Basic math for air pressure?

  • #1

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Hiya! I'm an ameture "maker", and recently I've become interested in air pressure and such. I thought a fun project to help me learn about it would be to make one of those small air cannons that are powered by bicycle pumps (safety first and all that). One thing confuses me though: how exactly does one calculate pressure in psi from volume? I've been thinking about this in terms of air (from a bike pump?) being added to a container of X volume in Y cubic inches per cycle/pump/etc. Is there a set of formulas for how pressure changes this way, or is it supposed to be inferred? Thanks in advance for any help you guys can give me - for the record, I think PF is a lot better than the Stack Exchange :)
 

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  • #2
scottdave
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The Hyperphysics website would be a good place to start. Here is a page discussing the Ideal Gas Law.
 
  • #3
BvU
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For air you can use the ideal gas law . For those obsolete units you are on your own or left to a converter (1 psi = 0.0689475729 bar )

Doesn't help you much at this point, but nevertheless.

With a bike pump you compress the contents of your pump and (if the pressure in the container is lower) send it to the container.

Suppose your pump piston has an area of 10 cm2 and your can compress it from 40 to 5 cm. Assume the temperature doesn't change (in fact it does!) with pV = nRT you calculate (nRT the same, V 8 times smaller) that you can increase the pressure to 8 bar. 8 bar (800000 N/m2) times 0.001 m2 means the force you need at the end of the stroke is 800 N -- rather hefty !

In a container of 1 dm3 (0.001 m3) at 8 bar (800000 N/m2) and 300 K you have pV = 800000 * 1/1000 and PV / RT = 0.32 mol of air (molweight 29 g/mol) , so 9.3 grams.

May I suggest you build a pet bottle rocket instead of an air cannon ? You'll still need the bike pump ...
 
  • #4
For air you can use the ideal gas law .

With a bike pump you compress the contents of your pump and (if the pressure in the container is lower) send it to the container.

Suppose your pump piston has an area of 10 cm2 and your can compress it from 40 to 5 cm. Assume the temperature doesn't change (in fact it does!) with pV = nRT you calculate (nRT the same, V 8 times smaller) that you can increase the pressure to 8 bar. 8 bar (800000 N/m2) times 0.001 m2 means the force you need at the end of the stroke is 800 N -- rather hefty !

In a container of 1 dm3 (0.001 m3) at 8 bar (800000 N/m2) and 300 K you have pV = 800000 * 1/1000 and PV / RT = 0.32 mol of air (molweight 29 g/mol) , so 9.3 grams.
Interesting; From some some "back of the envelope" math:
(please excuse my terrible and convoluted science/math skills - I totally get it if you don't want to read through all this ↓)
Estimating, the dimensions of my little pump are 2" wide by some 4" long. From the cylinder volume formula that makes about 12.57"3.
Then we take this "Ideal gas law" (I'd heard of it but never bothered to learn :frown:) pV = nRT and rearrange it to get:
p = nRT / V
now I looked at that wiki page you gave me, and from what I can see of that.. oh. Now that's a wild goose chase if I ever saw one. OK from what I gather:
p = pressure; that'll be my solution
n = "absolute substance in moles" -> ??
R = "Universal gas constant" -> Ya' know what, I'll use my calculator for that
T = "Absolute temperature of the gas" -> I googled it and from what I understand absol. temp. is just in Kelvin. You said this changed, but I'll ignore that for now?
...So far I have:
p = ? x R x T / V
Soooo for the pump itself, I'd have:
p = ? x R x T / 12.56, and I'll substitute T for the average room temperature (295 Kelvin, per Google) to get p = ? x R x 295 / 12.56
But I still don't know what n is.. I do have this equation though :biggrin: so I'll try "calibrating" it:
(the wiki page you gave me said that p should be in pascals -> 1 atm = 101325 pascals)
(the page also said V should be in meters3) -> 1 in3 = 0.000016387)
101325 = n x R x 295 x 0.000016387
we use some algebra to get:
n = RT / V * P -> n = R(295) / 0.000016387 * 101325
So assuming I did ALL that right, for 1 in3 at average room temperature, n = 1.516609771e13
Golly, that's a large number! But now, armed with that unit rate, I can find the value of n for my pump system with a proportion:
(1/15,166,097,710,000) = (12.56 / X) -> n = 1.904861872e14
So after ALL THAT, we can finally find the quasi-answer- first I find the pressure (in pascals) when the pump is fully "open":
p = 190,486,187,200,000 * R * 295 / 12.56 = 3.719893795e16 pascals which means 5.395249806366171875e12 psi
but that doesn't make sense, as I doubt anything in the known universe short of a nuclear warhead can make pressures that high o_O
So if you've read all the way to down here, thank you, and I'm sorry you wasted your valuable time for a wrong answer.
I should just not post this, but hopefully someone can help me out?
Thanks again!
 
  • #5
BvU
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p = ? x R x T / V
101325 = n x R x 295 x 0.000016387
Oops, you switch from / to x !
n is the number of moles. 1 mole of air weighs 29 gram (main component is N2 ).

Kudos for the attempt to use SI units !
 
  • #6
scottdave
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Just as a check, note that an ideal gas at Standard Temperature and Pressure (0°C or 273 Kelvin and 1 atmosphere): 1 mole will occupy 22.4 liters volume.
 
  • #7
Oops, you switch from / to x !
n is the number of moles. 1 mole of air weighs 29 gram (main component is N2 ).
Kudos for the attempt to use SI units !
Thanks! :smile:
I always tend to make silly mistakes like that. I wont't dump a ton of math here again, but once again I totally get it if you don't want to read it:
Hey...it might be easier if I used metric units to start with: 12.56 in3 = 205.82152 cm3
switching like you suggested to P = nRTV and using @scottdave 's suggestion
(1mole = 22.4L @ 1atm and 273K - > convert Liter to in3 and solve by proportion for 205.82152 in3 to get ≈0.0092 moles)
I'll get the following for the "open" pump (P is atm converted to pascals):
101325 = 0.0092 x R x 273 x 205.82152 -> prove it by: 0.0092 x R x 273 x 205.82152 = 4298.094211 ≠ 101325
Wrong again..
 

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