Answer to the Question 06/01
MINIMAL CONTAINERThe 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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