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mahzad November 6, 2012 02:51

Wall shear stress
 
Hi every one
As you know, there are two kind of forces acting on the cylinder surface,
pressure force and viscous force.
pressure forces act inward normal to the surface ,
but my question is about the direction of viscous forces, I know that they are tangential to the surface but I want to know about the direction? I have this question even about the airfoil case.
Does anybody know the answer to my question? or does anybody know any book or references which can help me with my problem?

Thanks in advance

mb.pejvak November 7, 2012 00:17

total viscous force is outcome of all friction force in all cells. you can not certainly say exact direction of the force and I think it differ from case to case.
I think you can find your answer in " fundamental of aerodynamics" by J.D. Anderson in section 1-5.
If you can not find it, let me know to sed it to you.

mahzad November 7, 2012 03:50

1 Attachment(s)
Quote:

Originally Posted by mb.pejvak (Post 390716)
total viscous force is outcome of all friction force in all cells. you can not certainly say exact direction of the force and I think it differ from case to case.
I think you can find your answer in " fundamental of aerodynamics" by J.D. Anderson in section 1-5.
If you can not find it, let me know to sed it to you.

Dear Pejvak

Thank you for replying my question, I checked this book, it was a really good suggestion, and came in handy. But as you see from the picture which I've attached from this book, page17 third edition, you can see that the author have specified a certain direction for shear stress both on upper and lower surface of the cylinder. So now, my question is about the way the author determined the direction of the tangential forces on the airfoil surface, I can not understand why this is so?
also I have the same question about a circular cylinder when the flow is unsteady over it. I mean when it has natural shedding...how can we specify the direction of shear stress on the cylinder surface?
I hope I expressed what I mean clearly.

Thanks for your time and help
regards,

mb.pejvak November 7, 2012 04:24

Dear Mahzad;
I can not understand what you mean, but if you mean how the direction of friction force is determined? first of all, this figure you mentioned is schematically shows the friction and pressure force and these forces depend on coordination (x,y) and due to this reason, we called CF(x). lift and drag forces have specific directions (parallel and perpendicular to Vinf), but for friction and pressure forces are tangential and normal to the surface in each points.
I hope I can convey my purpose.

michujo November 7, 2012 04:34

Hi, the total force due to skin friction on the body surface is the "sum" of all the friction force acting on each surface element of the body (so you have to integrate the friction force vector over the whole surface to obtain the resultant friction force). The direction of each elementary skin friction force contribution is, as you just said, tangential to the body. If you know the geometry of your body you know what is the direction of a vector tangential to the body at every point on its surface, therefore you know what are the components in the x and y axis (cos(angle) and sin(angle) respectively, right?).

\vec{F}_{friction}=F_x \cdot \vec{i}+F_y \cdot \vec{j}=\int{\tau \cdot d\vec{A}} \approx \sum{blabla}

Is this what you were asking?

Cheers,
Michujo.

mahzad November 7, 2012 06:25

Quote:

Originally Posted by michujo (Post 390738)
Hi, the total force due to skin friction on the body surface is the "sum" of all the friction force acting on each surface element of the body (so you have to integrate the friction force vector over the whole surface to obtain the resultant friction force). The direction of each elementary skin friction force contribution is, as you just said, tangential to the body. If you know the geometry of your body you know what is the direction of a vector tangential to the body at every point on its surface, therefore you know what are the components in the x and y axis (cos(angle) and sin(angle) respectively, right?).

\vec{F}_{friction}=F_x \cdot \vec{i}+F_y \cdot \vec{j}=\int{\tau \cdot d\vec{A}} \approx \sum{blabla}

Is this what you were asking?

Cheers,
Michujo.

Dear Michujo
I know that the friction force is tangential to the body but what I mean, is that how can we specify the sign of this direction? as you see from the picture which I have attached the friction forces are tangential to the surface, but what about the direction? why there are not in the reverse direction tangential to the surface? how can we determine this direction?
that's my question!

thanks for your time and help
Regards,

mahzad November 7, 2012 06:31

Quote:

Originally Posted by mb.pejvak (Post 390732)
Dear Mahzad;
I can not understand what you mean, but if you mean how the direction of friction force is determined? first of all, this figure you mentioned is schematically shows the friction and pressure force and these forces depend on coordination (x,y) and due to this reason, we called CF(x). lift and drag forces have specific directions (parallel and perpendicular to Vinf), but for friction and pressure forces are tangential and normal to the surface in each points.
I hope I can convey my purpose.

dear Pejvak

my question is that these forces are in what direction tangential to the surface? how can we determine this direction?

Regards,

michujo November 7, 2012 06:42

Hi, sorry I misunderstood you. The direction of the friction force is completely determined by the product of the stress tensor and the vector normal to the surface.

I hope it helps.

Cheers,
Michujo.

mahzad November 7, 2012 07:13

Quote:

Originally Posted by michujo (Post 390763)
Hi, sorry I misunderstood you. The direction of the friction force is completely determined by the product of the stress tensor and the vector normal to the surface.

I hope it helps.

Cheers,
Michujo.

Dear michujo

Assume we are calculating the lift force for an airfoil, we must consider both the contribution of pressure and shear stresses. the pressure in each point is given, and as result Cp is easily calculated. We can use the Tw=gama*(du/dy)_y=0, Cf=Tw/q formulations to calculate Cf. so now we don't have any stress tensor here. how can we specify this direction?

Again thanks for your help and time
Regards,

michujo November 7, 2012 07:43

Hi, have a look at this thread. A similar question was posed (for a channel flow though), but I think the same explanation applies.

http://www.cfd-online.com/Forums/mai...lus-minus.html

You know the value of du/dy at every point on your surface, right? you also know the direction of the vector normal to the surface.

mahzad November 7, 2012 08:47

Quote:

Originally Posted by michujo (Post 390776)
Hi, have a look at this thread. A similar question was posed (for a channel flow though), but I think the same explanation applies.

http://www.cfd-online.com/Forums/mai...lus-minus.html

You know the value of du/dy at every point on your surface, right? you also know the direction of the vector normal to the surface.

Dear michujo

Thank you for your good guidance, it was really helpful and now I understand how can we specify the direction of shear FORCES on the body surface.
but regarding what one of the member has said in the thread you introduced, a question arised to me.
FMDenaro: "The wall stress is proportional to vorticity at wall, therefore is correct to have same magnitude and opposite sign"

I understand this sentence, because I myself have plotted the vorticity and skin friction coefficient on a cylinder surface, the magnitude of them at the upper and lower surface of the cylinder are the same, but with opposite sign.

As you know we can calculate the shear stress from the following formulation,
Tw=gama*du/dy

there is no contribution of normal vector of the surface in this formulation, so the reason for the difference in the sign of skin friction coefficient(or vorticity) at the upper and lower surface of the cylinder, might be another thing, and do you think the normal vector of the surface is a logical reason for the difference in sign?!!!


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