Session 1
Stack Effect Fundamentals: Hydrostatic Pressure Difference and the Neutral Pressure Plane
Objectives
- Derive the hydrostatic pressure difference that drives stack effect from the density difference between warm indoor air and cool outdoor air.
- Locate the neutral pressure plane in a building section and explain how its position sets inflow below and outflow above.
- Explain why stack effect grows with building height and with the indoor-outdoor temperature difference.
- Predict the direction of air movement through a given opening in winter and in summer from its position relative to the neutral plane.
Buoyancy-driven ventilation is powered by a simple fact: air is a fluid, and its density depends on its temperature. When the air inside a building is warmer than the air outside, the indoor air is lighter, so it rises. That rising motion has to be fed from below and vented from above, and so a continuous circulation establishes itself: cool outdoor air enters low in the building, warms up from occupants and equipment, rises through the occupied volume, and leaves through openings near the top. No fan is involved. The energy that moves the air is the heat already present in the building.
The driving force can be quantified exactly from hydrostatics. Consider two vertical columns of air standing side by side, one outdoors at temperature To and one indoors at temperature Ti, both extending from a datum at height H above an opening to that opening. The pressure at the base of each column is its weight per unit area, which for a uniform column is rho * g * H. Because the warm column is less dense, it exerts less pressure at its base than the cold column does. The difference between the two base pressures is the stack pressure, and it is what pushes air through the opening. Written per unit height, the stack pressure available at a vertical distance H from the level where indoor and outdoor pressures are equal is:
DeltaPstack = rhoo * g * H * (Ti - To) / Ti
- DeltaPstack
- hydrostatic pressure difference between outdoors and indoors at height H, in pascals (Pa)
- rhoo
- outdoor air density, kg/m3 (commonly taken as 1.2 kg/m3 at near-standard conditions)
- g
- gravitational acceleration, 9.81 m/s2
- H
- vertical distance from the neutral pressure plane to the opening considered, in meters (m)
- Ti
- indoor absolute temperature, kelvin (K)
- To
- outdoor absolute temperature, kelvin (K)
A useful equivalent form replaces the density ratio with a direct density difference: DeltaPstack = g * H * (rhoo - rhoi), which says the same thing, that the available pressure equals the weight per unit area of the density difference over the height H. Both forms make two things obvious: the driving pressure is proportional to height, and it is proportional to the temperature difference normalized by the absolute temperature. The absolute-temperature denominator matters. A 10 K difference between inside and outside produces the same stack pressure in absolute terms whether the outdoor temperature is -10 degrees C or +20 degrees C, as long as the indoor temperature is the same, because the denominator Ti barely changes. What changes between winter and summer is the sign and the typical magnitude of the difference.
The level where indoor and outdoor static pressures are equal is the neutral pressure plane (NPP). Below the NPP, outdoor pressure exceeds indoor pressure, so air flows inward; above the NPP, indoor pressure exceeds outdoor pressure, so air flows outward. In the simplest case, a building with one opening low and one opening high, the NPP sits somewhere between them, its exact position determined by the relative sizes of the openings (Session 3 develops this quantitatively). If the two openings have equal effective area, the NPP sits roughly midway between them. If the lower opening is much larger than the upper one, the NPP drops toward the lower opening, meaning most of the available stack pressure is spent across the small upper outlet.
Tall buildings and winter amplify the effect for two independent reasons. Height enters directly: double the vertical separation between openings and you double the stack pressure. Winter enters through the temperature difference: a building held at 22 degrees C against an outdoor temperature of 0 degrees C sees a 22 K difference, roughly four times the difference available on a 30 degrees C day with the indoor space at 24 degrees C. Together, a tall building in cold weather can generate tens of pascals of driving pressure, enough to ventilate aggressively through modest openings, and enough to cause real problems: cold drafts at ground-floor entries, difficult-to-open doors, and unwanted infiltration through the envelope. In summer the difference reverses sign and shrinks; with indoor air only slightly cooler than outdoor air, stack pressures drop to a few pascals or less, and buoyancy alone often cannot ventilate adequately, which is why wind assistance or mechanical backup is usually considered for warm-climate design.
One practical consequence follows immediately. Any opening's flow direction can be read off a section drawing once the NPP is placed: openings below it are inlets, openings above it are outlets, and an opening exactly at the NPP has no driving pressure and contributes nothing. This is why a window left ajar at mid-height does little for stack ventilation, while a low vent and a high vent of the same size move a useful volume of air. It also explains why ground-floor lobbies in tall buildings feel drafty in January: the lobby sits far below the NPP, where the inward driving pressure is largest.
Worked example: stack pressure in a 4-storey office in winter
Given: indoor temperature Ti = 21 degrees C = 294 K; outdoor temperature To = 1 degree C = 274 K; vertical distance from the neutral plane to the roof vent H = 6 m; outdoor density rhoo = 1.24 kg/m3 (air at about 1 degree C and standard pressure).
Calculation: DeltaP = rhoo * g * H * (Ti - To) / Ti = 1.24 * 9.81 * 6.0 * (294 - 274) / 294 = 1.24 * 9.81 * 6.0 * 20 / 294 = 4.97 Pa. Roughly 5 Pa of driving pressure across the roof vent.
What it means: 5 Pa is a healthy driving pressure for natural ventilation, comparable to what a small fan stage might provide through low-resistance openings. Repeating the calculation for a July day with To = 32 degrees C and Ti = 24 degrees C (interior slightly cooler than exterior) reverses the sign and yields about -3 Pa with a comparable magnitude scaled by the smaller 8 K difference; the building now exhausts low and draws air in high, and the small pressure tells you buoyancy alone will not cool the space. This is the seasonal swing a designer must plan for.
Common mistakes
A frequent error is assuming the NPP is always at mid-height. It is not; it shifts toward the larger opening. Another is applying the stack equation to a volume that is not well mixed: in a stratified atrium the local driving pressure at a high outlet depends on the actual temperature at that level, not the room-average temperature. Finally, remember that the stack equation gives the maximum pressure difference across a single opening at height H from the NPP; the actual flow through a pair of openings depends on how that total pressure is divided between them, which Session 3 takes up.
Key takeaways
- Stack pressure comes from the weight of the density difference between indoor and outdoor air over a vertical height.
- Below the neutral pressure plane air flows in; above it air flows out; the plane's position depends on the relative opening areas.
- Driving pressure scales with height and with the absolute temperature difference, so tall buildings in cold weather see the strongest effect.
- Summer reverses the direction and shrinks the magnitude, which limits what buoyancy alone can do in warm conditions.
Session 2
Wind Plus Buoyancy: Pressure Coefficients and Combined Driving Pressure
Objectives
- Interpret wind pressure coefficients and use them to estimate the wind-induced pressure at any facade point.
- Compare the magnitudes of wind pressure and stack pressure for typical building geometries and weather conditions.
- Decide, for a given design case, whether wind or buoyancy dominates the ventilation driving force.
- Combine the two effects into a total driving pressure and account for opposing cases where wind fights the stack.
In practice, buoyancy almost never acts alone. Wind striking a building reshapes the pressure field around it: pressure rises on the windward face, drops along the side walls and over the roof, and settles somewhere in between on the leeward face. These local pressures can assist or oppose the stack effect, and on a breezy day they often dominate it entirely. Designing natural ventilation therefore means handling both drivers together.
The standard way to express wind-induced pressure is through the wind pressure coefficient, Cp. The coefficient is defined as the local pressure rise above the undisturbed static pressure, normalized by the dynamic pressure of the approaching wind:
Pw = Cp * rho * V2 / 2
- Pw
- wind-induced static pressure at a point on the building surface, relative to undisturbed ambient pressure, in pascals (Pa); positive means above ambient
- Cp
- wind pressure coefficient, dimensionless; typically positive on windward faces and negative (suction) on leeward faces, side walls, and roofs
- rho
- air density, kg/m3 (commonly taken as 1.2 kg/m3)
- V
- approach wind speed at a reference height (often roof height or 10 m meteorological height), in m/s
Published Cp values come from wind-tunnel tests and full-scale measurements compiled in handbooks and standards (for example, the ASHRAE Handbook of Fundamentals tabulates surface-averaged coefficients for common building shapes). Typical surface-averaged values for a simple rectangular block: about +0.6 to +0.8 on the windward face, about -0.3 to -0.5 on the leeward face, and about -0.6 to -0.8 on the roof. Local peak values near corners and roof edges are more extreme, but ventilation design uses area-averaged coefficients over the opening zones. What matters for airflow through an opening is the difference in Cp between the inlet zone and the outlet zone, multiplied by the dynamic pressure.
A quick comparison of magnitudes tells you which driver runs the show. Take a moderate breeze of V = 5 m/s: dynamic pressure is 0.5 * 1.2 * 25 = 15 Pa. With a windward-to-leeward Cp difference of about 1.0, wind alone can drive roughly 15 Pa across the building. Compare that with the 5 Pa stack pressure from Session 1's four-storey winter example. At 5 m/s the wind dominates; the stack contributes but does not set the flow. Now drop the wind to 1.5 m/s: dynamic pressure is only 1.35 Pa, and the same stack example gives nearly 4 times that. On calm days buoyancy rules; on breezy days wind rules. The crossover sits roughly where the two driving pressures are equal, which for common low-rise buildings happens at wind speeds of a few meters per second.
Because pressure is additive, the total driving pressure across a ventilation path is the algebraic sum of the wind and stack components, provided each is evaluated with a consistent sign convention. Define positive driving pressure as acting in the intended flow direction (outdoor to indoor at the inlet, indoor to outdoor at the outlet). Windward openings see positive wind pressure that assists inflow; leeward openings see suction that assists outflow; roof vents sit in suction that strongly assists stack exhaust. The dangerous case is wind on the face where the stack expects to exhaust, or more precisely, wind pressure that opposes the stack direction. If wind pushes into the high outlet with enough force to exceed the stack pressure there, the exhaust reverses and the building pressurizes from the top down, potentially pushing warm air out through unintended leaks low in the building. This is the classic "wind overpowering the stack" failure, and it is why stack exhausts are rarely placed on windward faces.
There is one more subtlety worth stating plainly: wind is unsteady. Gusts arrive in bursts, so the instantaneous driving pressure fluctuates around the mean, and so does the flow. Buoyancy, by contrast, changes slowly with temperature. A design that just barely works on mean pressures may stall or reverse during lulls and gusts. Conservative practice sizes openings against the weaker expected condition (for example, a hot, nearly calm day for a cooling-driven design) and uses adjustable openings, dampers, or wind cowls to keep control when conditions are favorable. A wind cowl or turbine vent on a roof exhaust turns wind from a variable nuisance into a reliable assistant: whatever direction the wind blows, the device presents suction to the exhaust stream.
Finally, note the different spatial character of the two drivers. Stack pressure varies linearly with height and is the same on all faces at a given level; it is a global, vertically distributed effect. Wind pressure varies with position around the building and is roughly uniform over the height of a low-rise block; it is a local, facade-dependent effect. This is why the two combine best when the ventilation path is short in the wind direction and tall in the vertical direction: cross-ventilation paths exploit facade-to-facade Cp differences, while stack paths exploit height. A building that uses both, with low windward inlets and high leeward or roof outlets, gets the two drivers pulling in the same direction, which is the most robust natural ventilation strategy there is.
Worked example: wind and stack compared on a two-storey school
Given: two-storey classroom block, vertical separation between low inlet and roof outlet H = 7 m; indoor 24 degrees C (297 K), outdoor 10 degrees C (283 K), rho = 1.23 kg/m3; wind V = 4 m/s; windward inlet Cp = +0.7, roof outlet Cp = -0.6.
Stack component: DeltaPstack = rho * g * H * (Ti - To) / Ti = 1.23 * 9.81 * 7.0 * 14 / 297 = 3.98 Pa, about 4.0 Pa.
Wind component: dynamic pressure = 0.5 * 1.23 * 16 = 9.84 Pa. DeltaCp = 0.7 - (-0.6) = 1.3. Wind driving pressure = 1.3 * 9.84 = 12.8 Pa.
Combined: approximately 4.0 + 12.8 = 16.8 Pa in the intended direction (both drivers assist, since the inlet is windward and the outlet is in roof suction).
What it means: wind contributes about three quarters of the total here. If the wind swung around so the outlet sat on the windward side, the wind term could oppose the stack; at 4 m/s the opposition (on the order of 10 Pa or more) would overwhelm the 4 Pa stack and reverse the intended flow. Orientation of openings relative to prevailing wind is therefore as important as their size.
Design note: never let wind fight the stack
The single most avoidable failure in combined natural ventilation is placing the exhaust where the wind pushes in. Keep high exhausts in zones of reliable suction (leeward slopes, ridge lines, roof monitors oriented with the prevailing wind, or rotating cowls), keep low inlets on windward faces where positive pressure assists entry, and check the opposing-wind case explicitly: if the design wind pressure against the exhaust exceeds the stack pressure, either relocate the exhaust, add a wind-driven assist device, or accept that the mode only works in calm conditions and provide an alternative. Designing only for the favorable wind direction is designing for half the year.
Key takeaways
- Wind pressure at a facade point is Cp times the dynamic pressure; the difference in Cp between inlet and outlet zones drives cross flow.
- Wind dominates buoyancy once wind speeds reach a few meters per second for typical low-rise buildings; buoyancy dominates on calm days.
- Total driving pressure is the algebraic sum of the wind and stack terms, so wind can assist or oppose the stack depending on orientation.
- Place inlets windward and exhausts in suction zones so both drivers pull the same way, and always check the opposing-wind case.
Session 3
Sizing Natural Ventilation Openings: The Orifice Equation and Effective Area
Objectives
- Apply the orifice equation to compute airflow through a ventilation opening from the pressure difference across it.
- Explain what the discharge coefficient represents and select appropriate values for common opening types.
- Combine an inlet and an outlet in series into a single effective area.
- Size an inlet/outlet pair for a target airflow given the available stack or combined driving pressure.
Driving pressure means nothing until it is converted into airflow, and the conversion device is the opening. Every vent, window, louver, and grille behaves, to a first approximation, like an orifice: flow through it is proportional to the opening area and to the square root of the pressure difference across it. This square-root relationship is the single most important quantitative fact in natural ventilation design. It means that doubling the driving pressure increases flow by only about 41 percent, while doubling the area doubles the flow. When a design falls short on air, adding area helps far more than chasing more pressure.
The orifice equation comes from Bernoulli's principle with a correction for real-world losses. An ideal frictionless opening would pass Q = A * sqrt(2 * DeltaP / rho). Real openings suffer a vena contracta (the jet necks down as it enters) plus friction and turbulence, so the actual flow is lower by a factor Cd, the discharge coefficient:
Q = Cd * A * sqrt(2 * DeltaP / rho)
- Q
- volumetric airflow rate through the opening, m3/s
- Cd
- discharge coefficient, dimensionless; accounts for contraction and friction losses in the opening
- A
- free (unobstructed) area of the opening, m2; for louvers and grilles this is the net free area, not the gross frame size
- DeltaP
- static pressure difference across the opening, Pa
- rho
- air density, kg/m3
The discharge coefficient typically ranges from about 0.6 to 0.7 for a sharp-edged orifice-like opening. A plain open window or a well-designed vent approaches the upper end, around 0.6 to 0.65 in many published tabulations. Louvers, insect screens, and grilles add resistance and lower the effective Cd, sometimes substantially; a louvered vent with a screen can behave as if its coefficient were 0.4 or less. Because flow is linear in Cd, misjudging it is the same as misjudging the area. When in doubt, use the net free area published by the louver or grille manufacturer together with the manufacturer's tested coefficient, and treat textbook 0.6 values as appropriate only for clean, unobstructed openings.
Real ventilation paths have at least two openings in series: an inlet and an outlet. The same air passes through both, so the flow is limited by the combination, and the total available pressure is split between them. The pair can be represented by a single effective area Aeff defined so that the orifice equation applied once, with the total pressure difference, gives the same flow. For two openings with areas A1, A2 and coefficients Cd1, Cd2:
1 / (Cd * Aeff)2 = 1 / (Cd1 * A1)2 + 1 / (Cd2 * A2)2
- Aeff
- effective area of the opening pair in series, m2
- Cd
- discharge coefficient applied to the effective area (often taken equal to a representative opening coefficient), dimensionless
- A1, A2
- free areas of the inlet and outlet openings, m2
- Cd1, Cd2
- discharge coefficients of the inlet and outlet openings, dimensionless
When both openings share the same coefficient, this simplifies to the familiar form 1/Aeff2 = 1/A12 + 1/A22. Two consequences deserve emphasis. First, the effective area is always smaller than the smaller of the two openings; the bottleneck rules. If one opening is much smaller than the other, the effective area is approximately the smaller area, and enlarging the larger one buys almost nothing. Second, for a fixed total opening area budget, the maximum flow comes from splitting the area equally between inlet and outlet (with equal coefficients). A balanced pair is the most area-efficient pair.
The vertical separation between inlet and outlet is what creates the stack pressure in the first place (Session 1), so it enters sizing twice: once in the driving pressure and once in the geometry. For pure stack ventilation with a balanced pair, the flow can be written directly in terms of the separation H between the openings' mid-heights:
Q = Cd * Aeff * sqrt(2 * g * H * (Ti - To) / Ti)
- Q
- volumetric airflow rate, m3/s
- Cd
- representative discharge coefficient, dimensionless
- Aeff
- effective area of the inlet/outlet pair in series, m2
- g
- gravitational acceleration, 9.81 m/s2
- H
- vertical separation between the mid-heights of the inlet and outlet openings, m
- Ti, To
- indoor and outdoor absolute temperatures, K
This combined form is the workhorse of stack-ventilation sizing. Notice what it rewards: taller separation, larger temperature difference, larger effective area, all under a square root except the area. If the target flow doubles, the required area roughly doubles (for fixed conditions), but getting twice the flow from height alone would require four times the separation, which is why generous vertical separation is cheap to provide at the drawing stage and nearly impossible to add later. This is the argument for high-level exhausts, open stairwells used as stack paths, and clerestories: height is the one parameter that costs nothing in energy.
Two practical cautions round out the picture. First, large openings at similar heights short-circuit: a pair of big windows at the same level on opposite walls gives fine cross-ventilation from wind but negligible stack flow, because H is near zero. Second, the neutral pressure plane (Session 1) divides the total stack pressure between the inlet and the outlet in inverse proportion to their effective areas; the equations above already account for this through Aeff, but the physical reading matters for comfort. The bulk of the pressure drop, and hence the highest local jet velocities, occurs across the smaller opening. If the smaller opening is the low inlet beside occupied desks, occupants will feel it as a draft even when the total flow is exactly as designed. Put the flow restriction where its jet will not strike people, typically at the high outlet.
Worked example: airflow through a balanced louver pair
Given: inlet louver net free area A1 = 0.8 m2, outlet louver net free area A2 = 0.8 m2, both with Cd = 0.5 (louver plus screen); vertical separation H = 5 m; Ti = 25 degrees C = 298 K, To = 15 degrees C = 288 K.
Effective area: 1/Aeff2 = 1/0.82 + 1/0.82 = 3.125, so Aeff = 1/sqrt(3.125) = 0.566 m2.
Airflow: Q = 0.5 * 0.566 * sqrt(2 * 9.81 * 5.0 * 10 / 298) = 0.283 * sqrt(3.291) = 0.283 * 1.814 = 0.513 m3/s, about 513 L/s.
What it means: a balanced pair of modest louvers delivers over 500 L/s under a mild 10 K difference with 5 m of separation. If the outlet were choked to 0.4 m2 while the inlet stayed at 0.8 m2, the effective area would fall to 0.357 m2 and the flow to about 324 L/s; the small outlet, not the large inlet, would be setting the flow, and the fast jet through it would be felt as a draft if it discharged near occupants.
Common mistakes
The most common sizing error is using gross frame area instead of net free area for louvers and grilles, which overstates the flow by the blockage fraction. Next is assuming Cd = 0.6 for a louvered, screened vent; use the manufacturer's tested data. Third is unbalanced pairs: a huge inlet with a tiny outlet (or vice versa) wastes most of the large opening's area. Finally, do not forget that insect screens, filters, and dampers left half-closed all count as area reductions; the as-built effective area is what matters, not the as-drawn one.
Key takeaways
- Flow through an opening follows Q = Cd * A * sqrt(2 * DeltaP / rho); area matters linearly, pressure only as a square root.
- Discharge coefficients run about 0.6 for clean openings and lower for louvers, screens, and grilles; use net free area, not gross.
- An inlet/outlet pair in series is governed by its effective area, which is always smaller than the smaller opening; balanced pairs are the most area-efficient.
- Vertical separation drives the stack pressure and is cheapest to secure at the drawing stage; keep the restriction away from occupied zones to avoid drafts.
Session 4
Atria and Tall Spaces: Stratification, Exhaust Placement, and Smoke-Control-Adjacent Thinking
Objectives
- Describe how warm air stratifies in tall volumes and why a single average temperature misrepresents the driving pressure.
- Explain how the neutral pressure plane shifts in tall, open-connected spaces and what that means for doorway flows.
- Select exhaust locations in atria and tall halls that work with stratification instead of against it.
- Distinguish everyday buoyancy ventilation from life-safety smoke control, and identify when a fire engineer must be engaged.
Everything in the first three sessions assumed a well-mixed room with one indoor temperature. Atria, auditoria, airport halls, and industrial high bays violate that assumption routinely. Warm air rising from occupants, lighting, and equipment does not mix evenly through a 15-meter volume; it stratifies, forming a warm layer near the ceiling that grows downward over time while cooler air remains near the floor. The temperature at the roof vent can be several kelvin warmer than the occupied zone, and it is the vent-level temperature, not the thermostat reading, that sets the stack pressure at the exhaust.
Stratification strengthens the stack effect at the top of the space. Recall from Session 1 that the driving pressure at an opening is proportional to the local density difference at that height. In a stratified atrium, the air leaving through a roof vent is warmer and lighter than the room average, so the effective (Ti - To) at the vent is larger than a mixed-room calculation would give. Using the occupied-zone temperature underestimates the available pressure and leads to oversized exhausts. The correct approach, even in hand calculations, is to work with a representative exhaust-air temperature: estimate or measure the temperature at the vent level under design conditions and use that in the stack equation. In energy models and CFD this happens automatically; in a spreadsheet it requires a deliberate choice, and the conservative direction for ventilation sizing is to use the lower, mixed temperature (which underestimates flow), while the conservative direction for overheating analysis is to use the higher, stratified temperature.
Tall connected spaces also move the neutral pressure plane problem from a single room to a whole building section. In an atrium open to several floors, the atrium acts as a vertical duct coupling them all. Warm air from every floor feeds the rising column, the NPP of the combined system can sit well above the mid-height of any individual floor, and doorways between floors and the atrium see cross-flows driven by the floor-by-floor temperature differences relative to the atrium. The practical symptoms are familiar: atrium-adjacent corridors that are mysteriously warm, doors that are hard to open against the pressure, and stack-driven air spilling from the atrium into upper floors. Mitigation is architectural as much as mechanical: vestibules or lobbies at atrium entries, doors kept closed (or held closed by closers rated for the pressure), and exhaust capacity at the atrium head sized for the combined load, not just the atrium's own gains.
Exhaust placement in tall spaces follows from the stratification physics. The warmest air is at the top, so exhausts belong at the highest practical point, at the apex of a skylight monitor or in the upper sidewalls just below the roof. Low or mid-height exhausts in a stratified volume discharge cooler air, wasting driving pressure and leaving the hot layer in place; they ventilate the wrong air. Inlets, conversely, belong low, where the coolest air sits, so that incoming air sweeps the occupied zone before joining the rising plume. The ideal tall-space section is therefore simple: low inlets, high exhausts, and nothing in between that short-circuits the column. Skylight vents that open at the ridge, roof monitors with operable sides, and high clerestories all serve this pattern. One caution: a high exhaust with a large area but no rain or wind protection will admit weather; gravity dampers, wind cowls, and rain-sensing actuators are part of the exhaust design, not accessories.
This is also where everyday buoyancy ventilation meets the neighboring discipline of smoke control, and the boundary needs to be stated carefully. The physics is the same physics: hot smoke rises, stratifies, and can be exhausted from high points, which is exactly what smoke exhaust systems do. But life-safety smoke control is governed by building and fire codes (for example, the International Building Code provisions and NFPA 92 on smoke control systems), requires engineered analysis, acceptance testing, and coordination with the fire alarm and sprinkler systems, and must be designed by qualified fire protection engineers. A natural ventilation scheme that happens to exhaust from the atrium head is not a smoke control system, must never be presented as one, and cannot be substituted for code-required smoke management. Where a project needs both, the right move is early engagement with the fire protection engineer so the architectural openings can serve both purposes under proper design: for instance, the same roof vents sized for daily buoyancy ventilation can be specified with fire-rated actuators and controls that the smoke control design takes over in alarm mode. The ventilation engineer provides the openings and the airflow intent; the fire engineer owns the life-safety design.
A final operational point: tall spaces with operable high exhausts need a control strategy that respects stratification. Opening everything on a mild day can over-ventilate the upper floors while the occupied level barely changes, because the stack is fed mostly by the warm upper layer. Staged opening (low inlets first, then progressive exhaust opening), temperature sensors at two or three heights, and simple rules such as "exhaust follows the upper sensor, inlets follow the occupied sensor" keep the system serving comfort rather than just moving air. Commissioning should verify the actual temperature profile under load, not just that the actuators move.
Worked example: stratified vs mixed temperature at an atrium vent
Given: atrium vent 12 m above the neutral plane; outdoor 18 degrees C (291 K); occupied-zone temperature 24 degrees C (297 K); measured vent-level temperature 29 degrees C (302 K) under design internal gains; rhoo = 1.21 kg/m3.
Mixed-room estimate: DeltaP = 1.21 * 9.81 * 12 * (297 - 291) / 297 = 1.21 * 9.81 * 12 * 6 / 297 = 2.88 Pa.
Stratified estimate: DeltaP = 1.21 * 9.81 * 12 * (302 - 291) / 302 = 1.21 * 9.81 * 12 * 11 / 302 = 5.19 Pa.
What it means: accounting for stratification nearly doubles the estimated driving pressure at the vent, from 2.9 Pa to 5.2 Pa. A design sized on the mixed temperature would oversize the exhaust by roughly the square root of the ratio (about 1.34 times the needed area). More importantly, it would misjudge where the system operates: the real building exhausts more air than the mixed model predicts, which is welcome for ventilation but can surprise an energy model that assumed less.
Design note: the atrium as an accidental chimney for the whole building
An atrium open to occupied floors couples their air systems together whether you planned it or not. Before finalizing any atrium ventilation scheme, trace the stack path floor by floor: check door pressure differentials against closer ratings, confirm that upper-floor spaces will not overheat from atrium spill, and verify that makeup air has a deliberate low-level path rather than leaking through the envelope. If the project triggers code smoke-control requirements, bring the fire protection engineer in during schematic design, when vent locations and areas can still serve both everyday ventilation and the engineered smoke strategy.
Key takeaways
- Warm air stratifies in tall volumes; the temperature at the exhaust, not the room average, sets the stack pressure.
- Atria couple connected floors into one stack system, shifting the neutral plane and driving inter-floor flows through doorways.
- Exhausts belong at the highest practical point, inlets low; mid-height exhausts ventilate the wrong air.
- Everyday buoyancy ventilation and code-required smoke control share physics but not design authority; never present one as the other.
Session 5
Solar Chimneys and Windcatchers: Enhancing Buoyancy with Solar and Wind
Objectives
- Explain how a solar chimney boosts stack flow by adding solar heat to the exhaust air stream.
- Explain how a windcatcher captures wind pressure and directs it into or out of a building.
- Identify the key sizing parameters for each device: collector area, chimney height, inlet and outlet areas.
- Judge climate suitability for each device from solar availability, wind regime, and diurnal temperature range.
Solar chimneys and windcatchers are both amplifiers. Neither creates airflow from nothing; each takes one of the natural drivers and strengthens it at the point where it matters. The solar chimney strengthens buoyancy by heating the air in the exhaust path. The windcatcher strengthens wind ventilation by reaching up into faster, cleaner air and presenting a favorable pressure to the building. Both are old ideas with modern engineering behind them, and both fail when applied to the wrong climate.
A solar chimney is a vertical shaft, usually on the sun-facing side of the building, with a glazed solar collector at its base or along its sunlit face. Air from the building enters the bottom of the chimney, the sun heats the collector and the air in contact with it, and the now-warmer air rises faster, pulling more air through the building behind it. The governing equations are the same stack and orifice equations from Sessions 1 and 3, but the temperature difference is no longer set by the building's internal gains alone; it includes the solar temperature rise in the chimney. This decouples ventilation from occupancy to a useful degree: a solar chimney can ventilate a building even when internal heat gains are low, as long as the sun is shining.
The key sizing parameters are the collector area, the chimney height, and the flow areas at entry and exit. The collector area sets how much solar energy enters the air stream: roughly, the useful heat gain is the incident solar irradiance times the collector area times the collector efficiency, and that heat raises the chimney air temperature according to Qsolar = m_dot * cp * DeltaTchimney. The chimney height sets the stack pressure for a given temperature rise, exactly as in Session 1. The entry and exit areas set the flow for the resulting pressure via the orifice equation. In practice these are solved together, often iteratively: assume a flow, compute the temperature rise, compute the stack pressure, compute the flow the openings pass, and iterate to consistency. Tall, narrow chimneys with large collectors on the equator-facing side perform best; a rule of practical experience is that the chimney should be the tallest element the architecture can accommodate, because height multiplies every kelvin of solar temperature rise into pressure.
The climate suitability of the solar chimney follows directly from its energy source. It needs sun during the hours ventilation is wanted, so it suits climates with strong daytime solar radiation and a cooling or ventilation need that coincides with sunshine: hot-arid regions, sunny temperate summers, and industrial buildings with daytime occupancy. It is weakest exactly when many buildings need ventilation most critically at night, during overcast spells, and in winter when the sun is low and the building needs heating rather than exhaust. A solar chimney can also over-ventilate on a hot sunny afternoon, exhausting air the building would rather keep; adjustable dampers at the chimney base are essential, not optional.
The windcatcher works on the other driver. It is a tower rising above the roof, open on one or more sides, that intercepts wind at a height where it is stronger and less turbulent than at street level. In its simplest traditional form, the tower has openings facing the prevailing wind and internal partitions dividing it into shafts; windward shafts drive air down into the building (positive pressure) while leeward shafts exhaust it (suction), so one device provides both inlet and outlet. Modern variants add dampers, and some combine the tower with evaporative cooling or with a solar-heated exhaust shaft so the device works across more conditions.
Sizing a windcatcher starts from the wind pressure equation of Session 2: the tower's height determines the reference wind speed (wind speeds increase with height above ground, following the local terrain profile), the opening areas and their Cp values determine the pressure presented to each shaft, and the internal shaft resistance, bends, and dampers determine how much of that pressure survives to move room air. The critical parameter is the reliable wind resource: a windcatcher in a district with a steady prevailing breeze is a different device from the same tower in a sheltered courtyard with fickle winds. Unlike the solar chimney, the windcatcher's output is inherently gusty, so the building needs enough thermal mass or a large enough volume to ride through lulls, and the tower needs closable dampers for storms and for winter, when an open tower becomes an uncontrolled infiltration path.
Both devices share a design discipline: they must be integrated with the building's overall ventilation path, not bolted on. A solar chimney with no low-level inlet path starves; a windcatcher discharging into a sealed corridor pressurizes it and stalls. Each needs a continuous, low-resistance air path from entry to exit, and each needs controls that close it down when its driver becomes a liability. The most robust schemes pair the devices with the building's own stack: a windcatcher tower that also serves as the high exhaust for buoyancy ventilation, or a solar chimney whose collector preheats the exhaust of an atrium. In those hybrid roles the devices extend the hours that natural ventilation works rather than replacing the fundamental design.
Worked example: temperature rise in a solar chimney
Given: chimney collector area 8 m2, equator-facing; peak solar irradiance on the collector 700 W/m2; collector efficiency 0.4 (a commonly used design assumption for a simple glazed absorber); target ventilation flow 0.5 m3/s; air density 1.15 kg/m3, cp = 1005 J/(kg K).
Useful heat: Qsolar = 700 * 8 * 0.4 = 2240 W.
Mass flow: m_dot = 0.5 * 1.15 = 0.575 kg/s.
Temperature rise: DeltaT = Qsolar / (m_dot * cp) = 2240 / (0.575 * 1005) = 3.9 K.
What it means: the sun adds roughly 4 K to the chimney air at this flow. Added to a 6 K building-to-outdoor difference, the total driving difference becomes about 10 K, and since stack pressure is linear in the difference, the solar contribution boosts the available pressure by roughly two thirds. But note the feedback the simple calculation hides: more temperature rise drives more flow, which reduces the temperature rise, so the true operating point sits below this first estimate. A proper sizing iterates to consistency, and the iteration converges because the feedback is stabilizing.
Design note: match the device to the climate, not the fashion
The failure mode of both devices is the same: installing them where their driver is absent when needed. Before committing to a solar chimney, check the solar resource during the actual ventilation season and the hours of need, not the annual average; a site with cloudy summers or a nighttime-dominated load is a poor candidate. Before committing to a windcatcher, check the local wind rose at tower height for the ventilation season, and confirm the tower will stand clear of surrounding buildings that could shelter it or shed turbulence onto it. Either device without closable dampers is a liability in the off season. And in both cases, verify the full air path through the building: the device is only as good as the inlets and outlets it connects.
Key takeaways
- A solar chimney adds solar heat to the exhaust stream, boosting the stack temperature difference independently of internal gains.
- A windcatcher reaches into stronger winds aloft and can provide both pressurized supply and suction exhaust through divided shafts.
- Key sizing parameters are collector area, chimney or tower height, and the flow areas of entries, exits, and internal shafts.
- Solar chimneys suit sunny daytime ventilation needs; windcatchers suit sites with a reliable prevailing breeze; both need dampers and a continuous building air path.
Session 6
Night Flushing (Night Purge): Coupling Thermal Mass to Cool Night Air
Objectives
- Explain how night flushing stores "coolth" in a building's thermal mass for use the following day.
- Identify the climate condition that makes night flushing effective: a sufficient diurnal temperature swing.
- Describe the openings and controls strategy that delivers high nighttime air change rates securely.
- Recognize the building characteristics (mass, internal gains, occupancy schedule) that decide whether night flushing is worth pursuing.
Night flushing turns the building itself into a cooling battery. During the night, large volumes of cool outdoor air are passed through the building, and the exposed thermal mass (concrete slabs, masonry walls, dense floors) absorbs the coolth, dropping several degrees below its daytime temperature. The next day, that chilled mass absorbs heat from the room air, the occupants, and the sun, holding indoor temperatures down without mechanical cooling or with a much smaller mechanical system. The ventilation does not cool the people directly; it cools the mass, and the mass cools the people later.
The physics is straightforward energy storage. The heat a slab can absorb overnight is its mass times its specific heat times the temperature swing it undergoes: E = m * c * DeltaTslab. A 200 mm concrete slab weighs about 480 kg per square meter of floor; with a specific heat near 1000 J/(kg K), each square meter stores roughly 480 kJ per kelvin of temperature drop. Dropping the slab 4 K overnight banks nearly 2 MJ per square meter, which the next day offsets a substantial fraction of typical office or classroom gains. The night air is the heat sink that makes this possible, and the airflow rate sets how much of the sink the building can actually use: the heat removed overnight is m_dot * cp * (Texhaust - Toutdoor) integrated over the flushing hours. High air change rates, often 5 to 10 per hour or more during the flush window, are the norm, because the driving temperature differences at night are modest and the mass needs many air changes to be cooled through.
Whether the strategy works at all is decided by the climate's diurnal swing: the difference between the daytime maximum and the nighttime minimum outdoor temperature. Night flushing needs nights cool enough to pull the mass down to a useful temperature, commonly taken as nights that fall several kelvin below the desired daytime indoor temperature. As a commonly used design screening rule, a diurnal swing on the order of 10 K or more marks climates where the strategy is worth serious study, while swings under about 5 K rarely repay the effort; between those values, detailed analysis decides. Hot-arid and high-desert climates are the classic winners. Humid climates can still work if nights are cool, but moisture management then matters: flushing with humid night air loads the mass and furnishings with moisture that the next day's warmth may release as discomfort.
The building itself must cooperate. The mass has to be exposed to the room air: carpeted slabs, suspended ceilings, and insulated wall linings all throttle the heat transfer between air and mass, sometimes to the point where the strategy fails. Exposed concrete soffits are the textbook good case. Internal gains must be moderate; a data-heavy space generating far more heat than the mass can absorb will overwhelm the stored coolth by mid-morning. And the occupancy schedule matters: night flushing suits buildings occupied by day and empty by night (schools, offices), because the flush needs unoccupied hours with secure openings and no comfort constraints on air speed or temperature.
Openings strategy is where night flushing differs most from daytime natural ventilation. The flush wants maximum airflow for a limited number of hours, so openings are sized generously and arranged for a clear low-to-high path: secure low inlets (louvered vents, trickle paths, or ground-floor windows with security grilles) and high exhausts (roof vents, clerestories, atrium heads). Because the building is unoccupied, larger openings are acceptable, but security and weather protection are non-negotiable: automated windows with rain sensors, security-rated louvers, and insect screening are standard. Controls run the show. A typical sequence starts the flush when the outdoor temperature drops a set margin below the slab temperature, modulates or stages openings to hold a target slab or exhaust temperature, and closes everything before occupancy, with a morning lockout that prevents the flush from overcooling the space on an unusually cold night. Temperature sensors embedded in or attached to the slab, not just air sensors, give the truest control signal, because the slab is the battery being charged.
Buoyancy is the natural driver for the flush: at night the building mass is warmer than the outdoor air, so the stack effect (Sessions 1 and 3) pulls cool air in low and exhausts warmed air high, exactly the pattern night flushing wants. Wind assists when present. Where buoyancy alone cannot reach the target air change rate, typically in deep-plan or single-sided spaces, low-energy fan assistance (mixed-mode operation) is the honest answer; the fans run only during the flush window and still cost far less energy than daytime mechanical cooling. Commissioning should verify the actual overnight slab temperature pull-down under real conditions, because models are sensitive to the assumed convective heat transfer at the slab surface, and a shortfall there quietly halves the benefit.
Worked example: overnight coolth stored in an office slab
Given: open office, 200 m2 floor plate with exposed 200 mm concrete soffit (density 2400 kg/m3, c = 1000 J/(kg K)); flush window 8 hours; average night airflow 6 air changes per hour in a 600 m3 volume; mean outdoor temperature 17 degrees C; slab starts the night at 26 degrees C and the flush pulls it to 22 degrees C (DeltaTslab = 4 K).
Stored coolth: slab mass per m2 = 0.2 * 2400 = 480 kg/m2; total mass = 480 * 200 = 96,000 kg. E = 96,000 * 1000 * 4 = 3.84 x 108 J = 384 MJ.
Check against airflow capacity: airflow = 6 * 600 / 3600 = 1.0 m3/s; m_dot = 1.2 kg/s. If exhaust air leaves 3 K warmer than outdoor on average, heat removed = 1.2 * 1005 * 3 * 8 * 3600 = 104 MJ.
What it means: the airflow as assumed can remove only about 104 MJ, well short of the 384 MJ needed for the full 4 K pull-down, so the assumed 6 ACH is insufficient for this target; the design needs roughly three to four times the airflow, a longer flush window, or a smaller pull-down target. This is the central sizing check in night-flush design: the mass sets the storage target, but the night airflow sets what is actually achievable, and the two must be reconciled before promising next-day comfort.
Design note: the mass must see the air
The most common reason night flushing disappoints is buried mass. Every layer between the room air and the concrete (carpet, raised access floors with insulation, suspended acoustic ceilings, drywall linings) adds thermal resistance that stretches the charging time beyond the night window. Audit the finishes early: if the soffit must be covered for acoustics, use thermally conductive or partial-coverage treatments and account for the penalty in the energy model. And never value-engineer away the slab sensors; without them the controls are flying blind and the strategy degrades into fixed-schedule window opening, which overcools on cold nights and underperforms on warm ones.
Key takeaways
- Night flushing pre-cools exposed thermal mass with cool night air so the mass absorbs next-day heat gains passively.
- It needs a real diurnal swing; a commonly used screening threshold is on the order of 10 K day-night difference.
- Exposed mass, moderate internal gains, and daytime occupancy are the building-side prerequisites.
- Size openings for high nighttime air change rates, automate them with rain and security provisions, and control on slab temperature.
- Always reconcile the storage target of the mass with what the night airflow can actually remove.
Session 7
Worked Example: Sizing Stack-Ventilation Openings for a Classroom
Objectives
- Set up a complete stack-ventilation sizing calculation from stated design assumptions.
- Compute the required effective opening area from a target airflow and the available stack pressure.
- Split the effective area into inlet and outlet openings and translate the result into real window sizes.
- Interpret the result: judge feasibility, identify sensitivities, and state what the designer should check next.
This session pulls the whole course together in a single end-to-end sizing exercise: a naturally ventilated classroom cooled by stack effect alone. The method is the one developed across Sessions 1 and 3. First, establish the heat gains and convert them into a required airflow using the allowable indoor-outdoor temperature difference. Second, compute the stack pressure available from the vertical separation between inlet and outlet. Third, solve the orifice equation for the effective area that passes the target flow under that pressure. Fourth, split the effective area into inlet and outlet, and translate the numbers into operable windows a designer can actually draw. Every assumption is stated explicitly, because a sizing calculation is only as honest as its inputs.
Step 1: Given data and design assumptions
Consider a single-storey classroom, 8 m by 7.5 m in plan with a 3 m ceiling, giving a floor area of 60 m2 and a volume of 180 m3. The class holds 30 occupants. The design conditions and assumptions are:
Heat gains (design assumptions, stated as such): 30 occupants at 100 W sensible each gives 3000 W; lighting and teaching equipment at a commonly used allowance of 15 W/m2 over 60 m2 gives 900 W; solar gain through glazing on the design afternoon, taken as 1500 W for a partly shaded facade. Total sensible gain: 5400 W. (Latent gains are excluded here; this is a sensible-cooling sizing.)
Temperatures (design assumptions): outdoor design temperature 21 degrees C (294 K); target indoor temperature 26 degrees C (299 K), so the air absorbs heat over a 5 K rise as it passes through the room. Air density 1.2 kg/m3, specific heat cp = 1005 J/(kg K).
Target airflow: the airflow that removes the 5400 W sensible gain over the 5 K rise, from Qheat = rho * V_dot * cp * DeltaT. Solving, V_dot = 5400 / (1.2 * 1005 * 5) = 5400 / 6030 = 0.896 m3/s, rounded to 0.90 m3/s (900 L/s). For reference, this is about 30 L/s per occupant, well above the roughly 10 L/s per person commonly used as a ventilation minimum, which confirms that cooling, not indoor air quality, sets the airflow here. It also equals 18 air changes per hour, a high but normal figure for buoyancy-driven cooling design.
Geometry (design assumption): low operable windows with mid-height at 1.0 m above the floor; high clerestory windows with mid-height at 4.5 m. Vertical separation H = 3.5 m. Openings are plain operable windows, so the discharge coefficient is taken as Cd = 0.6, a commonly used design value for unobstructed openings.
Step 2: Available stack pressure
Using the stack equation from Session 1 with the separation H between the opening mid-heights:
DeltaPstack = rho * g * H * (Ti - To) / Ti
- DeltaPstack
- total stack pressure available between inlet and outlet levels, Pa
- rho
- air density, 1.2 kg/m3
- g
- gravitational acceleration, 9.81 m/s2
- H
- vertical separation between opening mid-heights, 3.5 m
- Ti, To
- indoor and outdoor absolute temperatures, 299 K and 294 K
DeltaPstack = 1.2 * 9.81 * 3.5 * (299 - 294) / 299 = 1.2 * 9.81 * 3.5 * 5 / 299 = 0.689 Pa, about 0.69 Pa. This is a small driving pressure, typical of single-storey stack ventilation on a mild day, and it is the number everything else hangs on.
Step 3: Required effective area
Rearrange the orifice equation (Session 3) to solve for the effective area of the inlet/outlet pair:
Aeff = Q / (Cd * sqrt(2 * DeltaPstack / rho))
- Aeff
- effective area of the inlet/outlet pair in series, m2
- Q
- target volumetric airflow, 0.90 m3/s
- Cd
- discharge coefficient, 0.6
- DeltaPstack
- available stack pressure, 0.689 Pa
- rho
- air density, 1.2 kg/m3
sqrt(2 * 0.689 / 1.2) = sqrt(1.148) = 1.072 m/s. Aeff = 0.90 / (0.6 * 1.072) = 0.90 / 0.643 = 1.40 m2.
Step 4: Split into inlet and outlet, and translate to windows
For equal discharge coefficients, the most area-efficient split is balanced: A1 = A2 = Aeff * sqrt(2) = 1.40 * 1.414 = 1.98 m2, about 2.0 m2 of net free area each for inlet and outlet. In window terms: the low inlet could be four 1.0 m wide by 0.5 m high awning windows opened fully (2.0 m2 gross, and with plain open windows the gross-to-net correction is small), and the high outlet could be a continuous clerestory band, for example 4.0 m long by 0.5 m high of operable panels, also 2.0 m2. If insect screens or louvers are added, their lower discharge coefficient and reduced net free area must be applied: roughly, dropping Cd from 0.6 to 0.4 increases the required gross area by a factor of 1.5, pushing each side toward 3.0 m2.
Step 5: Interpretation
What does 2.0 m2 per side mean for the designer? First, it is feasible: 2 m2 of operable window per side is an ordinary amount of glazing operation for a 60 m2 classroom, roughly 3 percent of the floor area per side. Second, the design is sensitive to the temperature difference because the driving pressure is small: on a hotter afternoon when the indoor-outdoor difference shrinks to 2.5 K, the stack pressure halves and the flow falls by a factor of sqrt(2), to about 0.64 m3/s, which no longer removes the full 5.4 kW. The honest conclusion is that this scheme covers mild cooling days on buoyancy alone and needs either wind assistance (a favorable breeze adds driving pressure per Session 2), larger openings, or a mixed-mode backup for peak conditions. Third, comfort needs a check: the inlet jet enters at 21 degrees C beside seated occupants, so the low windows should be detailed to direct incoming air upward or across the ceiling (high-level inward-opening hoppers do this well), and the design should confirm with the neutral-plane logic of Session 1 that the inlets sit well below the NPP so they truly act as inlets. Finally, controls matter: on cold mornings the same 2 m2 openings would over-ventilate dramatically, so the windows need staged operation tied to indoor temperature, not just open/closed.
Worked example: summary of the classroom sizing
Given (design assumptions): 60 m2 classroom, 30 occupants; sensible gains 5400 W (3000 W occupants + 900 W equipment/lighting + 1500 W solar); To = 21 degrees C, Ti = 26 degrees C; H = 3.5 m between opening mid-heights; Cd = 0.6.
Steps: (1) Target airflow from heat balance: V_dot = 5400/(1.2 * 1005 * 5) = 0.90 m3/s. (2) Stack pressure: DeltaP = 1.2 * 9.81 * 3.5 * 5/299 = 0.69 Pa. (3) Effective area: Aeff = 0.90/(0.6 * sqrt(2 * 0.69/1.2)) = 1.40 m2. (4) Balanced split: 1.98 m2 net free area per side, about 2.0 m2 each.
Result: roughly 2.0 m2 of operable window at low level and 2.0 m2 at high level (e.g., four 1.0 x 0.5 m low awnings plus a 4.0 x 0.5 m clerestory band), feasible within normal classroom glazing, adequate for mild-day cooling on buoyancy alone, and needing wind assistance or mixed-mode backup for peak hot afternoons.
Design note: size for the worst credible calm, hot hour, then add control
This example sized openings for a mild 5 K day. Real design repeats the calculation for the hottest calm hour the scheme is expected to handle, because that is when buoyancy is weakest and the load is highest; the mild-day result then becomes the minimum, not the answer. Whatever area that worst-case sizing produces, pair it with staged, temperature-controlled operation: big openings sized for peak buoyancy will massively over-ventilate on cool mornings. The professional deliverable is therefore not just an area but an area plus a control sequence, plus a stated backup mode for conditions beyond the natural scheme's reach.
Key takeaways
- Start from heat gains and the allowable temperature rise to set the target airflow; cooling usually governs over ventilation minimums.
- Compute the stack pressure from the real vertical separation between opening mid-heights, then solve the orifice equation for effective area.
- Split the effective area equally between inlet and outlet for the most area-efficient pair, and translate net free area into real window sizes.
- Small stack pressures make the design sensitive to temperature difference; check the hot calm hour and provide wind assistance, larger openings, or mixed-mode backup.
- Deliver area plus controls plus a stated backup mode; openings sized for peak buoyancy over-ventilate on cool days without staged control.
References
References
Every source below is an established engineering textbook, an ASHRAE or ISO standard or handbook, or a peer-reviewed paper from a Q1/Q2 journal. Inline citations in the sessions above refer to these entries.
- Etheridge, D. and Sandberg, M., Building Ventilation: Theory and Measurement, Wiley, 1996.
- Allard, F. (ed.), Natural Ventilation in Buildings: A Design Handbook, James and James, 1998.
- Awbi, H. B., Ventilation of Buildings, 2nd ed., Routledge, 2003.
- Linden, P. F., "The fluid mechanics of natural ventilation," Annual Review of Fluid Mechanics, Vol. 31, pp. 201-238, 1999.
- Li, Y. and Delsante, A., "Natural ventilation induced by combined wind and thermal forces," Building and Environment, Vol. 36, pp. 59-71, 2001.
- ASHRAE Handbook: Fundamentals, American Society of Heating, Refrigerating and Air-Conditioning Engineers (natural ventilation chapter).
