Compressed Air Reservoir Fill Time

I'd say do a few calculation based on 'empty', '1/4 full' 'half full' '3/4 full' plot 'em, join up the dots...

Everyone knows I hate maths! Unless you are into the differential stuff this will get you close enough for jazz. I'll bet any amount of calculation won't get you very close.

I'd say use the biggest convenient pipe and choke it down if you want to slow up the fill. Probably as quick to try as to calculate!

Have fun (I reserve the right to be of no help whatsoever)

Del

Ha... Thanks nonetheless Del,

You're actually right up with my line of thinking - I hate maths and I have tried joining the dots to avoid it !!!

Cheers,

Kim...

Great idea to use a large diameter pipe and choke it. Use a couple of ANSI flanges on the pipe and bolt the choke between them. Makes changing the choke very easy.

I don't do Windows, and I don't do Diff. Eq.

You're using 100 psi Air to fill a 5.6 CF bottle to 99 psi every 2 seconds.

This is about 175 cfm of 100 psi air.

Looking in my Crane Engineering Data booklet from 1976, 179CFM of 100 psig, 60 F air is equivalent to 1400 cfm of free air at 60 F and 14.7 psia.

Based on the table, I would suggest trialling 2", 2-1/2" and 3" pipe to see if any of them work to your satisfaction.

Not a pretty solution, but the best I could do in under 5 minutes.

Scan CR4 mechanical section. Ducon as I recall did give a conversion factor of free air to compressed inclusive other formulas. Maybe this may help.

"Aside from using differential equations can anyone recommend any other possible means of doing this or perhaps suggest how I should use the above formula ?"

The problem you pose is why there is a need for such mathematical things as differential equations, although sufficiently accurate answers may be gotten otherwise.

A similar problem is the charging of a capacitor with a fixed voltage source through a resistor. The peak flow rate will occur when the tank to be filled is empty, and decreases from that point on. Calculating the flow rate for "free air discharge" will give you an idea of resistance to the highest air flow rate. Thereafter, the flow rate decreases exponentially.

The time constant of the system is determined by the resistance of flow in the piping and the capacity of the tank to be filled. The tank will be essentially at full pressure after 5 time constants. It will be approximately at 63% of full pressure at one time constant.

Given what you already know, you may be able to figure out the rest. Keeping the units straight is up to you.

P. S. A short treatise on Time Constant may be found on Wikipedia at http://en.wikipedia.org/wiki/Time_constant

Here's a little help with my last post.

The equation which describes your system is P_t(t) = P(1 - e^-t/RV), where P_t(t) is tank pressure at time t, P is manifold pressure, R is resistance of pipe of a given length and diameter to air flow, and V is volume of tank.

And the plotted curve (tau is RV) is:

This still doesn't give you the system time constant, but I hope you get the idea.

Thanks to all who contributed - I suprised myself and knuckled down with the Math and turned up some pleasing results (even if they look waaaay to basic for the amount of work input !!)

Here is the resulting curve (plotted against the flowrate from the vessel).

For anyone wishing to do similar I highly recommend the following works:

http://www.ijee.dit.ie/articles/999982/experime.htm

Nice curve, about what I would guess would happen. I could not read the units on the chart though.

What size hose or pipe will you end up using for your tanks? How close is your number to my S.W.A.G.?