Equilibrium Temperature Distribution with Mixed Boundary Conditions and using EnsembleProblems

For this tutorial, we consider the following problem:

\[\begin{equation} \begin{aligned} \grad^2 T &= 0 & \vb x \in \Omega, \\ \grad T \vdot \vu n &= 0 & \vb x \in \Gamma_1, \\ T &= 40 & \vb x \in \Gamma_2, \\ k\grad T \vdot \vu n &= h(T_{\infty} - T) & \vb x \in \Gamma_3, \\ T &= 70 & \vb x \in \Gamma_4. \\ \end{aligned} \end{equation}\]

This domain $\Omega$ with boundary $\partial\Omega=\Gamma_1\cup\Gamma_2\cup\Gamma_3\cup\Gamma_4$ is shown below.

Let us start by defining the mesh.

using DelaunayTriangulation, FiniteVolumeMethod, CairoMakie
A, B, C, D, E, F, G = (0.0, 0.0),
(0.06, 0.0),
(0.06, 0.03),
(0.05, 0.03),
(0.03, 0.05),
(0.03, 0.06),
(0.0, 0.06)
bn1 = [G, A, B]
bn2 = [B, C]
bn3 = [C, D, E, F]
bn4 = [F, G]
bn = [bn1, bn2, bn3, bn4]
boundary_nodes, points = convert_boundary_points_to_indices(bn)
tri = triangulate(points; boundary_nodes)
refine!(tri; max_area=1e-4get_area(tri))
triplot(tri)

mesh = FVMGeometry(tri)

FVMGeometry with 8160 control volumes, 15990 triangles, and 24149 edges

For the boundary conditions, the parameters that we use are $k = 3$, $h = 20$, and $T_{\infty} = 20$ for thermal conductivity, heat transfer coefficient, and ambient temperature, respectively.

k = 3.0
h = 20.0
T∞ = 20.0
bc1 = (x, y, t, T, p) -> zero(T) # ∇T⋅n=0
bc2 = (x, y, t, T, p) -> oftype(T, 40.0) # T=40
bc3 = (x, y, t, T, p) -> -p.h * (p.T∞- T) / p.k # k∇T⋅n=h(T∞-T). The minus is since q = -∇T
bc4 = (x, y, t, T, p) -> oftype(T, 70.0) # T=70
parameters = (nothing, nothing, (h=h, T∞=T∞, k=k), nothing)
BCs = BoundaryConditions(mesh, (bc1, bc2, bc3, bc4),
    (Neumann, Dirichlet, Neumann, Dirichlet);
    parameters)

BoundaryConditions with 4 boundary conditions with types (Neumann, Dirichlet, Neumann, Dirichlet)

Now we can define the actual problem. For the initial condition, which recall is used as an initial guess for steady state problems, let us use an initial condition which ranges from $T=70$ at $y=0.06$ down to $T=40$ at $y=0$.

diffusion_function = (x, y, t, T, p) -> one(T)
f = (x, y) -> 500y + 40
initial_condition = [f(x, y) for (x, y) in DelaunayTriangulation.each_point(tri)]
final_time = Inf
prob = FVMProblem(mesh, BCs;
    diffusion_function,
    initial_condition,
    final_time)

FVMProblem with 8160 nodes and time span (0.0, Inf)

steady_prob = SteadyFVMProblem(prob)

SteadyFVMProblem with 8160 nodes

Now we can solve.

using OrdinaryDiffEq, SteadyStateDiffEq
sol = solve(steady_prob, DynamicSS(Rosenbrock23()))

retcode: Success
u: 8298-element Vector{Float64}:
 70.0
 53.10321550648938
 40.0
 40.0
 44.13396494556966
  ⋮
 65.60739642042194
 42.990857540603706
 66.53334753207265
 44.71982985096343
 50.874999641340054

fig, ax, sc = tricontourf(tri, sol.u, levels=40:70, axis=(xlabel="x", ylabel="y"))
fig

Just the code

An uncommented version of this example is given below. You can view the source code for this file here.

using DelaunayTriangulation, FiniteVolumeMethod, CairoMakie
A, B, C, D, E, F, G = (0.0, 0.0),
(0.06, 0.0),
(0.06, 0.03),
(0.05, 0.03),
(0.03, 0.05),
(0.03, 0.06),
(0.0, 0.06)
bn1 = [G, A, B]
bn2 = [B, C]
bn3 = [C, D, E, F]
bn4 = [F, G]
bn = [bn1, bn2, bn3, bn4]
boundary_nodes, points = convert_boundary_points_to_indices(bn)
tri = triangulate(points; boundary_nodes)
refine!(tri; max_area=1e-4get_area(tri))
triplot(tri)

mesh = FVMGeometry(tri)

k = 3.0
h = 20.0
T∞ = 20.0
bc1 = (x, y, t, T, p) -> zero(T) # ∇T⋅n=0
bc2 = (x, y, t, T, p) -> oftype(T, 40.0) # T=40
bc3 = (x, y, t, T, p) -> -p.h * (p.T∞- T) / p.k # k∇T⋅n=h(T∞-T). The minus is since q = -∇T
bc4 = (x, y, t, T, p) -> oftype(T, 70.0) # T=70
parameters = (nothing, nothing, (h=h, T∞=T∞, k=k), nothing)
BCs = BoundaryConditions(mesh, (bc1, bc2, bc3, bc4),
    (Neumann, Dirichlet, Neumann, Dirichlet);
    parameters)

diffusion_function = (x, y, t, T, p) -> one(T)
f = (x, y) -> 500y + 40
initial_condition = [f(x, y) for (x, y) in DelaunayTriangulation.each_point(tri)]
final_time = Inf
prob = FVMProblem(mesh, BCs;
    diffusion_function,
    initial_condition,
    final_time)

steady_prob = SteadyFVMProblem(prob)

using OrdinaryDiffEq, SteadyStateDiffEq
sol = solve(steady_prob, DynamicSS(Rosenbrock23()))

fig, ax, sc = tricontourf(tri, sol.u, levels=40:70, axis=(xlabel="x", ylabel="y"))
fig

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