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

Table of contents
- Solution d

Solution

The energy-conserving full-depth lock-exchange flow is symmetric between the cold lower current and warm upper return current, so

h = H /2 .

Part (b) then gives

u_{2} = g^{'} H

and the laboratory front speed

U_{f} = \frac{1}{2} g^{'} H .

During one opening, the cold current displaces the volume

V_{ex} = Wh U_{f} δ t = \frac{W H δ t}{4} g^{'} H .

The same volume of warm air exits in the upper layer.

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