1. 2D Transient Conduction Calculator Using Matlab
Greg Teichert
Kyle Halgren

2. Assumptions Use Finite Difference Equations shown in table 5.2
2D transient conduction with heat transfer in all directions (i.e. no internal corners as shown in the second condition in table 5.2)
Uniform temperature gradient in object
Only rectangular geometry will be analyzed

3. Program Inputs The calculator asks for Length of sides (a, b) (m)
Outside Temperatures (T_inf 1-T_inf 4) (K)
Temperature of object (T_0) (K)
Thermal Convection Coefficient (h1-h4) (W/m^2*K)
Thermal Conduction Coefficient (k) (W/m*K)
Density (ρ) (kg/m^3)
Specific Heat (Cp) (J/kg*K)
Desired Time Interval (t) (s)

4. Transient Conduction Example problem
suppose we have an object with rectangular cross-section with these boundary conditions:
Origin
T_inf 1, h1
T_inf 2, h2
T_inf 3, h3
T_inf 4, h4 a b

5. Conditions %Userdefined h values h(1) = 10; h(2) = .1; h(3) = 10;
%Boundary conditions
%Userdefined T infinity values in kelvin
T_inf(1) = 293;
T_inf(2) = 293;
T_inf(3) = 353;
T_inf(4) = 353;
%Initial condition (assume uniform initial temperature)
%Userdefined initial temperature value
T_0 = 573;
%Material properties
%Userdefined material values
k = .08;
rho = 7480;
c_p = .460;
%Userdefined physical variables
a = 1; %height of cross section
b = 1.3; %width of cross section
t = 3600; %time at which results are given

6. Time Step (Δt) We assumed a value of Δx = Δy = gcd(a, b)
Using each of the conditions (except the second) in the table 5.2, we calculate the Δt and choose the smallest value
Using that Δt we calculate Fo
Our outputs for delta_x, delta_t, Fo respectively
0.0500, ,

7. Method
Using the Finite Difference Method, matlab generates a matrix of temperature values that are represented in the graph shown on the next slide
This method allows for the calculation of every node in any 2D direction

8. Results

9. Solution to different Problem
%Userdefined h values
h(1) = 0;
h(2) = 1000;
h(3) = 1000;
h(4) = 100;
%Boundary conditions
%Userdefined T infinity values in kelvin
T_inf(1) = 273;
T_inf(2) = 150;
T_inf(3) = 590;
T_inf(4) = 273;
%Initial condition (assume uniform initial temperature)
%Userdefined initial temperature value
T_0 = 250;
%Material properties
%Userdefined material values
k = .8;
rho = 1000;
c_p = .460;
%Userdefined physical variables
a = 1; %height of cross section
b = 1.3; %width of cross section
t = 20; %time at which results are given

10. Conclusion and Recommendations
Works only in rectangular geometry
High values of h and t>1 causes errors to occur due to lack of memory
Use a better method to find Δx and Δt

11. Appendix-References
Incropera, Frank P. DeWitt, DaviD P. Fundamentals of Heat and Mass Transfer Fifth Edition, R. R. Donnelley & Sons Company John Wiley & Sons, Inc

12. Appendix-hand work

13. Appendix-hand work