Why Isn't dv/dy Zero in Stoke's First Problem?

[member 428835]

hey pf! please read this, it's honestly pretty easy although long to read.

in stoke's first problem, we have parallel flow initially at rest and then, instantaneously (i know, unrealistic) one wall starts to move in the x-direction with no acceleration (yes, it magically becomes a velocity). the y-direction is perpendicular to the x direction and is directing toward the other wall. the z-direction is coming "out of the page".

okay, now time for the question.

conservation of mass for incompressible flows yields: $$ \nabla \cdot \vec{U}=0 \Rightarrow \frac{dv}{dy}=0$$ since \frac{du}{dx}=\frac{dw}{dz} are identically zero. my question is, why isn't \frac{dv}{dy} identically zero? i know it equals zero from the above, but how does it exist at all? how is there velocity in the y direction?

it makes sense to me that the other two are zero by inspection, but why not y? to me it seems \frac{du}{dy} would not be zero but that all others would be. please help me understand how there exists y-direction velocity (y )

thanks ahead of time!

 

[njoysci]

Your intuition and majority of statements are accurate, just a little variable terminology confusion:
U = u i + v j + w k
For your example: 0<= u <= V where V = wall velocity; v = 0; w = 0
Therefore ∂v/∂x = ∂v/∂y = ∂v/∂z = 0 and ∂w/∂x = ∂w/∂y = ∂w/∂z = 0
∇ dot U = ∂u/∂x + ∂v/∂y + ∂w/∂z = 0 (conservation of mass for incompressible flow)
Since flow is constant in the x direction for a given y, then ∂u/∂x = 0
So indeed ∇ dot U = 0 based upon the above partials.
In summary, since v is the fluid velocity in the y direction, it is zero. Hence ∂v/∂y = 0.
The only fluid flow is in the x direction, and since u varies as a function of y and is not constant, then ∂u/∂y is not zero. The molecules of each horizontal layer collide and change the molecular velocity in the adjacent horizontal layer. For instance for a typical Newtonian fluid
∂u/∂y = - fluid shear stress / viscosity