dumb question on compressor displacement cfm

HHmmnnnn . . .

Dead space. It too is filled with the CFM flow. We call that "clearance > >

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.

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>>

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

Rounding errors do not constitute a variation in the fundamental units of physics.

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.

Well, *something's* going on here. My Quincy compressor requires an 18hp gas engine to run it. But it can be run by a 7.5hp electric motor.

If it didn't take an 18hp gas engine to run it, why would they have built this machine with one on it?

Your own rule of thumb (4 cfm / hp) bears out 25.4 cfm from 7.5hp electric.

I know that theoretically a horse is a horse.

So how do *you* explain it?

I'm not trying to diss you, Richard. I very much enjoyed your CO2 page and made a similar setup with which I taught my children to make their own soda pop. The result of that was that after they saw cup after cup of sugar going in, they stopped drinking pop. I also learned really a lot from you in the big air compressor thread from a few years ago.

Grant Erwin

A "rating" is not a physical unit. It is a performance prediction or expectation As such, it is always associated with a set of conditions that are explicit, implied or understood. Ratings depend upon definitions to have any meaning, whether the definitions are explicit, implied or understood.

A motor that can deliver 1 hp (745.7 Newton-meters/sec) continuously (at rated voltage, current, temperature, service life, etc etc) but can deliver 5 hp for one second per minute could be considered to be sufficient for a matching load -- one that requires 1 hp continuously with peaks of up to 5 hp lasting no longer than 1 second no more often than once per minute. Is that a 1 hp motor or a 5 hp motor? (Answer: yes)

Nevertheless that's how electric motor HP is given, and the time dependence of power output that electric motors have but gas ones don't is part of the reason for the non-equivalence. Another may be because small gas engines have been found to rarely meet spec. If you want to compare gas HP to Sears HP they may be equal. Watt's original definition of the 24/7 horse power rating of his steam engines was extremely indefinite; at any instant several of the horses might be sleeping.

I'm 62 and found out last weekend that I can't quite manage 1 HP any more, while unloading a pallet of cement bags. Lifting an 80 Lb bag 4 feet in 1/2 second is a bit over 1 HP.

jsw

Don, waxing cryptic, goes, "one that requires 1 hp continuously with peaks of up to 5 hp lasting no longer than 1 second no more often than once per minute. Is that a 1 hp motor or a 5 hp motor? (Answer: yes)"

Say what, Don? Are you trying to enter a ratings war with Dr. Kinch, or what?

Bob Swinney

A "rating" is not a physical unit. It is a performance prediction or expectation As such, it is always associated with a set of conditions that are explicit, implied or understood. Ratings depend upon definitions to have any meaning, whether the definitions are explicit, implied or understood.

A motor that can deliver 1 hp (745.7 Newton-meters/sec) continuously (at rated voltage, current, temperature, service life, etc etc) but can deliver 5 hp for one second per minute could be considered to be sufficient for a matching load --

Of course. My point is simply that the unit of horsepower is a specific and well-defined physical quantity, and that the word "horsepower" on a gasoline engine is not that unit, and there is no conversion between the two because the latter usage can mean anything the manufacturer whimsically chooses. Contrast this to an automobile they are honest in specifying "brake horsepower" to reflect useful rpm x torque delivered to the wheels.

In the case of gasoline-powered air compressors, the situation is difficult to characterize in terms of true horsepower units, because the crankshaft power delivered by the engine is a complicated function of many factors, and the crankshaft sink at the piston compressor is likewise varying through factors like the cycle hysteresis.

No, I'm agreeing with him.

correct

and that the word "horsepower" on a

Blather

and there is no conversion between the

As long as they choose it but that does not change the definition

Contrast this to an automobile they are honest in specifying

REAL blather. There is NO vehicle manufacturer that talks about "brake hp to the rear wheels.

torque at engine RPM

a complicated function of many factors,

Complete blather.

The real point is that manufacturer's either publish a spec sheet and meet those specs. If they don't they are guilty of misrepresentation or outright fraud.

But you have to realize that a gasoline engine has NO OVERLOAD capacity, unlike an electric motor.

At 900 rpm when the piston is at bottom dead center, the pressure inside the cylinder is not 1 atmosphere. It is somewhat less because of the drop in pressure due to the filter and intake manifold.

When the piston is at top dead center, there is still a little volume inside the cylinder. The compression ratio on compressors is quite high because they use reed valves. But it is still not infinite. Otherwise with just a slight change due to manufacturing tolerances or temperature changes, the piston would be hitting the head.

And then those reed valves on both the intake and the exhaust, need some pressure across them to cause them to open.

And then there is some leakage by the piston rings.

None of these are biggies but they all affect how much air actually gets pumped.

So you calculated the swept volume correctly, but the efficiency can only be determined by an actual test.

=20 Dan

Just to throw another (largely irrelevant) definition of HP into this discussion,