User:Dry.eats/Numerical Solution: Spoon in Hot Tea

The solution of the temperature profile of a spoon in hot tea can be approximated by a rod. It is assumed that half of the rod is in one medium at temperature, $$T_0$$, of 80 C and the other half of the rod is in another medium at temperature, $$T_\infty$$, of 20 C. The spoon is approximated by a rod that is 10 cm long, has a cross sectional area, $$A_c$$, of approximately 0.2 cm^2 and a perimeter, $$P$$, of approximately 2 cm.

The governing equations are

$$A_c k \frac{d^2T}{dx^2} - h_c P (T-T_0) = 0$$ from x = 0 to L/2

$$A_c k \frac{d^2T}{dx^2} - h_c P (T-T_\infty) = 0$$ from x = L/2 to L

with adiabatic boundary conditions defined by

$$\left.\frac{dT}{dx}\right|_{x=0}=0$$

and

$$\left.\frac{dT}{dx}\right|_{x=L}=0$$

where $$T$$ is the temperature, $$k$$ is the conductivity, $$h_c$$ is the convective heat transfer coefficient.

The difference in materials can be seen by calculating the temperature profile for metals with different thermal conductivity $$k$$ such as silver and stainless steel. The thermal conductivity of silver is 400 W/m^2/K while the thermal conductivity of stainless steel is 15 W/m^2/K. With silver having the greater thermal conductivity, one would suppose that it would conduct heat better, thus resulting in a hotter spoon handle. The results and MATLAB code for this analysis are found below.

Numerical Solution with MATLAB
clear all; close all; clc;

%Geometric Parameters l = 0.1; %m Ac = 0.2/10000; %m^2 P = 0.02; %m

%Computational Parameters n = 10; %number of points h = l/n; %cell size %Locations of points + boundaries i = 1:(n); x(i) = 0.5*h + h*(i-1); x = [0 x l];

%Thermal Parameters k(1) = 15; %W/(m^2*K) - Stainless Steel k(2) = 400; %W/(m^2*K) - Silver hc(1:n/2) = 1000; %W/(m^2*K) hc(n/2+1:n) = 10; %W/(m^2*K) Tinf(1:n/2) = 80; %353; %K Tinf(n/2+1:n) = 20; %293; %K for i = 1:2 %Compute coefficients of super-diagonal, sub-diagonal, and diagonal l(1:n) = k(i)*Ac/(h^2); d(1:n)= -(2*k(i)*Ac/(h^2)+hc*P); u(1:n) = k(i)*Ac/(h^2); %Implement zero flux boundary conditions d(1) = -(k(i)*Ac/(h^2)+hc(1)*P); d(n) = -(k(i)*Ac/(h^2)+hc(n)*P); %Compute Source term b = -hc'.*P.*Tinf'; %Create tridiagonal matrix A = spdiags([l;d;u]',-1:1,n,n); %Solve linear system for temperature t = zeros(n,1); t = A\b; %Boundary conditions at end of fin tTip0 = 1/8*(9*t(1)-t(2)); tTipN = 1/8*(9*t(n)-t(n-1)); t = [tTip0 t' tTipN]; T(i,:) = t; end

%Plot Statements plot(x,T(1,:),'-d',x,T(2,:),'--*') xlabel('x position (m)'); ylabel('Temperature (C)'); title('Temperature Profile Along Length of Spoon in Hot Tea'); legend('Stainless Steel','Silver'); Dry.eats 18:04, 24 November 2009 (UTC)