Numerical method on heat diffusion problem

In summary, the conversation is about solving a heat diffusion problem using a differential equation and two different methods. The person is comparing the results from both methods and noticing some differences in the first few seconds. The suggested solution is to use more points along the length of the rod and reduce the time step to accurately represent the rapid change in temperature near the ends. The series solution should be exact with enough terms.
  • #1
loukoumas
15
0
Hello there!

I am solving a heat diffusion problem which is described by the differential equation: θΤ/θt=a*(θ^2Τ/ΘΤ^2). I will apply two diffrent temperatures on the ends of a metal rod. Solving the equation i conclude, according to the boundary conditions of course, at the function:
Τ(x,t)=sum i=1-infinity [e^-i^2 * Pi^2 *a*t*(160*(-1)^i /(i*Pi) + 40/(i*Pi))*sin(i*Pi*x)] +100 x.( I gues its not very readable!sorry)

i also use a numerical method, crank-nicolson. It uses the previous(T') accordinng to time temperatures and previous(Ti-1) and next(Ti+1) temperatures according to length
Ti=A*T'(i) + B[T(i-1) +T(i+1)+T'(i-1)+T'(i+1)] A,B are known and also temperatures on the ends are known( 0,100 degrees respectively) so if i cut the rod on n+1 i get nXn system to solve! So far so good. Where is my question?
Comparing the temperatures that i get from both methods are satisfactorily close,but only after 60-70 seconds. On the first tens of seconds, on the points next to the ends of the rod(10cm and 90 cm if the hole rod is 1m), i have 3-5 degrees difference between these two methods.Wich results are the most relyable? why do i find these differences?
thank you very much!
 
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  • #2
loukoumas said:
On the first tens of seconds, on the points next to the ends of the rod(10cm and 90 cm if the hole rod is 1m), i have 3-5 degrees difference between these two methods.

So you are only using 10 (or 9 or 11) points along length of the rod?

Try using more points (for example twice as many) and see what happens to the results. If temperature against time at each mesh point is oscillating up and down at each time step, you need to reduce the time step as well.

In the first part of the solution, you probably have a rapid change of temperature along the length of the rod near to the ends, and you don't have enough points in the finite difference mesh to represent that accurately.

Assuming you did the math right, your series solution should be exact, so long as you include enough terms to ensure it has converged to the accuracy you want.
 

1. What is a numerical method for solving heat diffusion problems?

A numerical method for solving heat diffusion problems is a mathematical technique used to approximate the solution of a heat diffusion equation. It involves dividing the problem into smaller parts and using iterative methods to find the solution at each step.

2. What is the most commonly used numerical method for solving heat diffusion problems?

The most commonly used numerical method for solving heat diffusion problems is the finite difference method. This method divides the problem into a grid and uses the values at each grid point to approximate the solution at that point.

3. How does the finite difference method work?

The finite difference method works by approximating the derivatives in the heat diffusion equation using the values at neighboring grid points. This results in a system of equations that can be solved iteratively to find the solution at each grid point.

4. What are the advantages of using numerical methods for heat diffusion problems?

Numerical methods offer a more efficient and accurate way of solving heat diffusion problems compared to analytical methods. They also allow for the solution of more complex problems that cannot be solved analytically.

5. What are the limitations of numerical methods for heat diffusion problems?

Numerical methods are limited by the accuracy of the grid used and the computational power available. They also require careful consideration of boundary conditions and initial conditions to ensure an accurate solution. Additionally, they may not provide a physical understanding of the problem compared to analytical methods.

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