Air Compressor Air Delivery

Are you saying 10psi over the 14.7psi, 24.7psi, or 10psi period?

68torino

14.7 PSIA = 0 PSIG. The change in CFM would be equal to the change in the density of the inlet air. The air compressor would simply use less energy to produce the amount of air discharged at the 60 PSIG.

Replying to #2 - I'm fairly sure N6377B means 10 PSIG inlet pressure = 24.7 PSIA. Then as you say change in SCFM = change in density = change in abs pressure, and this agrees with his figure - 49.1*24.7/14.7 = 82.5 SCFM.

Compressor draws a little more power at higher inlet pressure. Assuming 60% compressor efficiency I make shaft power

10.3 hp at 14.7 PSIA

11.1 hp at 24.7 PSIA

But the compressor is set to produce 60 psig, therefore it "raises" the pressure from 10 psig to 60 psig, rather than 0 psig to 60 psig. How about pump raising pressure from 0 psig to 60 psig vs 20 psig(static head) to 60 psig. "Fan" laws similar to "pump" laws except air (gas) is compressable and water (liquid) is not.

Reply to #4 OLD F**T - it's the compressibility that makes the difference.

Formula for shaft power is P1*Q1/n/η*(((P2/P1)^n)-1)) where P1 and P2 are inlet and outlet absolute pressures, Q1 = actual inlet flow, n = polytropic exponent, usual figure 0.23, η = compressor efficiency. Actual inlet flow doesn't vary with inlet pressure, so SCFM changes proportional to pressure.

Power in watt if SI units used, there's a factor needed if you use CFM, psi etc and want power in HP (I used Mathcad which looks after the units for you, so I don't know offhand what the factor is, if you want I'll work it out).

Depending on P2, power rises with increasing P1 up to a point, then falls. A quick bit of calculus shows (if I've done it right) that for a given P2, max power is when P1 = 0.321*P2. 71.4PSIA*0.321 = 24PSIA, so going from 14.7 to 24.7PSIA is mostly on the rising part of the curve hence higher power at 24.7. Also explains why when pulling a vacuum on a vessel, starting at atmospheric, the vac pump starts off running light, then can be heard to labour, and finally runs light again.

Repeating much of what has already been said but adding something extra - SCFM is used in a traditional way because it relates to ambient atmospheric pressure and temperature conditions in which the compressor operates. The 'standard' differs slightly from country to country around the world (mainly temperature). The difference does not matter too much for the purpose of an explanation.

In other words, your compressor sucks in 49.1 scfm of the atmosphere and compresses it to 60 psig (it could be any pressure depending on compressor mechanical design and appropriate motor size).

The compressed air (now in a much smaller volume) then passes through the system. It does it' s design job and then expands back to atmosphere when it leave the system (and assuming no leaks) to re-occupy 49.1 scfm.

If you start with a source of air at higher pressure (assuming the same temperature) and that the compressor is capable of handling the extra workload (as explained by others) then this air will expand to occupy more than 49.1 scfm by the simple ratio of the absolute pressures you are working with.

Which in you case will be (14.7 + 10)/14.7 = 16.8. Thus multiply 49.1 x 16.8 and you get your new flow rate = 82.50 scfm

Which is pretty close to your own calculation.

Replying to #6 Horace40

I think you meant 1.68, not 16.8, but what's a factor of 10 at Christmas?

Your dead right. It should be 1.68. Sorry about that, it was a typing error. Luckily the answer is the same.

To calculate the FAD of a compressor, we must know the pressure and temperature of the inlet gas and then we measure the discharged volume at the outlet conditions ( temperatue and presssure). The discharged volume is then worked back to the inlet conditions in accordance with the Ideal Gas Law. This volume is the free air delivery of the the compressor. V1 = V2xP2xT1/T2xP1

Hence by definition , if the outlet pressure and volume are unchanged and the inlet pressue is increased ,the FAD will actually be decreased by a factor of 1.88. However in this case , the compressor wil be using less power because it has to do less work on compression - starting from 10 PSIG instead of 0 PSIG provided the discharged volumetric flow rate is unaltered.

Hello Guest

...Hence by definition , if the outlet pressure and volume are unchanged....

I cannot see how the outlet volume V2 is 'unchanged'.

V2 comes from V1, where the original V1 is the free air volume sucked in by the compressor and is equal to the (per minute) volume of the actual compressor cylinder. This remains the same.

So regardless of any intermediate pressure (or actual volume/temperature), this air will re-occupy a free air volume of V1 back in the atmosphere in equilibrium with atmospheric pressure and temperature when pushed out of the cylinder.

If we start with the same cylinder V1 full of air at a pressure of 10 psig, then regardless of any intermediate pressure (or actual volume/temperature) this air will then expand to occupy a free air volume greater than the original V1.

Assuming temperatures are the same before and after, then the free-air volume is simply a ratio of the two starting pressures - 10 psig and 0 psig. Which in terms of absolute pressure comes out at 1.68.

Unless you disagree. If so would you be good enough to say which actual values from the original question should be used so as to clarify how you reached your conclusion.

"Hence by definition , if the outlet pressure and volume are unchanged and the inlet pressure is increased ,the FAD will actually be decreased by a factor of 1.88."

Why FAD will decrease when you say outlet volume is unchanged 'by definition'(which definition)? By increasing the suction pressure, suction air density is increased by ratio 24.7/14.7, so volume or volumetric flow rate or mass flow rate or FAD or SCFM (whatever you call) will increase by the same ratio compared to earlier.

"in this case , the compressor Will be using less power because it has to do less work on compression - starting from 10 PSIG instead of 0 PSIG provided the discharged volumetric flow rate is unaltered."

You are right if volumetric flow rate is unaltered but unfortunately it is not true as I have stated above. So, compressor will use more power (as also commented by Codemaster at post 3), because incease in power due to more mass flow rate is much higher compared to decrease in power due to lower pressure ratio (5.08 to 3.02). This decrease in pressure ratio will bring down the power requirement by about 14%. But 68% increase in mass flow rate will require 68% more power.