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Thermal expansion over a large temperature range

  1. Mar 14, 2016 #1
    So I've been working at a steel mill where we deal with billets cooling from temperatures around 1200 C to between 10-25 C. I have access to average thermal expansion coefficients over this temperature range.

    First question: Over a large temperature range as aforementioned am I correct in assuming it is better to integrate the thermal expansion formula rather than taking an average value for the thermal expansion coefficient?

    Second question: When integrating the thermal expansion formula am I correct in assuming (as is usually the case) that I should use temperatures in Kelvin?
     
  2. jcsd
  3. Mar 14, 2016 #2
    I should clarify that the average expansion coefficients are averaged over various steels, not the temperature range.
     
  4. Mar 15, 2016 #3
    Most thermodynamics formulas were worked out in Kelvin and hence you should either look for the modified version in Celsius or simply convert your temperatures to Kelvin. As for the expansion coefficient, the accuracy of the values becomes more important (over the averaged one) the smaller the volume. It is not clear from your post what sizes are you dealing with exactly, but if you wants absolute accuracy, don't use the averaged value, or do a solid scientific experiment and run a couple of identical ones (one with an averaged value for C and one with an integrated value) and compare the two.
     
  5. Mar 15, 2016 #4
    Please specify the exact forms of the thermal expansion equations that you are talking about here.
     
  6. Mar 16, 2016 #5
  7. Mar 21, 2016 #6
    I'm only concerned with length so its the linear equation of thermal expansion: (ΔL/L0)=α(T)ΔT
     
  8. Mar 21, 2016 #7
    Eric, in this equation, alpha depends on both T and T0, and the equation is valid only for small values of ΔT. If you want to write the equation correctly, and you want it to apply over a large range of temperatures, you have to write $$\frac{1}{L}\frac{dL}{dT}=\alpha(T)$$and integrate this differential equation with respect to T. Otherwise, you will get the wrong answer.
     
  9. Apr 4, 2016 #8
    Ok perfect that answers my first question, thanks! And you would use temperatures in Kelvin after integrating correct?
     
  10. Apr 4, 2016 #9
    I would use degrees C. There is no need to convert to Kelvin in this calculation.
     
  11. Apr 4, 2016 #10
    What if α(T) takes the form of a third order polynomial for instance? Wouldn't units have an effect then?
    For instance (3254-2754)≠(524-24)
     
  12. Apr 4, 2016 #11
    The relationship for alpha has to specify whether the temperature in the equation is expressed in C or whether it is expressed in K.
     
  13. Apr 4, 2016 #12
    Makes total sense, so since in my case since I plotted α vs. T in Kelvin to get an equation for α(T) the units for the integrated equation would also be Kelvin. But had I plotted α vs. T in Celcius I would then use Celcius in the integrated equation as well.
     
  14. Apr 4, 2016 #13

    Nidum

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    Worth noting that if the top temperature puts the metal in the plastic range then contraction on cooling is partly dependent on the cooling procedure .
     
  15. Apr 5, 2016 #14
    Interesting, could you elaborate? I wasn't aware different cooling mechanisms could affect how much a material contracts. If you could give a few examples I'd be interested to hear.
     
  16. Apr 5, 2016 #15
    Is this true, even though the expansion occurs at negligible stress (and isotropically)?
     
    Last edited: Apr 5, 2016
  17. Apr 7, 2016 #16

    Nidum

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    Thank you for your interest . I am preparing a note on this subject .
     
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