Quick thermal contraction question 1

[DRWeig]

Help an EE who knows the methods but not enough of the principle when an object has a hole in it?

My issue is a hollow cylinder (pipe) that will be chilled. I was taught by an old geezer long ago that the outer diameter of the pipe will expand in linear fashion according to the linear coefficient of expansion when heated, and will contract when cooled.

However, he didn't mention what happens to the inner diameter of the pipe. I am not sure whether the thickness of the walls will contract toward the center of the wall, or if the whole pipe contracts toward the center of the circle. In other words, does the inner diameter get smaller or larger when cooled?

Many thanks.

Best to you,

Goober Dave

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[MintJulep]

Imagine it were solid instead of a tube.

What would happen to the middle part?

 

[rb1957]

if you put a solid shaft in an oven i think it takes a little time before it expands. i think initially the outer surface wants to expand ('cause it's been heated by the oven) but the inner material doesn't ('cause it hasn't been heated yet); so the inner material resisits the expansion of the outer. but i don't see it contracting !?

if you put a tube in an oven (and let's say it's heated on the OD only) i think you've "just" removed some of the restraining material so the outer surface is freer to expand as it wants, and the ID will expand too. there might be some goemetry effect (maybe some material properties too) such that the thickness wants to expand more than ID which would (i think) cause the ID to shrink.

if you heat the tube from both surfaces i think it works similar to above, only now you're adding heat to the ID, it wants to expand similar to the OD.

at the end of the day, once the tube is up to heat the cross-section will have expanded by the coeff of expansion of the material, something like (1+mu)^3. intuitively, the diamter is (1+mu)*original. if the cross-section area is (1+mu)^2*original, then both ID and OD would be (1+mu)*original, which means that the thickness has increased by a factor of (1+mu) as you'd (ok, i'd) expect.

 

[rb1957]

now i did hear (?) that heating a plate will make a hole in it smaller, which sort of makes sense (as the material around the hole expands)

 

[SNORGY]

thread391-246558

 

[hydtools]

The ID will contract as it is cooled. Think in terms of the circumference. With cooling the length of the circumference contracts, therefore the diameters contract. s = pi*D

Ted

 

[IRstuff]

"However, he didn't mention what happens to the inner diameter of the pipe"

You probably already know the answer to that. When a jar lid is stuck you heat it up, which means the ID decrease as temperature decreases for a positive coefficient of expansion.

Another way to look at it is to consider the circumference as a linear entity, which it is, piecewise, and as temperature decreases, the circumference must shrink, which requires the inner diameter to shrink

TTFN
faq731-376

 

[DRWeig]

Thank you all! Mint made it clearest by forcing me to think. All the other confirmations are appreciated too.

It think it comes with getting old. I have done shrink-fit (interference) inserts and tube-in-tube stuff a number of times over the years, but I didn't make the connection.

Best to you,

Goober Dave

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