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The Heat Equation. Temperature Distribution in a Bar with Radiating Ends. MA428 Class Project. Boundary Conditions. Boundary Conditions.
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The Heat Equation Temperature Distribution in a Bar with Radiating Ends MA428 Class Project
Boundary Conditions 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.
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.
At the other end of the bar, The problem for X(x) is therefore, Apply Boundary Conditions
To have a nontrivial solution we must have , and this requires that To have a nontrivial solution we must have and this requires that Case 2 Case 1 Applying the boundary conditions we get, After converting to and solving for characteristic equation we have, After converting to and solving for characteristic equation we have, Since this equation has a sum greater than zero Applying the boundary conditions we get Applying the boundary conditions we get We get only the trivial solution from this case. 0 is not an eigenvalue of this problem. We get only the trivial solution from this case. 0 is not an eigenvalue of this problem. Solving the ODEsSturm-Liouville
Since k=z/L, then And We can now solve our second ode for the variable T Eigenvalue and Eigenfunction
The equation for T is so Solving the Second ODE
To Satisfy the initial conditions, let We must choose the ‘s so that Applying Initial Conditions
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.
Before we can solve for all values of we must first find all values of .
> 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 Utilizing Maple 7 to find all z
All C C1=8.074886200 C2=2.573922021 C3=-.6295352004 C4=.6105802995 C5=-.2798584694 C6=.2531477337 C7=-.1484813629 C8=.1360552034 C9=-.09083927767 C10=.08444279860