Heat Conduction in a Heat Conduction in a Hollow Cylinder Hollow Cylinder

(Pipe)(Pipe)Robby PantellasRobby Pantellas

Millersville UniversityMillersville UniversityDr. BuchananDr. Buchanan

Math 467.01 Partial Differential EquationsMath 467.01 Partial Differential Equations

Problem StatementProblem Statement Consider heat conduction in a cylindrical (hollow) Consider heat conduction in a cylindrical (hollow)

pipe of length L and radius a. Let the initial pipe of length L and radius a. Let the initial temperature distribution on the pipe be given by temperature distribution on the pipe be given by the function the function

f(f(θθ, z) = 1 + (z/L) cos , z) = 1 + (z/L) cos θθ. Suppose the ends of the . Suppose the ends of the pipe are insulated. Find the solution to the heat pipe are insulated. Find the solution to the heat equation in this situation.equation in this situation.

Since the wall of the pipe is so thin, heat is not Since the wall of the pipe is so thin, heat is not going to flow radially outward, so we can neglect going to flow radially outward, so we can neglect the thickness of the pipe’s wall. We will see this the thickness of the pipe’s wall. We will see this come into play in the separation of variables.come into play in the separation of variables.

Starting the ProblemStarting the Problem First, we have to change coordinate systems from First, we have to change coordinate systems from

Cartesian coordinates to cylindrical coordinates, Cartesian coordinates to cylindrical coordinates, then once we do that we will plug this conversion then once we do that we will plug this conversion into the Laplacian operator.into the Laplacian operator.

Below is the heat equation and the conversion of Below is the heat equation and the conversion of the Laplacian operator from Cartesian to cylindrical the Laplacian operator from Cartesian to cylindrical coordinates: coordinates:

uutt = k u = k uxx xx (for this problem we will assume the (for this problem we will assume the thermal diffusivity is 1 since the material of the pipe thermal diffusivity is 1 since the material of the pipe is not specified)is not specified)

∆ ∆² = ∂² = ∂22/∂x/∂x22 + ∂ + ∂22/∂y/∂y2 2 + ∂+ ∂22/∂z/∂z2 2 (in Cartesian coordinates)(in Cartesian coordinates) ΔΔ² = ∂² = ∂22/∂r/∂r22 + (1/r)∂/∂r + (1/r + (1/r)∂/∂r + (1/r22)∂)∂22/∂/∂θθ22 + + ∂∂22//∂∂zz22 (in (in

cylindrical coordinates)cylindrical coordinates)

Separation of VariablesSeparation of Variables As usual, separation of variables still works for this problem. As usual, separation of variables still works for this problem.

We want to look for a solution in the form: We want to look for a solution in the form: T = R(r) T = R(r) θθ((θθ) Z(z) ) Z(z) ФФ(t)(t) Taking the partial derivatives according the Laplacian operator Taking the partial derivatives according the Laplacian operator

we get the following:we get the following: R’’R’’θθZZФФ + (1/r) R’ + (1/r) R’ θθZZФФ + (1/r + (1/r22)R)Rθθ’’ Z’’ ZФФ + R + RθθZ’’ Z’’ ФФ = R = RθθZZФФ’’ and dividing both sides by Rand dividing both sides by RθθZZФФ and canceling we get: and canceling we get:

R’’/R + (1/r)R’/R + (1/rR’’/R + (1/r)R’/R + (1/r22))θθ22//θθ + Z’’/Z = + Z’’/Z = ФФ’/’/ФФ Recall that we are going to neglect the thickness of the pipe’s 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 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 move the Z’s to the right side and substitute a for r for the radius and we get the following:radius and we get the following:

(1/a(1/a22) ) θθ’’/’’/θθ = = ФФ’/’/ФФ – Z’’/Z and – Z’’/Z and θθ’’/’’/θθ = a = a22 ( (ФФ’/’/ФФ – Z’’/Z) = - – Z’’/Z) = -λλ22

Finding the SolutionFinding the Solution Now we want to get the different parts of our Now we want to get the different parts of our

solution, so we will start with our second order solution, so we will start with our second order homogeneous ODE homogeneous ODE θθ’’ + ’’ + λλ²²θθ = 0. This ODE has a = 0. This ODE has a solution in the form solution in the form θθ((θθ) = A) = Ann cos n cos nθθ + B + Bnn sin n sin nθθ where n = 0,1,2…. where n = 0,1,2….

Similarly, we have another second order Similarly, we have another second order homogeneous ODE Z’’ + homogeneous ODE Z’’ + μμ²Z = 0, and we need ²Z = 0, and we need to satisfy the boundary conditions Z’(0) = 0 = to satisfy the boundary conditions Z’(0) = 0 = Z’(L), since the ends of the pipe are insulated. Z’(L), since the ends of the pipe are insulated. This ODE has a solution in the form This ODE has a solution in the form

Zm(Z) = cos mZm(Z) = cos mππ/L where m = 0,1,2…../L where m = 0,1,2…..

Finding the Solution Finding the Solution (cont’d)(cont’d)

Next, we have a first order homogeneous ODENext, we have a first order homogeneous ODE ФФ’ + [(n’ + [(n22/a/a22) + (m) + (m22ππ22/L/L22)])]ФФ = 0. This ODE has a = 0. This ODE has a

solution of the form solution of the form ФФn,mn,m(t) = e(t) = e--ααn,mtn,mt

Putting all our solutions together, we get the Putting all our solutions together, we get the following solution to the heat equation:following solution to the heat equation:

u (u (θθ,z,t) = ,z,t) = ΣΣ ΣΣ e e--ααn,mt n,mt cos mcos mππ/L (4/n/L (4/nππ) (-1)) (-1)nn cos n cos nθθ where the summations are n, m from 0 -> ∞.where the summations are n, m from 0 -> ∞.

Finishing the ProblemFinishing the Problem We are not done yet, we still need to satisfy the We are not done yet, we still need to satisfy the

initial conditions. We need our solution to equal the initial conditions. We need our solution to equal the initial temperature distribution f(initial temperature distribution f(θθ, z) = 1 + (z/L) cos , z) = 1 + (z/L) cos θθ. So, we get the following: . So, we get the following: u ( u (θθ,z) = ,z) = ΣΣ ΣΣ A An n CCmm

cos mcos mππ/L cos n/L cos nθθ = 1 + (z/L) cos = 1 + (z/L) cos θθ.. We will let AWe will let A00 = 1, and we get: = 1, and we get:

u ( u (θθ,z) = 1 + ,z) = 1 + ΣΣ ΣΣ A An n CCmm cos mcos mππ/L cos n/L cos nθθ

We will factor out the An, get rid of the one We will factor out the An, get rid of the one summation, and let Asummation, and let A11 = (1/L) and we get the = (1/L) and we get the following:following:u (u (θθ,z) = 1 + (1/L)cos ,z) = 1 + (1/L)cos θθ ΣΣ C Cmm

cos mcos mππ/L/L Evaluating Cm using the Fourier Integral Formula we Evaluating Cm using the Fourier Integral Formula we

get Cget Cnn = - 4zL/n = - 4zL/n22ππ22. When n is even, C is 0, and when . When n is even, C is 0, and when n is odd we get (-4/nn is odd we get (-4/n22ππ22).).

Mathematica AnalysisMathematica Analysis

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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.

ConclusionConclusion We are now done solving the problem. We We are now done solving the problem. We

satisfied both the boundary conditions satisfied both the boundary conditions (slide 5) and the initial conditions (slide 7).(slide 5) and the initial conditions (slide 7).

Going further, if the wall of the pipe was of Going further, if the wall of the pipe was of significant thickness, then our solution significant thickness, then our solution would have another part to it to would have another part to it to compensate for the heat flowing radially compensate for the heat flowing radially through the thickness of the pipe.through the thickness of the pipe.