### [example][freeflowchannel] Stokes not Navier Stokes.

parent fab3fb50
 ... ... @@ -10,7 +10,7 @@ This example contains a stationary free flow of a fluid through two parallel sol ## Model description The Stokes model without gravitation and without sources or sinks for a stationary, incompressible, laminar, single phase, one-component, isothermal ($T=10^\circ C$) flow is considered assuming a Newtonian fluid of constant density $ \varrho = 1~\frac{\text{kg}}{\text{m}^3} $ and constant kinematic viscosity $ \nu = 1~\frac{\text{m}^2}{\text{s}} $. The momentum balance math \nabla \cdot (\varrho\boldsymbol{u} \boldsymbol{u}^{\text{T}}) - \nabla\cdot\left(\mu\left(\nabla\boldsymbol{u}+\nabla\boldsymbol{u}^{\text{T}}\right)\right)+ \nabla p = 0 - \nabla\cdot\left(\mu\left(\nabla\boldsymbol{u}+\nabla\boldsymbol{u}^{\text{T}}\right)\right)+ \nabla p = 0  with density $\varrho$, velocity $\boldsymbol{u}$, dynamic viscosity $\mu=\varrho\nu$ and pressure $p$ and the mass balance math ... ...
 ... ... @@ -10,7 +10,7 @@ This example contains a stationary free flow of a fluid through two parallel sol ## Model description The Stokes model without gravitation and without sources or sinks for a stationary, incompressible, laminar, single phase, one-component, isothermal ($T=10^\circ C$) flow is considered assuming a Newtonian fluid of constant density $ \varrho = 1~\frac{\text{kg}}{\text{m}^3} $ and constant kinematic viscosity $ \nu = 1~\frac{\text{m}^2}{\text{s}} $. The momentum balance math \nabla \cdot (\varrho\boldsymbol{u} \boldsymbol{u}^{\text{T}}) - \nabla\cdot\left(\mu\left(\nabla\boldsymbol{u}+\nabla\boldsymbol{u}^{\text{T}}\right)\right)+ \nabla p = 0 - \nabla\cdot\left(\mu\left(\nabla\boldsymbol{u}+\nabla\boldsymbol{u}^{\text{T}}\right)\right)+ \nabla p = 0  with density $\varrho$, velocity $\boldsymbol{u}$, dynamic viscosity $\mu=\varrho\nu$ and pressure $p$ and the mass balance math ... ...
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