Lines Matching refs:phi

10 Denote $\langle \phi \rangle$ as the Reynolds average of $\phi$, which in this case would be a aver…
13 \langle \phi \rangle(x,y) = \frac{1}{L_z + (T_f - T_0)}\int_0^{L_z} \int_{T_0}^{T_f} \phi(x, y, z, …
22 The function $\langle \phi \rangle (x,y)$ is represented on a 2-D finite element grid, taken from t…
34 \langle \phi \rangle_z(x,y,t) = \frac{1}{L_z} \int_0^{L_z} \phi(x, y, z, t) \mathrm{d}z
37 where the function $\phi$ may be the product of multiple solution functions and $\langle \phi \rang…
44 Substituting the spanwise average of $\phi$ for $u$, we get:
47 \bm M [\langle \phi \rangle_z]_N = \int_0^{L_x} \int_0^{L_y} \left [\frac{1}{L_z} \int_0^{L_z} \phi…
53 \bm M [\langle \phi \rangle_z]_N = \frac{1}{L_z} \int_\Omega \phi(x,y,z,t) \psi^\mathrm{parent}_N(x…
63 At the beginning of each simulation, the time integral of a statistic is set to 0, $\overline{\phi}…
66 $$\overline{\phi}_\mathrm{new} = \overline{\phi}_{\mathrm{old}} + \phi(t_\mathrm{new}) \Delta T$$
67 where $\phi(t_\mathrm{new})$ is the statistic at the current time and $\Delta T$ is the time since …
73 \bm M [\langle \phi \rangle]_N = \frac{1}{L_z + (T_f - T_0)} \int_\Omega \int_{T_0}^{T_f} \phi(x,y,…
75 where the integral $\int_{T_0}^{T_f} \phi(x,y,z,t) \mathrm{d}t$ is calculated on a running basis.
135 \mean{\phi' \theta'} = \mean{\phi \theta \rangle - \langle \phi} \mean{\theta}
175 \overline{\phi} - \nabla \cdot (\beta (\bm{D}\bm{\Delta})^2 \nabla \overline{\phi} ) = \phi
178 for $\phi$ the scalar solution field we want to filter, $\overline \phi$ the filtered scalar soluti…
182 \int_\Omega \left( v \overline \phi + \beta \nabla v \cdot (\bm{D}\bm{\Delta})^2 \nabla \overline \…
183 - \cancel{\int_{\partial \Omega} \beta v \nabla \overline \phi \cdot (\bm{D}\bm{\Delta})^2 \bm{\hat…
184 \int_\Omega v \phi \, , \; \forall v \in \mathcal{V}_p
187 …assume that $(\bm{D}\bm{\Delta})^2 = \bm{0} \Leftrightarrow \overline \phi = \phi$ at boundaries (…
227 The filtering at the wall may also be damped, to smoothly meet the $\overline \phi = \phi$ boundary…