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

praveen · November 21, 2011, 10:31

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

jessy · November 21, 2011, 21:08

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, 06:03

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

$\frac{m}{s^2}$ . Then we first determine

by multiplying by H and dividing by

.

For instance, let's say we took H = 0.4 m and V = 2 m/s, this gives that

is equal to 9.8*0.4/4 is 0.98.

Good luck!

Regards,

Vincent

VincentD · November 22, 2011, 06:40

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

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 :

non-dimensional parameters which we use here:

for r-component we have:

asal · May 3, 2012, 10:39

for theta-component, we have:

and finally for z-component, we have:

o_mars_2010 · January 24, 2013, 22:03

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

spark999 · November 25, 2013, 06:00

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

engahmed9221 · August 21, 2015, 12:39

Quote:

Originally Posted by asal

Dimensionless Navior-Stoks in cylindrical coordinate :

non-dimensional parameters which we use here:

for r-component we have:

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.

FMDenaro · August 21, 2015, 17:18

Quote:

Originally Posted by engahmed9221

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.

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

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

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

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

I GOT IT FINALLY !! Thanks a lot.

AdarshC · July 10, 2017, 04:55

How to non dimensionalise if we have to include surface tension and obtain Capillary number in the final equation

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.

November 21, 2011, 21:08	the force term problem	#3
jessy New Member wuzj Join Date: Mar 2011 Posts: 6 Rep Power: 15	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???

November 22, 2011, 06:40		#5
VincentD New Member Vincent Join Date: Jul 2011 Posts: 29 Rep Power: 14	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 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 randolph likes this.

May 3, 2012, 08:28		#6
asal Senior Member Astio Lamar Join Date: May 2012 Location: Pipe Posts: 186 Rep Power: 13	Dimensionless Navior-Stoks in cylindrical coordinate : non-dimensional parameters which we use here: for r-component we have: onya and masoud.k like this. Last edited by asal; May 3, 2012 at 10:35.

May 3, 2012, 10:39		#7
asal Senior Member Astio Lamar Join Date: May 2012 Location: Pipe Posts: 186 Rep Power: 13	for theta-component, we have: and finally for z-component, we have: onya and masoud.k like this.

November 21, 2011, 10:31		#2
praveen Super Moderator Praveen. C Join Date: Mar 2009 Location: Bangalore Posts: 342 Blog Entries: 6 Rep Power: 18	Hi Vincent, why dont you post this into the wiki.

November 22, 2011, 06:03		#4
VincentD New Member Vincent Join Date: Jul 2011 Posts: 29 Rep Power: 14	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 $\frac{m}{s^2}$ . Then we first determine by multiplying by H and dividing by . For instance, let's say we took H = 0.4 m and V = 2 m/s, this gives that is equal to 9.8*0.4/4 is 0.98. Good luck! Regards, Vincent

January 24, 2013, 22:03		#8
o_mars_2010 Member Osman Join Date: Oct 2012 Location: Japan Posts: 53 Rep Power: 13	Hi VincentD, could you tell me how to choose V and H ???

November 25, 2013, 06:00		#9
spark999 New Member teja Join Date: Nov 2013 Posts: 2 Rep Power: 0	Hi, I need to non dimensionalize energy equation of navier stokes. Can you help me out

August 21, 2015, 13:14		#11
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	theta is not dimensionally homogeneous to a lenght that needs to be non-dimensionalized but it is an angle position (rad).

August 21, 2015, 17:08		#12
engahmed9221 New Member Ahmed Aql Join Date: Aug 2015 Location: Kuwait Posts: 4 Rep Power: 10	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.

July 10, 2017, 04:55	Surface tension	#17
AdarshC New Member Adarsh Choudhury Join Date: Jul 2017 Posts: 1 Rep Power: 0	How to non dimensionalise if we have to include surface tension and obtain Capillary number in the final equation

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