Sizing Air Tank: Estimating Polytropic Efficiency 2

[erk1313]

Problem: Starting with an air compressor with a known free air displacement, how long would it take to fill a 5 gallon air tank to 95psig? See below for more detail.

Method: Use (PV^n = constant) to convert free air displacement (in standard atmosphere units SCFM) into resulting pressure built up in the fixed volume of the tank. Because the free air displacement changes based on the outlet pressure (see manufacturer's

http://www.eng-tips.com/threadminder.cfm?pid=391

chart), take small iterative steps until the final pressure is reached.
P = Patm*Vatm^n/V^n
Vatm = Vinitial + Vcompressor; The resulting air volume in the tank, but expressed as (SCF) standard atmosph. pressure and temperature.

Question: In using (PV^n = constant), how would I estimate the polytropic efficiency so I am not stuck assuming adiabatic expansion (n=1.4) (all heat generated is absorbed as increased gas temperature) or isothermal expansion (n=1) (all heat generated is removed so that the gas is kept at a constant temperature). Looking over manufacturer's pages, I typically see the assumption of isothermal expansion (n=1) being made when estimating fill time. Is this because inefficiency is already built into their free air displacement spec?

Knowns (assume STP conditions):
Air Tank Volume: V = 5 Gal = 0.67CF
Flow Rate: Vatm ~ 0.6SCFM (See manufacturer's table for Free Air Displacement chart) or approximate with best-fit line (Q = 1.117*Poutlet^-0.169)
Pressure (1atm): Patm = 14.7psia
Target Pressure: Pf = 95psig = 109.7psia
Compressor: Oil-lubed, reciprocating compressor
Specific Heat of Air: k = 1.4

Notes:
-Other threads mention taking an assumption of 85% isothermal efficiency for reciprocating compressors, but how would this figure into the polytropic exponent ?
(thread1036-238662)
-The following thread mentions a way to calculate the polytropic exponent , but I am unclear at how to conduct this (or may be missing information from the manufacturer) (no discharge temperature or motor rpm).
thread391-151370 (6 April 06 post)

 

[erk1313]

Found another helpful reference from Montemayor, who states that a reciprocating cylinder is very close to an isentropic cycle (thus adiabatic). If this is the case, why do some compressor manufacturers estimate fill time based on isothermal assumptions? I ask because setting k=1 gets me the closest to experimental results (assuming manufacturer-published free air displacement curve).

Free Air Discharge (pg-12): Q = (P2 - P1)/Patm * V/t, Basically derived from P1*V1 = P2*V2

thread798-161670

06 Aug 06
The process taking place inside a reciprocating cylinder is so close to the isentropic cycle that it always converts a neophyte into a believer that the cooling jackets do absolutely nil or no cooling. I’ve run cylinder dry just to prove this point and it worked. The process is not polytropic. The only people who think so are those who don’t understand recips and haven’t operated them for years on a variety of gases (such as college profs). I have, with oxygen, nitrogen, CO2, acetylene, nitrous oxide, air, carbon monoxide, Hydrogen, natural gas, ammonia, most of all the Freons, propylene, propane, ethane, off gas, syn gas, and many more. And in all of these cases the compression was identical to Isentropic – never “polytropic”. I’ve also operated centrifugal compressors (which are the subject at hand here) and these are strictly polytropic in nature.

 

[erk1313]

A colleague pointed out that there are two compression stages taking place, that inside the compressor and that inside the air tank. However, it would seem to me that both would fall under adiabatic compression. Unless the air cooled in the tank and heat was absorbed by the tank wall which would bias it toward isothermal.

Thoughts? Should I post this in another section. Not many responses here....

 

[erk1313]

A colleague pointed out that there are two compression stages taking place, that inside the compressor and that inside the air tank. However, it would seem to me that both would fall under adiabatic compression. Unless the air cooled in the tank and heat was absorbed by the tank wall which would bias it toward isothermal.

Thoughts? Should I post this in another section. Not many responses here....

 

[crshears]

To me, this post smacks an awful lot of homew_erk...

CR

 

[EnergyMix]

Crshears

That's why he's not getting any responders (because it sure seems like homework to me, too). This sort of question, he should be posting his results and asking what what we think of it. We might then nudge him in the right direction, if needed. But doing it for him? I have a job to do, I don't have time to do someone's homework.

Want to know the do's and don'ts of Eng-Tips? Read FAQ731-376.
English not your native language? Looking for some help in getting your question across to others or understanding their answers? Go to forum1529.

 

[erk1313]

What gave you the impression this is homework? Because I tried to lay out the problem clearly?

This project is for a phase I prototype of an inflatable hospital bed. Assuming adiabatic expansion, I am off by a factor of 2 in estimating tank fill time. I cannot simply go up in size because noise is an issue in a hospital environment. As such, air compressor capacity comes at a price and I would like advice on how to estimate this accurately. I don't need a theoretical answer for homework, I need a real-life estimate that I can be confident in.

Thank you to any who can chime in on this.

 

[crshears]

Hello erk1313,

In my humble opininon you did not lay out your problem at all clearly...

I have been a member of these fora for less than a year, and in that time I have repeatedly seen original posts asking theoretical questions without providing any background as to the application. In some of these cases I have undertaken to dig a little deeper, particularly where I thought I might be of some help but where the OP was sadly lacking in specifics.

Again, imho, your OP falls into this category.

If you had instead begun your OP with the last paragraph of your latest post you might have had a much more positive response; speaking only for myself, if I see a post that begins with "Problem: Starting with an air compressor with a known free air displacement, how long would it take to fill a 5 gallon air tank to 95 psig?" the highest probability is that I'll simply click the 'back' button.

End of scolding.

Now:

What kind of 'inflatable' bed? Are you looking to develop some kind of rapid-deployment bed for emergency use that would take up a minimum of storage space? In that case, you could construct a unit with an on-board air receiver, maybe even of the quick-swappable type, that could be recharged elsewhere, obviating the noise concerns.

Or are you looking to have a soft variable-pressure bladder that can be adjusted to differing levels of firmness for patient comfort? In that case the paramaeters change considerably from Case One.

Where did the 95 psig come from?

Maybe it's just me, but when I read a sentence like "Assuming adiabatic expansion, I am off by a factor of 2 in estimating tank fill time" I think, "Huh? Adiabatic expansion occurs when gas expands to a lower pressure; is this guy talking about adiabatic compression instead and just got his murds wixed up?"

CR

 

[erk1313]

Sure, I'll take the abuse, as I am a novice when it comes to pneumatics and I highly respect the engineers that lend their expertise here.

For this problem, it does not matter what the air is used for as the first stage is filling an air tank. But yes, it will be later used to inflate the bed and adjust pressure. Nor does the specific tank pressure matter. I'm looking for a fundamental approach to estimate fill time with an oil lube reciprocating compressor. I've seen several references that assume isothermal compression. Why is that? I would think both the compressor and tank fill would more approximate adiabatic compression.

 

[EnergyMix]

I'm doing a bit of a head scratch here. As an airbed owner, none of the 5 or 6 airbeds that I have, including the portable camping non-electric versions, require a tank. Compressor + air mattress tubes = airbed. From experience, to fill a raised twin with a .33 hp compressor motor takes about 2 minutes with most of the last minute spent adjusting the firmness. A five gallon tank would be rather unwieldy. If you're talking about a design for burn patients where the bed is keeping them on a layer of air rather than a mattress, then I would think you would want to fill your bed through a constant air supply line, not a tank. Those are relatively quiet, as the machinery is elsewhere.

So you ran an experiment? More than one? How long did it take you to fill your tank? What size compressor did you use? What was the final pressure inside the tank (i.e., how did you decide it was full?) Did you measure it? How did it compare to your analytical results? What papers have you researched? What textbooks have you consulted?

Not that I'm really interested in the answers ... but those are the sort of things I would expect to see if it weren't a theoretical problem. I'm not trying to trivilize your issue, just saying I'm not going to do it for you.

Want to know the do's and don'ts of Eng-Tips? Read FAQ731-376.
English not your native language? Looking for some help in getting your question across to others or understanding their answers? Go to forum1529.

 

[erk1313]

Aborting this thread. Disappointed that I asked a simple question and was barraged with unrelated advice about inflatable beds and accusations of posting homework. Will later document my measurement of tank fill time and predictions based on the manufacturers FAD curve in case it helps another.

 

[quark]

For receiver filling, you need to consider Isothermal Process (PV = k). Check the total volume in the receiver at atmospheric pressure in the given time and divide it with compressor FAD. The resulting value is your fill time. This is fairly accurate. Compressor manufacturers can give you more accurate formula, considering the temperatures and nozzle efficiencies etc.

 

[crshears]

Post aborted...hmmm...was it just me, or did we have a simple failure to communicate?

Paranoia that someone might scoop the idea, so playing cards close to the vest?

I definitely do not feel I was being abusive.

 

[EnergyMix]

You weren't.

Want to know the do's and don'ts of Eng-Tips? Read FAQ731-376.
English not your native language? Looking for some help in getting your question across to others or understanding their answers? Go to forum1529.

 

[crshears]

Thanks, EnergyMix; I didn't think I was, but it's good to have this affirmed.

CR

 

[quark]

The OP has done research fairly well before asking this question and I don't see any issue with the quality of the question. Many engineers doubt as to why compression is adiabatic (almost) at one end and isothermal at the other end, though the flow is continuous. The key is the speed of compression. Adiabatic processes are fast processes. Receiver filling happens at a slower rate and there is no temperature rise.

 

[crshears]

Hello quark,

I agree with you fully...at lower pressures; there are limits, however, such as in the case of filling diving cylinders to ~3300 psi. Depending on the capacity of the compressor, there may be [a] no issues at all, some cyinder warming, or [c] so much warming due to adiabatic compression that the cylinder must be placed in a water bath to control and limit its temperature rise.

Those who are divers will be able to speak to this much more authoritatively than I, but it is my understanding that when relatively small cylinders are being filled from much larger ones using the cascade method, the danger rises to a very serious level indeed; the only recourse, even with a water bath, is to limit the rate at which the bottles are filled.

CR

 

[quark]

You are right on money.In compressors, air is squeezed in short space superfast and when it comes to air receivers, of industrial size, it is at a slow rate. If the receiver size is small then temperture rises. Nevertheless, during factory acceptance testing, FAD calculations are done with modified isothermal equation (modified not for process but other inefficiencies).


The key is speed of pressure rise.

 

[racookpe1978]

Well, in addition to speed of filling (1200 compressions a minute in a cylinder space of 1.5 inch dia x 2 inch tall ?), you that little bit of compressed gas 91/2 x 1/2 x 1/2 cubic inch ?) from each compression going into a long 1/4 inch diameter steel line (all of which will cool it) going into a large tank with large heavy steel walls.

So the heat load into the large tank is trivial compared to the thermal inertia of the tank.

Now, coming out of the tank, ALL of that cooling happens right at the point where the air is expanding, so that tip may get very cold very quickly. But even on the hose itself? Little cooling from the (still high pressure) flowing gas.

 

[erk1313]

Great replies. This is helpful!
May I walk through my assumptions and see if they are correct?

So the rapid action of the compressor results in the gas quickly heating up. Before heat is transferred to the cylinder walls, the gas is sent downstream (hence adiabatic). As the pulse of gas travels into the large volume of the receiver, it compresses slowly, and the small amount of added heat is absorbed by the walls of the receiver.

However, quark mentioned that FAD specs are based on isothermal equations--is this because compressor manufacturers are anticipating the gas would eventually arrive at a receiver under isothermal conditions? Assuming isothermal, for a FAD of 0.65SCFM, this would result in an actual flow at 50psia of 0.2ACFM:

​

V = Patm*Vatm/P = 14.7psia*0.67SCF/50psia = 0.20CF (assumes STP conditions)

This actually matches up well with my observations:
Equipment:
[ul][li]Werther PTC15 Compressor (flow chart)[/li]
[li]36" of tubing (.16"ID)[/li]
[li]2.5G steel receiver[/li]
[li]36" of tubing (.32"ID)[/li]
[li]2.5G steel receiver[/li]
[li]Electronic pressure gauge (logging numbers by eye so easily 5% off)[/li]
[/ul]

Method for predicting tank fill over time:
[ul][li] Converted compressor FAD flow table to best fit line (R2 = 0.999)[/li]
[li] For small increment of time (0.1min), found the amount of air added to the receiver (as SCF). This is based on flow at the current outlet pressure[/li]
[li] Converted SCF to ACF using isothermal equation (Pactual = Patm*Vatm/Vactual)[/li]
[/ul]