Heat Equation Boundary Conditions

In summary, Chet found two expressions for u'(c) and solved for u(0) and u(c) using the boundary conditions and the properties of the heat transfer equation.
  • #1
Vector1962
61
8

Homework Statement


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

Homework Equations


[itex]U_t=K\nabla^2U[/itex]

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 [itex]u(0)=0[/itex] and [itex]u(c) = T[/itex] 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 [itex]u(x)=c_1x+c_2[/itex] comfortably which leads to [tex]u(x)=u_x(x)x+u(0) [/tex] [tex]u(c)=u_x(c)c+u(0) [/tex] but from there I'm lost.
 
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  • #2
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
 
  • #3
geoffrey159 said:
1. Find that u ′ (x) is constant. Find it's expression
[tex] u''(x)=0 [/tex] [tex] \int u''(x)\,dx = u'(x)+c_1[/tex] [tex]u'(x)=c_1[/tex]
geoffrey159 said:
Deduce that the general expression for u is u(x)=u(0)+xu ′ (0) . Improve that with 1.
[tex] \int u'(x)\,dx = \int c_1\,dx[/tex] [tex] u(x)= c_1x + c_2 [/tex]
at [itex]x=0 [/itex] yields [itex] u(0)=c_2 [/itex]
[tex] u(x)= x u'(x) + u(0) [/tex] [itex] u(x) = xu'(0) + u(0)
[/itex] because [itex] u'(x)=u'(0) = u'(c)=c_1 [/itex]
from the boundary values ; [itex] u'(0)=\frac {H} {K} u(0) [/itex] and [itex] u'(c)=\frac{H}{K} [T-u(c)] [/itex]

geoffrey159 said:
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?
 
  • #4
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.
 
  • #5
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?
 
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  • #6
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
 
  • #7
geoffrey159 said:
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

Chestermiller said:
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.
 
  • #8
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?
 
  • #9
Vector1962 said:
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.
 
  • #10
[tex]u(x)=u(0)+\frac{u(c)-u(0)}{c}x[/tex]
[tex]u'(x)=u'(0)=u'(c)=\frac{u(c)-u(0)}{c}[/tex]
[tex]K\frac{u(c)-u(0)}{c}=Hu(0)[/tex]
[tex]K\frac{u(c)-u(0)}{c}=H(T-u(c))[/tex]
Solve the previous two equations for u(0) and u(c)

Chet
 
  • #11
##u(0)=\frac{u(c)}{1+hc}## and ##u(c)=\frac{hcT+u(0)}{1+hc}##
insert these into [tex]u(x)=u(0)+\frac{u(c)-u(0)}{c}x[/tex] ?
 
  • #12
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.
 
  • #13
Vector1962 said:
##u(0)=\frac{u(c)}{1+hc}## and ##u(c)=\frac{hcT+u(0)}{1+hc}##
insert these into [tex]u(x)=u(0)+\frac{u(c)-u(0)}{c}x[/tex] ?
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
 
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  • #14
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:
[tex]u(x)=\frac{T}{ch+2}(hx+1)[/tex]
 
  • #15
Vector1962 said:
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:
[tex]u(x)=\frac{T}{ch+2}(hx+1)[/tex]
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
 

FAQ: Heat Equation Boundary Conditions

1. What is a boundary condition in the heat equation?

A boundary condition in the heat equation is a set of conditions that must be specified at the boundaries of a system in order to solve the equation and determine the temperature distribution within the system. These conditions can include the temperature or heat flux at the boundaries, as well as the material properties of the boundary itself.

2. How does the choice of boundary conditions affect the solution of the heat equation?

The choice of boundary conditions can greatly affect the solution of the heat equation. Different boundary conditions can lead to different temperature distributions within the system, and can also impact the stability and accuracy of the solution. Therefore, it is important to carefully consider and select appropriate boundary conditions for a given system.

3. What are the most common types of boundary conditions in the heat equation?

The most common types of boundary conditions in the heat equation include Dirichlet boundary conditions, where the temperature at the boundary is specified, and Neumann boundary conditions, where the heat flux at the boundary is specified. Other types of boundary conditions include Robin boundary conditions, which are a combination of Dirichlet and Neumann conditions, and periodic boundary conditions, where the temperature at one boundary is equal to the temperature at the opposite boundary.

4. Can the heat equation be solved without specifying boundary conditions?

No, the heat equation cannot be solved without specifying boundary conditions. The boundary conditions are necessary for determining the temperature distribution within a system and are an integral part of the heat equation.

5. How do different boundary conditions impact the behavior of the heat equation over time?

Different boundary conditions can result in different behavior of the heat equation over time. For example, some boundary conditions may lead to a stable and predictable temperature distribution, while others may result in oscillations or instability. The choice of boundary conditions should be carefully considered in order to achieve the desired behavior of the heat equation over time.

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