# A Is there an analytical solution to the following problem?

1. May 20, 2016

### deepakphysics

Differential equation
\begin{align}
\frac{{\partial}^2 T (x,t)}{\partial x^2} = \frac{1}{\alpha}\frac{\partial T(x,t)}{\partial t},
\end{align}

boundary conditions

\begin{aligned}
T(x, 0) = T_\infty \\
-k\frac{\partial T(0,t)}{\partial x}= h(T_{max} - T(0,t)) \\
-k\frac{\partial T(l,t)}{\partial x}= h(T(l,t) - T_\infty).
\end{aligned}

It is 1D transient heat diffusion problem. x[0,L] represents a 1D spatial domain and t[0,tao] represents a time domain. T(x,t) is temperature distribution over the spatial and temporal domain. \alpha = k/(cp*rho), where k is thermal conductivity, cp heat capacity and \rho thermal conductivity.

Last edited by a moderator: Jun 6, 2016
2. May 20, 2016

### wrobel

what is $T_{\max}$

3. May 20, 2016

### deepakphysics

T_max is a constant value. T_infinity is also a constant value.

Last edited: May 20, 2016
4. May 21, 2016

### wrobel

then try to use the Laplace transform in the variable t

5. May 21, 2016

### deepakphysics

Yes doing that we will arrive at solution T(x,t) = (A*cos(w*x)+Bsin(w*x))*T_inf*e^(-alpha*w^2*t).
I have problem now finding values of A, B and W with given boundary conditions.

6. May 21, 2016

### wrobel

I have got the following Laplace transform
$$\hat T(x,p)=\frac{T_\infty}{p}+C_1e^{x\sqrt{p/\alpha}}+C_2e^{-x\sqrt{p/\alpha}}$$
then we have two equations:
$$k\hat T_x(0,p)=\frac{hT_\max}{p}-h\hat T(0,p),\quad k\hat T_x(l,p)=h\hat T(l,p)-\frac{h T_\infty}{p}$$ and two unknown functions $C_j=C_j(p),\quad j=1,2$ I do not see what is the problem

Last edited: May 21, 2016
7. May 21, 2016

### deepakphysics

Thank you for the solution. It is possible to find C1 and C2. I got as follows (picture attached) a = alpha Tm = T_max, Ti = T_inf. I am stuck again..how can we find inverse laplace ?

File size:
40.2 KB
Views:
150
8. Jun 3, 2016

### Staff: Mentor

There is a much easier way of solving this problem analytically. The first step is to solve for the steady state solution. What do you get for the steady state solution?

Chet

9. Jun 4, 2016

### deepakphysics

Hi
I have following analytical solution (Please find it in the attachment). T1 and T2 are temperature at the surface of inner and outer wall.

#### Attached Files:

File size:
104.3 KB
Views:
209
• ###### Screen Shot 2016-06-05 at 00.16.59.png
File size:
16.9 KB
Views:
201
10. Jun 4, 2016

### Staff: Mentor

OK. Let $T(x,\infty)$ represent this final steady state solution. Now express the temperature at an arbitrary time t as follows:
$$T(x,t)=T(x,\infty)-\theta (x,t)$$
Now, if you substitute this into the differential equation, boundary conditions and initial conditions, what do you get (in terms of $\theta$)?

11. Jun 5, 2016

### deepakphysics

This is what I got. One more thing, T_infty is not the temperature corresponding to steady state solution. It is a ambient temperature. For some reason it should be taken as a variable. The results need to be expressed in terms of T_infty and T_max and other material parameters (h, l, k ).

#### Attached Files:

• ###### Screen Shot 2016-06-05 at 11.39.38.png
File size:
160.9 KB
Views:
198
12. Jun 5, 2016

### Staff: Mentor

I understand all this. Now please answer my question in the previous post.

13. Jun 5, 2016

### deepakphysics

Thank you. I have tried to answer your question. I have attached a screen shot in the previous reply (forgot to mention about the attached screen shot).

14. Jun 5, 2016

### Staff: Mentor

Oh, sorry. I missed that.

There is a mistake in your equation for $T(x,\infty)$. The x should be an x/l. Please make this correction in your initial condition and boundary conditions, and also reduce these equations to a common denominator.

15. Jun 5, 2016

### deepakphysics

Thank you for the correction. Please find the updated results in the attachments.

#### Attached Files:

• ###### Screen Shot 2016-06-05 at 17.09.46.png
File size:
180.5 KB
Views:
202
16. Jun 5, 2016

### Staff: Mentor

My solution for $T(x,/infty)$ does not match yours. Based on the boundary condition equations you gave in your first post, I get:
$$T(x,\infty)=\frac{-T_{max}+\beta (T_{max}+T_{\infty})+(T_{max}-T_{\infty})x/l}{(2\beta -1)}$$
Note the minus sign in the denominator. Try this and see if it satisfies you boundary conditions.

I suggest that, when you do the formulation in terms of $\theta$, you continue to use $\beta$ in the equations.

17. Jun 6, 2016

### deepakphysics

I think I did little mistake in the first post. There should have been a minus sign in the left hand side of both equations representing spatial boundary conditions. Also I did mistake in my previous reply. I think I have corrected those mistake this time. Please find the outcome in the attachments I) in terms of \beta and ii) interms of (k,h and l).

#### Attached Files:

File size:
57.5 KB
Views:
130
File size:
59.5 KB
Views:
125
18. Jun 6, 2016

### Delta²

We can prove mathematically that the steady state solution exists? That is if $T(x,t)$ is a function that satisfies (1) then the limit $\lim_{t \to \infty}T(x,t)=f(x)$ exists and it is finite?

OR we assume that it exists because we have to do with a physical process that will reach in some sort of equilibrium and that nothing can become infinite in a physical process?

19. Jun 6, 2016

### Staff: Mentor

Yes. I missed this. I'm editing your original post to correct this.
I'll check it out in a little while.

20. Jun 6, 2016

### Staff: Mentor

I'm not a mathematician so I couldn't answer whether it can be proven mathematically that there is a stable steady state solution. I let the physics of the situation tell me this, based on an abundance of heat transfer experience. This problem obviously involves convective heat transfer at the two ends of the rod, with the external temperatures being Tmax and T.

21. Jun 6, 2016

### Staff: Mentor

I couldn't read what you had, but here's what I got (for the corrected boundary conditions):
$$T(x,\infty)=\frac{T_{max}+\beta (T_{max}+T_{\infty})-(T_{max}-T_{\infty})x/l}{(1+2\beta)}$$

Initial condition: $$\theta (x,0)=\frac{(T_{max}-T_{\infty})(\beta + x/l)}{(1+2\beta)}$$

Boundary condition at x = 0:$$\frac{\partial \theta}{\partial (x/l)}=\frac{\theta}{\beta}$$

Boundary condition at (x/l) = 1:$$\frac{\partial \theta}{\partial (x/l)}=-\frac{\theta}{\beta}$$