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Manufacturing question for thermal expansion disagreement

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adammclean1
#1
Mar12-14, 06:54 AM
P: 2
Hello, my colleague and I are having a disagreement about the amount of thermal expansion in a particular part we are working on (manufacturing environment).

It is a piece of structural steel, which has a thermal expansion value of 12 (10^-6/K according to the chart at http://www.owlnet.rice.edu/~msci301/...lExpansion.pdf - please check out to confirm ) and we are wondering how much linear expansion would take place over a distance of 2.794 meters with every 1 degree (K/C) change in temperature. I will not bias the answers by posting what each of us think. Please help us solve this debate. Thanks!
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sgb27
#2
Mar12-14, 07:49 AM
P: 46
Isn't the answer just 12*10^-6 times 2.794? About 34 microns per degree.
See http://en.wikipedia.org/wiki/Thermal...near_expansion.
adammclean1
#3
Mar12-14, 08:02 AM
P: 2
I believe that 12 is the coefficient of thermal expansion on that chart, we are looking for linear expansion at every degree temp change.

Simon Bridge
#4
Mar12-14, 08:12 AM
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Manufacturing question for thermal expansion disagreement

Welcome to PF;
The figure given in the reference p11 for (structural) steel is for linear thermal expansion.

If the sheet starts out with length L, then is heated by ΔT, the change in length is given by:

##\Delta L = \alpha L \Delta T## (1)

A 1m sheet will increase it's length by 12 microns per degC
A 2m sheet will increase by 24 microns per degC

If 2.794m is the initial length of the sheet, then sgb27 has it right :)
But I think you are after something else...

After heating by ##\Delta T##, the new length is (from equ 1) given by:

(2) ##L=L_0(1+\alpha\Delta T)##

... lets compare this with the case we increment by 1deg at a time:

If you start at ##L_0## at 20degC then ##L_1=L_0(1+\alpha)## at 21degC
the next degree takes you to ##L_2=L_1(1+\alpha)=L_0(1+\alpha)^2## at 22degC and so on...

...so a whole N degrees heating to 20+N degC gives final length

(3) ##L_N=L_0(1+\alpha)^N \simeq L_0(1+N\alpha)##... provided ##\alpha<<1##.

i.e. eqn(2) is actually an approximation.
I suspect this is the heart of the dispute.
sgb27
#5
Mar12-14, 01:05 PM
P: 46
Simon is of course correct, however I suspect that a larger source of error when using eqn(2) with large temperature changes is that the coefficient of expansion is not constant over temperature.
Simon Bridge
#6
Mar12-14, 06:29 PM
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Indeed - and the reference cited says:
, (CTE) Approximate Ranges at Room Temperature to 100 C (212 F),
-- Table 2.1 p11
... which is why I was careful that my heating example started at 20degC ;)

If OP is thinking of ΔL=2.794m, I don't really know what is intended by that figure, then the sheet must start out over a km long for the number to be valid.

For 100 degrees temp change:
A 1m sheet of that steel would increase it's length by 1.2007mm, predicted by equ(3).
That would be a 1.2000mm increase by equ(1) - a difference of 0.7 microns in the predictions.
... about the wavelength of dark red light.

If you got anything different, then it is because the coefficient is only valid for the particular kind of steel and under specific conditions.

Need to hear from OP now :)
Chestermiller
#7
Mar12-14, 10:42 PM
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Quote Quote by sgb27 View Post
Simon is of course correct, however I suspect that a larger source of error when using eqn(2) with large temperature changes is that the coefficient of expansion is not constant over temperature.
Actually, the more technically correct and precise definition of the coefficient of thermal expansion is:
[tex]α=\frac{d\ln L}{dT}[/tex]
So, [tex]\frac{L}{L_0}=e^{αΔT}[/tex]
or, if the coefficient varies with temperature:
[tex]\frac{L}{L_0}=\exp \left(\int_{T_0}^T{α(T')dT'}\right)[/tex]
Chet


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