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Unread 12-01-2015, 15:24
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Re: Number of Pneumatic Air Tanks

Quote:
Originally Posted by Nate Laverdure View Post
By the combined gas law the quantity (P*V)/T remains constant throughout a closed system in steady state. If we can also assume that the temperature of the system remains approximately constant over the duration of interest, then the quantity P*V remains approximately constant too.

P*V ≈ constant

This is a pretty powerful tool. We can use it to equate the energies of two different volumes of gas at two different pressures.

P1*V1 ≈ constant ≈ P2*V2
P1*V1 ≈ P2*V2

The quantity P*V will be in units of energy. This number will in fact be a estimate of how much stored energy is contained in the pressurized system. (But it's not quite a measure of how much energy is available in the system to do useful work!)

Here's an example. First let's look at how much energy is contained by the storage tanks.

P1 = 120 psig storage pressure
V1 = two 16 in^3 tanks = 32 in^3
P1*V1 = (120 psig)(32 in^3) = 3840 in*lbf

Now, how many times can you use that energy to actuate an example cylinder?

P2 = 60 psig point-of-use pressure
V2 = (n actuations)(5/16" bore x 14" stroke) = n(0.342 in^3)
P2*V2 = n(60 psig)(0.342 in^3) = n(20.5 in*lbf)

P1*V1 = P2*V2
3840 in*lbf = n(20.5 in*lbf)
Solve for n...
n = 187 actuations
Nate, can you go over how you calculate V2 in a little more detail? Less than 1 cubic inch for one of those cylinders seems really small to me, and doesn't match my simple math (but I'm not a much e, so I could very well be missing something). 187 actuations also seems to be really high, based on practical experience with FRC pneumatic systems.

Does it take into account what happens when storage pressure drops below 60 PSI, it does it assume you can go down to 0 psi storage while maintaining 60 psi working? When your storage drops below 60 PSI, the equations have to change a bit, as you can no longer have the desired working pressure.

Perhaps a better approach would be to do the math from the other direction - what volume of your stored air (at 120 psi) would be needed for a single actuation (at 60 psi)? Since we know PV is roughly a constant, we know we have roughly half our total storage volume available to us before we drop below 60 psi storage. If it takes 2% of the total storage volume for one actuation, then we would get 25 actuations before dropping below 60 psi, for example.
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