Non-dimensionalizing Navier Stokes

VincentD · November 21, 2011, 08:29

This post is in response to a message I recieved of someone that wanted to have a more detailed explanation on how to obtain the non-dimensional NS equations. Since others might be interested (and the fact that I like this tex compiler) I posted on this forum.

For starters we introduce the Navier Stokes equations and look at the x-direction component:
$\rho \Big(\frac{\partial v_x}{\partial t} + v_x\frac{\partial v_x}{\partial x} + v_y\frac{\partial v_x}{\partial y} + v_z\frac{\partial v_x}{\partial z}\Big) = - \frac{\partial p}{\partial x} + \mu \Big(\frac{\partial^2 v_x}{\partial x^2} + \frac{\partial^2 v_x}{\partial y^2} + \frac{\partial^2 v_x}{\partial z^2}\Big)$
We assume no additional body forces. The components of the velocity field v are denoted by subscripts (x,y,z). The pressure is given by p and $\rho$ is the density.

First we divide by the density $\rho$ in order to simplify the equation.
$\frac{\partial v_x}{\partial t} + v_x\frac{\partial v_x}{\partial x} + v_y\frac{\partial v_x}{\partial y} + v_z\frac{\partial v_x}{\partial z} = - \frac{1}{\rho} \frac{\partial p}{\partial x} + \nu \Big(\frac{\partial^2 v_x}{\partial x^2} + \frac{\partial^2 v_x}{\partial y^2} + \frac{\partial^2 v_x}{\partial z^2}\Big)$
The reader should check that indeed each terms has a dimension equal to $\frac{m}{s^2}$ . Now in order to obtaint the non-dimensional equation we make the following substitutions.

$p = p^* {V^2 \rho}$

Quantities with a superscript star are dimensionless quantities and the capital letters V and H have dimensions $\frac{m}{s}$ and

respectively.
Please check that these substitutions are indeed valid. For instance the physical length is 4 m. We decide to take the length H equal to 1 m, giving a non-dimensionalised length of 4.

Using the above substitutions we end up with the following equation:
$\frac{\partial v_x^* V}{\partial t^* T} + v_x^* V\frac{\partial v_x^* V}{\partial x^* H} + v_y^* V\frac{\partial v_x^* V}{\partial y^* H} + v_z^* V\frac{\partial v_x^* V}{\partial z^* H}$ $= - \frac{1}{\rho} \frac{\partial p^* V^2 \rho}{\partial x^* H } + \nu \Big(\frac{\partial^2 v_x^* V}{\partial {x^*}^2 H^2} + \frac{\partial^2 v_x^* V}{\partial {y^*}^2 H} + \frac{\partial^2 v_x^* V}{\partial {z^*}^2 H^2}\Big)$

We simplify this equation and use that T = H/V.

$\frac{\partial v_x^*}{\partial t^*} + v_x^* \frac{\partial v_x^*}{\partial x^*} + v_y^* \frac{\partial v_x^*}{\partial y^*} + v_z^*\frac{\partial v_x^*}{\partial z^*}$ $= - \frac{\partial p^*}{\partial x^*} + \nu \frac{1}{V H} \Big(\frac{\partial^2 v_x^*}{\partial {x^*}^2} + \frac{\partial^2 v_x^*}{\partial {y^*}^2} + \frac{\partial^2 v_x^*}{\partial {z^*}^2}\Big)$

Here $\frac{\nu}{V H}$ is our inverse reynolds number.

$\frac{\partial v_x^*}{\partial t^*} + v_x^* \frac{\partial v_x^*}{\partial x^*} + v_y^* \frac{\partial v_x^*}{\partial y^*} + v_z^*\frac{\partial v_x^*}{\partial z^*}$ $= - \frac{\partial p^*}{\partial x^*} + \frac{1}{Re} \Big(\frac{\partial^2 v_x^*}{\partial {x^*}^2} + \frac{\partial^2 v_x^*}{\partial {y^*}^2} + \frac{\partial^2 v_x^*}{\partial {z^*}^2}\Big)$

Where Re is defined as $\frac{V H}{\nu}$ . Now we have non-dimensionalised the NS equations. Please rememeber that if you solve this equation you will end up with the nondimensionalised quantities. In order to revert them to physical quantities you will need to use the equation we proposed above when we non-dimensionalised the quantities.

Good luck!

Regards,

Vincent

November 21, 2011, 08:29	Non-dimensionalizing Navier Stokes	#1
VincentD New Member Vincent Join Date: Jul 2011 Posts: 29 Rep Power: 14	This post is in response to a message I recieved of someone that wanted to have a more detailed explanation on how to obtain the non-dimensional NS equations. Since others might be interested (and the fact that I like this tex compiler) I posted on this forum. For starters we introduce the Navier Stokes equations and look at the x-direction component: $\rho \Big(\frac{\partial v_x}{\partial t} + v_x\frac{\partial v_x}{\partial x} + v_y\frac{\partial v_x}{\partial y} + v_z\frac{\partial v_x}{\partial z}\Big) = - \frac{\partial p}{\partial x} + \mu \Big(\frac{\partial^2 v_x}{\partial x^2} + \frac{\partial^2 v_x}{\partial y^2} + \frac{\partial^2 v_x}{\partial z^2}\Big)$ We assume no additional body forces. The components of the velocity field v are denoted by subscripts (x,y,z). The pressure is given by p and $\rho$ is the density. First we divide by the density $\rho$ in order to simplify the equation. $\frac{\partial v_x}{\partial t} + v_x\frac{\partial v_x}{\partial x} + v_y\frac{\partial v_x}{\partial y} + v_z\frac{\partial v_x}{\partial z} = - \frac{1}{\rho} \frac{\partial p}{\partial x} + \nu \Big(\frac{\partial^2 v_x}{\partial x^2} + \frac{\partial^2 v_x}{\partial y^2} + \frac{\partial^2 v_x}{\partial z^2}\Big)$ The reader should check that indeed each terms has a dimension equal to $\frac{m}{s^2}$ . Now in order to obtaint the non-dimensional equation we make the following substitutions. $p = p^* {V^2 \rho}$ Quantities with a superscript star are dimensionless quantities and the capital letters V and H have dimensions $\frac{m}{s}$ and respectively. Please check that these substitutions are indeed valid. For instance the physical length is 4 m. We decide to take the length H equal to 1 m, giving a non-dimensionalised length of 4. Using the above substitutions we end up with the following equation: $\frac{\partial v_x^* V}{\partial t^* T} + v_x^* V\frac{\partial v_x^* V}{\partial x^* H} + v_y^* V\frac{\partial v_x^* V}{\partial y^* H} + v_z^* V\frac{\partial v_x^* V}{\partial z^* H}$ $= - \frac{1}{\rho} \frac{\partial p^* V^2 \rho}{\partial x^* H } + \nu \Big(\frac{\partial^2 v_x^* V}{\partial {x^}^2 H^2} + \frac{\partial^2 v_x^ V}{\partial {y^}^2 H} + \frac{\partial^2 v_x^ V}{\partial {z^}^2 H^2}\Big)$ We simplify this equation and use that T = H/V. $\frac{\partial v_x^}{\partial t^} + v_x^ \frac{\partial v_x^}{\partial x^} + v_y^* \frac{\partial v_x^}{\partial y^} + v_z^\frac{\partial v_x^}{\partial z^}$ $= - \frac{\partial p^}{\partial x^} + \nu \frac{1}{V H} \Big(\frac{\partial^2 v_x^}{\partial {x^}^2} + \frac{\partial^2 v_x^}{\partial {y^}^2} + \frac{\partial^2 v_x^}{\partial {z^}^2}\Big)$ Here $\frac{\nu}{V H}$ is our inverse reynolds number. $\frac{\partial v_x^}{\partial t^} + v_x^ \frac{\partial v_x^}{\partial x^} + v_y^* \frac{\partial v_x^}{\partial y^} + v_z^\frac{\partial v_x^}{\partial z^}$ $= - \frac{\partial p^}{\partial x^} + \frac{1}{Re} \Big(\frac{\partial^2 v_x^}{\partial {x^}^2} + \frac{\partial^2 v_x^}{\partial {y^}^2} + \frac{\partial^2 v_x^}{\partial {z^*}^2}\Big)$ Where Re is defined as $\frac{V H}{\nu}$ . Now we have non-dimensionalised the NS equations. Please rememeber that if you solve this equation you will end up with the nondimensionalised quantities. In order to revert them to physical quantities you will need to use the equation we proposed above when we non-dimensionalised the quantities. Good luck! Regards, Vincent DoHander, nimav, shahrooz.omd and 7 others like this.

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