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 8185 control volumes, 16031 triangles, and 24215 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 8185 nodes and time span (0.0, Inf)
steady_prob = SteadyFVMProblem(prob)
SteadyFVMProblem with 8185 nodes
Now we can solve.
using OrdinaryDiffEq, SteadyStateDiffEq
sol = solve(steady_prob, DynamicSS(Rosenbrock23()))
retcode: Success
u: 8320-element Vector{Float64}:
70.0
53.10293622451229
40.0
40.0
44.128418439597596
⋮
61.652468942477334
59.31965243753265
61.44777462514571
55.36594253951101
67.99988351815033
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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