Transient heat conduction of a semi-infinite solid

In summary: This is a problem concerning transient heat conduction in an undefined semi-infinite solid, initially at a temperature T0 whose surface temperature is suddenly raised to a new constant level at Ts. We want to:Derive the Energy EquationState initial and boundary conditionsDefine two new variables η=Cxtmθ=(T-T0)/(Ts-T0)Determine C and m (the constants?) of η and θ from Part 3Solve for η and θ themselves.In summary, the student is trying to solve for θ and C from Part 3 of the homework problem, but is having trouble
  • #1
mudweez0009
46
1

Homework Statement



This is a problem regarding transient heat conduction in an undefined semi-infinite solid, initially at a temperature T0 whose surface temperature is suddenly raised to a new constant level at Ts.
I also supplied the problem as an attachment for ease in explaining the problem.
We want to:
  1. Derive the Energy Equation
  2. State initial and boundary conditions
  3. Define two new variables
    1. η=Cxtm
    2. θ=(T-T0)/(Ts-T0)
  4. Determine C and m (the constants?) of η and θ from Part 3
  5. Solve for η and θ themselves.
I understand and completed Parts 1 and 2 of this problem. Unfortunately I don't understand how to begin Part 3... I assume its a method of Separation of Variables?

Transient Heat Conduction.PNG


Homework Equations


There are a decent amount of equations for transient heat conduction, but I don't think Parts 3, 4, and 5 relate to any specific equation. I think its a method (I forget the name) of where you define new equations so that you can re-define the variables, then plug into the original Energy equation (shown in Part 1).

The Attempt at a Solution


Here is what I've tried so far. I am getting very lost on what variables to solve for, which ones to take derivatives of, etc... I would appreciate it if someone could help me get started if I'm wrong, or point me in the right direction if I am on the right track. Thanks in advance.

upload_2015-4-2_10-34-27.png
 
Physics news on Phys.org
  • #2
This is not a separation of variables problem. The approach they are trying to lead you through is called a similarity transformation. They are trying to help you transform the differential equation from a partial differential equation to an ordinary differential equation by means of a change of independent variable. The original independent variables are combined in such a way that, when you apply the transformation, they form a single independent variable. In part 3, what they want you to try is to transform the problem from a PDE to an ODE by expressing the dimensionless temperature θ uniquely in terms of the combined variable η of the form:

θ = θ(η)

If you can do this without any t's or x's remaining in the transformed equation, then you can solve the resulting ODE for the dimensionless temperature profile in terms of η. You need to choose a value of m to get rid of the t in the transformed equation. And you need to choose a value of C to work the equation into the form that they have provided. That's what they want you to do in part 4.

Chet
 
  • #3
Okay so I get that we need to get rid of the "x's" because the final equation we want in Part 3 does not have an x. However, it does have a "t". So I'm not sure I understand the reason why I would guess an "m" that would eliminate the "t".

Is the initial approach I took correct?
Do I need to do anything with θ? I'm not sure how the final equation has dθ/dη and d2θ/dη2 in it, when θ is only in terms of T, not η.
 
  • #4
Your equation for ##\frac{\partial ^2 θ}{\partial x^2}## is incomplete. You need to apply the chain rule here too. After you do this, you get:

$$\frac{\partial θ}{\partial t}=\frac{dθ}{dη}mCxt^{m-1}=\alpha C^2t^{2m}\frac{d^2θ}{dη^2}$$

But, ##η=Cxt^m##

So you can substitute ##x=\frac{η}{Ct^m}##.

Substitute that, and see what this leaves you with.

Chet
 
  • #5
Sorry for the late reply, I was actually in the midst of studying for the PE Exam, which I took on Friday, so I didnt have time to get to this until now. I did however get through Parts1 through 4, and am fairly confident in my answer. I am stuck on solving Part 5, where you have to solve d2θ/dη2+2η(dθ/dη)=0, to obtain θ as a functino of η.
I believe this is somewhat of a common diff. eq. problem, but my diff. eq. is rusty to say the least!
 
  • #6
Hint: Substitute u=dθ/dη

Chet
 

1. What is transient heat conduction of a semi-infinite solid?

Transient heat conduction refers to the transfer of thermal energy through a solid material that has a finite size, but is considered to be infinitely large in one direction. This means that heat is only being transferred in one direction, while the other sides of the solid are well insulated.

2. How is transient heat conduction different from steady-state heat conduction?

In steady-state heat conduction, the temperature throughout the solid remains constant over time. In contrast, transient heat conduction involves changes in temperature over time as heat is transferred through the material.

3. What factors affect the rate of transient heat conduction in a semi-infinite solid?

The rate of heat conduction in a semi-infinite solid is affected by several factors, including the thermal conductivity of the material, the temperature difference between the two ends of the solid, and the length of time that heat is being transferred.

4. How is the temperature distribution in a semi-infinite solid affected by transient heat conduction?

During transient heat conduction, the temperature distribution in a semi-infinite solid changes over time. Initially, the temperature at the surface of the solid will change rapidly, while the interior of the solid will take longer to reach the new temperature. As time passes, the temperature throughout the solid will eventually become uniform.

5. How is transient heat conduction of a semi-infinite solid used in real-world applications?

Transient heat conduction is an important concept in various fields, including thermodynamics, materials science, and engineering. It is used to study and predict the behavior of materials in situations where heat transfer is important, such as in the design of heating and cooling systems, or in the analysis of thermal insulation materials.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
20
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
13
Views
1K
  • Advanced Physics Homework Help
Replies
11
Views
4K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
13
Views
4K
Back
Top