1. Heat Conduction in a Hollow Cylinder (Pipe) Robby Pantellas Millersville University Dr. Buchanan Math 467.01 Partial Differential Equations

2. Problem Statement Consider heat conduction in a cylindrical (hollow) pipe of length L and radius a. Let the initial temperature distribution on the pipe be given by the function Consider heat conduction in a cylindrical (hollow) pipe of length L and radius a. Let the initial temperature distribution on the pipe be given by the function f(θ, z) = 1 + (z/L) cos θ. Suppose the ends of the pipe are insulated. Find the solution to the heat equation in this situation. f(θ, z) = 1 + (z/L) cos θ. Suppose the ends of the pipe are insulated. Find the solution to the heat equation in this situation. Since the wall of the pipe is so thin, heat is not going to flow radially outward, so we can neglect the thickness of the pipe’s wall. We will see this come into play in the separation of variables. Since the wall of the pipe is so thin, heat is not going to flow radially outward, so we can neglect the thickness of the pipe’s wall. We will see this come into play in the separation of variables.

3. Starting the Problem First, we have to change coordinate systems from Cartesian coordinates to cylindrical coordinates, then once we do that we will plug this conversion into the Laplacian operator. First, we have to change coordinate systems from Cartesian coordinates to cylindrical coordinates, then once we do that we will plug this conversion into the Laplacian operator. Below is the heat equation and the conversion of the Laplacian operator from Cartesian to cylindrical coordinates: Below is the heat equation and the conversion of the Laplacian operator from Cartesian to cylindrical coordinates: u t = k u xx (for this problem we will assume the thermal diffusivity is 1 since the material of the pipe is not specified) u t = k u xx (for this problem we will assume the thermal diffusivity is 1 since the material of the pipe is not specified) ∆² = ∂ 2 /∂x 2 + ∂ 2 /∂y 2 + ∂ 2 /∂z 2 (in Cartesian coordinates) ∆² = ∂ 2 /∂x 2 + ∂ 2 /∂y 2 + ∂ 2 /∂z 2 (in Cartesian coordinates) Δ² = ∂ 2 /∂r 2 + (1/r)∂/∂r + (1/r 2 )∂ 2 /∂θ 2 + ∂ 2 /∂z 2 (in cylindrical coordinates) Δ² = ∂ 2 /∂r 2 + (1/r)∂/∂r + (1/r 2 )∂ 2 /∂θ 2 + ∂ 2 /∂z 2 (in cylindrical coordinates)

4. Separation of Variables As usual, separation of variables still works for this problem. We want to look for a solution in the form: As usual, separation of variables still works for this problem. We want to look for a solution in the form: T = R(r) θ(θ) Z(z) Ф(t) T = R(r) θ(θ) Z(z) Ф(t) Taking the partial derivatives according the Laplacian operator we get the following: Taking the partial derivatives according the Laplacian operator we get the following: R’’θZФ + (1/r) R’ θZФ + (1/r 2 )Rθ’’ ZФ + RθZ’’ Ф = RθZФ’ R’’θZФ + (1/r) R’ θZФ + (1/r 2 )Rθ’’ ZФ + RθZ’’ Ф = RθZФ’ and dividing both sides by RθZФ and canceling we get: and dividing both sides by RθZФ and canceling we get: R’’/R + (1/r)R’/R + (1/r 2 )θ 2 /θ + Z’’/Z = Ф’/Ф R’’/R + (1/r)R’/R + (1/r 2 )θ 2 /θ + Z’’/Z = Ф’/Ф Recall that we are going to neglect the thickness of the pipe’s wall, so we can get rid of the R’s in our equation. Also, we can move the Z’s to the right side and substitute a for r for the radius and we get the following: Recall that we are going to neglect the thickness of the pipe’s wall, so we can get rid of the R’s in our equation. Also, we can move the Z’s to the right side and substitute a for r for the radius and we get the following: (1/a 2 ) θ’’/θ = Ф’/Ф – Z’’/Z and (1/a 2 ) θ’’/θ = Ф’/Ф – Z’’/Z and θ’’/θ = a 2 (Ф’/Ф – Z’’/Z) = -λ 2 θ’’/θ = a 2 (Ф’/Ф – Z’’/Z) = -λ 2

5. Finding the Solution Now we want to get the different parts of our solution, so we will start with our second order homogeneous ODE θ’’ + λ²θ = 0. This ODE has a solution in the form θ( θ ) = A n cos nθ + B n sin nθ where n = 0,1,2…. Now we want to get the different parts of our solution, so we will start with our second order homogeneous ODE θ’’ + λ²θ = 0. This ODE has a solution in the form θ( θ ) = A n cos nθ + B n sin nθ where n = 0,1,2…. Similarly, we have another second order homogeneous ODE Z’’ + μ²Z = 0, and we need to satisfy the boundary conditions Z’(0) = 0 = Z’(L), since the ends of the pipe are insulated. This ODE has a solution in the form Similarly, we have another second order homogeneous ODE Z’’ + μ²Z = 0, and we need to satisfy the boundary conditions Z’(0) = 0 = Z’(L), since the ends of the pipe are insulated. This ODE has a solution in the form Zm(Z) = cos mπ/L where m = 0,1,2….. Zm(Z) = cos mπ/L where m = 0,1,2…..

6. Finding the Solution (cont’d) Next, we have a first order homogeneous ODE Next, we have a first order homogeneous ODE Ф’ + [(n 2 /a 2 ) + (m 2 π 2 /L 2 )]Ф = 0. This ODE has a solution of the form Ф n,m (t) = e -αn,mt Ф’ + [(n 2 /a 2 ) + (m 2 π 2 /L 2 )]Ф = 0. This ODE has a solution of the form Ф n,m (t) = e -αn,mt Putting all our solutions together, we get the following solution to the heat equation: Putting all our solutions together, we get the following solution to the heat equation: u (θ,z,t) = Σ Σ e -αn,mt cos mπ/L (4/nπ) (-1) n cos nθ where the summations are n, m from 0 -> ∞. u (θ,z,t) = Σ Σ e -αn,mt cos mπ/L (4/nπ) (-1) n cos nθ where the summations are n, m from 0 -> ∞.

7. Finishing the Problem We are not done yet, we still need to satisfy the initial conditions. We need our solution to equal the initial temperature distribution f(θ, z) = 1 + (z/L) cos θ. So, we get the following: u (θ,z) = Σ Σ A n C m cos mπ/L cos nθ = 1 + (z/L) cos θ. We are not done yet, we still need to satisfy the initial conditions. We need our solution to equal the initial temperature distribution f(θ, z) = 1 + (z/L) cos θ. So, we get the following: u (θ,z) = Σ Σ A n C m cos mπ/L cos nθ = 1 + (z/L) cos θ. We will let A 0 = 1, and we get: u (θ,z) = 1 + Σ Σ A n C m cos mπ/L cos nθ We will let A 0 = 1, and we get: u (θ,z) = 1 + Σ Σ A n C m cos mπ/L cos nθ We will factor out the An, get rid of the one summation, and let A 1 = (1/L) and we get the following: u (θ,z) = 1 + (1/L)cos θ Σ C m cos mπ/L We will factor out the An, get rid of the one summation, and let A 1 = (1/L) and we get the following: u (θ,z) = 1 + (1/L)cos θ Σ C m cos mπ/L Evaluating Cm using the Fourier Integral Formula we get C n = - 4zL/n 2 π 2. When n is even, C is 0, and when n is odd we get (-4/n 2 π 2 ). Evaluating Cm using the Fourier Integral Formula we get C n = - 4zL/n 2 π 2. When n is even, C is 0, and when n is odd we get (-4/n 2 π 2 ).

8. Mathematica Analysis  This plot shows the temperature distribution throughout the pipe, where the red, oranges, and yellows are the cold areas and the blues, indigos, and violets are the hot areas.

9. Conclusion We are now done solving the problem. We satisfied both the boundary conditions (slide 5) and the initial conditions (slide 7). We are now done solving the problem. We satisfied both the boundary conditions (slide 5) and the initial conditions (slide 7). Going further, if the wall of the pipe was of significant thickness, then our solution would have another part to it to compensate for the heat flowing radially through the thickness of the pipe. Going further, if the wall of the pipe was of significant thickness, then our solution would have another part to it to compensate for the heat flowing radially through the thickness of the pipe.