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Non-dimensionalizing Navier Stokes

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Old   November 21, 2011, 08:29
Default Non-dimensionalizing Navier Stokes
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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.
v_x = v_x^* V

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

t = t^* T

x = x^* H

y = y^* H

z = z^* H

Quantities with a superscript star are dimensionless quantities and the capital letters V and H have dimensions \frac{m}{s} and m 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
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Old   November 21, 2011, 10:31
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Hi Vincent, why dont you post this into the wiki.
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Old   November 21, 2011, 21:08
Default the force term problem
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if there are force term in the NS equation, how to deal with it??

for example, when use the immersed boundary method to solve the flow past fixed circular, what should i do with the force term nondimension???
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Old   November 22, 2011, 06:03
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In the case of an additional body force (like gravity) the equation changes to include an additional term.
\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) + \rho g_x

Again dividing by \rho yields this term indeed as an acceleration \Big(\frac{m}{s^2}\Big).

If we then substitute in the following way:

g_x = g_x^* \frac{V^2}{H}

We end up with the following 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{\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) + g_x^*

So for instance if we want to simulate gravity where g_x = 9.8 \frac{m}{s^2}. Then we first determine g_x^* by multiplying by H and dividing by V^2.

For instance, let's say we took H = 0.4 m and V = 2 m/s, this gives that g_x^* is equal to 9.8*0.4/4 is 0.98.

Good luck!

Regards,

Vincent
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Old   November 22, 2011, 06:40
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A simple immersed boundary method can be implemented in the following way. The philosophy behind the idea is that we will determine the fluid flow without the obstacle and the in a next step force the fluid flow to zero using a body force.

Introducing a phase indicator function \alpha that can either be one (solid-phase) or zero (fluid-phase). By multiplying our body force with this phase indicator function we only imply a force where our obstacle is.

However we can't use a simple constant for the forcing term g_x^* since we want this opposing force to set the velocity exactly to zero and not a positive or negative value.

So we need a parametrization for our body force. The forcing in a certain cell can be expressed in the following form:

f = \frac{\alpha_{i,j,k}}{\Delta t}({v_x^*}_{i,j,k}-{v_d^*}_{i,j,k})

Here \alpha_{i,j,k} is our phase indicator function as defined above. \Delta t is our numerical timestep {v_x^{*}}_{i,j,k} the velocity in a certain cell at the current timestep before forcing. Our desired velocity is {v_d^*}_{i,j,k} which is equal to zero in the case of a solid. Combining these elements gives that our full 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{\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) + \frac{\alpha_{i,j,k}}{\Delta t}({v_x^*}_{i,j,k}-{v_d^*}_{i,j,k})

Now we need to supply a geometry field (\alpha) and the desired velocity field (which for simple cases will be zero everywhere).

Good luck!

Regards,

Vincent
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Old   May 3, 2012, 08:28
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Dimensionless Navior-Stoks in cylindrical coordinate :



non-dimensional parameters which we use here:



for r-component we have:




Last edited by asal; May 3, 2012 at 10:35.
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Old   May 3, 2012, 10:39
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for theta-component, we have:



and finally for z-component, we have:

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Old   January 24, 2013, 22:03
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Hi VincentD, could you tell me how to choose V and H ???
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Old   November 25, 2013, 06:00
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Hi,
I need to non dimensionalize energy equation of navier stokes.
Can you help me out
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