Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 65 / 2

Write $\theta =T-T_e$ and use the Boussinesq approximation for air, ${\rho }_e-\rho \simeq \rho \theta /T_e$. A uniformly heated, well-mixed room has the buoyancy pressure difference $\Delta p_b=\rho gH\theta /T_e$ between its openings. Let $C_o$ denote the combined hydraulic coefficient of the two equal openings, so the effective opening area for pressure-driven ventilation givesFor example, if each orifice has conventional discharge coefficient $C_d$ in $V=C_{d}A\sqrt{2\Delta p/\rho }$, the two pressure losses add and $C_o=C_d$. The question's coefficient is the prefactor in the final heat-load law; define $c=(2C_o^2{)}^{1/3}$. This explicitly states the numerical loss-coefficient convention. Steady heat balance is $Q=\rho C_{p}V\theta$. Eliminating $\theta$ givesHere $Q$ is total heating power and $C_p$ is specific heat capacity at constant pressure. Writing $c=C_d$ without this conversion would incorrectly change the power of the discharge coefficient.

After the reduction, cooler newly heated air occupies the bottom and displaces the old warmer air upward. Use the ideal two-layer displacement ventilation model: each layer is well mixed, the upper layer remains at ${\theta }_0$, walls have negligible heat capacity, and the sharp interface is advected without exchange between the layers. Let the lower layer have depth $h$ and excess temperature $\theta$. Integrating the hydrostatic buoyancy over the full height givesThe room's excess thermal energy is $E_T=\rho C_p{A}_r[h\theta +(H-h){\theta }_0]$. The floor supplies $\lambda Q$, incoming air has zero excess temperature, and the ceiling removes $\rho C_{p}V{\theta }_0$. Thus the required heat balance isThis moving-interface energy term must include the loss of old upper-layer volume; balancing only the lower-layer temperature would miss it.

For a genuine reduction assume $0<\lambda <1$. SetThe integrated buoyancy factor is $H{\theta }_0(1-x)$, so $V/V_0=\sqrt{1-x}$. Since $Q=\rho C_p{V}_0{\theta }_0$, the thermal balance becomes the ventilation after a heating reduction equationThe material interface also moves with the upward displacement velocity, $A_r\dot{h}=V$. Therefore ${\eta }_s=\sqrt{1-x}$, and subtracting the two equations givesThe second relation is also the lower-layer energy balance: its total excess energy increases at rate $\lambda Q$. It does not assume its temperature stays constant. Since $x_s>0$ until $x=1-{\lambda }^2$, the solution remains $0\le x<1-{\lambda }^2$, and $V/V_0\ge \lambda$. Hence ${\eta }_s\ge \lambda$, so the lower layer reaches $\eta =1$ in finite time $s_H\le 1/\lambda$. Also ${\eta }_s\le 1$, so $s_H\ge 1$.

An explicit implicit solution is useful. Put $y=\sqrt{1-x}$, which decreases from $1$ toward $\lambda$. Separation givesFilling occurs when $1=1-y_H^2+\lambda s_H$, so $\lambda s_H=y_H^2$ and ${\theta }_H/{\theta }_0=y_H^2$. These relations determine the exit time and the continuous temperature at that time. For complete switch-off $\lambda =0$, the limiting solution is $y=1-s/2$, $x=\eta =s-s^2/4$ for $0\le s\le 2$, so the cold lower layer fills the room at $s_H=2$ even though its final ventilation rate is zero.

After the interface exits, the whole room is well mixed. For $z=\theta /{\theta }_0$, the natural ventilation rate is $V=V_0\sqrt{z}$, and the well-mixed ventilation temperature balance becomesEquivalently, $\rho C_p{A}_{r}H\dot{\theta }=\lambda Q-\rho C_p{C}_{o}A\sqrt{gH/T_e}{\theta }^{3/2}$. The heat-removal term is strictly increasing in $\theta \ge 0$, so every nonnegative solution tends to the unique stable equilibriumFor $\lambda =0$ the final temperature is $T_e$. The unchanged-load case $\lambda =1$ has no distinct new temperature layer and is outside the reduction scenario.