Compressor Air Delivery Results ASME PTC-9 Test Vs. Pump Up Time.

Volumetric efficiency over 100% is achieved by 'supercharging'. With a recip. compressor the low pressure high volume cylinder adds more volume and some intial pressure to the high pressure cylinder, which in turn compresses the charge to a higher pressure. The low pressure high volume side can not acheive high pressure efficiently and the high pressure low volume side can not achieve high volume efficiently. So they work together to produce more effiency. Similar to a superchrger on a car motor. If the motor is 85% volumetric efficient by adding a supercharger you can go well over 100% volumetric efficiecy.

Thanks for your imput, however this is not a supercharged application.

You have ACFM and SCFM which one is 22CFM? You are going to have specific heat created which hurts air density. A compressor capable of pumping 10,000AFCM @ 460psi and 300 degrees will have a SCFM of 6800 SCFM due to heat. It takes alot more air pressure to overcome the heat issue that causes the initial drop in air density, thus reducing CFM capacity. Supercharging could be a way, or by raising the air pressure enough to overcome the effect of heat. 68Torino would think of this as positive boost over atmospheric. The sound of a Kenne Bell supercharger is alot sweeter than the sound of any air compressor in my opinion.

Thanks for your input TLGengrco.

Once again this has nothing in common with a supercharged application. It is simply a difference in the measure of air delivery expressed in CFM under the same actual conditions. Since the air sucked into the inlet is not 68 F, 36% RH and at 14.7 PSIA, it would be defined as ACFM.

I am not attempting various methods to achieve the identical results. I am simply trying to account for the why the big difference between the two methods of measuring the Recip Compressor air delivery.

I cant help with the ASME PTC-9 test because I know nothing about it, but pumping up a tank from empty has to take account of the fall off in volumetric efficiency as the pressure rises in the tank. If you have allowed for this, then I can't help.

I would be interested myself, so I will watch the response.

Thanks for your input horace40. The problem is that computing air delivery based on pump up time results in a volumetric efficiency over 100%. (NOTE: This is not a supercharged application).

I am interested in your opion based on your knowledge to other questions, which I have agreed with.

Thanks,

I think we need to know more about it to comment properly. I realise some might be covered by ASME PTC-9 test requirements.

You say the figures are ACFM at compressor suction conditions - does this apply to both the 28 and 22 CFM?

How was the CFM into the 80-gal receiver estimated? You need to know final pressure and temp, and it would be usual to correct it to SCFM.

Need to correct the ACFM at compressor suction to SCFM for comparison.

How accurate is the 80-gal? Is pipe volume included?

How is the 22 CFM compressor flow measured?

How accurate is the swept volume used in calculating vol. eff?

Another thought on this - after filling the 80-gal receiver you need to know the air temp in it to calculate air mass (high temp just after filling). Even if there's an thermometer in a thermowell, temp might not be uniform.

So I suggest you do it indoors where room temp is constant (and known):-

1. Charge the receiver to max pressure compressor will deliver. Note time of running. Actual receiver pressure not important at this stage. Isolate the receiver, valves must be leak-tight.

2. Leave it overnight to cool to room temp. Measure pressure.

3. You can then calculate mass from volume, pressure and temp., hence SCFM.

Thanks to everybody for your commemts, they were all very helpful. Thanks again

N6377B

Please advise what did you conclude in the end babout your problem.

horace40,

Unfortunately, I failed to find an explanation to satisfy the question. Supercharged application explanations were introduced, which was not a factor. Perhaps it is my misfortune of submitting an ambiguous question. Now assume all the calculations are based on actual conditions. (Since I am not at 68 F, 36% RH and 14.7 PSIA conditions) Computing Compressor air delivery based on pump up time is a fairly easy computation. CFM = (final tank press-intial tank press) x V (tank size in gal.)/7.48 x atmos. press (psia) x time (in mins.). The ASME Power Test Code 9 computes air delivery as follows:

(2.552 x Dn squared x c X t1/Pa) x (square root of PB x delta P/t2):

Where:

Dn = Nozzle Diameter (inches)

c = Nozzle Coefficient

t1 = Absolute Fahrenheit tempertaure at which volume of flowing air is to be expressed; usually compressor intake temperature (Degrees F + 460)

Pa = Atmospheric Pressure Absolute ( Inches of Mercury x .491)

PB = Atmospheric Pressure ( Inches of Mercury)

Delta P = Differential pressure ( The pressure drop through the nozzle, expressed in inches of water)

t2 = Absolute Fahrenheit temperature, upstream side of nozzle (degrees F + 460)

The two air delivery results are quite different. When you consider that the air delivery expressed in CFM based on pump-up time results in a volumretic efficiency over 100% which is not possible without the presence of supercharging which does not apply to this application.

Hello N6377B.

Sorry you have not solved your problem. But to be honest I do not fully understand what supercharging means - other than having a source of already partly pressurised air - which must be fairly obvious if you have.

My approach has been to assume you pump the tank up from zero - and if you did that might explain the error. This is because a compressor can pump out more air at low pressure - ie, the free air delivered (FAD).

FAD goes down as the pressure goes up - more or less depending on the design of the compressor (or to be more honest - the precision and skill and the permitted tolerances in machining the components to build a compressor).

A perfect compressor would deliver every ounce of air as the piston moves up the cylinder until top dead centre. This being so you can calculate the output from precise measurements of bore and stroke (not so easy with a screw comp) and speed. Or with a water test. Measure precisely how much you put in to fill the cylinder at BDC and then measure precisely how much is pumped out at TDC - and the difference is the clearance volume.

In practice it is nigh on impossible to machine parts to such close tolerances and as such you end up with 'slackness' that added together create a 'clearance' volume that will be full of air (at pressure) at the end of the stroke. The air in this clearance volume will never leave the compressor.

It then has a marked effect because it expands back into the cylinder on the suction stroke - thus to prevent air being being drawn in - which reduces the amount pumped out - and so on.

It gets worse as the working pressure rises. In the extreme (assuming mechanical strength and enough power and no leakage past the piston rings) the pressurised air in the clearance volume will be enough to completely fill the cylinder on the suction stroke, thus to prevent any output at all. You simply re-pump the contents of the cylinder.

Remember pumping up bike tyres - you pump away harder and harder until no air goes into the tyre. But it is still hard work - and the pump gets hot.

The point of this, is that the time taken to pump up a tank from empty give a results likely to be on the 'high' side. Based on a precise volume, you need to accurately time the precise pressure change between two close values near the working pressure and correcting for temperature.

If you have done this then I can't help any more - but the difference needs to be explained because an awful lot of people do it your way.

HORACE40

As usual your explanations are right on the money. As you stated I do not use the air delivery base on pimp-up time from 0 psig to final pressure due to the exact reason you provided. I usually compute pimp up time based on the recovery time from usually say a 20 to 30 psig (145 to 175 PSIG). Also addressing the clearance volume issue, it is for this exact reason why it is impossible to achieve over 100% volumetric efficiency. The supercharged issue was introduced by someone else attempting to explain how to achieve a volumetric efficieny greater than 100%. which they are correct in the event this was a supercharged application. My application is not supercharged, therefore it does not apply. As you mentioned, alot of people use pump-up time to determine the air delivery for Compressors. However, I wonder if they ever went a step further and calculated the volumetric efficiency of the compressor based on pump up time. I think they would be in for a big surprise. Pump up time is basically a rough approxiamte of air delivery. They arrive at a inflated air delivery.

Thanks for your comments

N6377B

Thanks for the feedback.

People buying compressors based on quoted 'displacement' volumes will be in for an even bigger surprise. In practice these people will never know. Lack of air to run the pneumatic equipment will be put down to under-estimating the demand.

One thing I did not mention when estimating output based on displacement volume is the additional effect of 'slowing' the motor. Motors are assumed to run at constant speed but in actual fact will slow down a little as pressure increases and thus the flow will decrease.

Obviously when using pump-up time this 'slowness' is automatically included. However, even here the nominal 'volume' of the tank is questionable - this should include the volume of all the associated pipework and equipment.

There are so many variables in pump-up time calcs that confuse the issue that standards call up proven flow meter methods instead.

Having said that, pump up time is 'near enough' for most purposes.

N6377B - reply to #11, couple of comments.

For compressor air delivery based on vessel pump-up time, the pressure on the bottom of the fraction should be standard pressure to get answer in SCFM. Need to put answer in SCFM for the different measurements for comparison. In practice, atmos. press not likely to differ much from standard. More importantly, should x by T_s/T_a (obvious terminology). If you're straight off a compressor T_a could be a lot higher than T_s and this could explain your discrepancy. Also it might not be too easy to measure the temp at all accurately, so you could try doing what I suggested in #8.

In the ASME test formula quoted, are you sure the square root is in the right place? It's hard to be sure, but from definitions it looks like t1 = std. temp, t2 = actual temp, Pa = std pressure, PB = actual pressure. My calc is

q_a = C.A.√(2.ΔP/ρ_a) - basic flow formula, based on any consistent system of units. q_a = actual flow, ρ_a = actual density.

Converting q_a to q_s - (ρ_s = std density)

q_s = C.A.√(2.ΔP/ρ_a).ρ_a/ρ_s = C.A.√(2.ΔP/ρ_s).√(ρ_a/ρ_s) =

C.A.√(2.ΔP/ρ_s).√(PB.t1/Pa.t2)

Strictly speaking, need to x by expansion factor, but this is usually close to 1. And the ASME formula uses arbitrary units so needs a correction factor, so the expansion factor might be in there, along with the 2 and ρ_s.

Cheers