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Couette flow with the thermal wall entropy balance

Example

Let $x$ be periodic, $0 < y < L$ , and fix $p_{0} > 0$ , $T_{w} > 0$ , $a \neq 0$ , $μ > 0$ , and $κ > 0$ . Define

$u (x, y) = (a y, 0), T (y) = T_{w} + \frac{μ a^{2}}{2 κ} y (L - y), ρ (y) = \frac{p_{0}}{R T (y)},$

and $m = ρ u$ , $E = p_{0} / (γ - 1) + ρ a^{2} y^{2} / 2$ . This is a smooth positive steady solution with constant pressure $p_{0}$ , no-slip wall velocities $u (x, 0) = (0, 0)$ and $u (x, L) = (a L, 0)$ , and wall temperature $T_{w}$ at both walls. Per unit periodic length, the outward physical-entropy flux through the two thermal walls equals

${[- κ \frac{T^{'}}{T}]}_{0}^{L} = \int_{0}^{L} (\frac{μ a^{2}}{T} + \frac{κ (T^{'})^{2}}{T^{2}}) d y > 0.$

This is a continuous wall-balance example; it specifies no discrete wall flux.

Verification

Given: The two-dimensional ideal-gas Navier–Stokes–Fourier model, $Ω$ periodic in $x$ and equal to $(0, L)$ in $y$ , and the displayed fields.

[F1] Ideal-gas compressible Navier–Stokes–Fourier model: $p = (γ - 1) (E - ∣ m ∣^{2} / (2 ρ))$ , $τ = 2 μ D^{0} + ζ (tr D) I$ , and $F_{j}^{v i s c} = (0, τ_{j \cdot}, u \cdot τ_{j \cdot} + κ \partial_{j} T)$ .

[F2] Smooth ideal-gas Navier–Stokes–Fourier entropy identity: physical entropy $ρ s = - η$ satisfies $\partial_{t} (ρ s) + \nabla \cdot (ρ s u - κ \nabla T / T) = τ : \nabla u / T + κ ∣ \nabla T ∣^{2} / T^{2}$ .

1.1F1givenalgebra

The profile has $T (y) \geq T_{w} > 0$ , so $ρ > 0$ . Its internal-energy density is $E - ∣ m ∣^{2} / (2 ρ) = p_{0} / (γ - 1)$ , hence [F1] gives $p = p_{0}$ . The velocity gradient has only $\partial_{y} u_{1} = a$ , so $\nabla \cdot u = 0$ , $τ_{12} = τ_{21} = μ a$ , and all entries of $τ$ are constant.

2.1F1step 1.1algebra

The mass divergence is zero because the field is independent of $x$ and $u_{2} = 0$ . The momentum divergence is zero because $p$ is constant, $τ$ is constant, and every $y$ -velocity is zero. For total energy, the Euler flux has zero divergence; the only nonzero viscous energy flux is $F_{2, E}^{v i s c} = u \cdot τ_{2 \cdot} + κ T^{'} = μ a^{2} y + κ T^{'}$ . Since $T^{''} = - μ a^{2} / κ$ , its derivative is zero. Thus all conservative equations hold.

3.1F1F2step 1.1step 2.1algebra∎

The stress production is $τ : \nabla u = μ a^{2}$ . The physical entropy law [F2], no normal velocity at either wall, and periodicity in $x$ give the boundary-flux identity after integration in $y$ . Directly, $(κ T^{'} / T)^{'} = κ T^{''} / T - κ (T^{'})^{2} / T^{2} = - μ a^{2} / T - κ (T^{'})^{2} / T^{2}$ , so integrating its negative proves the equality. The integrand is strictly positive because $a \neq 0$ , $μ > 0$ , and $T > 0$ .

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Sources

Pablo Fernandez, Ngoc-Cuong Nguyen, and Jaime Peraire, Entropy-stable hybridized discontinuous Galerkin methods for the compressible Euler and Navier-Stokes equations, complete author preprint (standard reference, not scraped)