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Heat Conduction Equation - BVP

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Folks, I am self studying through a heat conduction problem involving a 2nd order linear homogenous differential equation which has the solution of the form

    ##\theta (x)=C_1\cosh mx+ C_2\sinh mx## (1)

    where ##m \equiv \sqrt \frac{c}{a}= \sqrt{\frac{\beta P}{k A}} ##

    The constants are dertermined via the BC's ##\theta(0)=\theta_0## and

    ##[\theta_x+\frac{\beta}{k} \theta]_{x=l}=0## using ##sinh x =(e^x-e^{-x})/2## etc etc.

    I can determine ##C_1=\theta(0)## but I dont know how ##C_2## is determined using the hyperbolic expression....

    3. The attempt at a solution
    I attempted to rearrange ##\theta_x=-\frac{\beta}{k} \theta## from the BC given and equate that to the derivative of the general form of solution and then subsitute x=l in order to find C_2....ie

    ##\theta'(x)= m C_1 \sinh mx +C_2 \cosh mx## therefore at x=L

    ##\theta'(L)=m \theta_0 \sinh m L +C_2 \cosh m L=-\frac{\beta}{k} \theta##....?
     
  2. jcsd
  3. Jun 26, 2012 #2

    TSny

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    Looks like the correct approach. But you dropped a factor of m in one of the terms when you took the derivative of theta(x). It shouldn't be hard to solve for C2.
     
  4. Jun 27, 2012 #3
    That was a typo. I calculate ##C_2## to be

    ##\theta'(L)=m \theta_0 \sinh m L +m C_2 \cosh m L=-\frac{\beta}{k} \theta##

    Thus ## \displaystyle C_2=\frac{-\frac{\beta}{k} \theta - \theta_0 m \sinh m L}{m \cosh m L}##

    but the book gives it as

    ## \displaystyle C_2=-\theta_0[ \frac{\sinh mL+(\beta/mk) \cosh mL}{\cosh mL+(\beta/mk) \sinh mL}]##..?
     
  5. Jun 27, 2012 #4

    gabbagabbahey

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    Don't you mean [itex]\theta'(L)=m \theta_0 \sinh m L +m C_2 \cosh m L=-\frac{\beta}{k} \theta(L)[/itex] ? :wink:
     
  6. Jun 27, 2012 #5
    Thanks for that typo. Still doesn't clarify how the book arrives at their answer..?
     
  7. Jun 27, 2012 #6

    gabbagabbahey

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    Well, what is [itex]\theta(L)[/itex]?:wink:
     
  8. Jun 27, 2012 #7
    ##\theta(L) = \theta_0 \cosh mL+C_2 \sinh mL## obtained from the first equation in post#1....therefor substituting in equation in the last post we get

    [itex]\theta'(L)=m \theta_0 \sinh m L +m C_2 \cosh m L=-\frac{\beta}{k} (\theta_0 \cosh mL+C_2 \sinh mL)[/itex]...

    not sure how you can get ##C_2## from this...thanks
     
  9. Jun 27, 2012 #8

    TSny

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    Rearrange the terms so that the two terms with C2 are on the same side of the equation and the other terms on the other side. Then factor out the C2. For example if you had

    a + b*C2 = d + e*C2

    then

    b*C2-e*C2 = d - a

    C2*(b-e) = d-a

    C2 = (d-a)/(b-e)
     
  10. Jun 27, 2012 #9
    Very good guys...should have spotted that.

    Thank you.
     
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