# Conjugate heat transfer modelling of reciprocating flow crashes for long flow times

The following transient problem is essentially the reciprocating (i.e., fully reversing) flow of a fluid over a thick heated block until the system reaches a cyclic steady-state (i.e., the system temperature oscillates around a mean when this point is reached). The following code simulates the system for a fluid velocity boundary condition $$u=3*\sin(2*\pi*0.5*t)$$, i.e., a frequency of $$0.5Hz$$ and a heat flux input of $$1000 W/m^2$$. tflow refers to the total time till which the flow oscillation occurs and t0 is the time-step. The code uses the flow solver developed by Alex Trounev and described in this page.

Needs["NDSolveFEM"]

{f = 0.5; L = 0.040, d = 0.003, e = 0.005, kf = 0.026499, ks = 16,
rho = 1.1492, rhos = 7860, mu = 18.923*10^-6, cp = 1.0069*10^3,
cps = 502.4}; u0 = 3; nu = mu/rho; om = 2 Pi f;
tflow = 400;
t0 = .3;
NV = 2 f tflow;
nn = Round[NV \[Pi]/(om t0)]
Ti = 307; q = 1000/Ti;

reg1 = ImplicitRegion[0 <= x <= L && 0 <= y <= d, {x, y}]; reg2 =
ImplicitRegion[0 <= x <= L && -e <= y <= d, {x, y}];
UX[x_, y_] := 0;
VY[x_, y_] := 0;
P[x_, y_] := 0;
Tfs[x_, y_] := 307/Ti; appro =
With[{k = 2. 10^6}, ArcTan[k #]/Pi + 1/2 &];
ade[y_] := (ks + (kf - ks) appro[y])
rde[y_] := (cps rhos + (cp rho - cps rhos) appro[y]);

Monitor[Do[{UX[i], VY[i], P[i]} =
NDSolveValue[{{Inactive[
u[x, y], {x, y}]), {x, y}] + D[p[x, y], x] +
UX[i - 1][x, y]*D[u[x, y], x] +
VY[i - 1][x, y]*D[u[x, y], y] + (u[x, y] - UX[i - 1][x, y])/
t0, Inactive[

v[x, y], {x, y}]), {x, y}] + D[p[x, y], y] +
UX[i - 1][x, y]*D[v[x, y], x] +
VY[i - 1][x, y]*D[v[x, y], y] + (v[x, y] - VY[i - 1][x, y])/
t0, D[u[x, y], x] + D[v[x, y], y]} == {0, 0, 0} /. \[Mu] ->
nu, {DirichletCondition[{u[x, y] == u0*Sin[om*i*t0],
v[x, y] == 0},
x == L (1 - Sign[Sin[om*i*t0]])/2 && 0 < y < d],
DirichletCondition[{u[x, y] == 0, v[x, y] == 0},
y == 0 || y == d]},
DirichletCondition[p[x, y] == 0,
x == L (1 + Sign[Sin[om*i*t0]])/2 && 0 < y < d]}, {u, v,
p}, {x, y} \[Element] reg1,
Method -> {"FiniteElement",
"InterpolationOrder" -> {u -> 2, v -> 2, p -> 1},
"MeshOptions" -> {"MaxCellMeasure" -> 0.0000005}}];
ux = If[y <= 0, 0, UX[i][x, y]]; vy = If[y <= 0, 0, VY[i][x, y]];
Tfs[i] =
NDSolveValue[{rde[
y] ((ux*D[T[x, y], x] +
vy*D[T[x, y], y]) + (T[x, y] - Tfs[i - 1][x, y])/t0) -
Inactive[Div][
NeumannValue[q, y == -e],
DirichletCondition[{T[x, y] == 1},
x == L (1 - Sign[Sin[om*i*t0]])/2 && 0 <= y <= d]},
T, {x, y} \[Element] reg2,
Method -> {"FiniteElement", "InterpolationOrder" -> {T -> 2},
"MeshOptions" -> {"MaxCellMeasure" -> 0.0000001}}] //
Quiet;, {i, 1, nn}],ProgressIndicator[i,{1,nn}]]; // AbsoluteTiming

ListLinePlot[
Table[{i t0, Tfs[i][0.5*L, -e/2]*Ti - 273.16}, {i, 0, nn}],
AxesLabel -> {"t(s)", "T(in deg. C)"}, PlotRange -> Full]


If one plots the temperature profile at a point $$(0.5L,-e/2)$$ in the solid with time, it turns out to be: As can be seen, the temperature of the system initially increases fast to gradually slow down and reach a cyclic steady-state (marked with red), where it oscillates around a mean.

However, keeping all other parameters same, if I increase the frequency to say f=2 Hz, I need to give a smaller time-step which should be at-least $$<\frac{1}{2*2}=0.25s$$, to resolve the flow reversal in each half-period correctly. In this scenario, the tflow required to reach a cyclic steady-state increases.

When I run the above code with f=2, t0=0.1,tflow=800, it crashes each and every time with the error Wolfram Kernel has stopped working. My question is:

1. Can this code be improved so that it does not crash for high f, which require small t0? For example: f=2, t0=0.1

To answer first question we can improve code by separating fluid part and using mesh with QuadElement only as follows

Needs["NDSolveFEM"]
{f = 2;
L = 0.040, d = 0.003, e = 0.005, kf = 0.026499, ks = 16,
rho = 1.1492, rhos = 7860, mu = 18.923*10^-6, cp = 1.0069*10^3,
cps = 502.4}; u0 = 3; nu = mu/rho; om = 2 Pi f;
tflow = 800;
t0 = .1;
NV = 2 f tflow;
nn = Round[NV \[Pi]/(om t0)]
Ti = 307; q = 1000/Ti;

reg1 = ImplicitRegion[0 <= x <= L && 0 <= y <= d, {x, y}]; reg2 =
ImplicitRegion[0 <= x <= L && -e <= y <= d, {x, y}];

mesh = ToElementMesh[FullRegion, {{0, L}, {0, d}}];
mesh1 = ToElementMesh[FullRegion, {{0, L}, {-e, d}}];
UX[x_, y_] := 0;
VY[x_, y_] := 0;
P[x_, y_] := 0;
Tfs[x_, y_] := 307/Ti; appro =
With[{k = 2. 10^6}, ArcTan[k #]/Pi + 1/2 &];
ade[y_] := (ks + (kf - ks) appro[y])
rde[y_] := (cps rhos + (cp rho - cps rhos) appro[y]);
eqs = {Inactive[
Div][({{-\[Mu], 0}, {0, -\[Mu]}} .
Inactive[Grad][u[x, y], {x, y}]), {x, y}] + D[p[x, y], x] +
UX[i - 1][x, y]*D[u[x, y], x] +
VY[i - 1][x, y]*D[u[x, y], y] + (u[x, y] - UX[i - 1][x, y])/t0,
Inactive[
Div][({{-\[Mu], 0}, {0, -\[Mu]}} .
Inactive[Grad][v[x, y], {x, y}]), {x, y}] + D[p[x, y], y] +
UX[i - 1][x, y]*D[v[x, y], x] +
VY[i - 1][x, y]*D[v[x, y], y] + (v[x, y] - VY[i - 1][x, y])/t0,
D[u[x, y], x] + D[v[x, y], y]};
bc[i_] := {DirichletCondition[{u[x, y] == u0*Sin[om*i*t0],
v[x, y] == 0}, x == L (1 - Sign[Sin[om*i*t0]])/2 && 0 < y < d],
DirichletCondition[{u[x, y] == 0, v[x, y] == 0}, y == 0 || y == d],
DirichletCondition[p[x, y] == 0,
x == L (1 + Sign[Sin[om*i*t0]])/2 && 0 < y < d]};

Monitor[Do[{UX[i], VY[i], P[i]} =
NDSolveValue[{eqs == {0, 0, 0} /. \[Mu] -> nu, bc[i]}, {u, v,
p}, {x, y} \[Element] mesh,
Method -> {"FiniteElement",
"InterpolationOrder" -> {u -> 2, v -> 2, p -> 1}}];, {i, 1,
nn}], ProgressIndicator[i, {1, nn}]]; // AbsoluteTiming


It takes about 1871s on my laptop. Visualization last 20 frames

Table[DensityPlot[UX[i][x, y], {x, y} \[Element] mesh,
ColorFunction -> "Rainbow", PlotRange -> All,
AspectRatio -> Automatic, Frame -> False], {i, nn - 19, nn}] With this velocity field we can compute temperature as follows

Monitor[Do[ux = If[y <= 0, 0, UX[i][x, y]];
vy = If[y <= 0, 0, VY[i][x, y]];
Tfs[i] =
NDSolveValue[{rde[
y] ((ux*D[T[x, y], x] +
vy*D[T[x, y], y]) + (T[x, y] - Tfs[i - 1][x, y])/t0) -
Inactive[Div][
NeumannValue[q, y == -e],
DirichletCondition[{T[x, y] == 1},
x == L (1 - Sign[Sin[om*i*t0]])/2 && 0 <= y <= d]},
T, {x, y} \[Element] mesh1,
Method -> {"FiniteElement",
"InterpolationOrder" -> {T -> 2}}] // Quiet;, {i, 1, nn}],
ProgressIndicator[i, {1, nn}]]; // AbsoluteTiming


{ListLinePlot[
Table[{i t0, Tfs[i][0.5*L, -e/2]*Ti - 273.16}, {i, 0, nn}],
AxesLabel -> {"t(s)", "T(in deg. C)"}, PlotRange -> Full],
ListLinePlot[
Table[{i t0, Tfs[i][0.5*L, -e/2]*Ti - 273.16}, {i, nn - 40, nn}],
AxesLabel -> {"t(s)", "T(in deg. C)"}, PlotRange -> Full]} • Thankyou for these modifications. Right now I am running a example and it has not crashed till now. I am optimistic that this has done the trick. Dec 13, 2022 at 5:50
• I ran the code for f=2, tflow=1000, q=5000/Ti. It took a total of 8hr 51min to execute on my 8-core AMD Ryzen 7 5800H laptop. But it did not crash. This solution surely answers the first question. I will accept it in a day if no answer to the second part is forthcoming. Thankyou again Alex. Dec 14, 2022 at 4:50
• @Avrana It could be better to ask second question in separate post. Since answer is not simple. There are several papers about this problem as I remember from discussion on mathematica.stackexchange.com/questions/262999/… Dec 14, 2022 at 5:53
• I think you are right. I have accepted the answer and will post a separate question for the second part. Dec 14, 2022 at 6:17
• I have posted a separate question here. Dec 15, 2022 at 6:53