Volume of air from psi and diameter of pipe?

  1. I need to find the volume of air (ft^3) for this sample. The air is discharged at 40 psi for 1 minute and the diameter of the pipe is 0.5 inches (Area = 0.001363538 ft^2). I think I need to solve for the velocity of the air (vel = sqrt(2P/rho), where P is the pressure and rho is the density). For the density I will assume the air is at 70°F and atmospheric pressure, so the density will be 0.074887 lbm/ft^3. I think I keep messing up with my units so I am not sure if I have the right answer or not. In US units I get about 32 ft^3 (I may be using the gravitational constant incorrectly). In SI units I get about 5.14 m^3 (181.67 ft^3). I am not really sure what I can expect for the volume of air, for air being discharged at 40 psi for 1 minute through a 0.5 inch diameter pipe. Any help would be greatly appreciated. Thank you.
     
  2. jcsd
  3. aWhy not just use the standart 'p-v-t' relationships??

    in one second, yoiu have a known volume of gas at a known temperature and pressure.

    That volume is then released to occupy a much larger (unknown) volume at a different pressure-volume-temperature condition...

    simple calculations ought to give you X ft^3 per unit time

    Hopefully this might give you 'enough' of an acceptable answer.
     
    Last edited: Nov 28, 2011
  4. Let me give you a little bit more information of the background. This is a large compressed air system. The samples are to be taken at several use points. Each use point has it owns regulator, which during the sample taking process is set to 40 psi.

    So how would I have a known volume? Would I get that from the max volume capacity of the regulator (I don't think that would be correct)? If I were to use the p-v-t relationship would my initial pressure be 40 psi and my final be atmospheric? I would imagine the temperature would not change too much, but I don't know for sure and I would not be able to measure the initial temperature of the air in the compressed air system.

    Why can't I just solve for the velocity of the air, which would seem to be a lot simpler?
     
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