Answer to the Question 06/01
MINIMAL CONTAINER
The question was:
A given amount of gas is held in a spherical container. For what gas
pressure the weight of the container will be minimal?
(7/01) The problem has been answered correctly by
Fred Goesmann (6/6/2001) from Max-Planck-Institut fuer Aeronomie
in Katlenburg-Lindau (Germany) (e-mail
goesmann@linmpi.mpg.de) - see remarks at the bottom of the page,
and by Larry Weinstein (5/7/2001) from Old Dominion University (Norfolk, VA, USA)
(e-mail
weinstei@physics.odu.edu)
Answer: The weight of the container is almost independent of gas pressure.
The solution:
The weight of the container is proportional to HR2,
where H is the thickness of the wall and R is the radius
of the sphere. (This assumes that the thickness of the walls is much smaller
than the radius of the sphere.)
The thickness should be such that the container does not rupture
under pressure p. If the critical stress that can be sustained by the
metal from which the container is made is S, then the critical thickness
is proportional to (p/S)R. This result can be obtained from a direct analysis
of the forces, since the stress in the sphere must create a resulting force
opposing the internal pressure.* Alternatively, the same answer can be obtained from
pure dimensional analysis: Abviously, the H must be proportional to pressure
since all the forces are proportional to it; the only combination
of pressure, critical stress and radius that has dimensions of length is (p/S)R.
Since the pressure is inversely proportional
to the volume of the gas (i.e. proportional to 1/R3), the
critical thickness H is simply proportional to 1/R2.
Consequently, the weight of the container is independent of R !
As we saw, the decrease of thickness with increasing radius is exactly
compensated by increasing area, and consequently the weight of the
container is really independent of the radius of the container, or
of the gas pressure.
(Of course, if you do not neglect the ratio between the thickness and
the radius, then you will discover that larger radia (smaller pressures)
are slightly more advantageous.)
* Comment: A simple way to get that relation is as follows: Consider half of the sphere
exactly at a critical pressure;
the pressure p applies force p{pi}R2 to the hemisphere; it is opposed
by the stress S which is applied to the the container walls, which have cross section area
2{pi}RH, creating force 2{pi}RHS. By equating the two forces, you will get
the desired relation.
Remark: Fred Goesmann wrote us (slightly edited):
In our department we are building a gas-chromatograph/mass-spectrometer
for space applications and we need to take
carrier gas with us. And as always one has to
minimise mass, volume, power, everything.
A description can be found
here.
The surprising result is that the mass of a
spherical container is independent of the pressure.
... The result holds also for long cylindrical containers.
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