How much energy is there in a filled scuba tank?

We debated this and I was wrong to suggest it was relatively low so here is the derivation and the formula to work it out. Skip over the maths and get to the stuff at the end if you just want the results and don't want to see how I get to the numbers.

Revised Version 2
Thanks to Jorge Carballeira who found the error
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Derivation of a formula

Consider a box. It is is h high and has a top and bottom area of A units.
It is full of gas at pressure p and the height h is variable so as h varies so p varies.
The outside pressure is Pa

Let's be simplistic and use Charles's law that states p*v is constant for a given volume of gas provided the temperature stays constant.
So for a given starting pressure P0 and volume V0 we can deduce any later pressure as p = P0*V0 / v

But since the volume is v=h*A we can reason that
p = P0*V0/(h*A)

Now if we let the gas expand just a teeny bit (a distance dh) the work done (dW) is the distance moved multiplied by the force applied
Force is (p - Pa) * A so
dW = ((P0*V0)/(h*A) - Pa) * A * dh
dW = (P0*V0/h - Pa*A) * dh

So now we want to sum all the dW for all the dh for a range of h from our initial pressure/volume until the pressure reaches Pa.
So the initial limit is based on v=h*A giving h = V0/A
and the closing limit is based on p = P0 * V0 / (h * A) so for p=Pa we get h = P0*V0/(Pa * A)

W = integrate((P0*V0/h - Pa*A) * dh) using limits above

W = P0*V0 * integrate(h-1*dh) - A * Pa * integrate(dh)

W = P0*V0 * [logeh] - A * Pa * [h]

substitute in the initial and final values for h

W = P0*V0 * (loge(P0*V0/(A * Pa)) - loge(V0/A)) - A * Pa * (P0*V0/(A * Pa) - V0/A)

since log(a)-log(b) = log(a/b) we can simplify this a bit

W = P0*V0 * (loge((P0*V0/(A*Pa))/(V0/A)) - A*Pa*(P0*V0/(A*Pa)-V0/A)

now it all comes together with basic algebra

W = P0*V0 * loge(P0/Pa) - P0*V0 + Pa*V0

W = P0*V0 * (loge(P0/Pa)-1) + Pa*V0

Well I could use this but it's messy and I only want a feel for the numbers. P0/Pa is the tank pressure divided by normal air pressure and
loge of 200 is 5.3 and loge of 300 is 5.7 so I'll call it 5.5 also Pa*V0 is pretty trivial beside the first term so it all becomes:
W = P0*V0* 4.5

Great. Just need to sort out the units.
The world in general seems to have settled for MKS (Meters/Kilograms/Seconds) as a way to do business in physics but I'd prefer litres and bar here.
1 bar is 105 Newtons/square meter and 1 litre is 10-3 cubic meters so our formula becomes
W (in Joules) = P0*V0*450

E = P0*V0*450

So for a 230 bar 12L tank we have 230*12*450 Joules. 1242000 joules!

You're not impressed? You don't have a feel for a joule perhaps?
A joule is one watt for one second. so 1242000 joules is 3 kilowatts for just under 7 minutes.
That would boil 3.5 litres of water and so make coffee for everybody on the boat?
Still doesn't seem much does it? The trick is to release it in an instant.
In a previous life (see CV) I used a unit of energy that was 4.184x1012 joules representing the energy released by 1000 tons of TNT.
That works out at about 1866000 joules per pound.
In metric units the tank contains the energy in 300 grams of TNT. A normal hand grenade has about 150 grams.

Hum. That works out at 650 grams for my 10L twins at 300 bar.
Just behind my head? Now I see why people worry about it.

Second problem: How much energy stored in elastic deformation of the tank?

This seems to be very dependent on tank geometry but it can be estimated.
As before skip the analysis if you just want the numbers.

Derivation of a formula

Consider a bit of tank wall. The radius is R and the thickness is T. We will work on a slice of this W long rather than do the whole tank.
We want the force that pulls left and right at the line A. I contend that this is half the total force pulling the left hand side to the left plus the force pulling the right hand side to the right. Half because the piece of wall opposite A is also being stretched.

The force on A can therefore be the sum of all the right hand components of the pressure on the wall for all values of θ from 0 to π/2

Well for a vanishingly small slice of θ ie. dθ we have an area W*R*dθ
so if the pressure difference across the wall is P a force of W*R*P*dθ

The horizontal component is then W*R*P*sinθ*dθ
so integrating from 0 to π/2 we get W*R*P*(-cos(π/2)+cos(0))) = 2*W*R*P

We want to use the Young's modulus values for the tank material and this is the ratio of the stress to the strain. ie: m = (force/unit area)/fractional increase in length

Now the piece of than wall this acts on is a the a area at A which is W*T so the stress is 2*W*R*P/W*T giving 2*R*P/T
so the stretch is 2*R*P/(T*m)

Now the energy in a even stretch is ½ * force * distance
The distance stretched is the circumference * the strain ie: D = 4*R2*π*P/(T*m)
giving an energy of E = 4*R3*π*P2*W/(T*m)

OK so lets try some numbers in MKS units again using one of my 10L tanks.
R = 0.09 meters
P = 300x105N/sq meter (300 bar)
W = 0.45 meters
m = 2x1011 for steel

T is a problem but to a first order guess the tanks weigh 16Kg so that is virtually 2 Litres = 0.002 cu meters to make a cylinder 0.5m long by 0.5m circumference leaving 0.008m (0.8cms for the wall)

Bang all the numbers in and we get 2319 joules
less than 0.2% of the air. So that's all right then I suppose.


by Nigel Hewitt
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