Lesson 9: Diffusion of Heat (discrete rod) and ode45

Transcription

1 Lesson 9: Diffusion of Heat (discrete rod) and ode Applied Problem. Consider the temperature in a thin rod such as a wire with electrical current. Assume the ends have a fixed temperature and the current generates heat in the rod. Then the temperature will not be uniform and must depend both on space and time, u(x,t). The problem is to approximate the temperature given the initial temperature and the temperature at the ends of the rod. Partitioning the rod into a finite number of segments can form a system of ordinary differential equations, whose solution can be approximated using the Matlab command ode Differential Equation Model. The model for the diffusion of heat in space is based on empirical observations. Discrete Fourier Heat Law: (a). heat flows from hot to cold, (b). the change in heat is proportional to the cross sectional area, change in time and (change in temperature)/(change space normal to the cross section). The last term is a good approximation provided the change in space is small. The proportionality constant, K, is called the thermal conductivity. The K varies with the particular material and with the temperature. Here we will assume the temperature varies over a smaller range so that K is approximately a constant.

2 2 Consider a thin wire so that there is diffusion in only one direction, x. The wire will have a current going through it with a source of heat, f, which is from the electrical resistance of the wire. The f has units of (heat)/(volume*time). We will assume the ends of the wire are kept at zero temperature, and the initial temperature is given. The goal is to be able to predict the temperature inside the wire for any future time and space location. In order to develop a model to predict the temperature = u(x,t), we will discretize space and let u(ih,t) be approximated by u i where h = L/n and L is the length of the wire. The model will have the general form change in heat content (heat from the source) + (heat from diffusion through the right end) + (heat from diffusion through the left end). This is depicted in the figure below where the time step has not been indicated. For time we can choose either t or t+ t. u i-1 * u u i i+1 * * h A h is the change in length A is the cross section area u i is the approximate temperature. Figure: Heat Diffusion in a Thin Wire The heat diffusing in the right face (when (u i+1 - u i ) /h > 0) is A t K (u i+1 - u i ) /h.

3 3 The heat diffusing out the left face (when (u i - u i-1 ) /h > 0) is A t K (u i - u i-1 ) /h. Therefore, the heat from diffusion is A t K (u i+1 - u i ) /h - A t K (ui - u i-1 ) /h. The heat from the source is Ah t f. The heat content of the volume Ah at time t is ρc u i Ah where ρ is the density and c is the specific heat. By combining these we have the following approximation of the change in the heat for the small volume Ah: ρc u i (t+ t) Ah - ρc u i (t) Ah = Ah t f + A t K (u i+1 - u i )/h - A t K (u i - u i-1 )/h. Now, divide by ρcah, define α = (K/ρc) /h 2 and let t go the zero to the get the system of ordinary differential equations where the left side gives the time derivative, u i '. ODE Model for Heat Diffusion. u i ' = 1/ρc f + α(u i+1 + ui-1 ) - 2α u i where (1) i = 1,...,n-1, u i (0) = 0 for i = 1,...,n-1 and (2) u 0 (t) = u n (t) = 0 for t > 0. (3) Equation (2) is the initial temperature set equal to zero, and (3) is the temperature at the left and right ends set equal to zero.

4 4 Equation (1) may be put into the matrix version of a system of ODEs. For example, if the wire is divided into four equal parts, then n = 4 and (1) may be written as three scalar equations u 1 ' = 1/ρc f + α(u 2 + 0) - 2α u1 u 2 ' = 1ρc f + α(u 1 + u3 ) - 2α u 2 u 3 ' = 1/ρc f + α(0 + u 2 ) - 2α u 3 Or, using a vector equation and a matrix u' + A u = b where ' u 1 u ' u' = u2, u u2, b (1/ ρc) f 1 and A α = = =. ' u 3 u A steady state solution is given by setting the derivative vector equal to the zero vector and solving the algebraic system Au = bwhere ' u 1 0 u 1 ' u' = u 2 0 = and u = u2 is constant vector. ' u 3 0 u 3 Thus, the steady state solution is inv(a)*b or A\b. u -1 = A b. This can be done in Matlab by either Another model uses partial derivatives with respect to time and space. PDE Model for Heat Diffusion. ρcu t - (Ku x ) x = f u(x,0) = given and u(0,t), u(1,t) = given, a partial differential equation, an initial condition, boundary conditions.

5 5 The steady state solution is given by setting u t = 0 so that the temperature depends only on space and - (Ku x ) x = f and u(0), u(1) are given. For example, if K = 0.001, f = 1, u(0) = 0 and u(1) = 0, then u(x) must be a quadratic function of x u(x) = x 2 /2 + a x + b. The boundary conditions imply the choice of a and b so that u(x) = -500x(1 x). 9.3 Method of Solution. In this lesson we will use the Matlab command ode45 to solve our systems of differential equation. This command is a robust implementation for systems of differential equations, which uses a variable step size method and the fourth and fifth order Runge-Kutta method. 9.4 Matlab Implementation. Since there are three unknowns and three differential equations and we wish to use Matlab's ode45 scheme, the file, ypheatode.m, must contain the three right sides of the differential equations where y(1) = temperature in the left segment, y(2) = temperature in the center segment and y(3) = temperature in the right segment. The heatode.m file contains the call to ode45, and the Matlab command plot generates the graphs of the three population groups. The initial temperatures are 1.0 and stored in the uo array in the heatode.m file.

6 6 m-files ypheatode.m function ypheatode = ypheatode(t,u) K =.001; rho = 1.; c = 1.; L = 1.; dx = L/4; const = (K/(rho*c))/(dx*dx); f = [1 1 1]/(rho*c); ypheatode(1) = -const*(2*u(1) - u(2)) + f(1); ypheatode(2) = -const*(-u(1) + 2*u(2) - u(3)) + f(2); ypheatode(3) = -const*(2*u(3) - u(2)) + f(3); ypheatode = [ypheatode(1) ypheatode(2) ypheatode(3)]'; m-files heatode.m. clear; clf K =.001; rho = 1.; c = 1.; L = 1.; dx = L/4; const = (K/(rho*c))/(dx*dx); f = [1 1 1]/(rho*c); uo = [1 1 1]; to = 0; tf = 400; [t u] = ode45('ypheatode',[to tf],uo); maxk = size(t,1); x = 0:.25:1; figure(1) for k = 1:5:maxk plot(x,[0 u(k,:) 0]) title('temperatures at various times') xlabel('position in space') axis([ ]) hold on pause end t(1:5:maxk) A = const*[2-1 0;-1 2-1;0-1 2]; d = [1 1 1]'/(rho*c); sssol = A\d plot(x, [0 sssol' 0]) figure(2) plot(t,u(:,2)) title('temperature in the center') xlabel('time')

7 7 9.5 Numerical Experiments. The first graphical output has multiple curves each associated with a particular time as given in the numerical output times = sssol = Note the last curve approximates the steady state solution found by the command A\d. The second graph is for the temperature at the center segment versus time. This graph indicates the approximation of 125.0, which is the steady state solution at the center node.

8 8 temperatures at various times position in space 140 temperature in the center time

9 9 9.6 Additional Calculations. Consider the ODE heat diffusion model with K = 0.001, rho = 1.0, c = 1.0, L = 1.0 and variable heat source f. (a). (b). (c). (d). State the system of differential equations. Modify the ypheatode.m and heatode.m files. For f = [ ]' execute the heatode.m file. Use the hold on command and repeat (c) using f = [ ]' and f = [ ]' (e). Examine the three sets of three curves and numerical outputs. What happens as the heat source f increases?