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Old   August 8, 2006, 12:47
Default boundary layer
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jj
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dear cfd researchers, in boundary layer equation (flat plate) the pressure gradient is zero , then how the flow is sustained?

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Old   August 8, 2006, 14:51
Default Re: boundary layer
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haifa
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only the pressure gradient perpendicular to the wall!!!
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Old   August 9, 2006, 03:33
Default Re: boundary layer
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Tom
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It's sustained by the relative motion of the plate (stationary) to the freestream (moving). The motion in the freestream is sustained simply because it is inviscid, uniform and unidirectional (Newtons first law).

NOTE: there are visous corrections to the pressure at higher order in the infinite Reynolds number expansion.
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Old   August 10, 2006, 02:34
Default Re: boundary layer
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Ram Dayal
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Hi, Pressure gradient is assumed to be zero in transverse direction only (perpendicular to plate), and it is not zero in streamwise direction. Bye Ram Dayal
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Old   August 10, 2006, 03:57
Default Re: boundary layer
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Tom
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"Pressure gradient is assumed to be zero in transverse direction only (perpendicular to plate), and it is not zero in streamwise direction."

No! In the case of the flat plate boundary layer (Blasius) the pressure gradient IS zero in the streamwise direction. The inviscid outer flow has v=0(=w) and u constant => (Bernoulli) that p is constant. Within the boundary-layer it then follows that dp/dx=0 to lead order in the inverse Reynolds number expansion.
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Old   August 10, 2006, 20:50
Default Re: boundary layer
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Ram Dayal
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Hi Tom, Would you mind explaining it more elaborately? Thanks in advance. Bye Ram Dayal
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Old   August 11, 2006, 03:56
Default Re: boundary layer
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Tom
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The inviscid flow past a flat plate at zero incidence is simply u=U(const) v=0 => p=P(const) (i.e. there is zero pressure gradient in the inviscid flow). Now consider the viscous equations at high Reynolds number R. Since the Reynolds number is large we can seek a solution as a power series in the small parameter 1/R. Now this type of series can never satisfy the noslip condition on the surface since the lead order term fails to satisfy it and adding in smaller terms cannot rectify it. The reason for this is is simply that the noslip conditions arise from the viscous terms which have been neglected to lead order and must be reinstated. To do this rescale the normal coordinate Y=y/a and observe that the coefficient of (1/R)d^2u/dy^2 becomes unity when a = R^{-1/2}. Noting that continuity requires v = O(R^{-1/2}) we are led to Prandtl's boundary layer equations after again seeking a power series solution in 1/R of the new scaled equations:

uu_x + vu_Y = -P_x + u_YY, p_Y =0, u_x + v_Y =0.

The second equation shows that p is a function of x alone (i.e. the normal pressure gradient is zero).

The solution of these equations must satisfy appropriate boundary conditions (and one initial condition) the noslip condition u=v=0 on y=0 are two obvious ones. For the third boundary condition we require that the inviscid outer solution and viscous boundary layer solution agree in some region of common overlap. If we write the outer solution in terms of Y instead of y and expand as R->infinity we see that u->U, p->P as Y-> infinity from within the boundary layer. However since p_Y=0 we must have p=P throughout the boundary layer and hence p_x=0. This closes the problem and allows u and v to be determined within the boundary layer once an initial condition at some x station has been supplied => Blasius solution.

If you proceed to higher order in the 1/R expansion you will obtain a pressure gradient correction due to the displacement of fluid form the wall due to boundary layer growth (v is nonzero in the boundary layer).

Note that the above argument is quite general and can be applied to problems other than the flat plate where p_x is nonzero.

There's a section in "Elementary fluid dynamics" by D.J. Acheson that discusses this type of analysis.

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Old   August 11, 2006, 04:00
Default Re: boundary layer
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Ram Dayal
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Hi, Tom Thanks for Details. Bye Ram Dayal
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