In steady inviscid two-dimensional flow the spanwise vorticity obeys
$$
\mathbf u\mathbin{\cdot}\nabla\omega
=\left[\nabla(\rho'/\rho_0)\times\mathbf g\right]_y.
$$
Integrating this equation over the front-frame control volume converts the left side to vorticity flux through its upstream and downstream faces. For a sharp interface, the baroclinic source integrates to the circulation generated by the hydrostatic pressure jump, $g'h$. With plug flow downstream, the resulting balance is
$$
\frac12u_2^2=g'h,
\qquad
\boxed{u_2=\sqrt{2g'h}}.
$$
Volume conservation in the front frame gives $u_1H=u_2(H-h)$. The undisturbed indoor air is stationary in the laboratory, so $u_1$ is the front speed:
$$
\boxed{U_f=\frac{H-h}{H}\sqrt{2g'h}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 345 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-345/3/b/solution