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Expansions in thermodynamics using differentials

  1. Jul 14, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A welded railway train, of length 15km, is laid without expansion joints in a desert where the night and day temperatures differ by 50K. The cross sectional area of the rail is 3.6 x 10-3m2.
    A)What is the difference in the night and day tension in the rail if it kept at constant length?
    B)If the rail is free to expand, by how much does its length change between night and day? (linear expansion coefficient, ##\alpha = 8 \times 10^{-6} K^{-1}##, Young's modulus ##Y = 2 \times 10^{11} N m^{-2}##)

    2. Relevant equations

    Total differential, cyclical relation

    3. The attempt at a solution
    I have the correct answer to A) and know how the answer to B) is attained, but I don't quite get it.

    Expressing the tension F = F(L,T) as a total differential; $$dF = \left(\frac{\partial F}{\partial L}\right)_T dL + \left(\frac{\partial F}{\partial T}\right)_L dT$$

    In A), I can cancel the first term on RHS. For B), I cannot.

    First attempt

    I rearranged the above to obtain: $$\int \left(\frac{\partial F}{\partial L}\right)_T dL = \int dF - YA\alpha(T_2 - T_1),$$using the expression for ##Y##.

    Simplifying further gives: $$L_2 - L_1 = \exp\left(\frac{\Delta F - YA\alpha(T_2 - T_1)}{YA}\right)$$
    I don't know ##\Delta F## so I was unable to proceed here.

    Second attempt

    Write the differential for L = L(F,T} to give ##dL = \frac{A}{YL}dF + \alpha L dT## This gives the correct answer for B) provided the answer to part A) is used as dF in the above eqn.
    Why is it okay to use this dF here? (In part A), that dF is derived assuming constancy in the length of the rail - part B) is concerned with the contrary - i.e dropping that assumption so I don't see how that dF is applicable)
    Secondly, why does method 1) fail?

    Many thanks.
     
  2. jcsd
  3. Jul 14, 2013 #2
    Thermal expansion causes the zero-stress length of an object to increase. When the object is heated, the new zero stress length is equal to [itex]l_o(1+αΔT)[/itex], where l0 is the length with no temperature change and no loading stress. When stress is also added to the object, the strain must be calculated relative to this thermally expanded length:
    [tex]strain=\frac{l-l_0}{l_0}-αΔT[/tex]
    This relation assumes that the strain is small, and neglects second order interactions between thermal expansion and loading stress. With this strain, the stress is given by
    [tex]σ=\frac{F}{A}=Y\left(\frac{l-l_0}{l_0}-αΔT\right)[/tex]
    If the rail is kept at constant length, then l = l0, and
    [tex]σ=\frac{F}{A}=Y(-αΔT)[/tex]
    If the rail is free to expand, then σ = 0, and
    [tex]\left(\frac{l-l_0}{l_0}-αΔT\right)=0[/tex]
    A good way of thinking about this is to envision it as a two step process...first, heating at zero stress to a new length, and then applying a stress to the thermally expanded object (with strain relative to its thermally expanded length).
     
    Last edited: Jul 14, 2013
  4. Jul 14, 2013 #3

    SteamKing

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    Shouldn't (L - Lo)/Lo = αΔT ?
     
  5. Jul 14, 2013 #4
    Oops. Yes. My mistake. Thanks for pointing that out. If I can, I'm going to go back and correct it in my earlier response. Thanks again.

    Chet
     
  6. Jul 15, 2013 #5

    CAF123

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    Thank you Chestermiller, but this problem comes from the second chapter of the book I am referring to, before stress/strain details are considered.
    Could you tell me what is wrong with my method 1) and why using dF in B) as found in A) is appropriate?
     
  7. Jul 15, 2013 #6

    SteamKing

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    In your method 1, you have assumed that F is a function of T and L. However, if T is held constant, then F must equal 0, since there will be no thermal stresses created in the material. I don't follow the rest of your derivation where you calculate L2 - L1 = exp (mess).
     
  8. Jul 15, 2013 #7
    One thing to note is that the notation ## \left( \frac {\partial X} {\partial Y} \right)_Z ## is highly misleading and is best avoided altogether. ## \frac {\partial X} {\partial Y} ## already means that all variables except ## Y ## are held constant while differentiating, so wrapping that in ## \left( \ \right)_Z ## is meaningless mathematically, yet leaves the impression that it is somehow different, but it is not.
     
  9. Jul 15, 2013 #8

    CAF123

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    Hi SteamKing,
    Why is T being held constant here? The difference in T being night and day is a given 50K.
    I can write it out more explicitly (and following voko's advice above): $$\int \frac{\partial F}{\partial L} dL = \Delta F - YA\alpha(T_2 - T_1),$$ Using ##\frac{\partial F}{\partial L} = YA/L## this becomes $$YA\int \frac{1}{L} dL = \Delta F - YA\alpha(T_2 - T_1)$$ Solving gives the result in OP.
     
  10. Jul 15, 2013 #9

    CAF123

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    Hi voko,
    Yes, I understand this, I was just following the notation in my book.

    P.S Was my last post in my other thread fine?
     
  11. Jul 15, 2013 #10
    dF = X dL + Y dT. In case A, dL = 0, dT is given, find dF. In case B, dF = 0, dT is given, find dL.
     
  12. Jul 15, 2013 #11

    CAF123

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    dF is zero because the rail is allowed to freely change length as a result of the change in T and since there is no restriction on the final length of the rail, the tension is zero?
     
  13. Jul 15, 2013 #12
    Correct.
     
  14. Jul 15, 2013 #13

    CAF123

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    Using dF = 0 in the second attempt in the OP gives the correct answer, while in the first attempt, I should set ##(\partial F/\partial L) = 0## but, by doing this, I consequently cancel out the dL and left with 0 = YAα(T2-T1), which is nonsense.
     
  15. Jul 15, 2013 #14
    I am not even sure how you set up "integration" in the first attempt. It does not look right at all. dF is a full differential, so its integral does not depend on the path. The other terms are not full differentials, so they are path dependent, and you do not specify the path.
     
  16. Jul 15, 2013 #15

    CAF123

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    Do you mean specify the initial and final points of dL and dT in my integration limits?
     
  17. Jul 15, 2013 #16
    I mean specify the path. The functions X and Y (as in dF = X dL + Y dT) depend on (L, T), and it takes more than specifying a couple of points to integrate them.
     
  18. Jul 15, 2013 #17

    CAF123

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    I don't think I understand exactly what you are saying, but I think what you said stems from the fact I am expressing dL ( not a full differential given that F = F(L,T)) in terms of a full differential (dF) and another non-full differential (dT). Is this so? Expressing L =L(F,T) was easier as dL in this instance was a full differential.

    Is my first method perhaps more complicated then? What would I need to do to specify explicitly the path?
     
  19. Jul 15, 2013 #18
    As I said, I do not quite follow the method. And the reason is that F and its derivatives are functions of two variables. Integration of such functions generally requires a particular path - one and the same path on both sides of the equation, too. It is not clear at all how you set up your integration, but it seems that ## \frac {\partial F} {\partial L} ## is integrated along the L-direction, while the thing on the right along the T-direction, which is not good.
     
    Last edited: Jul 15, 2013
  20. Jul 15, 2013 #19
    I don't understand what you did mathematically in your first attempt, but your starting equation is correct, so let's go from there.

    $$dF = \left(\frac{\partial F}{\partial L}\right)_T dL + \left(\frac{\partial F}{\partial T}\right)_L dT$$

    Hooke's law assumes that the temperature is held constant while the object is stretched, so
    $$\left(\frac{YA}{L}\right)dL = \left(\frac{\partial F}{\partial L}\right)_T dL $$ and so
    $$\left(\frac{\partial F}{\partial L}\right)_T =\left(\frac{YA}{L}\right)$$

    The coefficient of thermal expansion is defined by the change in length of the object as a function of temperature at zero constant tensile force (or any other constant tensile force). Thus, with pure thermal expansion, you have dF = 0, and thus
    $$0 = \left(\frac{\partial F}{\partial L}\right)_T dL + \left(\frac{\partial F}{\partial T}\right)_L dT$$
    or, equivalently,
    [tex]\left(\frac{\partial L}{\partial T}\right)_F=-\frac{\left(\frac{\partial F}{\partial T}\right)_L }{\left(\frac{\partial F}{\partial L}\right)_T}[/tex]
    The left hand side of this equation is αL. So,
    [tex]\left(\frac{\partial F}{\partial T}\right)_L =-αL{\left(\frac{\partial F}{\partial L}\right)_T}=-YAα[/tex]
    Substituting back into your original equation gives:
    $$dF = YA(d\ln L -αdT) $$
    I hope this helps.
    Chet
     
  21. Jul 16, 2013 #20

    CAF123

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    This was derived using the cyclical relation, right?

    This means ##F = YA(\ln(L_2 - L_1) - \alpha(T_2 - T_1))##. For part A), the first term is zero on the RHS and so ##F = -YA\alpha(T_2 - T_1)##. For part B), the LHS is zero but when I solve for ##L_2 - L_1##, I obtain the incorrect answer. Why is this so?
     
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