Conservation of volume flux through a circular arc of radius $r$ gives
$$
\alpha r(-u_r)=\alpha Q,
$$
so the outer flow is the <sink flow in a sector>
$$
\boxed{u_r=-\frac Qr,
\qquad u_\theta=0,
\qquad \psi=-Q\theta}.
$$
This is a <potential flow> and already satisfies the <no-penetration boundary condition> on the radial walls. It has nonzero tangential speed $Q/r$ there, however, and therefore violates the <no-slip boundary condition>. Thin vorticity-bearing <boundary layer>[boundary layers] are needed to reduce that speed to zero at each wall.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 38B / b / i / Solution

A line sink of strength $\alpha Q$ at the vertex of a sector $0<\theta<\alpha$ produces the irrotational outer flow
$$
u_r=-\frac Qr,
\qquad u_\theta=0,
\qquad \psi=-Q\theta.
$$
It satisfies no penetration at the radial walls but violates no slip there, so viscous boundary layers form along both walls.

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