The energy-conserving full-depth <lock-exchange flow> is symmetric between the cold lower current and warm upper return current, so
$$
\boxed{h=H/2}.
$$
Part (b) then gives $u_2=\sqrt{g'H}$ and the laboratory front speed
$$
\boxed{U_f=\frac12\sqrt{g'H}}.
$$
During one opening, the cold current displaces the volume
$$
\boxed{V_{\rm ex}=WhU_f\,\delta t
=\frac{WH\,\delta t}{4}\sqrt{g'H}}.
$$
The same volume of warm air exits in the upper layer.

Solution

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

 Solution

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