2. MA428 Class Project The Heat Equation Temperature Distribution in a Bar with Radiating Ends

3. Boundary Conditions

4. The boundary conditions at L assumes that the energy radiates from this end at a rate proportional to the temperature at that end of the bar. A is a positive constant called the transfer coefficient.

5. Finding the General Solution

6. Finding the General Solution Cont. Put u(x,t)=X(x)T(t) into the differential equation to get XT’= X”T. Which we will obtain two ODEs.

7. Apply Boundary Conditions At the other end of the bar, The problem for X(x) is therefore,

8. Solving the ODEs Sturm-Liouville Case 1 We get only the trivial solution from this case. 0 is not an eigenvalue of this problem. Applying the boundary conditions we get, Case 2 After converting to and solving for characteristic equation we have, Applying the boundary conditions we get To have a nontrivial solution we must have and this requires that Since this equation has a sum greater than zero We get only the trivial solution from this case. 0 is not an eigenvalue of this problem. After converting to and solving for characteristic equation we have, Applying the boundary conditions we get To have a nontrivial solution we must have, and this requires that

9. Graphs of y=tan(z) & y=-z/AL

10. Eigenvalue and Eigenfunction Since k=z/L, then And We can now solve our second ode for the variable T

11. Solving the Second ODE The equation for T is so

12. Combining the Two ODEs

13. Applying Initial Conditions To Satisfy the initial conditions, let We must choose the ‘s so that

14. The General Solution

15. Example Problem A thin, homogeneous bar of thermal diffusivity 4 and length 6 cm with insulated sides, has it’s left end at temperature zero. Its right end is radiating (with transfer coefficient 1/2 ) into the surrounding medium, which has temperature zero. The bar has an initial temperature given by f(x)=x(6- x). Approximate the temperature distribution u(x,t) by finding the partial sum of the series representation.

17. Before we can solve for all values of we must first find all values of.

18. Recall that

19. Utilizing Maple 7 to find all z > fsolve(tan(z)=-z/3,z,z=0..Pi/2); -0. > z1:=fsolve(tan(z)=-z/3,z,z=Pi/2..3*Pi/2); z1 := 2.455643863 > z2:=fsolve(tan(z)=-z/3.0,z,z=3*Pi/2..5*Pi/2); z2 := 5.232938454 > z3:=fsolve(tan(z)=-z/3.0,z,z=5*Pi/2..7*Pi/2); z3 := 8.204531363 > z4:=fsolve(tan(z)=-z/3.0,z,z=7*Pi/2..9*Pi/2); z4 := 11.25604301 > z5:=fsolve(tan(z)=- z/3.0,z,z=9*Pi/2..11*Pi/2); z5 := 14.34335079 > z6:=fsolve(tan(z)=- z/3.0,z,z=11*Pi/2..13*Pi/2); z6 := 17.44902434 > z7:=fsolve(tan(z)=- z/3.0,z,z=13*Pi/2..15*Pi/2); z7 := 20.56520794 > z8:=fsolve(tan(z)=- z/3.0,z,z=15*Pi/2..17*Pi/2); z8 := 23.68792106

20. All C C1=8.074886200 C2=2.573922021 C3=-.6295352004 C4=.6105802995 C5=-.2798584694 C6=.2531477337 C7=-.1484813629 C8=.1360552034 C9=-.09083927767 C10=.08444279860

21. Final Answer