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Thread: temperature problem

  1. #1
    Junior Member pirateboy's Avatar
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    temperature problem

    Problem Statement

    The temperature, $\displaystyle T$, of a cooling fin of a finite length, $\displaystyle L$, satisfies
    $\displaystyle \frac{d^2T}{dx^2} = -k^2\left( T_L-T \right),\quad 0\leq x\leqL,\quad T(0) = 0,\quad T(L) = T_L$
    where $\displaystyle k$, $\displaystyle T_0$, and $\displaystyle T_L$ are constants


    a) Find the complimentary function$\displaystyle T_c(x)$, stating your answer in terms of unknown constants, $\displaystyle c_1$ and $\displaystyle c_2$


    b)Complete the following identity
    $\displaystyle \sinh(kl-kx)=$


    c) write down your starting point for the particular solution, $\displaystyle T_p(x)$


    d) find $\displaystyle T_p(x)$ and the coefficients$\displaystyle c_1$, and $\displaystyle c_1$ and hence show that hte solution of the above problem is
    $\displaystyle T(x) = T_L + (T_0 - T_L)\frac{\sinh(k(L-x))}{\sinh(kL)}$

    My attempt

    a)
    The first thing I do here (which may be where I'm going wrong) is that I rearrange the equation. I take

    $\displaystyle T''(x) = -k^2(T_L-T)$

    $\displaystyle T''(x) -k^2T(x) = -k^2T_L$

    characteristic equation:
    $\displaystyle r^2 - k^2 = 0$

    $\displaystyle \therefore r = \pm k$

    So

    $\displaystyle T_c(x) = c_1 \cosh(kx) + c_2\sinh(kx) + c_3 k^2T_L$

    With this, I jump straight to

    $\displaystyle T_c''(x) = k^2c_1\cosh(kx) +k^2c_2\sinh(kx)$

    I can then verify that $\displaystyle T''(x) -k^2T(x) + c_3k^2T_L = -k^2T_L$ is a solution, with $\displaystyle c_3=-1$. But this helps me very little when I try to solve for my initial values and helps me less when I try to complete part (d).

    Grinding through a bit further I can solve for $\displaystyle c_1$ with initial values with

    $\displaystyle T(0) = T_0 = c_1\cosh(k\cdot 0) + c_2\sinh(k\cdot 0) -k^2T_L$

    $\displaystyle c_1 - k^2T_L = T_0$

    $\displaystyle \therefore c_1 = T_0 + k^2T_L$

    So my new $\displaystyle T(x)$

    $\displaystyle T(x) = (T_0 + k^2T_L)\cosh(kx)+c_2\sinh(kx)-k^2T_L$

    But then attempting to solve for $\displaystyle c_2$ I get stuck. I end up with

    $\displaystyle T(L) = T_L = (T_0+k^2T_L)\cosh(kL)+c_2\sinh(kL)-k^2T_L$


    And I think I'm stuck.
    Am I missing some simple algebra trick here, or have I gone down the wrong road?



    b) $\displaystyle \sinh(kL-kx) = \sinh(kL)\cosh(kx)-\cosh(kL)\sinh(kx)$
    I'm pretty sure that's right, at least.

    c)???

    d)???


    So, I'm thinking I'm starting this wrong, but I can't think of another way to attempt this. Any help would be greatly appreciated.
    Last edited by pirateboy; Aug 3rd 2011 at 03:21 PM. Reason: oops forgot something.
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  2. #2
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    Re: temperature problem

    Everything in your equation for $\displaystyle c_2$ is constant, so you should just be able to rearrange it to get $\displaystyle c_2$ on its own.
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  3. #3
    Junior Member pirateboy's Avatar
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    Re: temperature problem

    Quote Originally Posted by TheBigH View Post
    Everything in your equation for $\displaystyle c_2$ is constant, so you should just be able to rearrange it to get $\displaystyle c_2$ on its own.
    Ah, right, but solving that I get a monster. How do I get that to look like (d), is the question.
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  4. #4
    Junior Member pirateboy's Avatar
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    Re: temperature problem

    ok, got it. I was starting off wrong. If anyone is interested, I'll post the solution when I get a chance.

    hint:

    $\displaystyle T(x) = c_1\cosh(kx) + c_2\sinh(kx) +c_3 T_L$

    $\displaystyle c_3 = 1$

    The rest is algebra.
    Last edited by Ackbeet; Aug 3rd 2011 at 05:02 PM.
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  5. #5
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    Re: temperature problem

    Quote Originally Posted by pirateboy View Post
    Problem Statement

    The temperature, $\displaystyle T$, of a cooling fin of a finite length, $\displaystyle L$, satisfies
    $\displaystyle \frac{d^2T}{dx^2} = -k^2\left( T_L-T \right),\quad 0\leq x\leqL,\quad T(0) = 0,\quad T(L) = T_L$
    where $\displaystyle k$, $\displaystyle T_0$, and $\displaystyle T_L$ are constants


    a) Find the complimentary function$\displaystyle T_c(x)$, stating your answer in terms of unknown constants, $\displaystyle c_1$ and $\displaystyle c_2$


    b)Complete the following identity
    $\displaystyle \sinh(kl-kx)=$


    c) write down your starting point for the particular solution, $\displaystyle T_p(x)$


    d) find $\displaystyle T_p(x)$ and the coefficients$\displaystyle c_1$, and $\displaystyle c_1$ and hence show that hte solution of the above problem is
    $\displaystyle T(x) = T_L + (T_0 - T_L)\frac{\sinh(k(L-x))}{\sinh(kL)}$

    My attempt

    a)
    The first thing I do here (which may be where I'm going wrong) is that I rearrange the equation. I take

    $\displaystyle T''(x) = -k^2(T_L-T)$

    $\displaystyle T''(x) -k^2T(x) = -k^2T_L$

    characteristic equation:
    $\displaystyle r^2 - k^2 = 0$

    $\displaystyle \therefore r = \pm k$

    So

    $\displaystyle T_c(x) = c_1 \cosh(kx) + c_2\sinh(kx) + c_3 k^2T_L$

    With this, I jump straight to

    $\displaystyle T_c''(x) = k^2c_1\cosh(kx) +k^2c_2\sinh(kx)$

    I can then verify that $\displaystyle T''(x) -k^2T(x) + c_3k^2T_L = -k^2T_L$ is a solution, with $\displaystyle c_3=-1$. But this helps me very little when I try to solve for my initial values and helps me less when I try to complete part (d).

    Grinding through a bit further I can solve for $\displaystyle c_1$ with initial values with

    $\displaystyle T(0) = T_0 = c_1\cosh(k\cdot 0) + c_2\sinh(k\cdot 0) -k^2T_L$

    $\displaystyle c_1 - k^2T_L = T_0$

    $\displaystyle \therefore c_1 = T_0 + k^2T_L$

    So my new $\displaystyle T(x)$

    $\displaystyle T(x) = (T_0 + k^2T_L)\cosh(kx)+c_2\sinh(kx)-k^2T_L$

    But then attempting to solve for $\displaystyle c_2$ I get stuck. I end up with

    $\displaystyle T(L) = T_L = (T_0+k^2T_L)\cosh(kL)+c_2\sinh(kL)-k^2T_L$
    That's a simple linear equation:
    $\displaystyle c_2\sinh(kL)= T_L+ k^2T- (T_0+ k^2T_L)\cosh(kL)$
    $\displaystyle c_2= \frac{T_L+ k^2T- (T_0+ k^2T_L)\cosh(kL)}{\sinh(kL)}$


    And I think I'm stuck.
    Am I missing some simple algebra trick here, or have I gone down the wrong road?



    b) $\displaystyle \sinh(kL-kx) = \sinh(kL)\cosh(kx)-\cosh(kL)\sinh(kx)$
    I'm pretty sure that's right, at least.

    c)???

    d)???


    So, I'm thinking I'm starting this wrong, but I can't think of another way to attempt this. Any help would be greatly appreciated.
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