Calculating the force provided by an opened compressed air tank

  • Thread starter AussieDave
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  • #1
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Hey. Quick question.

If we have a compressed air tank of 550 kPa with a opening area of 0.25 cm^2, would the force provided just be P*A so it'd just be 550 * 0.25 (with appropriate conversions to metres and pascals etc). We are looking for the force that'd be provided to a model car that the tank (from a paintball gun) is attached to.

???

Or is it something a little bit more complicated than that. This isn't a homework question where we need an exact answer, we only need to estimate it.
 

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  • #2
I was having a similar problem actually. Any ideas?
 
  • #3
Q_Goest
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Hey. Quick question.

If we have a compressed air tank of 550 kPa with a opening area of 0.25 cm^2, would the force provided just be P*A so it'd just be 550 * 0.25 (with appropriate conversions to metres and pascals etc). We are looking for the force that'd be provided to a model car that the tank (from a paintball gun) is attached to.

???

Or is it something a little bit more complicated than that. This isn't a homework question where we need an exact answer, we only need to estimate it.
If this is an estimate, then yes it's a perfectly acceptable method for calculating force. The force determined using this method will be a bit higher than actual.
 
  • #4
minger
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Will you not also have a component due to mass flow?
 
  • #5
Q_Goest
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Hi minger,
Will you not also have a component due to mass flow?
Good question. One can either analyze this assuming conservation of momentum OR assuming a pressure over an area. For example, consider a rocket. The total thrust is accurately determined using conservation of momentum alone - the pressure inside the combustion chamber is not additive, it is simply used to determine the flow and velocity.

Using just pressure times area is often done just to get a worst case (maximum force) value. Because that pressure has to be converted to kinetic energy (momentum), there are losses in this conversion for a simple orifice, so the force determined using pressure x area will always be higher than the force determined using conservation of momentum.
 
  • #6
minger
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OK OK, basically yea it depends on where the CV is drawn. If its drawn right at the exit, then there is negligible pressure difference as the pressure has already been converted to velocity, whereas if it's across a larger portion, then velocities are small.

I recently did a problem with a nozzle. With a nozzle because there is a dP and change in velocity across the nozzle, both effects need to be considered.
 
  • #7
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The total thrust is accurately determined using conservation of momentum alone - the pressure inside the combustion chamber is not additive, it is simply used to determine the flow and velocity.
I think this depends on your definition of accurate. For a small CO2 canister you can probably neglect the internal pressure but for something like a large diameter ducted fan, probably not.

Here's a good source to learn about thrust.
http://www.grc.nasa.gov/WWW/K-12/airplane/thrsteq.html
 
  • #8
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Hey fellas, thanks for your help.

I tried using conservation of momentum because I use that all the time in rocket equations. This is, however, just an estimate based on a gas tank that we'd buy online (we don't actually buy it...we just do the design) and there isn't any info on the mass flow rate.

I'll just use the Pressure x Area because it doesn't have to be a great estimate and the thng will never actually be tested for accuracy.

Thanks again.
 

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