# temperature problem

• August 3rd 2011, 03:09 PM
pirateboy
temperature problem
Problem Statement

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

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

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

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

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

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

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

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

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

$\therefore r = \pm k$

So

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

With this, I jump straight to

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

I can then verify that $T''(x) -k^2T(x) + c_3k^2T_L = -k^2T_L$ is a solution, with $c_3=-1$. 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 $c_1$ with initial values with

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

$c_1 - k^2T_L = T_0$

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

So my new $T(x)$

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

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

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

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

b) $\sinh(kL-kx) = \sinh(kL)\cosh(kx)-\cosh(kL)\sinh(kx)$
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.
• August 3rd 2011, 03:48 PM
TheBigH
Re: temperature problem
Everything in your equation for $c_2$ is constant, so you should just be able to rearrange it to get $c_2$ on its own.
• August 3rd 2011, 04:13 PM
pirateboy
Re: temperature problem
Quote:

Originally Posted by TheBigH
Everything in your equation for $c_2$ is constant, so you should just be able to rearrange it to get $c_2$ on its own.

Ah, right, but solving that I get a monster. How do I get that to look like (d), is the question.
• August 3rd 2011, 04:35 PM
pirateboy
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:

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

$c_3 = 1$

The rest is algebra.
• August 5th 2011, 08:26 AM
HallsofIvy
Re: temperature problem
Quote:

Originally Posted by pirateboy
Problem Statement

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

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

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

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

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

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

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

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

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

$\therefore r = \pm k$

So

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

With this, I jump straight to

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

I can then verify that $T''(x) -k^2T(x) + c_3k^2T_L = -k^2T_L$ is a solution, with $c_3=-1$. 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 $c_1$ with initial values with

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

$c_1 - k^2T_L = T_0$

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

So my new $T(x)$

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

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

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

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

Quote:

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

b) $\sinh(kL-kx) = \sinh(kL)\cosh(kx)-\cosh(kL)\sinh(kx)$
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.