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Linear thermal expansion problem

  1. Feb 16, 2014 #1
    Hey fellas plz consider helping me with this..
    There is a bar of some material that is heated from state 1 to 2 to 3.
    If l1 is the length at 1 then we have,
    l2 - l1 = l1(1 + a(t2-t1))
    l3 - l1 = l1(1 + a(t3-t1))
    If, t3-t2 = t2-t1,
    Then l3-l2 = l2-l1
    But if it is written
    l3 - l2 = l2(1+a(t3-t2))

    l3 - l2 > l2 - l1
    Why is this..???
  2. jcsd
  3. Feb 16, 2014 #2
    The difference between the two calculations is negligible. For a more accurate calculation you must take into account that the expansion coefficient is a function of temperature and the change in length is given by the integral [tex]\Delta L = \int {a(T)}dT.[/tex] If the expansion coefficient is considered constant (excellent approximation for most circumstances), we get [tex]\Delta L = \int {a(T)}dT=a \int dT = a\Delta T.[/tex] That way we recover the simpler expression you're used to.
  4. Feb 16, 2014 #3


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    You need to be careful here, if the temperature range is large. The expansion coefficient values should be given relative to a reference temperature, and may also be temperature dependent.

    In that case, if you want to find the expansion from temperature ##T_1## to ##T_2##, the correct way is to calculate $$\alpha_{T_2}(T_2 - {T_{\text{ref}}}) - \alpha_{T_1}(T_1 - {T_{\text{ref}}})$$
    For example the coefficient for Titanium alloys can vary by a factor of 4 over a temperature range from -150C to +200C.

    See http://www.kayelaby.npl.co.uk/general_physics/2_3/2_3_5.html for many examples of non-constant thermal expansion - and note the two ways it is specified. (The above method uses "mean expansion coefficients" which is how engineering data is usually tabulated).

    But for "homework" type questions, you can probably ignore all of this, and just use ##\alpha(T_2 - T_1)##.
    Last edited: Feb 16, 2014
  5. Feb 16, 2014 #4
    That depends on how the table was designed. It's always possible and sensible to use the temperature itself as the reference temperature. In other words [tex]\alpha = \frac{\partial L}{\partial T}[/tex]
  6. Feb 16, 2014 #5
    Because the absolute value of the thermal expansion depends not only on the temperature difference but on initial lengths.
    A constant thermal expansion coefficient tells you that for same temperature difference the relative expansion is the same.
    If you heat up a 1m bar and a 2m bar by the same ΔT, they will have the same relative expansion, Δl/l. But the 2m bar will expand twice as much.
    So your l2 being more than l1, you can expect l3-l2 to be more than l2-l1.
    Why would you think otherwise?
  7. Feb 16, 2014 #6


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    You seem to be worrying about the consequences of the arithmetic involved and not the Physics. As nasu says, if you interpret you answer correctly, in terms of the data you are inserting, there isn't a problem.
    The other posts are more about the Physics of the business and the practicalities. There is no such thing as truly linear expansion - expansion 'approaches' linear as the temperature change approaches zero ( as with all the 'Laws' that assume linearity and a single coefficient to calculate with)
  8. Feb 16, 2014 #7
    Thnk u all but the thing is yes it is true that 'a' coefficient of thermal expansion depends on temperature and thermal expansion is linear for small temperature changes, the problem i am facing is even if i am considering these two i am getting a diff result.

    For ex. If we cosider
    t3-t2 = t2-t1 = unity
    Then for expnasion from l2 to l1
    The base (reference) length is l1
    But for expansion from l3 to l2
    The base (reference) length is l2

    So, for same change in temp. Change in length is diff.

    One more thing is if we consider a thermometer working betweenqq
  9. Feb 16, 2014 #8
    If we consider a thermometer working between t3 and t1 then for some intermediate temperature say t2 are the equations of thermal expansion applicable there as it ...
  10. Feb 16, 2014 #9
    Yes, for same temperature change the change is different. There is no contradiction in this, in principle. I already gave you the example with two bars of different lengths and same temperature difference.

    Anyway the linear formula for thermal expansion is just an approximation.
    It works well if ε=α ΔT <<1.
    For your example,
    l3-l2=l2αΔT=l2ε=l1(1+ε)ε=l1ε +l1ε2
    But l1ε=l2-l1.
    So the difference between the two changes is of the order ε2.
    However when we decided to use the linear formula for thermal expansion we assumed that terms of higher order are negligible. Then in this model, the two differences are "the same".

    If the difference is not negligible, then the linear formula does not work so you have to use a better model.

    In practice the linear model is quite good and the two differences are the "same" with good approximation for temperature ranges of the order of 100 C, as found on most common thermometers.
    The values of α for solids are of the order 10-6. For a change in temperature of the order of 100 C, the ε in the above equation will be 10-4.
    By neglecting the second order terms the error will be of the order of 10-4 or 0.01%.
    So the the two differences, even though they are technically different they are practically the same. For mercury the coefficient is a little higher, of the order of 10-5 but still low enough to have a scale approximately linear for a thermometer with a range of 100 C.
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