1. Finite-Difference Solutions Part 2

2. Nodal Network
Assumptions:
2 dimensional temperature distribution
Constant thermal conductivity
Steady State
No heat source or sink
Temperature distribution is given by
Similarly to the 1 dimensional approach this differential equation
can be solve by finite-difference

3. Given a 2-D geometry we divide it into a nodal network as illustrated
y,n
m,n+1
x,m
m,n
m+1,n
m-1,n
m,n-1

4. Using central difference we have

5. Depending on the location of the node finite equations simulating
can be derive and yield the following results (for ).
Standard configurations
Interior Node
Finite-Difference
Equation
m,n+1
m,n
m+1,n
m-1,n
m,n-1

6. Node at an internal corner with convection
Finite-Difference
Equation
m,n+1
m,n
m+1,n
m-1,n
m,n-1

7. Node at a plane surface with convection
Finite-Difference
Equation
m,n+1
m,n
m-1,n
m,n-1

8. Node at an adiabatic plane surface (Symmetry)
Finite-Difference
Equation
m,n+1
m,n
m-1,n
m,n-1

9. Node at a plane surface with uniform heat flux
Finite-Difference
Equation
m,n+1
m,n
m-1,n
m,n-1

10. Node at an external corner with convection
Finite-Difference
Equation
m,n
m-1,n
m,n-1

11. One Dimensional Numerical Solution of Transient Conduction
Assumptions:
1 dimensional temperature distribution
Constant thermal conductivity
No heat source or sink
Temperature distribution is given by

12. Now using central difference on an interior node for the position
we find
And forward difference for the time component
Which yields
Solve for x than for t.
Note: system maybe unstable depending on and