Pressurised cylinders

[makeup]

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.

 

[GregLocock]

Its absolute pressure would drop. Its gauge pressure would increase.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[RPstress]

It would increase in size and the stresses in the rubber would increase. External pressure drop is about 770 mbar
(11 psi) at 35000 ft.

 

[makeup]

And this increase would give rise to a volume change, correct? Thus bike tyres are deflated before being allowed into the storage compartment.

 

[rb1957]

as RPstress notes, the tire would "feel" a higher pressure differential (as the outside pressure decreases). because the tire is an elastic membrane it would expand, as you note causing a volume change, but this would reduce the tire's internal pressure (as the air inside the tire is fixed), and somewhere an equilibrium would be achieved. as the tire would be more highly stressed than before, it is quite possible that it would explode, causing consternation to the crew and pax. i'm willing to bet this has happened in the past, causing, as you've noted, the airline policy to deflate the tires. if the tires are deflated and left uncapped, then there will be no load in the tire. if the tires are capped, the maximum pressure differeintial is less than 14.7psi, easily within the capacity of a tire.

 

[makeup]

Thanks everyone.

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?

 

[rb1957]

a mixture of statics, and chemistry !

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

 

[makeup]

Thats brilliant thanks.