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Accurate Formulas on Thermal Expansion

  1. Jun 20, 2012 #1
    1.The problem statement, all variables and given/known data.
    Can someone give me more accurate mathematical models on thermal expansion? for example, linear expansion.


    2.Relevant equations
    L=L0(1+α∆T)
    this equation is not accurate. for example.
    When an object 100m long at 0oC (example: aluminum, α=2.4x10-5) is heated to 100oC, then:

    L0=100 and ∆T=100
    L= 100.24

    When it is cooled back to 0oC, then:

    L0=100.24 and ∆T=-100
    L=99.999424

    There is a little discrepancy but it still is not accurate.

    another is this:

    First situation: aluminum (100m long at 0oC) is heated to 200oC, then measure the length.
    L=100.48

    Second situation: aluminum (100m long at 0oC) is heated to 100oC, the length is measured. then from 100oC it is heated again to 200oC then the length is measured again.
    at 0oC, L=100
    at 100oC, L=100.24
    at 200oC, L=100.480576

    Again, there is a dicrepancy at 200oC . one is 100.48 and another is 100.480576.


    3. The attempt at a solution
    I made a formula. can you test its validity?
    L=L0eα∆T
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 20, 2012 #2
    There is another question:
    If L=L0(1+α∆T)
    then it is possible for L to be zero?

    0=L0(1+α∆T)
    Suppose L0 is not equal to zero.
    0=1+α∆T
    ∆T= -1/α
    α is a very small number. for example, aluminum again.
    α = 2.4x10-5.
    then ∆T=-41666 2/3.

    It is a very low temperature and I'm not sure if it can be achieved physically but according to the computation, it is possible.
     
  4. Jun 21, 2012 #3
    Hi Keldric! Welcome to Physics Forums :smile:

    That is because the coefficient of linear expansion changes a tiny little bit as the temperature changes. The change is usually very very less, so it doesn't cause much trouble, and hence we use the equation [itex]L = L_o (1+\alpha \Delta T)[/itex]

    A more accurate formula can be derived through integration, knowing how α changes. But for most practical cases, the one we use is good enough.

    The lowest possible physical temperature is absolute zero, which is 0 Kelvin, or -273.16 Celcius. And even that has not been attained yet(or cannot be achieved at all) :wink:
     
  5. Jun 21, 2012 #4

    ehild

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    Lo is defined as the length at a standard temperature To, usually at 0 °C, and ΔT=T-To.
    If the length of the rod is 100.24 cm at 100°C its standard length is
    Lo=L(100)/(1+100 α).

    ehild


    Edited!
     
    Last edited: Jun 21, 2012
  6. Jun 21, 2012 #5
    t0=t(100)/(1+100α)?

    T0=100.24(100)/(1+100α)

    α=2.4*10^-5

    t=10048.0576 ?
     
  7. Jun 21, 2012 #6
    According to this site (http://www.wisegeek.com/what-is-the-highest-possible-temperature.htm), the highest possible temperature is 1.41679 x 1032 Kelvin.
    If we are going to use T0≥ 41666 2/3, then T≥0 if ΔT=-41666 2/3. The final temperature is now possible to be achieved (T≥0).

    Can you please show me the more accurate formula? I don't know much about higher math but I still want to see it.
     
  8. Jun 21, 2012 #7
    Of course, but at those very high temperatures, all known elements would be in their gaseous form. Tungsten probably has the highest boiling point, and that's a puny 5660 degrees. And our formula is -not- valid for gases or liquids.

    You cannot have the 'more accurate formula' without knowing exactly how the coefficient of expansion varies. I presume the variation would be different for different metals.
     
  9. Jun 21, 2012 #8

    ehild

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    Read my edited post #4 again. I mistakenly typed T instead of L in the last formula. It explains the error in your calculation in the original post.
    Lo is not the starting length at any temperature, but the length at a reference temperature.

    Apart from this, the coefficient alpha changes in broad temperature range. You can find it in tables.
    And of course, it changes at phase transitions.

    ehild
     
  10. Jun 22, 2012 #9
    in my previous post, I already used the value for L.

    L0=t(100)/(1+100α)?

    L0=100.24(100)/(1+100α)

    α=2.4*10^-5

    L0=10048.0576 ?

    Is there a list of α of a certain material at different temperatures? if there is one, please show me.
    Another thing, please show me your reference material. is it a book? an e-book?
     
  11. Jun 22, 2012 #10
    Ya, but I remember that this isn't Classical thermodynaimics.
     
  12. Jun 22, 2012 #11

    ehild

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    L(T) is function of the temperature. L(100) means the length at 100°C, alpha is the coefficient defined at 0°C. The length at 0°C is Lo=100 cm.

    L(100)=Lo(1+100α)=100.24 cm.

    If you know the length at 100 °C you get the length Lo by dividing 100.24 by (1+100α): Lo=100.24/1.0024 =100.

    The length is linear function of the temperature in first approximation. But is is fairly good in practice. It is not exponential as you supposed in your original post. Think of the mercury thermometer. The scale is equidistant instead of being logarithmic.

    Linear function means that L(T)=AT+B where and B are constants. At a reference temperature To, the length is Lo: Lo=ATo+B.
    The change of length is L(T)-Lo=A(T-To), that can be rewritten as L(T)=Lo+A(T-To). Introducing the coefficient of thermal expansion α as the relative change of length at 1 °C change of temperature, α=A/Lo:

    L(T)=Lo(1+α(T-To)).

    In Thermodynamics, the linear coefficient of thermal expansion is defined as the logarithmic derivative of L with respect to the temperature. It is not the same as α in the formula above. It changes with the temperature.
    Think of the volumetric thermal expansion of an ideal gas. You know that it is linear function of the temperature. It expands by 1/273-th of its volume at 0°C. But the thermal expansion coefficient defined as 1/V(dV/dT) at constant pressure is 1/T. It is not constant.


    ehild
     
  13. Jun 22, 2012 #12
    I think it would be helpful to discuss the precise definition of the coefficient of linear expansion. The starting point for the discussion is the specific volume of the material V (which is volume per unit mass). The specific volume is a function of temperature and pressure:

    V = V (T,p)

    The coefficient of volume expansion is equal to the fractional change in specific volume with respect to temperature at constant pressure (typically 1 atm):

    β (T,p) = [itex]\frac{1}{V}[/itex]([itex]\frac{\partial V}{\partial T}[/itex])p

    The coefficient of linear expansion is 1/3 the coefficient of volume expansion:

    α (T,p)= β/3 = [itex]\frac{1}{3V}[/itex]([itex]\frac{\partial V}{\partial T}[/itex])p

    The coefficient of linear expansion describes how straight line segments within the material increase in length with differential changes in temperature. Note that neither the coefficient of volume expansion nor the coefficient of linear expansion are constants. They are functions of temperature (and pressure). However, over small finite temperature intervals, these parameters and nearly constants. So the word "linear" does not mean that the length of a line segment increases "linearly" with temperature. The word "linear" implies "line segment."
     
  14. Jun 22, 2012 #13
  15. Jun 22, 2012 #14
    Thanks for the answers.
     
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