Working Note · Fluid Dynamics · Part IX
Bernoulli's Equation: Why Faster Flow Exerts Less Pressure
A counterintuitive fact on which half of aviation and the entire perfume industry relies: where a fluid or gas accelerates, the surrounding pressure drops, not rises.
Daniel Bernoulli derived the energy conservation equation for flowing fluid in 1738: along a streamline, the sum of pressure, kinetic, and potential 'energy per unit volume' remains constant. A direct consequence — where the flow accelerates (pipe narrows), static pressure must drop. This same principle explains atomizers, carburetors, and (with important caveats) the lift of a wing. In this note — derivation of the equation, Venturi tube with numbers, and an honest breakdown of what Bernoulli explains in aircraft flight, and what it does not.
The Equation in a Nutshell
| Term | Meaning |
|---|---|
| P | static pressure — what a pressure gauge moving with the flow would show |
| ½ρv² | 'dynamic pressure' — energy of motion per unit volume |
| ρgh | 'hydrostatic pressure' — energy due to height (as in Archimedes' principle, Part VIII) |
Three caveats in the definition — not fine print, but precisely the conditions where the law works accurately: steady flow (flow pattern does not change with time), inviscid (no internal friction), incompressible (density does not change). For water this is almost always a good approximation; for air at high subsonic speeds — also decent, but not perfect (§6).
Where It Comes From: Flow Energy
Bernoulli's equation is not a new law of nature, but the law of energy conservation (Part II) applied to a small fluid volume moving along a streamline. Each term is a form of energy per unit volume:
- ½ρv²: a direct analog of kinetic energy ½mv², only instead of mass m — density ρ (mass per unit volume).
- ρgh: analog of potential energy mgh, by the same substitution (m → ρ).
- P: 'pressure energy': the work done by adjacent fluid layers to push the volume forward.
If the fluid flows horizontally (h does not change) and accelerates, the kinetic term ½ρv² grows — and since the sum must remain constant, the pressure term P must drop. This is no exotic thing, but simple energy accounting: speed has 'borrowed' energy from pressure.
Pipe Narrowing: Venturi Tube
Before calculating pressure, you need the velocity in the narrow part — it comes from a separate, simpler law: continuity equation (the fluid cannot 'accumulate' inside the pipe; what flows in equals what flows out):
Numerical example: the wide part of the pipe is 4 times wider than the narrow part (by cross-sectional area), water (ρ = 1000 kg/m³) flows in the wide part at velocity v₁ = 2 m/s. The pipe is horizontal (h does not change).
Now the pressure drop according to Bernoulli (h unchanged, ρgh term cancels on both sides):
The pressure in the narrow part is lower than in the wide part by a full 30 kilopascals — nearly a third of atmospheric pressure! It is on this principle (Venturi tube) that carburetors, airbrushes, and medical nebulizers work: a narrow channel creates a local low-pressure zone that draws fluid in from a side opening.
Aircraft Wing — Not So Simple
The popular explanation of lift: 'the wing is curved on top, the air above it moves faster, so the pressure there is lower, the wing is pushed upward.' This is a qualitatively correct line of reasoning (faster on top — indeed lower pressure on top — indeed there is lift), but the specific, often repeated justification of 'why the air moves faster on top' is a myth.
It is often said: 'air particles above and below the wing must meet at the trailing edge at the same time, and the path on top is longer — so the speed there is greater.' This is incorrect: experiment shows particles do NOT meet simultaneously — the flow over the wing overtakes the 'below' pair by a large margin. The real reason for the higher speed on top is that the wing shape and angle of attack create circulation around the entire profile (a result of air viscosity at the surface and the Kutta condition at the trailing edge) — a significantly more subtle effect than 'longer path — faster.'
Bernoulli's equation here is a valid TOOL for calculation (since there is a speed difference — there will be a pressure difference), but it does not explain where the speed difference itself comes from. A complete explanation of lift also requires Newton's third law: the wing deflects the airflow downward (downwash), and by action-reaction, the air pushes the wing upward — both descriptions (energy-based via Bernoulli and momentum-based via Newton) are mutually consistent and together give the full picture.
A Running Example: Pitot Tube
Aircraft measure airspeed with a Pitot tube — it compares the pressure at two points: where the flow is brought to rest (at the tube's tip, 'stagnation point'), and where it flows freely.
Numerical example: the sensor shows a pressure difference ΔP = 250 Pa, air density ρ ≈ 1.25 kg/m³. What is the speed of the aircraft relative to the air?
≈72 km/h — a modest speed (light plane at takeoff), but the method is the same for any speeds up to near-sonic, where the formula must be corrected for air compressibility (§6).
Where It Leads: Limits of Applicability
The three assumptions from §1 are not formalities. Viscosity of real fluids dissipates some energy as heat along the pipe (hence pressure drop in long pipelines even WITHOUT narrowing — the pure Bernoulli equation does not describe this; friction corrections are needed). Compressibility becomes significant for gases at speeds comparable to the speed of sound (Mach number approaching 1) — then more complete gas dynamics is needed. Turbulence makes the flow unsteady on small scales — the equation holds for averaged, not instantaneous, quantities.
Bernoulli's equation is a special, highly simplified case of the far more general Navier-Stokes equations, which describe the flow of real viscous fluids and gases. Their exact analytical solution in the general case — one of the seven 'Millennium Problems' (Clay Mathematics Institute, $1,000,000 for proving existence and smoothness of solutions) — remains unsolved to this day, though engineers happily compute flows numerically without waiting for mathematicians.
On the site this is a separate law, if you'd like to dive deeper:
Home Experiment
Take two sheets of paper, hold them parallel to each other at a distance of 3-4 cm (you can just hold the top edges with two hands) and blow between them, right into the gap.
What to Notice sheets attract each other, rather than flying apart as one might intuitively expect. The air between the sheets accelerates — by Bernoulli, the pressure there drops below the outside atmospheric pressure, and the external pressure literally presses the sheets inward, toward each other.
Point a hairdryer (or a vacuum cleaner in blower mode) straight up and gently place a lightweight ping-pong ball in the airflow above it, then release.
What to Notice the ball hovers in the air, swaying slightly, and does not drift out of the jet to the sides — as soon as it shifts to the edge of the stream, the surrounding air there moves slower (normal atmospheric pressure), and the higher outside pressure 'pushes' the ball back toward the center of the fast flow, where pressure is lower. The same principle that holds two sheets of paper together in the previous experiment.
Check Your Understanding
First think on your own, then open the solution. The solution follows the pattern everywhere: Given → Law → Solution → Answer.
1. Water flows through a horizontal pipe that narrows by a factor of three in cross-sectional area. The speed in the wide part is 3 m/s. Find the speed in the narrow part.
Given A₁/A₂ = 3, v₁ = 3 m/s.
Law continuity equation (§3): A₁v₁ = A₂v₂.
Solution v₂ = v₁·(A₁/A₂) = 3·3 = 9 m/s.
Answer 9 m/s.
2. In the previous problem, find the pressure drop between the wide and narrow part (water, ρ = 1000 kg/m³).
Given v₁ = 3 m/s, v₂ = 9 m/s, ρ = 1000 kg/m³.
Law Bernoulli's equation for a horizontal pipe (§3): ΔP = ½ρ(v₂²−v₁²).
Solution ΔP = 0.5·1000·(81−9) = 500·72 = 36,000 Pa.
Answer 36 kPa.
3. A Pitot tube shows a pressure difference of 500 Pa in air with density 1.25 kg/m³. Find the speed.
Given ΔP = 500 Pa, ρ = 1.25 kg/m³.
Law Pitot tube formula (§5): v = √(2ΔP/ρ).
Solution v = √(2·500/1.25) = √800 ≈ 28.3 m/s.
Answer ≈28.3 m/s (≈102 km/h).
4. Is it true that 'air particles above and below an aircraft wing meet at the trailing edge at the same time'? How does this relate to the actual mechanism of lift?
Law §4 — debunking the equal transit time myth.
Solution No, this is a common misconception: experimentally, the particles do NOT meet at the same time; the flow on top arrives earlier. The real reason for the speed difference is circulation around the wing profile (an effect of viscosity and the Kutta condition at the trailing edge), not 'a longer path on top.'
Answer myth, incorrect — Bernoulli correctly CALCULATES the pressure from an already known speed difference, but does not explain where the difference itself comes from.
5. Why does the basic form of Bernoulli's equation not predict a pressure drop along a LONG pipe of CONSTANT cross-section, even though it is observed in practice?
Law limits of applicability (§6) — the assumption of 'inviscid fluid.'
Solution The basic Bernoulli equation is derived for an IDEAL (inviscid) fluid, where energy along a streamline is not lost. In real pipes, viscous friction against the walls gradually turns some of the flow's energy into heat — this loss is not included in the simple form P+½ρv²+ρgh=const, hence the pressure drop observed in practice even without a change in cross-section.
Answer viscosity — real fluids are not ideal; energy is dissipated as heat, which the basic equation does not account for.
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