Pressurised cylinders
Pressurised cylinders
(OP)
I am struggling to visualise the effect and apply the relevant laws to the following model and am wondering if anyone could help.
An inflatable such as a cycle inner-tube fully inflated at ground level to the correct pressure would react 'how' at altitude in say a non-pressured storage compartment of a passenger plane.
An inflatable such as a cycle inner-tube fully inflated at ground level to the correct pressure would react 'how' at altitude in say a non-pressured storage compartment of a passenger plane.





RE: Pressurised cylinders
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: Pressurised cylinders
(11 psi) at 35000 ft.
RE: Pressurised cylinders
RE: Pressurised cylinders
RE: Pressurised cylinders
rb1957 you are exactly where I am, to quote you: somewhere there will be an equilibrium achieved. What can be used to calculate this equilibrium position?
RE: Pressurised cylinders
how big is the tire on the ground ? R+dr ... R is the on-ground uninflated tire tube radius
circumferential strain=stress*E = (pR/t)*E
circumferential deflection = 2piR*(1+(pR/t)*E)
so dr = R*(pR/t)*E
at altitude p increases, to say p+11psi (from RPstress above)
which will cause an increase in radius (as above)
and an increase in volume, piR^2*L (L is the length of the rim, near enough)
then the chemistry part, pV = constant; the increase in volume means a reduction in pressure p2 = p1*V1/V2
which will cause a reduction in the pressure differential, a reduction in the volume, an increase (from the previous estimate) in pressure differential ...
iterate for awhile
note some approximations, pR/t assumes a racing tire (a rubber tube). for a standard tire, anchored on a fixed rim, the volume calculation would be slightly different (as a portion of the (cross-section) circumference won't strain (appreciably). i've based the volume of the tire on the rim length (L), more accurate would be (L/pi+2R)*pi = L+pi*2R
good luck
RE: Pressurised cylinders