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VincentD November 21, 2011 09:29

Non-dimensionalizing Navier Stokes
 
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

praveen November 21, 2011 11:31

Hi Vincent, why dont you post this into the wiki.

jessy November 21, 2011 22:08

the force term problem
 
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???

VincentD November 22, 2011 07:03

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

VincentD November 22, 2011 07:40

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

asal May 3, 2012 08:28

Dimensionless Navior-Stoks in cylindrical coordinate :

http://i49.tinypic.com/24o66wn.jpg

non-dimensional parameters which we use here:

http://i47.tinypic.com/5cbmuv.jpg

for r-component we have:

http://i49.tinypic.com/21oaid4.jpg

http://www.cfd-online.com/Forums/dat...AAAElFTkSuQmCC

asal May 3, 2012 10:39

for theta-component, we have:

http://i47.tinypic.com/av6879.jpg

and finally for z-component, we have:

http://i48.tinypic.com/2q19gcy.jpg

o_mars_2010 January 24, 2013 23:03

Hi VincentD, could you tell me how to choose V and H ???

spark999 November 25, 2013 07:00

Hi,
I need to non dimensionalize energy equation of navier stokes.
Can you help me out

engahmed9221 August 21, 2015 12:39

Non-dimensionalization of NV equations in cylindrical coordinates
 
Quote:

Originally Posted by asal (Post 359064)
Dimensionless Navior-Stoks in cylindrical coordinate :

http://i49.tinypic.com/24o66wn.jpg

non-dimensional parameters which we use here:

http://i47.tinypic.com/5cbmuv.jpg

for r-component we have:

http://i49.tinypic.com/21oaid4.jpg

http://www.cfd-online.com/Forums/dat...AAAElFTkSuQmCC

Please, Thank you for the great explanation. I have a question, how didn't you non-dimensionalized theta?! and is it okay to express the azimuthal (tangential) velocity with the same reference parameter as the axial and radial velocities? thank you!

FMDenaro August 21, 2015 13:14

theta is not dimensionally homogeneous to a lenght that needs to be non-dimensionalized but it is an angle position (rad).

engahmed9221 August 21, 2015 17:08

Thank You! but what will be wrong with the below dimensionless parameters?
I considered theta as a reference variable to dimensionalize the tangential coordinate

I also expressed the tangential dimensionless velocity considering the angular velocity as a reference variable.


http://s10.postimg.org/72y0uw6t5/dim...parameters.png

FMDenaro August 21, 2015 17:18

Quote:

Originally Posted by engahmed9221 (Post 560570)
Thank You! but what will be wrong with the below dimensionless parameters?
I considered theta as a reference variable to dimensionalize the tangential coordinate

I also expressed the tangential dimensionless velocity considering the angular velocity as a reference variable.


http://s10.postimg.org/72y0uw6t5/dim...parameters.png


No, theta ranges from 0 to 2*pi is already a non-dimensional number, furthermore, the velocity reference is unique

engahmed9221 August 21, 2015 17:26

Quote:

Originally Posted by FMDenaro (Post 560575)
No, theta ranges from 0 to 2*pi is already a non-dimensional number, furthermore, the velocity reference is unique

Thank you Prof.

I can see now how theta is already dimensionless. I am still a little bet confused about how the tangential velocity can be expressed in terms of a the unique reference velocity (Vr) which has [ L / T ] dimensions while omega r have a dimension [ T ^-1] ?

Thank you so much

FMDenaro August 21, 2015 17:33

Quote:

Originally Posted by engahmed9221 (Post 560577)
Thank you Prof.

I can see now how theta is already dimensionless. I am still a little bet confused about how the tangential velocity can be expressed in terms of a the unique reference velocity (Vr) which has [ L / T ] dimensions while omega r have a dimension [ T ^-1] ?

Thank you so much

but v_theta is dimensionally homogenous, remember that the velocity magnitude is omega*r

engahmed9221 August 21, 2015 17:40

Quote:

Originally Posted by FMDenaro (Post 560579)
but v_theta is dimensionally homogenous, remember that the velocity magnitude is omega*r

I GOT IT FINALLY !! Thanks a lot.

FMallaco September 14, 2015 08:45

It should be noted this seems to be a derivation for an incompressible flow.


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