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Determination of ID and OD at Cryogenic Temperatures

Determination of ID and OD at Cryogenic Temperatures

Determination of ID and OD at Cryogenic Temperatures

(OP)
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!

RE: Determination of ID and OD at Cryogenic Temperatures

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.

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

RE: Determination of ID and OD at Cryogenic Temperatures

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 ?

RE: Determination of ID and OD at Cryogenic Temperatures

(OP)
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?

RE: Determination of ID and OD at Cryogenic Temperatures

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 ?

RE: Determination of ID and OD at Cryogenic Temperatures

(OP)
rb1957

If the cross-section were to contract with the expansion in length, would the Poisson's ratio figure in the expression?

RE: Determination of ID and OD at Cryogenic Temperatures

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

RE: Determination of ID and OD at Cryogenic Temperatures

(OP)
EdStainless

Based on your statement, will the below expressions suffice? Thank you.
OD (Temp) = OD (293) + alpha (Temp)
ID (Temp) = ID (293) - alpha (Temp)

RE: Determination of ID and OD at Cryogenic Temperatures

'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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