De Vahl Davis natural convection test

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Intro

The classical de Vahl Davis benchmark test is defined for the natural convection of the air ($\Pr =0.71$) in the square closed cavity (${{\text{A}}_{\text{R}}}=1$). The only physical free parameter of the test remains the thermal Rayleigh number. In the original paper [1] de Vahl Davis tested the problem up to the Rayleigh number ${{10}^{6}}$, however in the latter publications, the results of more intense simulations were presented with the Rayleigh number up to ${{10}^{8}}$. Lage and Bejan [2] showed that the laminar domain of the closed cavity natural convection problem is roughly below $\text{Gr1}{{\text{0}}^{9}}$. It was reported [3, 4] that the natural convection becomes unsteady for $\text{Ra}=2\cdot {{10}^5}$. Here we present a MLSM solution of the case.

\begin{equation} \text{Ra}\text{=}\,\frac{\left| \mathbf{g} \right|{{\beta }_{T}}\left( {{T}_{H}}-{{T}_{C}} \right){{\Omega }_{H}}^{3}{{\rho }^{2}}{{c}_{p}}}{\lambda \mu } \end{equation} \begin{equation} \text{Pr}=\frac{\mu {{c}_{p}}}{\lambda } \end{equation}

[1] de Vahl Davis G. Natural convection of air in a square cavity: a bench mark numerical solution. Int J Numer Meth Fl. 1983;3:249-64.

[2] Lage JL, Bejan A. The Ra-Pr domain of laminar natural convection in an enclosure heated from the side. Numer Heat Transfer. 1991;A19:21-41.

[3] Janssen RJA, Henkes RAWM. Accuracy of finite-volume disretizations for the bifurcating natural-convection flow in a square cavity. Numer Heat Transfer. 1993;B24:191-207.

[4] Nobile E. Simulation of time-dependent flow in cavities with the additive-correction multigrid method, part II: Apllications. Numer Heat Transfer. 1996;B30:341-50.

Image.png. Figure 1: Scheme of the de Vahl Davis benchmark test

Code

Full examples can be found under the examples in the code repository Main Page).

Explicit ACM method with CBS looks

 1     v2[boundary] = vec_t{0.0, 0.0};
 2     T2[left] = O.T_cold;
 3     T2[right] = O.T_hot;
 4     //Time stepping
 5     for (int step = 0; step <= O.t_steps; ++step) {
 6         for (int i_count = 1; i_count < _MAX_ITER_; ++i_count) {
 7             // Navier Stokes
 8             for (auto c : interior) {
 9                 v2[c] = v1[c] + O.dt * (-1 / O.rho * op.grad(P1, c)
10                                         + O.mu / O.rho * op.lap(v1, c)
11                                         - op.grad(v1, c) * v1[c]
12                                         + O.g * (1 - O.beta * (T1[c] - O.T_ref)));
13             }
14 
15             //Speed of sound
16             Range<scal_t> norm = v2.map([](const vec_t& p) { return p.norm(); });
17             scal_t C = O.dl * std::max(*std::max_element(norm.begin(), norm.end()), O.v_ref);
18             // Mass continuity
19             Range<scal_t> div_v;
20             for (auto c:all) {
21                 div_v[c] = op.div(v2, c);
22                 P2[c] = P1[c] - C * C * O.dt * O.rho * div_v[c] +
23                         O.dl2 * C * C * O.dt * O.dt * op.lap(P1, c);
24             }
25             P1.swap(P2);
26         }
27 
28         //heat transport
29         for (auto c : interior) {
30             T2[c] = T1[c] + O.dt * O.lam / O.rho / O.c_p * op.lap(T1, c) -
31                     O.dt * v1[c].transpose() * op.grad(T1, c);
32         }
33         for (auto c : top) T2[c] = op.neumann(T2, c, vec_t{0, -1}, 0.0);
34         for (auto c : bottom) T2[c] = op.neumann(T2, c, vec_t{0, 1}, 0.0);
35     }

Explicit pressure correction

The solution of heat equation is the same as in above example

 1     for (int step = 0; step <= O.t_steps; ++step) {
 2 
 3         // Explicit Navier-Stokes computed on whole domain, including boundaries
 4         // without pressure
 5         for (int c:all) {
 6             v_2[c] = v_1[c] + O.dt (
 7                                    O.mu / O.rho * op.lap(v_1, c)
 8                                     - op.grad(v_1, c) * v_1[c]
 9                                     + O.g * (1 - O.beta * (T_1[c] - O.T_ref)));
10         }
11         // Pressure correction
12         VecXd rhs_pressure(N + 1, 0); //Note N+1, +1 stands for regularization equation
13         rhs_pressure(N) = 0; // = 0 part of the regularization equation
14         for (int i:interior) rhs_pressure(c) = O.rho / O.dt * op.div(v_2, c);
15         
16         for (int i: boundary) rhs_pressure(c) = O.rho / O.dt * v_2[c].dot(domain.normal(c));
17         VecXd solution = solver_p.solve(rhs_pressure);
18         alpha = solution[N];
19         VecXd P_c = solution.head(N);
20         for ( int i = interior) v_2[c] -=  O.dt / O.rho * op.grad(P_c, c);
21         
22         v_2[boundary] = 0; // force boundary conditions
23 
24         //heat transport
25         for (auto c : interior) {
26             T2[c] = T1[c] + O.dt * O.lam / O.rho / O.c_p * op.lap(T1, c) -
27                     O.dt * v1[c].transpose() * op.grad(T1, c);
28         }
29         for (auto c : top) T2[c] = op.neumann(T2, c, vec_t{0, -1}, 0.0);
30         for (auto c : bottom) T2[c] = op.neumann(T2, c, vec_t{0, 1}, 0.0);
31 }

Implicit

 1 for (int step = 0; step <= O.t_steps_i; ++step) {
 2         time_1 = std::chrono::high_resolution_clock::now();
 3         // NAVIER STOKES
 4         M_velocity = mat_t(2 * N, 2 * N);
 5         // system
 6         M_velocity.reserve(Range<int>(2 * N, O.n));
 7         for (int i : interior) {
 8             op.valuevec(M_velocity, i, 1 / O.dt_i);
 9             op.lapvec(M_velocity, i, -O.mu / O.rho);
10             op.gradvec(M_velocity, i, v_1[i]);
11         }
12         for (int i : boundary) op.valuevec(M_velocity, i, 1); //sets velocity to 0
13 
14         M_velocity.makeCompressed();
15         solver_v.compute(M_velocity);
16         // solution
17         Range <vec_t> rhs_vec(domain.size(), 0);
18         for (int i : interior) {
19             rhs_vec[i] = -1 / O.rho * op.grad(P, i) +
20                          v_1[i] / O.dt_i
21                          + O.g * (1 - O.beta * (T[i] - O.T_ref));
22         }
23         // for (int i:top) rhs_vec[i] = vec_t{0,1};
24 
25         v_2 = reshape<2>(solver_v.solveWithGuess(reshape(rhs_vec), reshape(v_1)));
26         // END OF NAVIER STOKES
27 
28         // PRESSURE CORRECTION
29         VecXd rhs_pressure(N + 1, 0); //Note N+1, +1 stands for regularization equation
30         rhs_pressure(N) = 0; // = 0 part of the regularization equation
31         double alpha;
32         for (int i : interior) rhs_pressure(i) = O.rho / O.dt_i * op.div(v_2, i);
33         for (int i : boundary) rhs_pressure(i) = O.rho / O.dt * v_2[i].dot(domain.normal(i));
34 
35         VecXd solution = solver_p.solve(rhs_pressure);
36         alpha = solution[N];
37         VecXd P_c = solution.head(N);
38         // apply velocity correction
39         for (int i : interior) {
40             v_2[i] -= O.dl * O.dt_i / O.rho * op.grad(P_c, i);
41         }
42         P += O.dl * P_c;
43         // enforce velocity BC
44         // v_2[boundary] = 0;
45         // END OF PRESSURE CORRECTION
46 
47         // HEAT TRANSPORT
48         M_temperature = mat_t(N, N);
49         Range<int> per_row(N, O.n);
50         M_temperature.reserve(per_row);
51         // outer boundary dirichlet BC
52         for (int i : top) op.neumann_implicit(M_temperature, i, domain.normal(i), 1);
53         for (int i : bottom) op.neumann_implicit(M_temperature, i, domain.normal(i), 1);
54         for (int i : left) op.value(M_temperature, i, 1.0);
55         for (int i : right) op.value(M_temperature, i, 1.0);
56         // heat transport in air
57         for (int i : interior) {
58             op.value(M_temperature, i, 1.0);                          // time dependency
59             op.lap(M_temperature, i, -O.dt_i * O.lam / O.rho / O.c_p);  //laplace in interior
60             op.grad(M_temperature, i, O.dt * v_2[i]);
61         }
62         M_temperature.makeCompressed();
63         solver_T.compute(M_temperature);
64 
65         VectorXd rhs = VectorXd::Zero(N);
66         for (int i : interior) rhs(i) = T(i);
67         for (int i : top) rhs(i) = 0;
68         for (int i : bottom) rhs(i) = 0;
69         for (int i : left) rhs(i) = O.T_hot;
70         for (int i : right) rhs(i) = O.T_cold;
71         T = solver_T.solveWithGuess(rhs, T);
72         // END OF HEAT TRANSPORT
73         v_1.swap(v_2);
74 }

Results

Comparison of MLSM solution with reference data

Following video shows evolution of temperature and velocity magnitude for the $Ra=10^8$ case.


In below galley you can find temperature contour plots, velocity magnitude contour plots, v_max and average hot side Nusselt number convergence behavior. The reference values are from:

  • [a] de Vahl Davis G. Natural convection of air in a square cavity: a bench mark numerical solution. Int J Numer Meth Fl. 1983;3:249-64.
  • [b] Sadat H, Couturier S. Performance and accuracy of a meshless method for laminar natural convection. Numer Heat Transfer. 2000;B37:455-67.
  • [c] Wan DC, Patnaik BSV, Wei GW. A new benchmark quality solution for the buoyancy-driven cavity by discrete singular convolution. Numer Heat Transfer. 2001;B40:199-228.
  • [d] Šarler B. A radial basis function collocation approach in computational fluid dynamics. CMES-Comp Model Eng. 2005;7:185-93.
  • [e] Kosec G, Šarler B. Solution of thermo-fluid problems by collocation with local pressure correction. Int J Numer Method H. 2008;18:868-82.