 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-345/3/c/solution

The weaker plume detrains into the middle layer at

h_{2}

, while the stronger plume passes through it and feeds the upper layer at

h_{1}

. Middle-layer volume balance is

Q_{2} (h_{2}) = Q_{1} (h_{1}) - Q_{1} (h_{2}) .

Using

Q_{i} (z) = C_{Q} F_{i}^{1/3} z^{5/3}

gives

\frac{h _{2}}{h _{1}} = [1 + (\frac{F _{2}}{F _{1}})^{1/3}]^{- 3/5} .

The middle layer receives buoyancy flux

F_{2}

and loses volume

Q_{2} (h_{2})

by entrainment into the strong plume. The upper layer ultimately receives both source buoyancy fluxes and loses volume

Q_{1} (h_{1})

through the ceiling. Hence

g_{10}^{'} = \frac{F _{2}}{Q _{2} ( h _{2} )}, g_{20}^{'} = \frac{F _{1} + F _{2}}{Q _{1} ( h _{1} )} .

The interior-minus-exterior pressure is constant below

h_{2}

, rises with slope

ρ_{0} g_{10}^{'}

in the middle layer, and with slope

ρ_{0} g_{20}^{'}

in the upper layer. Put

B = g_{10}^{'} (h_{1} - h_{2}) + g_{20}^{'} (H - h_{1}) .

Equal ideal openings split the total pressure difference equally, and the common ventilation rate is

Q_{1} (h_{1}) = C_{d} A_{f} B = C_{d} A_{c} B .

Thus the required equal vent areas are

A_{f} = A_{c} = \frac{Q _{1} ( h _{1} )}{C _{d} g _{10}^{'} ( h _{1} - h _{2} ) + g _{20}^{'} ( H - h _{1} )} .

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