From 4c43abd47981b2115e0deb8f586a5c588b4feee1 Mon Sep 17 00:00:00 2001 From: Wolfgang Bangerth Date: Mon, 17 Aug 2009 00:45:24 +0000 Subject: [PATCH] Fix more places where we referenced the Rayleigh number. git-svn-id: https://svn.dealii.org/trunk@19280 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-31/doc/intro.dox | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) diff --git a/deal.II/examples/step-31/doc/intro.dox b/deal.II/examples/step-31/doc/intro.dox index abe5a04005..00b948cbd5 100644 --- a/deal.II/examples/step-31/doc/intro.dox +++ b/deal.II/examples/step-31/doc/intro.dox @@ -29,7 +29,7 @@ can be neglected, the equations that describe such behavior are the Boussinesq equations that read as follows: @f{eqnarray*} -\nabla \cdot (2 \eta \varepsilon ({\mathbf u})) + \nabla p &=& - -\rho \beta \; T \mathbf{g}, + -\rho\; \beta \; T\; \mathbf{g}, \\ \nabla \cdot {\mathbf u} &=& 0, \\ @@ -54,13 +54,13 @@ The forcing term of the fluid motion is the buoyancy of the fluid, expressed as the product of the density $\rho$, the thermal expansion coefficient $\beta$, the temperature T and the gravity vector g pointing downward. (A -possibly more intuitive formulation would use $-\mathrm{Ra} \; (T-\bar T) +possibly more intuitive formulation would use $-\rho\; \beta \; (T-\bar T) \mathbf{g}$ as right hand side where $\bar T$ is the average temperature, and the right hand side then describes the forces due to local deviations from the average density; this formulation is entirely equivalent if the gravity vector results from a gravity potential $\phi$, i.e. $\mathbf{g}=-\nabla\phi$, and yields the exact -same solution except for the pressure which will now be $p-\mathrm{Ra} +same solution except for the pressure which will now be $p-\rho\; \beta \;\bar T \phi$.) While the first two equations describe how the fluid reacts to @@ -195,7 +195,7 @@ the previous time step. In other words, we first solve the Stokes system for time step n-1 as @f{eqnarray*} -\nabla \cdot (2\eta \varepsilon ({\mathbf u}^{n-1})) + \nabla p^{n-1} &=& - -\mathrm{Ra} \; T^{n-1} \mathbf{g}, + -\rho\; \beta \; T^{n-1} \mathbf{g}, \\ \nabla \cdot {\mathbf u}^{n-1} &=& 0, @f} @@ -315,7 +315,7 @@ by their discrete counterparts: - (\nabla \cdot {\mathbf v}_h, p^{n-1}_h) &=& - -({\mathbf v}_h, \mathrm{Ra} \; T^{n-1}_h \mathbf{g}), + -({\mathbf v}_h, \rho\; \beta \; T^{n-1}_h \mathbf{g}), \\ (q_h, \nabla \cdot {\mathbf u}^{n-1}_h) &=& 0, @f} @@ -334,7 +334,7 @@ that we implement is - (\nabla \cdot {\mathbf v}_h, p^{n-1}_h) &=& - -({\mathbf v}_h, \mathrm{Ra} \; T^{n-1}_h \mathbf{g}), + -({\mathbf v}_h, \rho\; \beta \; T^{n-1}_h \mathbf{g}), \\ (q_h, \nabla \cdot {\mathbf u}^{n-1}_h) &=& 0. @f} @@ -977,7 +977,7 @@ discussed here.

The testcase

The case we want to solve here is as follows: we solve the Boussinesq -equations described above with $\kappa=10^{-6}, \eta=1, \mathrm{Ra}=10$, +equations described above with $\kappa=10^{-6}, \eta=1, \rho=1, \beta=10$, i.e. a relatively slow moving fluid that has virtually no thermal diffusive conductivity and transports heat mainly through convection. On the boundary, we will require no-normal flux for the velocity -- 2.39.5