Need Help with an Air Tank Problem

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Jonny
Posts: 1
Joined: Tue Mar 19, 2019 10:13 pm

Need Help with an Air Tank Problem

Post by Jonny » Tue Mar 19, 2019 10:49 pm

Hi everyone
was wondering if you could help me with calculation for a small project of mine. I was trying to use some of the calculators but I'm struggling to figure out the maths involved, its a bit out of my depth.

I need to find the size of tank I would need to provide :
-150 psi for 10 reciprocations to two actuators (25mm bore and 100mm stroke)
-The flow rate does not have to be fast ( if the actuators took around one second to extend that would be optimal.)
-paintball tanks I was looking at can store air up to 3000psi just need to know the size
-I am using PU piping 5mm(ID) around 1 metre in length in total to connect a regulator, solenoid valve, manual cut off valve and a junction to the system.

If anyone could help me out it would be greatly appreciated, I'm doing my final exams in Ireland in school and don't understand most of the concepts Thanks.

admin
Site Admin
Posts: 366
Joined: Mon Feb 08, 2010 7:47 pm

Re: Need Help with an Air Tank Problem

Post by admin » Wed Mar 20, 2019 8:51 am

You should also know how many strokes you have in a min or hour?
So you calculate a needed flow as a volume of the cylinder * number of strokes; Q = V * n, where is Q - is a flow, V - is volume, n - is a number of strokes in a unit of time.
Another question is if you have a compressor to fill the tank? If so, you have a flow rate of air consumption (Q) and the flow rate of tank filling by a compressor (Qc). Depending on the actual consumption you can select to have Q > Qc, Q = Qc or Q < Qc. If Q > Qc than you might end up without air in the tank at some point.
Every air tank filled by the compressor has two pressure sets. Lower, when the compressor is turned on, and higher when the compressor is turned off. Lower pressure is the lowest pressure that your system needs to operate correctly. Higher is the highest that your tank and the system can tolerate.
Now, you need to calculate the difference of the air quantity in the tank between these two pressures using the equation for a perfect gas. That air quantity is the mass difference (G) in the tank between two pressures. Your Q also needs to be converted to mass flow, m = Q * density. From there you can calculate time in which air is consumed time t = G / m.
And that time to fill the tank is critical. If the tank is small the time will also be short and the compressor will need to start frequently which might lead to motor overheating and failure. If you set that time between two compressor starts is for example 60 sec, then from there you can calculate G - which is the mass of air between two compressor starts, and your tank needs to be able to accumulate G of air. To get a volume of tank you can use formula V = G / rho, and rho is the density on the higher pressure.

Hope you can follow this instruction...
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