, they attempt to minimize that dead space. But with a reed valve popping open, there must be some "blow back" as air in the reed valve area pushes back into the cylinder as the piston starts to fall and the valve closes. Or in other words, the area created by the opening of the valve, adds a little to the dead space.
At the same time, I wonder if too little dead space might be a safety concern? The compressor can only take so much pressure before blowing up or breaking a connecting rod, and one simple sure way to limit the max pressure is by intentionally including a little dead space. Like the numbers I used above the system was limited to 150 PSI. Maybe for a 120 PSI compressor they might limit the pressure to 200 PSI by adding a little dead space for safety reasons? They way, even if the pressure cutoff switch malfunctioned, the pressure could not go above 200 PSI even with the motor constantly running?
None the less, the only problem with Grant's numbers that I can see, is that he didn't include the effect of the dead space at the top of the cylinder. With the tank pressure at 0, his calculation should be close to correct. But as the tank pressure rises, the CFM numbers will drop. Without knowing the effective volume of the dead space, you won't be able to calculate how much the CFM drops.
But since he has a number for the CFM, we should be able to calculate the dead space.
Reply to
Curt Welch
Horsepower is a physical measure of power and as such it has NO reference or variation to the source. It is the same unit for waterwheels, horses, electric motors, bicyclists, turbines, IC engines, rockets, etc.
Of course we know that the word "horsepower" is applied to every fantastical characterization of power with no fixed relation to the physical unit. Like vacuum cleaners, air compressors, hi-fi amplifiers, etc.
Reply to
Richard J Kinch
True, but the rated HP is measured at the optimum speed and load. Less well matched real-world loads shift the operating point along the speed / torque curve.
jsw
Reply to
Jim Wilkins
Err, not really. for example:
# One mechanical horsepower of 550 foot-pounds per second is equivalent to 745.7 watts # A metric horsepower of 75 kgf-m per second is equivalent to 735.499 watts # A boiler horsepower is used for rating steam boilers and is equivalent to 34.5 pounds of water evaporated per hour at 212 degrees Fahrenheit, or 9809.5 watts # One horsepower for rating electric motors is equal to 746 watts
Cheers,
Bruce (bruceinbangkokatgmaildotcom)
Reply to
Bruce In Bangkok
It's often said that the rated HP power requirement difference is in the type of power source.. 60 cycle AC vs. 4 cycle IC.
A 5 HP 4 cycle gas engine adequately replaces a 2 HP electric motor.
Reply to
Wild_Bill
That 2 HP is (or should be) the continuous rating limited by temperature rise, and not nearly the short-time maximum.
jsw
Reply to
Jim Wilkins
HHmmnnnn . . .
Dead space. It too is filled with the CFM flow. We call that "clearance in steam engines.
Bob Swinney
"Pete C." wrote:
You create some of the more fun questions here Grant!
A few people seemed to have touched on the answer, but no one seems to have made it clear. Let me try.
The problem lies with the dead space at the top of the cylinder. You can't calculate the CFM unless you also have an good measure of the effective volume of that dead space (and I have no clue what is typical).
The problem Grant, is what Pete says here. Not all the air in the stroke actually gets pushed out into the tank. The worst case is that none of it gets pushed out. But even with the normal case, not all of the air which was in the stroke volume, will get pushed out.
The calculations that Pete posted yesterday I believe were invalid because they seemed to be based on the assumption that CFM was a measure of compressed volume (which is a good guess - but just not true).
Lets do an example with simple numbers. Lets say we have a stroke volume of 90 cubic inches, and 10 cubic inches of dead space (not trying to be realistic here). This means that the 100 ci of the cylinder gets compressed down to 10 ci at TDC. So the pressure will go from 15 PSI, to 150 PSI in this compressor. That means that the max it can do is 150 PSI. At 150 PSI, no air will go into the tank per the exmaple Pete gave.
At 140 PSI of tank pressure, the air won't start to flow until the cylinder pressure reaches 140 PSI. 140/15 is 9.3e so that's a 1 to 9.3 compression ratio we need. But because the piston must compress both the stroke volume AND the dead space volume, we need more than 1:9.3e of stroke compression. We need enough piston movement to create a 1:9.3e compression on that 110 ci of air before air will start to flow to the tank. The cylinder volume needs to be reduced from 110 to 110/9.33 or 11.79 ci before air flows. From that point, the stroke is only 1.79 ci away from TDC. So as it pumps the air out, the pressure effectively stays at 140 PSI, but only 1.79 ci of air gets pumped out. The rest stays in the dead space and doesn't go into the tank.
1.79 ci of air at 140 PSI, is the same as 1.79 * 9.33 = 16.7 ci of air at 15 PSI.
So where the stroke of this example was 100 ci per stroke, the air moved per stroke was only 16.7 ci when the pressure reached 140 PSI. So the efficiency of the compressor drops as the pressure rises. And it's all because of the dead space at the top of the cylinder. The larger the dead space, the faster the CFM numbers drop as tank pressure goes up.
Now I assume for air compressors, they attempt to minimize that dead space. But with a reed valve popping open, there must be some "blow back" as air in the reed valve area pushes back into the cylinder as the piston starts to fall and the valve closes. Or in other words, the area created by the opening of the valve, adds a little to the dead space.
At the same time, I wonder if too little dead space might be a safety concern? The compressor can only take so much pressure before blowing up or breaking a connecting rod, and one simple sure way to limit the max pressure is by intentionally including a little dead space. Like the numbers I used above the system was limited to 150 PSI. Maybe for a 120 PSI compressor they might limit the pressure to 200 PSI by adding a little dead space for safety reasons? They way, even if the pressure cutoff switch malfunctioned, the pressure could not go above 200 PSI even with the motor constantly running?
None the less, the only problem with Grant's numbers that I can see, is that he didn't include the effect of the dead space at the top of the cylinder. With the tank pressure at 0, his calculation should be close to correct. But as the tank pressure rises, the CFM numbers will drop. Without knowing the effective volume of the dead space, you won't be able to calculate how much the CFM drops.
But since he has a number for the CFM, we should be able to calculate the dead space.
Reply to
Robert Swinney
1 gas horsepower = 1 electric hp = 33,000 ft pounds of work per minute
But:
A gas engine has a max torque that is purely dependent on the burning fuel in the cylinder. What's there is there. If you overload it, it just kills. A big flywheel helps but that means the speed sag is quite long.
In contrast, an electric motor is quite capable of putting out much more **PEAK** HP (The Sears HP number) for a short period. The instantaneous peak can be anywhere from 3x to 10x rated HP depending on motor type and construction. The motor simply draws more current than rated for the short period. This matches nicely with real world loads that often have short period high loads.
The net is that for high peak load situations, you can indeed use a much smaller rated electric motor.
If you want to see where some of this falls apart, think about running a large water pump (sewage lift station) out at the end of a LONG power line. When you overload the electric motor on a hard start, the extra current drops the voltage down so low that the motor doesn't really get the extra power. In some cases you can get a full motor stall, the unit will never start. It lets the smoke out of a rather expensive piece of equipment in a few seconds.
Grant Erw>> Grant Erw>>
Reply to
RoyJ
The short version is that if the tank pressure is high enough, the compressed air in the dead space won't expand to below atmospheric pressure when the piston is down, and no more air will be sucked in.
jsw
Reply to
Jim Wilkins
At the age of 53 I am still myself capable of putting out more than 1 HP for brief periods (and they seem to become briefer with each birthday, soon to fall below zero). But surely even the strongest man alive still qualifies for only fractional "rated" horsepower. Prefixing "rated" to a specific physical unit makes it meaningless, and any argument in support of such flimsy talk is a definitional retreat.
Reply to
Richard J Kinch

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