## Need formula/help with mass air flow

Hello,

My name is Quinton and I am doing a small project that includes a small compressed air tank attached to a section of 2" PVC pipe. My problem is I need to find the force that the air creates when all the air from the tank is released through a valve (also 2"). Any help is appreciated.

-Quinton
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 You have to calculate the force when air is released by knowing what? I mean, i imagine you´ll be able to know the air's pressure and volume, is that the data that I can use to give you and answer? Thanks
 Correct. The tank has a volume of 603.18 inches3 and the tank would be at a pressure of 80psi.

## Need formula/help with mass air flow

You´ll see Quintonbs I've been doing some calculation and I have found a formulae that gives the force that will develop the air going out the tank. I'm quite unconfident of it beacuse it looks "odd" to me (too simple, too... I don't know), so please if anybody who can judge it read this post i ask him/her to do it.

If A is the section of the pipe inside which air flows and P is pressure inside of the tank and P0 outside it, the propulsion force in the opposite direction of the air flow in modulus is given by:

$$F_{prop}=2A(P-P_{0})$$

I´ll keep on this tomorrow so is probable i do some corrections

salutatios :)
 That kind of sounds right. But the variable that I can't seem to fit anywhere is time. The longer the mass of air takes to exit the nozzle the less force it will give. I don't know how to calculate that in.
 Yeh, as I said I've doing some more calculations and found that the problem is to obtain a P(t) since pressure inside the tank decreases as air is released of it.
 Here I show my calculations to get a P(t), so please if anybody see any mistake in my approach i'd be glad to know. Well lets start: 1st Consider the state equation for ideal gases $$PV=nRT\Rightarrow PV=\frac{m}{P_{m}}RT$$ then if m is the mass of air inside the tank and P its pressure (the rest of factors are constants), differenciate both sides of the eq. to get dP/dt $$\frac{dP}{dt}=\frac{dm}{dt}\frac{RT}{VP_{m}}$$ 2nd Using the expression I found for the above-calculated Fprop of dm/dt and simplifying we get the ODE $$\frac{dP}{dt}=\frac{Av}{V}\sqrt{P^2-PP_{0}}$$ here A is the section of the valve, V is the volume of the tank and v is a constant that depends on air's temperature as follows $$v=\sqrt{\frac{2RT}{P_{m}}}$$ where R is the ideal gasses constant (R=0.082 atm·L/mol·ºK), Pm air's molecuar mass and T air's temperature. 3rd Solve this 1st order-separable ODE (whith sightly heavy integration) with condition P(0)=Pi and get P(t) $$P=P_0sinh^{2}(\frac{Av}{2V}t-k)$$ where k is $$k=arcsinh\sqrt{\frac{P_i}{P_0}}$$ Just to make clear P is P(t), P0 is atmosferic pressure and Pi is initial pressure in the tank. That was the calculations I made for modelizing air flowing out a constant-volume tank, so to get the "desired" ;) F(t) just put P(t) in the formulae I wrote above: $$F_{prop}(t)=2AP_0(sinh^2(\frac{Av}{2V}t-k)-1)$$ Hope this helps, good science Quintonbs. :)

 Tags air, flow, mass, pipe, thermodynamics

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