2D Transient Conduction Calculator Using Matlab Greg Teichert Kyle Halgren 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 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 () (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 crosssection with these boundary conditions: Origin T_inf 2, h2 T_inf 1, h1 a b T_inf 4, h4 T_inf 3, h3
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 (t) We assumed a value of x = y = gcd(a, b) Using each of the conditions (except the second) in the table 5.2, we calculate the t and choose the smallest value Using that t 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 2D direction Results Transient conduction (the origin of the plot is the top left corner of the cross section) Transient conduction (the origin of the plot is the top left corner of the cross section) 550 550 500 Temperature (K)
1 a (m) b (m) 0 1 0.6 0.8 a (m) 0.4 0.2 0 Solution to different Problem
%Userdefined h values h(1) = 0; h(2) = 1000; h(3) = 1000; h(4) = 100; %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 600
500 Temperature (K) %Boundary conditions %Userdefined T infinity values in kelvin T_inf(1) = 273; T_inf(2) = 150; T_inf(3) = 590; T_inf(4) = 273; Transient conduction (the origin of the plot is the top left corner of the cross section) 400 300 200 100 1.5 1 0.5 b (m)
0 0 0.2 0.4 a (m) 0.6 0.8 1 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 x and t Appendix-References Incropera, Frank P. DeWitt, DaviD 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
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