Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2 / 36B (Fluid Dynamics II)

The outward flux through a spherical cap of radius $R$ inside the cone isThus the inward extraction flux is $Q$ at leading order. The thin no-slip wall layer is omitted from this approximation; its smaller displacement-flux effect supplies the later correction.

Using local speed $A/R^2$ and length $R$, the Reynolds number isInertia dominates and a thin viscous boundary layer is possible when this is large, namely $R\ll A/\nu$. The crossover scale is $R\sim Q/[2\pi (1-\cos \alpha )\nu ]$, or $Q/\nu$ up to the fixed-angle factor. The inverse dependence on $R$ explains why this inertial sink regime occurs near the tip rather than far away. The fixed cone angle must also allow the wall layer to be small relative to the local cone radius.

Let $x$ increase away from the tip along the wall and $y$ point normally into the fluid. Put $u=u_x$, $v=u_y$ for velocity components and $U(x)=-A/x^2$ for the outer tangential velocity. The local Cartesian momentum and pressure equations areFor a two-dimensional flat layer the usual continuity equation is ${\partial }_{x}u+{\partial }_{y}v=0$. For this axisymmetric cone, however, the circumference is proportional to $x$, so its leading continuity equation isThe circumference term is of the same order as ${\partial }_{x}u$ and cannot be omitted when deriving the cone's similarity equation. Conditions at the stationary wall are $u=v=0$; at the layer edge $u\to U$.

Balance $U^2/x\sim \nu |U|/{\delta }^2$. With $|U|=A/x^2$ this givesThe choice of multiplicative constant fixes the normalization of the similarity variable. To enforce axisymmetric continuity set $xu={\Psi }_y$, $xv=-{\Psi }_x$, withThenDifferentiation gives $u{\partial }_{x}u+v{\partial }_{y}u=-(A^2/x^5)[2F^{\prime 2}+F{F}^{\prime \prime }/2]$, whereas $U{U}^{\prime }=-2A^2/x^5$ and $\nu u_{yy}=-A^2{F}^{\prime \prime \prime }/x^5$. Consequently the conical sink boundary layer satisfiesThe first condition is no penetration, the second no slip, and the last matches the inward core flow. If one instead makes a strictly planar approximation and omits the circumference term, the streamfunction is $-\sqrt{A\nu }{x}^{-1/2}F$ and the equation becomes $F^{\prime \prime \prime }+F{F}^{\prime \prime }/2+2(1-F^{\prime 2})=0$. The thickness exponent is the same, but that planar equation is not the axisymmetric cone equation.

Across a cap near $R=x$, the inward flux in a layer truncated at $y=Y$ is, to leading order,To match to the core, use the finite displacement-flux deficit rather than integrate the uniform outer flow to infinite $Y$. If $I={\int }_0^{\infty }(1-F^{\prime })d\eta ={\lim }_{\eta \to \infty }(\eta -F)$, thenConservation of the fixed total extraction rate requires an extra core flux of this order. Dividing by cap area, proportional to $R^2$, provesThe correction is a relative $O(\sqrt{\nu R/A})$ contribution. Its angular structure requires outer matching; the requested radial power follows from the displacement flux alone. Its cap-average direction increases the inward core speed to compensate for the wall deficit.