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Determination of ID and OD at Cryogenic Temperatures

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Johnson_123

Mechanical
Sep 12, 2016
16
I have a 316 SS tube with an I.D. of 0.43″ and O.D. of 0.5 ″. Could someone please advise me on how I would calculate the I.D. and O.D. of the tube at cryogenic temperatures? The linear thermal expansion percent (L – L293/L293) at a temperature range (0 K – 300 K) is the only material property that I have access to. Thank you!
 
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What is wrong with starting with; original size - (CTE x delta T x original size)
NIST publishes good low temp CTE data. It isn't linear.

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P.E. Metallurgy, Plymouth Tube
 
it seems you have data to 0K. If CTE is not linear then start at room temperature and work down step-wise, working on the diameter (as the significant dimension).

another day in paradise, or is paradise one day closer ?
 
EdStainless and rb1957
I perhaps should have asked, if the percent change in the length of the tube for a given temperature difference is the same as the percent change in its diameters individually?
 
I "think" so ... that diameters change as ruled by CTE. But that does raise the question, "is there a CTE for volumes?" ... if a bar extends governed by CTE, does it's cross-section reduce (to keep volume constant) ?

another day in paradise, or is paradise one day closer ?
 
rb1957

If the cross-section were to contract with the expansion in length, would the Poisson's ratio figure in the expression?
 
every dimension changes by the same amount.
The tube will get larger in diameter, thicker wall, and longer as it gets hotter.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube
 
EdStainless

Based on your statement, will the below expressions suffice? Thank you.
OD (Temp) = OD (293) + alpha (Temp)
ID (Temp) = ID (293) - alpha (Temp)
 
'cept it's been said "It isn't linear"; so my suggestion of piecewise linear (or integration ??)

another day in paradise, or is paradise one day closer ?
 
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