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The volumetric coefficient of thermal expansion is the fractional volume increase per temperature increase at fixed pressure and composition:The mass density form requires fixed mass $m$. Its units are ${\mathrm{K}}^{-1}$; ${\alpha }_T$ is another notation for this coefficient. For small temperature changes $\theta$ at fixed pressure,For a finite change in a single thermodynamic phase, integrate the possibly varying coefficient: $\ln (V_2/V_1)={\int }_{T_1}^{T_2}\alpha (T,P,N)dT$. A constant-coefficient approximation gives an exponential, while the displayed linear formula additionally requires $|\alpha \theta |\ll 1$.

For an ideal gas at fixed particle number,Integrating $1/T$ recovers the exact fixed-pressure thermal expansion ratio $V_2/V_1=T_2/T_1$. Combining expansion with isothermal compressibility gives the local response of any smooth simple compressible system of fixed composition,The ratio has units $\mathrm{Pa}{\mathrm{K}}^{-1}$. For the one-mole ideal gas state in isothermal compressibility, it is $P_0/T_0=333.33\mathrm{Pa}{\mathrm{K}}^{-1}$: a $1\mathrm{K}$ increase at fixed volume raises the pressure by about $333.33\mathrm{Pa}$. Maxwell relations identify this same ratio as the entropy slope with respect to volume and give $C_P-C_V=TV{\alpha }^2/{\kappa }_T$.

Under the Boussinesq approximation, the small fractional mass density change produces buoyancy while mass density is otherwise treated as constant. A positive temperature excess $\theta$ gives reduced gravity $g^{\prime }=g{\alpha }_T\theta$ in the convention used in natural ventilation. The coefficient therefore converts a heat budget into a buoyancy budget. The same measurable expansion response supplies entropy changes and heat capacity differences through Maxwell relations, connecting fluid motion to equilibrium thermodynamics.

An effective opening area may absorb both the discharge coefficient and the factor of two in an orifice law. If $A_e=\sqrt{2}{C}_d{A}_{\mathrm{geom}}$, its discharge is $q=A_e\sqrt{\delta p/\rho }$. Two equal openings carrying the same flux share the total pressure drop, so $q^2=A_e^2\Delta p/(2\rho )$. They do not share equal drops when an internal volume source makes their fluxes unequal. State the effective-area convention before comparing numerical factors in natural ventilation models.

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.

Use the well-mixed natural ventilation model for a distributed floor source. Let the building height be $H_b$, its volume be $\mathcal{V}$, its temperature excess be $\theta$, and the exterior mass density be $\rho$. Let ${\alpha }_T$ be the coefficient of thermal expansion, so reduced gravity is $g^{\prime }=g{\alpha }_T\theta$. Convert the total heat input $Q$ to the temperature-volume flux $\mathcal{H}=Q/(\rho c_p)$, where $c_p$ is specific heat capacity. If the quoted heat flux is per floor area, first multiply it by that area.

For effective opening area for pressure-driven ventilation, define the equal-opening hydraulic coefficient $K_A$ such that the signed throughflow $q$ obeys $q|q|=K_A\Pi$, with $\Pi$ the total driving pressure divided by mass density. In the effective-area convention of question 6, each opening has discharge $A\sqrt{\delta p/\rho }$, so $K_A=A^2/2$. The usual physical-area orifice law instead gives $K_A=C_d^2{A}^2$; that prefactor does not affect the stability conclusions.

Set $W=\Delta P/\rho >0$ and $G_T=g{\alpha }_T{H}_b$. Positive $q$ denotes entry at the floor and exit at the roof. hydrostatic pressure and wind combine to giveBecause the inflowing air is at exterior temperature in either direction, the heat budget isThe well-mixed ventilation temperature balance at steady state is thereforeAll roots must satisfy $\theta >0$ and their appropriate flow-direction inequality; squaring a pressure relation without imposing those conditions could add spurious branches.

The assisting case has a unique positive solution and it is stable. For opposing wind, define ${\theta }_w=W/G_T$. There is always one solution with $\theta >{\theta }_w$, a buoyancy-dominated upward flow. Below ${\theta }_w$, the heat-removal curve is $R(\theta )=\sqrt{K_A}\theta \sqrt{W-G_T\theta }$. It is zero at both endpoints, and its maximum occurs at $\theta =2{\theta }_w/3$:For $0<\mathcal{H}<R_{\max }$ there are two downward-flow roots as well as the upward-flow root. Equivalently, setting $y=G_T\theta /W$ and $\epsilon =G_T\mathcal{H}/(\sqrt{K_A}{W}^{3/2})$, the equilibrium curve isFor linear stability analysis of the opposing-wind ventilation bistability, linearize the heat budget: the eigenvalue is $-R^{\prime }(\theta )/\mathcal{V}$. The cooler downward root has $0<\theta <2{\theta }_w/3$ and $R^{\prime }>0$, so it is stable. The warmer downward root has $2{\theta }_w/3<\theta <{\theta }_w$ and $R^{\prime }<0$, so it is unstable and separates the two basins. The upward root has $R^{\prime }>0$ and is stable. At $\mathcal{H}=R_{\max }$ the two downward roots merge at a saddle-node bifurcation; above that value only the upward equilibrium remains.

For ventilation switching under wind and heating changes, increasing wind pressure or decreasing heat input both reduce $\epsilon$, so both can create the pair of wind-driven equilibria. They nevertheless have different physical and transient effects. On the hot upward branch, implicit differentiation of $\mathcal{H}=\sqrt{K_A}\theta \sqrt{G_T\theta -W}$ shows that increasing $W$ increases $\theta$, while decreasing $\mathcal{H}$ decreases $\theta$. Both reduce the upward ventilation rate, but only a wind change shifts ${\theta }_w$.

Under slow parameter variation, a state follows its current stable branch while that branch persists; the upward branch does not disappear when wind increases or heating decreases. A sufficiently large abrupt wind increase can raise ${\theta }_w$ beyond the current temperature and place that state below the new unstable threshold, causing flow reversal and cooling towards the wind-dominated equilibrium. A jump insufficient to cross its basin boundary returns to the upward equilibrium.

At fixed $W$, reducing a still-positive heat input cannot force an initially upward-flow state through ${\theta }_w$: at that boundary the temperature balance has $\dot{\theta }=\mathcal{H}/\mathcal{V}>0$. Thus heat reduction alone does not cause that reversal in this model. Conversely, decreasing wind or increasing heat from the stable downward branch eventually destroys it at the fold and forces a switch to the upward branch, giving history dependence. Wind strengthening and heat reduction have the same effect on the steady dimensionless control parameter, but not the same temperature change or switching dynamics. These stability statements use quasi-steady opening flow and a single well-mixed temperature; other stratification or airflow-inertia models require their own stability analysis.

For an ideal two-layer natural ventilation transient following a heating change $Q\mapsto \lambda Q$, let $V_0$ and ${\theta }_0$ be the initial flow and temperature excess, $s=V_{0}t/(A_{r}H)$, $\eta =h/H$, and $x=\eta (1-\theta /{\theta }_0)$. The integrated buoyancy pressure gives $V/V_0=\sqrt{1-x}$, while heat balance gives $x_s=\sqrt{1-x}-\lambda$. A material interface satisfies ${\eta }_s=\sqrt{1-x}$, so $\eta =x+\lambda s$. For $0<\lambda <1$, $x$ increases toward $1-{\lambda }^2$ and ${\eta }_s\ge \lambda$, ensuring finite-time filling of the room. After filling, $y=\theta /{\theta }_0$ obeys $y_s=\lambda -y^{3/2}$ and tends to ${\lambda }^{2/3}$. This model assumes well-mixed individual layers, negligible wall heat capacity and a sharp advected interface.