Time Required to Fill a Tank with Air

I had a similar situation in which I had to produce a simulator to test a PLC based air control system.

Although I had not used an excel spreadsheet, the iterative process of a PLC is similar.

I found the air equation of PV =nRT to be useful. Assuming that T (Temperature) and V (Volume) are constant, calculation of P based on adding air mass (CFM) was fairly easy and was proven on two separate volumes.

I agree with the other post, with "T" time and "X" for tank volume size proportionate with "U" CFM

p911

as the tank fills, the cfm will drop unless the compressor can speed up to compensate. U and W are constantly dropping as X increases. I have found that a lot of simulations do not accomodate these parameters and seem to be inaccurate to actual figures.

This issue is complicated by the fact that the compressor specs only gives the output at a given working pressure, in terms of cfm of normal atmospheric conditions density.

Therefore, starting from 0 psi, until you reach X, the volume delivered will go diminishing until it reaches X and the specified cfm. This is due to the fact that there will be a backflow just before the Exhaust valve closes after the compression stroke (Piston...).

If this is neglected, then the time in minutes could be at best = X x V/U (that is T > X x V/U) V being the tank volume + Pipe volume.

If you can obtain curve characteristics for the compressor from the manufacturer, then you could be more accurate. Otherwise, theoretically, you should consider Integrating the estimated equation describing the diminishing performance (volume) of the compressor, until it reaches the X pressure.

This is the reason to work with "mass/time unit" and not with delivery in "volume/time unit".

The absorbed air by a compressor depends on the ratio between dead and stroke volume and ratio between final pressure at delivery and pressure before entry in compressor stage. This due to the fact that the compressor delivers when internal pressure>pressure in the delivery line and during the aspiration stroke first the compressed air in the dead volume expands till the internal pressure<pressure in front of aspiration valve. This is called the volumetric efficiency of the compressor. I consider a piston compressor and for further simplification with one stage .

The worst case is when the line pressure is the maximal one. For the computation considering this case will lead to a longer time but the differences are not very important since at start it is only an estimation of the filling time.

If you accept as order of magnitude Vd/Vs=0.05, ratio between delivery pressure and aspiration start =7.37 and the adiabatic exponent for air as 1.4 the volumetric efficiency will be in the worst case ηv=84.2 %. In the best case the same parameter will have a value around 99.5%. A good approximation can be done with 90% assumed as constant.

Now let us go further. The delivery is in volume/time unit for NORMAL conditions. This is usually the aspired volume at no load so that has to be corrected for the real conditions as temperature and atmospheric pressure and with the volumetric efficiency mentioned above. From this value one computes the air mass delivered per time unit and proceeds as mentioned above.

The same procedure can be used for multi-stage compressors analysing the behaviour of every stage independently and taking of course the intermediary cooling into consideration.

I have to give the definition of Vs and Vd:

Vs is the stroke volume = pi*d^2/4*Stroke

Vd, the dead volume is the volume between cylinder head and piston at stroke end.

One of the most important parameters for a compressor design is to maintain Vd as low as possible.

If it's a final accurate pressure you are looking at on a gauge, don't forget to calculate in the temperature difference of the air as it's compressed. IE: When i used to fill High Pressure SCBA tanks that were 3000 PSI directly from a compressor, we had to overfill the tank according to the gauges by approx. 300-400 psi (extra run Time)so when the tank & air cooled it would be the proper pressure. I might add ,this difference was when the tank was empty. A half full tank would not need to be overfilled as much. Another way we used to fill the tank was fill it to 3000 psi, let the tank cool, (which took hours) Psi dropped approx 300-400 then restart the compressor and top up the tank up to 3000 Psi.

Different size tanks and /or PSI would need different formulas.

If x tank takes 10 mins to fill to 2000, it will likely take 25-27 mins to fill the same tank to 4000 psi.

Rescue

Ok Heath

This is the KIS solution you are looking for.

V = volume of tank in cubic feet

U = SCFM (verify that the compressor output is Std Cu Ft / Min) CFM is another animal.

W = Final tank pressure in psig (gauge pressure)

1. Assume the compressor manufacture put a label on a compressor that indicated it would deliver 4 SCFM [U] @ 100 PSIG [W].

2. Determine the volume of the tank V in Cu Ft.

3. Then V x ((W + 14.7) / 14.7) / U = T (time in minute)

Just for fun say the tank in 2.5 Cu. Ft Then 2.5 x (114.7/14.7) / 4 = 4.877 min.

Simple as 1, 2, 3 KIS, Tom