Heat Equation Boundary Conditions

[Vector1962]

Homework Statement

Let a slab $0 \le x \le c$ be subject to surface heat transfer, according to Newtons's law of cooling, at its faces $x = 0$ and $x = c$, the furface conductance H being the same on each face. Show that if the medium $x\le0$ has temperature zero and medium $x=c$ has the constant temperature T then the boundary value problem for steady-state temperatures in the slab is $$u''(x)=0$$ $$Ku'(0)=Hu(0)$$ $$Ku'(c)=H[T-u(c)]$$ where K is the thermal conductivity of the material in the slab, write $h=\frac{H}{K}$ and derive the expression $$u(x)=\frac{T}{ch+2}(hx+1)$$

Homework Equations

$U_t=K\nabla^2U$

The Attempt at a Solution

.[/B]
I have the first part of the question complete. I'm not sure how to apply the boundary conditions? I've solved similar problems with boundary's $u(0)=0$ and $u(c) = T$ and I've solved various problems with heat flux boundary's. For some reason, applying the Newton boundary conditions is messing me up.
I can get to $u(x)=c_1x+c_2$ comfortably which leads to $$u(x)=u_x(x)x+u(0)$$ $$u(c)=u_x(c)c+u(0)$$ but from there I'm lost.

 

[geoffrey159]

1. Find that $u'(x)$ is constant. Find it's expression.
2. Deduce that the general expession for $u$ is $u(x) = u(0) + x u'(0)$. Improve that with 1.
3. From the text plus points 1. and 2. above, you should be able to find two expressions for $u'(c)$. Find it and deduce $u(0)$
4. Conclude

 

[Vector1962]

1. Find that u ′ (x) is constant. Find it's expression

$$u''(x)=0$$ $$\int u''(x)\,dx = u'(x)+c_1$$ $$u'(x)=c_1$$

Deduce that the general expression for u is u(x)=u(0)+xu ′ (0) . Improve that with 1.

$$\int u'(x)\,dx = \int c_1\,dx$$ $$u(x)= c_1x + c_2$$
at $x=0$ yields $u(0)=c_2$
$$u(x)= x u'(x) + u(0)$$ $u(x) = xu'(0) + u(0)$ because $u'(x)=u'(0) = u'(c)=c_1$
from the boundary values ; $u'(0)=\frac {H} {K} u(0)$ and $u'(c)=\frac{H}{K} [T-u(c)]$

From the text plus points 1. and 2. above, you should be able to find two expressions for $u'(c)$. Find it and deduce $u(0)$

I don't see it from here?
$u(0)=T-u(c)$ from the boundary conditions?

 

[geoffrey159]

Recall the fundamental theorem of calculus which says that $u(b) - u(a) = \int_a^b u'(x) dx$ if $u'$ is continuous.
From that theorem you should be able to put subscripts and superscripts on your integrals and get to the result easily.

 

[geoffrey159]

For point 1. : You have for whatever $0 \le x \le c$ that $u'(x) - u'(0) = \int_0^x u''(s) ds = 0$ so $u'(x) = u'(0)$ for any of these $x$. Which proves that $u'$ it is constant.

For point 2. :
$u(x) - u(0) = \int_0^x u'(s) ds = x u'(0) = x h u(0)$ because $u'$ is constant and because you have a constraint from your text.
Now you have $u(x) = u(0) (1+xh )$

Can you finish it?

 

[Chestermiller]

You were off to a great start when you wrote $u(x)=c_1x+c_2$. Now substitute u=u(0) at x = 0 and u = u(c) at x = c into the equation to determine the two constants of integration c₁ and c₂ in terms of u(0), u(c), and c. Next evaluate u'(0)=u'(c) in terms of u(0) and u(c). Substitute this into the two boundary condition equations. This will give you two equations that allow you to solve for u(0) and u(c) in terms of K, H, and T.

Chet

 

[Vector1962]

Can you finish it?

I can follow everything you did to get to $u(x)=u(0)(hx+1)$ but no clue how to finish

This will give you two equations that allow you to solve for u(0) and u(c) in terms of K, H, and T.

I've tried this route at least 20 times and always end up in algebra hell "dancing" around the solution but never getting there. After which, I thought I'd post it to the forum and see if I could get a hint to bring it to closure. Can't quite seem to get there.

 

[Vector1962]

Apparently, I'm applying the boundary conditions the right way, but failing miserably in the algebra. As I understand everything discussed this far, it appears $u'(x)=\frac{T}{c}$ correct?

 

[Chestermiller]

Apparently, I'm applying the boundary conditions the right way, but failing miserably in the algebra. As I understand everything discussed this far, it appears $u'(x)=\frac{T}{c}$ correct?

No.

 

[Chestermiller]

$$u(x)=u(0)+\frac{u(c)-u(0)}{c}x$$
$$u'(x)=u'(0)=u'(c)=\frac{u(c)-u(0)}{c}$$
$$K\frac{u(c)-u(0)}{c}=Hu(0)$$
$$K\frac{u(c)-u(0)}{c}=H(T-u(c))$$
Solve the previous two equations for u(0) and u(c)

Chet

 

[Vector1962]

$u(0)=\frac{u(c)}{1+hc}$ and $u(c)=\frac{hcT+u(0)}{1+hc}$
insert these into $$u(x)=u(0)+\frac{u(c)-u(0)}{c}x$$ ?

 

[geoffrey159]

You have two ways of expressing $u'(c)$.
First one is using the fact $u'$ is constant, so $u'(c) = u'(0) = h u(0)$.
Second expression of $u'(c)$ is in your text, it's a constraint, and says $u'(c) = h(T-u(c))$.
Now you have a general expression for $u$, which we wrote previously, so $u(c)= u(0) (1 + hc)$
Equating the two expressions of $u'(c)$ will give you $u(0)$ and final answer.

 

[Chestermiller]

$u(0)=\frac{u(c)}{1+hc}$ and $u(c)=\frac{hcT+u(0)}{1+hc}$
insert these into $$u(x)=u(0)+\frac{u(c)-u(0)}{c}x$$ ?

Do you know how to solve two simultaneous linear algebraic equations in two unknowns? We learned this in 1st year algebra in 8th grade.

Chet

 

[Vector1962]

Some people don't have the benefit of a formal high school or college education. I just get a math book every once in a while and work some problems. Math is not that hard. I get stuck every once in a while and need a "pointer" to get me out of the rut. Thanks so much for the help. Following both of your leads I arrive at:
$$u(x)=\frac{T}{ch+2}(hx+1)$$

 

[Chestermiller]

Some people don't have the benefit of a formal high school or college education. I just get a math book every once in a while and work some problems. Math is not that hard. I get stuck every once in a while and need a "pointer" to get me out of the rut. Thanks so much for the help. Following both of your leads I arrive at:
$$u(x)=\frac{T}{ch+2}(hx+1)$$

Sorry. I wasn't aware. You seemed to have some good math skills beyond HS, and I was surprised at your difficulty with algebra.

Chet