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Re: gas container



At 10:47 11/2/98 -0500, Ludwik wrote:
Suppose that a spherical shell of thickness d is used to contain a
gas whose pressure is given, for example, 100 or 100,000 atm.
What is the necessary value of d for steel?...
Any mechanical engineer (or a physicist, chemist, etc.) to help us?

I responded:

I would like to simulate some slight knowledge of this topic, when I have a
moment this evening....


A formalism used by students who wish to estimate the internal pressure
of a coke can with a strain gage ( a very popular project among a certain
class of students) is to suppose that the stress in a thin wall is constant
where the wall thickness is much less than the cylindrical wall radius.
The stress on the wall is calculated by imagining the can slit either
hoop wise or along its long axis. The force which holds the can together
is computed from the projected area either of its end caps, or of its
half walls, the internal pressure, and the wall length and thickness which
are restraining this explosive force.

In a ridiculously simple formula, we take the stress as p.r/t or p.r/2.t
(p pressure, r radius, t thickness) for the
hoop stress and the longitudinal stress respectively.

On a metalworking newsgroup recently, a contributor asked of a design for
a micro turbine being bruited by someone at MIT...

My question is how heavy would the container have to be to hold 7 grams
of hydrogen so it can run for an hour....

....to which I replied as follows with an estimate couched in a plaintive
mix of Imperial and SI, as the conversions came to mind, with corrections
included as provided by friendly hecklers....
(You are cordially invited to spot yet more glaring errors of elementary
arithmetic :-)

Density of H is 0.09 kg.m^3 (rounded)
From P.V = R.T
Start with what volume at stp?
0.007/0.09 m^3 = 0.0777... m^3 at P = 1 atmosphere,
or 78 ccs (rounded) at 1000 atmospheres.

A sphere is structurally efficient...
Volume of sphere is
4/3 pi.r cubed = 78 cc
r cubed = 78 . 3/4.pi = 18.6 cm^3
r = 2.65 cm = 1.043 in.

Now let's use carbon fiber at say 500 kpi allowable stress (!)
(the kpi unit is x1000 psi)
stress = P.R/2T so (pressure, radius, thickness)
T = P.R/2.Stress = 1000 x 14.7 x 1.043 /(2 x 500000) = 0.0154 in

Area of sphere is 4 pi r^2
so volume of material is 4 x pi x 1.043 x 1.043 x 0.0154 = 0.21 cu in

Density of graphite is 2300 kg / m3
No allowance for composite matrix (!)
Carbon fiber weight is 0.21/39.37^3 x 2300 kg = 0.0079 kg or
0.3 oz.
Well, waddaya know! <grin>
( Take this with several grains of salt...)