Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 345 / 3 / f

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- Solution f

Solution

Assume the two identical openings and their wall plumes behave symmetrically and do not interact before reaching the interface. If

Q_{V}^{(1)}

is the one-way rate through either vent, the global buoyancy balance is

B = 2 Q_{V}^{(1)} g_{H}^{'} .

Each wall plume therefore has buoyancy flux per unit span

B = B / (2 L)

. The steady volume balance now includes two descending plumes:

C_{P} B^{1/3} h^{5/3} = 2 Lα (z_{o} - h) (\frac{B}{2 αL})^{1/3} .

Hence the interface height is determined implicitly by

C_{P} h^{5/3} = 2^{2/3} α^{2/3} L^{2/3} (z_{o} - h), z_{o} = z_{V} + \frac{H ^{'}}{2 α} .

For completeness, if each opening retains the same single-opening exchange flow coefficient

K_{V}

, then

Q_{V}^{(1)} = (\frac{K _{V}^{2} B}{2})^{1/3}, Q_{V, total} = 2 Q_{V}^{(1)} = 2^{2/3} (K_{V}^{2} B)^{1/3} .

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