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 A.T. August 16, 1999 20:14

Non-slip Boundary condition

Hi everybody,

Can anybody explain why non-slip boundary condition from the physical (maybe atom) point of view please?

Say, we want to calculate the flow field of the mercury through a tube made of glass, can we still use this non-slip boundary condition? thanks.

 John C. Chien August 17, 1999 00:35

Re: Non-slip Boundary condition

(1). try to look up the properties of mercury in a handbook first. (2). If it says the viscosity is zero, then you can use slip boundary condition at the wall. (3). Otherwise, if it has a viscosity value, then use the non-slip boundary condition. (4). If you can't find it, then run two cases with slip and non-slip conditions. And compare the results with the test data.

 Tae-Chang Jo August 17, 1999 13:00

Re:Non-slip Boundary condition

Hi. Could anyone tell me what the slip boundary condition is? Thank you.

 Patrick Godon August 17, 1999 14:11

Re: Non-slip Boundary condition

For mercury the surface tension is extremely high and this has also to be taken into account. For example if you try to 'inject' mecury into a medium using a syringe, this one might just break. Also, a little amount of mercury in a tube would rather 'roll' rather than flow. So if you're talking about a small glass tube, you will certainly need a non-slip boundary condition (viscosity, friction) and some treatment for the surface tension. Check the viscosity of mercury, the velocity of its flow and find out (using simple books about the theory of the boundary layer) whether the boundary layer will be much smaller than the size of the glass tube (slip BC) or of the same scale (non-slip BC).

As to your question related to the microscopic scale of the friction, it is a whole field of Physics!

 John C. Chien August 17, 1999 20:38

Re:Non-slip Boundary condition

(1).For a stationary wall ( or ground ), a non-slip condition is that the fluid velocity at the wall point is zero.(because of the existence of the viscosity) (2). In some analyses, the fluid viscosity is removed from the governing equations, and the resultant equation is INVISCID. (that is non-viscous). (3). For inviscid flows, we can not apply the non-slip boundary condition at the wall because there is not viscosity by assumption. (4). In the inviscid flow case, the fluid velocity at the wall is allowed to slip along the contour of the surface (move on the surface and tangent to the surface). This condition is called "slip condition".

 Tae-Chang Jo August 18, 1999 16:05

Re:Non-slip Boundary condition

Thank you for your answering. I'd like to make sure that I understand well. For inviscid flows, the normal component of the velocity is zero at the boundary - this is what your saying, isn't it?

Then can the tangental velocity be any? Is there any condition such as the derivative in tangential direction is zero?

Thank you again.

 prishor November 28, 2012 00:26

Quote:
 Originally Posted by Tae-Chang Jo ;4652 Thank you for your answering. I'd like to make sure that I understand well. For inviscid flows, the normal component of the velocity is zero at the boundary - this is what your saying, isn't it? Then can the tangental velocity be any? Is there any condition such as the derivative in tangential direction is zero? Thank you again.
derivative of the tangential velocity will be zero for hydrodynamically(fully) developed flows only.it cannot be zero for boundary layer and entry length problems

regards,
prishor

 michujo November 28, 2012 03:19

Quote:
 Originally Posted by prishor (Post 394546) derivative of the tangential velocity will be zero for hydrodynamically(fully) developed flows only.it cannot be zero for boundary layer and entry length problems regards, prishor
Hi, I have to disagree with that. A null derivative of the tangential velocity component relative to the direction normal to the wall is a no-slip condition (zero shear stress at the wall). It does not correspond to a fully developed flow, check for instance the Poiseuille solution for laminar flow in a pipe and you'll readily see that it is not zero.

Cheers.

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