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Consider an incompressible viscous fluid of kinematic viscosity $ν$ , dynamic viscosity $µ$ and density $ρ$ . A viscous boundary layer is located over a solid surface at $y = 0$ and $x > 0$. The flow in the boundary layer must match with the constant inviscid bulk velocity $U$. The steady two-dimensional boundary-layer equations are $$\frac{∂ u}{ ∂ x} + \frac{∂ v}{ ∂ y} = 0$$ $$ρ \bigg( u \frac{∂ u }{∂ x} +v \frac{∂ u }{∂ y} \bigg)= µ \frac{∂ ^2u }{∂ y^2}$$ where $u(x,y)$ is the horizontal velocity and $v(x,y)$ the vertical velocity.

(i) State the boundary conditions on $u$ and $v$ along $y = 0$ and $x > 0$.

(ii) State the far-field condition on $u$ as $y→∞$.


It says "The flow in the boundary layer must match with the constant inviscid bulk velocity $U$" so does this mean that the resultant of the $u$ and $v$ velocity components should be equal to $U$?

Then for (i) $u(x,y=0)=v(x,y=0)=U/\sqrt2$?

For (ii), is the far field condition just $u \rightarrow 0$ as $y \rightarrow \infty$?

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  • $\begingroup$ If $y=0$ is at the solid surface, what are the velocities there (hint: no slip)? Likewise, at $y\rightarrow\infty$ the streamwise velocity is $U$ and the velocities must be continuous, so what are the values of $u$ and $v$? If still unsure, try to draw how you think the fluid velocity profile develops for $x\ge0$. $\endgroup$ – nluigi May 10 '16 at 20:00
  • $\begingroup$ @nluigi $u(x,0)=v(y,0)=0$? $\endgroup$ – snowman May 10 '16 at 20:13
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From your question, I guess that you don't really visualize what the flow geometry is. It's true that there's one unclear bit in the exercise: I guess it should read "The flow in the boundary layer must match with the constant inviscid bulk velocity $\vec{U} = (U, 0)$". The overall geometry is a flow meeeting a plate which is aligned with it (0 incidence).

In your answer to @nluigi, you got correctly $u(x,0)=v(x,0)=0$ for $x\geq 0$.

I think with the above phrasing, the answer to (ii) becomes obvious.

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