2 D Transient Conduction Calculator Using Matlab Greg

2 D Transient Conduction Calculator Using Matlab Greg Teichert Kyle Halgren

Assumptions Use Finite Difference Equations shown in table 5. 2 2 D transient conduction with heat transfer in all directions (i. e. no internal corners as shown in the secondition in table 5. 2) Uniform temperature gradient in object Only rectangular geometry will be analyzed

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 (h 1 -h 4) (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)

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

Conditions %Userdefined h values h(1) = 10; h(2) =. 1; h(3) = 10; h(4) =. 1; %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

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, 3. 7078, 0. 0345

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 2 D direction

Results

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

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

Appendix-References Incropera, Frank P. De. Witt, Davi. D P. Fundamentals of Heat and Mass Transfer Fifth Edition, R. R. Donnelley & Sons Company. 2002 John Wiley & Sons, Inc

Appendix-hand work

Appendix-hand work