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 8186 control volumes, 16033 triangles, and 24218 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 8186 nodes and time span (0.0, Inf)

steady_prob = SteadyFVMProblem(prob)

SteadyFVMProblem with 8186 nodes

Now we can solve.

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

retcode: Success
u: 8321-element Vector{Float64}:
 70.0
 53.10293700353859
 40.0
 40.0
 44.12841830740191
  ⋮
 61.65246886036293
 59.3196525701446
 61.44777453456248
 55.365941641800816
 67.99988355181776

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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