Energy Equations Driving Me Crazy

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The discussion focuses on designing an air chamber for a paintball marker using energy equations. The user aims to achieve a final velocity of 290 feet per second with an initial pressure of 150 psi, using the equations for kinetic and potential energy. There is confusion regarding the calculations, particularly in determining the required volume of air, with a friend's estimate of 6.4 in³ being questioned. Additional insights include the importance of accounting for potential energy and friction losses in the calculations. The user confirms they will use compressed air instead of CO2, which may affect the calculations.
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Alright, so I'm trying to design an air camber for a paintball marker. I know to get an approximation for the volume I can use energy equations. I'm using:

KE1 + PE1 = KE2 + PE2
and
KE = .5mv^2
PE = pressure(volume)

I want an initial pressure of 150 psi, and a final velocity of 290 feet per second. A paintball's mass is approximatly 3.19 grams or 0.007 lb.

I set up an equation like this:

(150 lb/in^2)(X)=(.5)(0.007lb)(3480 in/s)^2

A friend said it was 6.4 in^3, but I can't see what he did.

I don't know if I've just done conversions wrong, set up the problem wrong, or what.
 
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KE1 + PE1 = KE2 + PE2[/QUOTE said:
Is right how ever I can not re make your 2nd equation I need more information what is .5mv^ also KE1 should = 0. I also think that you might have a unit issue is mv mil volt? If so volts are not a measure of energy a measurement of energy would be like a joule.
 
I apologize: .5mv^2 is one-half mass times velocity squared. Or (1/2)(mass)(velocity)^2 if that is easier to see.

I realize that KE1 would be zero, and PE2 should be able to be ignored as well.
 
I think you are missing a term in potential energy. Maybe gravity ?
 
I found the following info, it might be of help for your calculations:

We know that 1.8 grams of CO2 == 61.02in3 so (12/1.8)(61.02)=406.7Vin3 at 1atm
We know that 1atm == 14.7 PSI so (406.7) * 14.7 / 150 = 39.86in3
So 12 grams of CO2 will fill 39.86in3 at 150 PSI at 77F.

Calculating the velocity:
V²rms = 3RT/M; where M is molecular mass in kg/mole and R is the gas constant with the units 8.314 J/mol K. These unusual units are used because we want the answer in m/s, and by definition, 1 joule equals 1 kg*m²/s². The molecular mass of CO2 is 44 g/mol, or 0.044 kg/mol.
The calculation on this:Vrms = √[(3) (8.314 J/mol K) (280 K) / (0.044 kg/mol)]
Vrms = 398.4 m/s. So the speed of molecules at 280K is 398,4 m/s.

You would need to calculate the loss due through friction (length pipe)
and the inertia of your paint ball.

good luck
 
The potential energy of a fluid can be expressed as the pressure times the volume if I'm not incorrect.

I'm going to use compressed air not co2, so I'm looking for the needed volume of air in a holding chamber at 150 psi (my desired pressure), that will propel a paintball up to 290 feet per second.
 
Ok, did'nt catch up that your using air.
Perhaps you don't need much more for calc the velocity by changing the mol weight?
 
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