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Need formula/help with mass air flow |
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| Sep4-09, 01:28 PM | #1 |
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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 |
| Sep4-09, 03:14 PM | #2 |
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You have to calculate the force when air is released by knowing what? I mean, i imagine youŽll be able to know the air's pressure and volume, is that the data that I can use to give you and answer?
Thanks |
| Sep4-09, 06:15 PM | #3 |
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Correct. The tank has a volume of 603.18 inches3 and the tank would be at a pressure of 80psi.
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| Sep4-09, 07:23 PM | #4 |
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Need formula/help with mass air flow
YouŽll 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: [tex]F_{prop}=2A(P-P_{0})[/tex] IŽll keep on this tomorrow so is probable i do some corrections salutatios :) |
| Sep6-09, 01:44 AM | #5 |
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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.
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| Sep6-09, 05:07 AM | #6 |
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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.
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| Sep6-09, 06:18 AM | #7 |
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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 [tex]PV=nRT\Rightarrow PV=\frac{m}{P_{m}}RT[/tex] 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 [tex]\frac{dP}{dt}=\frac{dm}{dt}\frac{RT}{VP_{m}}[/tex] 2nd Using the expression I found for the above-calculated Fprop of dm/dt and simplifying we get the ODE [tex]\frac{dP}{dt}=\frac{Av}{V}\sqrt{P^2-PP_{0}}[/tex] 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 [tex]v=\sqrt{\frac{2RT}{P_{m}}}[/tex] where R is the ideal gasses constant (R=0.082 atm·L/mol·șK), 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) [tex]P=P_0sinh^{2}(\frac{Av}{2V}t-k)[/tex] where k is [tex]k=arcsinh\sqrt{\frac{P_i}{P_0}}[/tex] 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: [tex]F_{prop}(t)=2AP_0(sinh^2(\frac{Av}{2V}t-k)-1)[/tex] Hope this helps, good science Quintonbs. :) |
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| Tags |
| air, flow, mass, pipe, thermodynamics |
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