Originally Posted By: paul
I said the nozzle area is 1 square inch.
I said the volue of the tank is 1000 sq ft.
I even gave you the pipe volume.
I said the fluid is air so SP = 1.29
the DP is 114.7 - 14.7 = 100


You do realize the the exact numbers are meaningless - regardless of how fast or slow the air comes out of the tank, or how much or little of it there there, the momentum of the tank will always be equal, but opposite, to that of the air expelled from the tank.

None-the-less, I did miss your specifications (although even here you didn't give them all). 60" diameter pipe (1.152m), 500' (152.5m) long. You haven't given a divergence angle for the nozzle, so I'll assume it is zero (i.e. there is no cone; just a striaght tube).

Volume of pipe = pi*r^2*h = 3.14*(1.152/2)^2*152.5m = 158.95m^3

I'm going to assume your volume of 1000 sq ft is actually cuft. 1000cuft = 28m^3, which is compressed to 10132kPa (100ATA).

The force generated upon the release of this air is:
F=[m*Ve]+[(Pt-Pa)*Ae]

Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is:
F = (Pt-Pa)*Ae
F = (10132kPa-101kPa)*0.00064516m^2
F = 6471N

As the tank empties and the pipe fills, that force will drop. Thrust ends when Pt = Pa. ASCII doesn't allow me to show my work, integrations and all that, but excel gives a total thrust of just over 60kN*s (kilo-newton seconds).

Originally Posted By: paul
you cant do the math , so you claim that I cant.


Just did the math. And I state again that you cannot; nor do I believe you understand it either. But as I stated at the beginning, the specific numbers do not matter - the force applied to the tank is equal, but opposite to the force applied to the air coming out of the tank. No matter how much force is generated the pipe sits still.

Unless you make enough force to rupture the pipe...

It isn't rocket science - oh wait, in this case it kinda is... smile

And here we are again - exactly where we were at the beginning of our "discussion" - you still do not have the faintest understanding of newtons laws of motion and law of conservation of energy. And once again I've provided yet another layer of mathematics showing the impossibility of your perpetual-motion engine.

Bryan
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