Circuler grid need to be solved by Finite difference method pls help me

In summary, the problem at hand is to solve a heat conduction problem for a circular grid using the finite difference method. The two-dimensional equation for this problem is given by 1/r * δ/δr (r * δT/δr) + 1/r^2 * ((δ^2 T)/(δΦ^2 )) = 0. The discretized values for this equation are given by (T_((i,j,k)-T_((i+1,j,k) ) )/Δr)+((T_((i+1.j.k) )+T_((i-1,j,k) )-2T_((i,j,k) ))/(Δr^2 )) +
  • #1
nafiz27me
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Circuler grid need to be solved by Finite difference method! pls help me...

hi this is the picture of the problem.. i have studied the rectangular grid but not the circular grid... now pls someone help me to find out the way to solve a heat conduction problem for circle using finite difference method.

1/r δ/δr (r δT/δr)+ 1/r^2 ((δ^2 T)/(δΦ^2 ))=0
this is the two dimensional equation and the discritise values are below...

1) 1/r δ/δr (r δT/δr)= 1/r (δT/δr)+(δ^2 T)/(δr^2 )=1/r (T_((i,j,k)-T_((i+1,j,k) ) )/Δr)+((T_((i+1.j.k) )+T_((i-1,j,k) )-2T_((i,j,k) ))/(Δr^2 ))

2) 1/r^2 ((δ^2 T)/(δΦ^2 ))= 1/r^2 ((T_((i.j+1.k) )+T_((i,j-1,k) )-2T_((i,j,k) ))/(ΔΦ^2 ))
 

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  • #2
so the final discretized form is: (T_((i,j,k)-T_((i+1,j,k) ) )/Δr)+((T_((i+1.j.k) )+T_((i-1,j,k) )-2T_((i,j,k) ))/(Δr^2 )) + (T_((i.j+1.k) )+T_((i,j-1,k) )-2T_((i,j,k) ))/(ΔΦ^2 )) =0or rearranging the terms, you get:T_((i,j,k)) = (T_((i+1,j,k) ) + (T_((i-1,j,k) )/(Δr^2 )) + (T_((i.j+1.k) )+T_((i,j-1,k) )/(ΔΦ^2 )) / (1/Δr + 1/Δr^2 + 1/ΔΦ^2 )Hope this helps!
 

Related to Circuler grid need to be solved by Finite difference method pls help me

1. What is a circular grid?

A circular grid is a type of grid used in computational methods, such as the Finite Difference Method, to solve problems involving circular or cylindrical geometries.

2. Why is the Finite Difference Method used to solve problems involving circular grids?

The Finite Difference Method is used because it is a numerical method that can handle complex geometries, such as circular grids, and can provide accurate solutions to differential equations.

3. How does the Finite Difference Method work?

The Finite Difference Method works by dividing the circular grid into a finite number of smaller cells and approximating the partial derivatives in the differential equations using the values at the neighboring grid points.

4. What are the advantages of using the Finite Difference Method to solve problems involving circular grids?

The Finite Difference Method is relatively easy to implement and can handle a wide range of complex geometries. It also provides accurate solutions and can be easily adapted to different boundary conditions.

5. Are there any limitations to using the Finite Difference Method for circular grids?

The Finite Difference Method may not be suitable for problems with irregular boundaries or highly non-linear equations. It also requires a fine grid resolution to accurately capture the behavior of the solution, which can be computationally expensive.

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