Finding speed of projectile propelled by pressurised air

Pharrahnox

I have a projectile of 4 grams, an air tank of 120 psi, a barrel length of 80 centimetres and a barrel diametre of 8 millimetres. I would like to find the ideal speed (muzzle velocity) that the projectile would be launched at, not taking into account the pressure drop, friction atmospheric pressure and stuff like that.

What I did was start by finding out how much force the air from the tank would exert on the projectile. So I used Pascal's Law: F (newtons) = Pressure (pascals) * Area (metres^2).
120 psi is roughly 827371 pascals
The area (m^2) - 0.004^2*pi = approx. 5*10^-5 (1/20000 or 0.00005)m^2
So F = 827371*0.00005
F = 41.37N

Then I put that into F = ma to find the acceleration:
a = F/m
a = 41.37N / 0.004kg
a = 10342.5 m/s

So I have the acceleration, but how can I find how lond it accelerates for? It travels (under pressure) down the barrel for 0.8 metres, is there a calculation I can use to find the speed?

Thanks for any help.

Simon Bridge

Fire it through a photogate.

If the volume of the barrel is small compared with the volume of your reservoir (so the pressure does not drop much with the increased volume) then you can just use kinematic equations and the length of the barrel as an approximate method.

If the projectile is light - you could figure that the projectile quickly reaches the same speed as the escaping air and work that out. If you have a pressure guage on the reservoir you can measure the pressure before and after firing once and work out the change in energy that suggests... guessing that most of that energy goes into the projectile. Be aware that anything you do will be approximate though.

You will be much better working empirically - since you've built the thing!

Pharrahnox

The air released in each shot is small compared to the tank's volume (24 litres). I was unfamiliar with the kinematic equations, so I had a look at them. None of them solve for time. Can one of them be rearranged to solve for time or speed with the known variables?

Simon Bridge

That's right.

JustinRyan

Acceleration is in metres per second squared. So if you re-arrange....

Seconds=sqrt(metres/acceleration)

Pharrahnox

@JustinRyan that would give me a value 89.44, which seems like a reasonable estimate.

Thankyou both for your help. I will refer to that method in the future, I had never even thought of that relationship before.

Simon Bridge

It's misleading:

##s=ut+\frac{1}{2}at^2 \Rightarrow t=\sqrt{2s/a}##
(s=barrel length, t=time in the barrel, a=acceleration, u=initial velocity)

But you don't need the time of the acceleration to get the muzzle velocity.
Since the initial velocity is 0, you have the two equations:

(1) ##s=\frac{1}{2}vt##
(2) ##v=at##

you know s, and a ... you want to know v.
solve for t in (2) and substitute into (1)... then make v the subject.

Pharrahnox

Sorry if this sounds like a stupid question, but what do v and s represent? If it is velocity and speed, then in equation (2), I still only know 1 of 3 variables.

Simon Bridge

... what is it you wanted to know? muzzle velocity right? That's v.

You told me earlier that you had looked up kinematic equations.
The standard set are called "suvat" equations - the standard notation is:

s: displacement
u: initial velocity
v: final velocity
a: acceleration
t: time

BTW: you know the relationship between area and diameter for a circle is ##A=\frac{1}{4}\pi d^2## ?

JustinRyan

[QUOTE=Simon Bridge;4134262]It's misleading:

Apologies.
Call it what it is...wrong. By a factor of two.

Had to look through the derivation of the suvat to where the 2 came from.

Simon Bridge

[QUOTE=JustinRyan;4134370]It was accurate dimensional analysis ;) Just remember that the displacement is the area under the v-t graph - and that, for u=0, that graph is a triangle: the 2 comes from the formula for the area of a triangle.

Pharrahnox

I just used pi*r^2 instead, which gives the same result.

Oh I see what you mean now. My brother helped me, I haven't done this sort of stuff yet in school...

So it would then be: s = 1/2*at^2
Which when rearranged to make t the subject gives: t = sqrt(2s/a), which I know see in the second part of the equation.

I then multiplied by the acceleration -approx. 10000- and that gave me 126.5 m/s.

Is this correct?

Simon Bridge

That would be it.

In general it is better to avoid lots of steps when you put the numbers in by doing the algebra earlier - like this:

Equations:
(1) v = at
(2) s = (1/2)at^2 (since you insist - but see eq1 post #7)

from (1) you get: t = v/a
put that into (2) to get: s = (1/2)a(v/a)^2 = (v^2)/2a
rearrange to give: v = sqrt(2as) <--<<< which is what you wanted.

The scientific bit would be to check this using an experiment.

cjl

That may be somewhat close, but that is a significant fraction (a bit under 40%) of the speed of sound in air, so the airflow down the tube is a factor that must be considered. My guess is that (assuming that your derivation and calculations are correct) your value should be used as an upper limit, when in reality, the exit velocity will probably be a bit slower.

Pharrahnox

Yeah, I was just looking for an ideal velocity, thankyou all for your help.