Recent content by Eric MIlburn

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    Thermal expansion over a large temperature range

    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.
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    Thermal expansion over a large temperature range

    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 celsius I would then use celsius in the integrated equation as well.
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    Thermal expansion over a large temperature range

    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)
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    Thermal expansion over a large temperature range

    Ok perfect that answers my first question, thanks! And you would use temperatures in Kelvin after integrating correct?
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    Thermal expansion over a large temperature range

    I'm only concerned with length so its the linear equation of thermal expansion: (ΔL/L0)=α(T)ΔT
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    Thermal expansion over a large temperature range

    I should clarify that the average expansion coefficients are averaged over various steels, not the temperature range.
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    Thermal expansion over a large temperature range

    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...