Equivalent of fully-developed boundary condition in polar coordinate

mostanad · February 22, 2019, 11:20

Hello Dear friends,
I am working on the flow over the computational domain similar to the attached file. In this regard, the polar coordinates has been defined for the field. I should implement the fully developed boundary condition ( $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}=0$ ) at the outlet ( $\Gamma_2$ ). The problem is that I can't find its equivalent in polar coordinates. Indeed, I should describe this BC based on derivatives of

and $u_\theta$ and radial and angular directions. Can anybody help me in this issue?

FMDenaro · February 22, 2019, 11:40

just consider that v = i*u+j*v=ir*ur+itheta*vtheta then project first along i and then along j and derive in x

mostanad · February 22, 2019, 14:16

Quote:

Originally Posted by FMDenaro

just consider that v = i*u+j*v=ir*ur+itheta*vtheta then project first along i and then along j and derive in x

Thanks, But it doesn't work for me. Could you please give me the exact output of this procedure? Please write it based on polar spatial parameters.

mostanad · February 22, 2019, 14:36

Quote:

Originally Posted by FMDenaro

just consider that v = i*u+j*v=ir*ur+itheta*vtheta then project first along i and then along j and derive in x

In fact, I obtained the following equations based on your suggested procedure. I have implemented these equations in output region ( $\Gamma_2$ ):

$\frac{\partial u_r}{\partial r}cos^2\theta-\frac{\partial u_\theta}{\partial r}sin\theta cos\theta-\frac{\partial u_r}{\partial \theta}\frac{sin\theta cos\theta}{r}+\frac{u_r sin^2\theta}{r}+\frac{\partial u_\theta}{\partial \theta}\frac{ sin^2\theta}{r}+u_\theta \frac{ sin\theta cos\theta}{r}=0$

$-\frac{\partial u_r}{\partial r}sin\theta cos\theta-\frac{\partial u_\theta}{\partial r}cos^2\theta+\frac{\partial u_r}{\partial \theta}\frac{sin^2\theta}{r}+\frac{u_r sin\theta cos\theta}{r}+\frac{\partial u_\theta}{\partial \theta}\frac{ sin\theta cos\theta}{r}-u_\theta \frac{ sin^2\theta}{r}=0$

FMDenaro · February 22, 2019, 14:40

Quote:

Originally Posted by mostanad

Thanks, But it doesn't work for me. Could you please give me the exact output of this procedure? Please write it based on polar spatial parameters.

Your issue is that the assumption of fully developed flow along x does not match the geometry of your described outlet.

To see that, you can work on the projection of the velocity gradient along the normal direction to the boundary, that is n= ir. Then insert the assumptions:

ir*Gradu =ir*( i du/dx+j du/dy)=ir*j du/dy

ir*Gradv =ir*( i dv/dx+j dv/dy)=ir*j dv/dy

You see that further assumptions on du/dy and dv/dy are needed.

mostanad · February 22, 2019, 14:46

Quote:

Originally Posted by FMDenaro

Your issue is that the assumption of fully developed flow along x does not match the geometry of your described outlet.

To see that, you can work on the projection of the velocity gradient along the normal direction to the boundary, that is n= ir. Then insert the assumptions:

ir*Gradu =ir*( i du/dx+j du/dy)=ir*j du/dy

ir*Gradv =ir*( i dv/dx+j dv/dy)=ir*j dv/dy

You see that further assumptions on du/dy and dv/dy are needed.

Why are you working with two coordinates systems simultaneously? Why do you think in this way? I only want to find a relationship between velocity gradients in two systems.

FMDenaro · February 22, 2019, 14:52

Quote:

Originally Posted by mostanad

Why are you working with two coordinates systems simultaneously? Why do you think in this way? I only want to find a relationship between velocity gradients in two systems.

The representation of the velocity gradient (as well as any vector field) in cartesian and polar coordinates can be used since they are equal. You expressed the assumptions d()/dx=0 only in the cartesian coordinates so that you need to link those to the polar representation.

mostanad · February 22, 2019, 14:57

Quote:

Originally Posted by FMDenaro

The representation of the velocity gradient (as well as any vector field) in cartesian and polar coordinates can be used since they are equal. You expressed the assumptions d()/dx=0 only in the cartesian coordinates so that you need to link those to the polar representation.

So what is the problem with du/dy?

FMDenaro · February 22, 2019, 15:05

Quote:

Originally Posted by mostanad

So what is the problem with du/dy?

In a Cartesian system you use a straight outlet line having as normal unit vector n=i and the assumption of fully developed flow is n.Grad () =0 on this line. How can you assume n.Grad () =0 on your line Gamma2??

mostanad · February 22, 2019, 15:08

Quote:

Originally Posted by FMDenaro

In a Cartesian system you use a straight outlet line having as normal unit vector n=i and the assumption of fully developed flow is n.Grad () =0 on this line. How can you assume n.Grad () =0 on your line Gamma2??

Perhaps by transforming gradient and

to polar coordinate. Am it right?

FMDenaro · February 22, 2019, 15:11

Quote:

Originally Posted by mostanad

Perhaps by transforming gradient and

to polar coordinate. Am it right?

Indeed you have to write the equivalence between the gradient expressed in Cartesian and polar coordinates. The you compute the component in polar coordinates after having inserted the conditions d()/dx in the gradient written in Cartesian coordinates. You end up to have the need to further assumptions

mostanad · February 22, 2019, 15:20

Quote:

Originally Posted by FMDenaro

Indeed you have to write the equivalence between the gradient expressed in Cartesian and polar coordinates. The you compute the component in polar coordinates after having inserted the conditions d()/dx in the gradient written in Cartesian coordinates. You end up to have the need to further assumptions

I think you can help me. I don't have any further idea on this issue. Could you please write the exact procedure for me?
Thanks

FMDenaro · February 22, 2019, 16:08

Quote:

Originally Posted by mostanad

I think you can help me. I don't have any further idea on this issue. Could you please write the exact procedure for me?
Thanks

It is a standard vector calculus

Grad f = i df/dx + j df/dy = ir df/dr + (itheta/r) df/dtheta

From that you have the relation between the two systems of coordinates

mostanad · February 22, 2019, 16:10

Quote:

Originally Posted by FMDenaro

It is a standard vector calculus

Grad f = i df/dx + j df/dy = ir df/dr + (itheta/r) df/dtheta

From that you have the relation between the two systems of coordinates

So what about projection?

FMDenaro · February 22, 2019, 16:16

First of all, you need to set the correct numerical outflow section at a certain distance along x. If you consider Gamma2 you cannot set n.Grad () =0 along it. The flow develop only along the streamwise direction x

mostanad · February 22, 2019, 16:34

Quote:

Originally Posted by FMDenaro

First of all, you need to set the correct numerical outflow section at a certain distance along x. If you consider Gamma2 you cannot set n.Grad () =0 along it. The flow develop only along the streamwise direction x

I can't understand.

FMDenaro · February 22, 2019, 16:44

See the sketch

mostanad · February 22, 2019, 16:59

Quote:

Originally Posted by FMDenaro

See the sketch

But the straight outlet cannot be imposed in polar coordinate. You said that $m.\nabla()\neq0$ at radial outlet. Why? We can sense it in physical approach. In every node over $\Gamma_2$ , we have a streamwise velocity that doesn't change along streamwise. Your first suggestion was so logical for me. Indeed, $v=u i+v j=u_r e_r+ u_\theta e_\theta$ . Then by $v.i=u=(u_r e_r+ u_\theta e_\theta).i=(u_r e_r+ u_\theta e_\theta).(-cos(\theta) e_r+ sin(\theta) e_\theta)$ , we can find $\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(-u_r cos(\theta)+u_\theta sin(\theta))$ . Afterward, by using chain rule. we can find a relation based on polar parameters. But it didn't work for me.

FMDenaro · February 23, 2019, 04:30

Quote:

Originally Posted by mostanad

But the straight outlet cannot be imposed in polar coordinate. You said that $m.\nabla()\neq0$ at radial outlet. Why? We can sense it in physical approach. In every node over $\Gamma_2$ , we have a streamwise velocity that doesn't change along streamwise. Your first suggestion was so logical for me. Indeed, $v=u i+v j=u_r e_r+ u_\theta e_\theta$ . Then by $v.i=u=(u_r e_r+ u_\theta e_\theta).i=(u_r e_r+ u_\theta e_\theta).(-cos(\theta) e_r+ sin(\theta) e_\theta)$ , we can find $\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(-u_r cos(\theta)+u_\theta sin(\theta))$ . Afterward, by using chain rule. we can find a relation based on polar parameters. But it didn't work for me.

in polar coordinates, the condition ir. Grad () = d()/dir= 0 would imply that the function does not vary along the radial direction. Do you think that such condition is physically acceptable along Gamma2? That strongly depends on the radial extension of the computational domain as well as on the flow problem you are simulating.

mostanad · February 23, 2019, 05:47

Quote:

Originally Posted by FMDenaro

in polar coordinates, the condition ir. Grad () = d()/dir= 0 would imply that the function does not vary along the radial direction. Do you think that such condition is physically acceptable along Gamma2? That strongly depends on the radial extension of the computational domain as well as on the flow problem you are simulating.

No, That is completely wrong. If you consider d()/dir= 0, by looking at theta=90, you will find out that the flow direction is not acceptable. As I said, d()/dx= 0 is completely correct. But I don't know how to transform it into polar coordinates?

February 22, 2019, 11:20	Equivalent of fully-developed boundary condition in polar coordinate	#1
mostanad Senior Member mohammad Join Date: Sep 2015 Posts: 274 Rep Power: 11	Hello Dear friends, I am working on the flow over the computational domain similar to the attached file. In this regard, the polar coordinates has been defined for the field. I should implement the fully developed boundary condition ( $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}=0$ ) at the outlet ( $\Gamma_2$ ). The problem is that I can't find its equivalent in polar coordinates. Indeed, I should describe this BC based on derivatives of and $u_\theta$ and radial and angular directions. Can anybody help me in this issue? Last edited by mostanad; February 22, 2019 at 11:26. Reason: wrong edition

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
sliding mesh problem in CFX	Saima	CFX	46	September 11, 2021 07:38
Radiation in semi-transparent media with surface-to-surface model?	mpeppels	CFX	11	August 22, 2019 07:30
Constant mass flow rate boundary condition	sahm	OpenFOAM	0	June 20, 2018 22:45
Wrong flow in ratating domain problem	Sanyo	CFX	17	August 15, 2015 06:20
several fields modified by single boundary condition	schröder	OpenFOAM Programming & Development	3	April 21, 2015 05:09

February 22, 2019, 11:40		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	just consider that v = iu+jv=irur+ithetavtheta then project first along i and then along j and derive in x

February 22, 2019, 16:16		#15
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	First of all, you need to set the correct numerical outflow section at a certain distance along x. If you consider Gamma2 you cannot set n.Grad () =0 along it. The flow develop only along the streamwise direction x