Fluids

Bernoulli’s Principle & Applications

Definition

Bernoulli’s principle states that in a steadily flowing fluid, an increase in speed occurs together with a decrease in pressure or a drop in height. In equation form, static pressure plus dynamic pressure (½ρv²) plus hydrostatic pressure (ρgh) stays constant along a streamline for an incompressible, low-friction flow.

Hold a strip of paper just below your lip and blow hard across its top. The strip does not flatten — it rises to meet the moving air, as if lifted by nothing at all.

That small mystery scales up spectacularly. The same physics helps hold a 400-tonne airliner in the sky, once fed petrol into every carburettor engine, and lets a perfume bottle turn liquid into mist. This guide unpacks how it works, where it applies — and, just as importantly, where it does not.

What Is Bernoulli’s Principle?

Picture water moving through a garden hose. Squeeze the end and the same amount of water must pass through a smaller opening every second, so it speeds up. Bernoulli’s principle answers the follow-up question most people never think to ask: what happens to the pressure?

The answer surprises almost everyone. Where the fluid moves faster, its pressure is lower — not higher. Where it moves slower, the pressure is higher.

Stated precisely: for a steady, incompressible flow with negligible friction, the sum of static pressure, kinetic energy per unit volume and gravitational potential energy per unit volume stays constant along a streamline. A streamline is simply the path a tiny parcel of fluid traces through the flow.

The idea comes from the Swiss mathematician Daniel Bernoulli, who published it in his 1738 book Hydrodynamica. The tidy equation we write today was set down soon afterwards by his colleague Leonhard Euler — but the name stuck to Bernoulli, and it has held for nearly three centuries.

Portrait of Daniel Bernoulli, who first published Bernoulli's principle in 1738
Daniel Bernoulli (1700–1782). His 1738 book Hydrodynamica applied energy conservation to moving fluids.

The Bernoulli Equation

The whole principle compresses into a single line:

P + ½ρv² + ρgh = constant

Every symbol has a precise meaning and an SI unit:

  • P — static pressure of the fluid, in pascals (Pa)
  • ρ (rho) — density of the fluid, in kilograms per cubic metre (kg/m³)
  • v — flow speed at that point, in metres per second (m/s)
  • g — gravitational field strength, approximately 9.81 m/s² on Earth
  • h — height above a chosen reference level, in metres (m)

Each of the three terms is an energy per unit volume, in joules per cubic metre — which works out to exactly the same unit as pressure, the pascal. That is no coincidence. Bernoulli’s equation is conservation of energy, written out for each cubic metre of moving fluid.

Term Name What it represents Grows when…
P Static pressure The squeeze the fluid exerts on its surroundings The fluid is pushed on harder
½ρv² Dynamic pressure Kinetic energy packed into each cubic metre The flow speeds up
ρgh Hydrostatic term Gravitational potential energy per cubic metre The fluid sits higher up

To compare two points along the same streamline, write the constant out twice:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

One partner equation almost always joins it. The continuity equation says what flows in must flow out, so for an incompressible fluid the cross-sectional area times the speed stays fixed:

A₁v₁ = A₂v₂

Continuity tells you how fast; Bernoulli tells you at what pressure. Nearly every exam problem uses the pair together. If you would rather skip the algebra, plug your numbers straight into our Bernoulli Equation Calculator and get the result in one step. NASA’s Glenn Research Center keeps a clear reference on the different forms of Bernoulli’s equation and the conditions attached to each one.

How Bernoulli’s Principle Works

Why should speeding up cost a fluid its pressure? The cleanest answer follows the energy.

Imagine a small parcel of water approaching a narrow section of pipe. The high-pressure fluid behind it pushes it forward — positive work done on the parcel. The lower-pressure fluid ahead pushes back, doing negative work. The push from behind wins, so the parcel accelerates into the constriction.

  1. The parcel moves from a wide, high-pressure region towards a narrow, low-pressure region.
  2. Net work done on it equals (pressure behind minus pressure ahead) × area × distance moved.
  3. By the work–energy theorem, that net work becomes extra kinetic energy — or potential energy, if the pipe climbs.
  4. Divide the whole ledger by the parcel’s volume, and the bookkeeping reads P + ½ρv² + ρgh = constant.

Notice the direction of cause and effect. The fluid does not speed up and then mysteriously lose pressure. A pressure difference already exists along the pipe, and that difference is the very force accelerating the fluid. High speed and low pressure arrive together because one is paid for by the other.

Georgia State University’s HyperPhysics puts it neatly: think of pressure as an energy density, and a constriction simply trades that energy for motion. For the full mathematical derivation — starting from the work–energy theorem and ending at the finished equation — OpenStax’s Bernoulli’s Equation chapter walks through every step with diagrams.

Higher pressure Lower pressure Slow flow high static pressure Fast flow low static pressure Slow again pressure recovers Same water per second everywhere — the throat pays for its speed with pressure

The Venturi effect: in the narrow throat the fluid moves faster and its static pressure falls — the manometer columns make the invisible visible.

Sliders beat words here. Squeeze the throat in the lab below and watch the speed jump and the pressure column drop in real time.

Bernoulli's Principle Lab

7 Real-World Applications of Bernoulli’s Principle

Once you know the signature — fast flow next to low pressure — you start spotting it everywhere. Here are seven places it earns its keep.

1. Aircraft wings

Air genuinely travels faster over the curved top of a lifting wing than beneath it, and measurements confirm the pressure up there is lower. Sum that pressure difference over the whole wing and you get a large upward force. The popular story of why the upper air is faster, however, is usually wrong — see the misconceptions below.

2. Atomisers and spray bottles

Squeezing the bulb of a perfume atomiser fires a fast air jet across the top of a thin tube. Pressure there drops below atmospheric, so ordinary air pressure inside the bottle pushes liquid up the tube and into the airstream, shredding it into mist.

3. Venturi meters and carburettors

A Venturi meter deliberately narrows a pipe and measures the pressure drop at the throat; that drop reveals the flow speed, so utilities can meter water or gas with no moving parts at all. Carburettors in older petrol engines ran on the same trick — the narrow throat lowers pressure just enough to draw fuel into the incoming air.

4. Pitot tubes on aircraft

An aircraft’s speedometer is really a pressure gauge. The pitot tube compares total pressure (a port facing straight into the flow, where air is brought to rest) with static pressure from side ports. The difference between them is the dynamic pressure ½ρv², and the airspeed follows directly.

5. Chimneys and prairie-dog burrows

Wind blowing across the top of a chimney lowers the pressure there and strengthens the upward draught. Prairie dogs exploit the same physics: one burrow entrance is built on a raised mound, wind moves faster over it, and the resulting pressure difference ventilates the tunnel with zero effort.

6. Curving balls in sport

A spinning football drags a thin layer of air around with it, so the airflow ends up faster on one side of the ball than the other. The faster side sits at lower pressure and the ball swerves towards it — the Magnus effect that bends a free kick around a defensive wall. Note that this one needs viscosity to grip the air, a reminder that real flows mix several effects.

7. Blood flow through narrowed arteries

Blood speeding through a constricted artery is a Venturi in miniature: pressure inside the narrowed section falls. In severe cases the surrounding pressure can briefly squash the vessel shut, then flow reopens it, over and over — the flutter behind some of the murmurs a doctor hears through a stethoscope.

Application The speed change The pressure change The payoff
Aircraft wing Air faster over the top surface Lower pressure above the wing Lift
Atomiser Air jet races over a tube Pressure at the tube top drops Liquid rises and turns to mist
Venturi meter / carburettor Flow accelerates in the throat Throat pressure falls Flow rate measured; fuel drawn in
Pitot tube Air brought to rest at the nose Pressure rises by ½ρv² Airspeed reading
Chimney / burrow Wind speeds over the opening Pressure at the top falls Natural ventilation draught
Spinning ball Air faster on one side Sideways pressure imbalance The ball curves in flight
Narrowed artery Blood accelerates in the constriction Pressure inside falls Vascular flutter, audible murmur

Common Misconceptions About Bernoulli’s Principle

Myth 1: “Air over the wing must catch up” — the equal-transit-time fallacy

The old textbook story claims air split at a wing’s leading edge must reunite at the trailing edge, forcing the top stream — with its longer path — to travel faster. It sounds plausible, and it is simply false. Wind-tunnel experiments show the two streams never reunite; the upper air typically reaches the trailing edge first.

The faster upper flow is real, but it comes from how the wing’s shape and angle of attack turn the entire flow field, not from any reunion appointment. NASA’s Glenn Research Center dissects this and the other incorrect theories of lift in detail. One quick sanity check: if equal transit time were the mechanism, aircraft could not fly upside down — and stunt pilots do it routinely.

Myth 2: “Bernoulli is wrong — lift is really Newton’s third law”

This one over-corrects. A wing does deflect air downward, and the reaction to that push is lift — perfectly true. But the pressure-difference account and the flow-turning account are two ledgers for the same transaction, and both give the right answer when applied properly. You do not have to pick a side.

Myth 3: “Moving air always has lower pressure than still air”

Bernoulli’s equation compares points along the same streamline, not any two parcels of air you fancy. The free jet from a hair dryer, for instance, sits at roughly the same atmospheric pressure as the room around it. Casually comparing a jet with unconnected still air — as many demo write-ups do — misuses the equation.

Myth 4: Mixing up static, dynamic and total pressure

“The pressure” is an ambiguous phrase in a moving fluid. Static pressure is the sideways squeeze the fluid exerts as it streams past; dynamic pressure ½ρv² is the extra you would register by bringing the flow to rest; their sum is the total pressure. Bernoulli says the total stays constant — it is the static share that falls as speed rises.

When Bernoulli’s Equation Breaks Down

Every clean equation carries small print. Bernoulli’s assumes four things, and real flows violate each of them somewhere.

  • Steady flow. The flow pattern must not change from moment to moment. Turn a tap on suddenly, or peer inside churning turbulence, and the equation loses its footing.
  • Incompressible fluid. Excellent for liquids, and fine for air below roughly 100 m/s — about a third of the speed of sound. Near the sound barrier, density changes and compressible-flow relations take over.
  • Negligible viscosity. Internal friction converts flow energy into heat. In long narrow pipes, in honey, or inside the thin boundary layer hugging every surface, viscous losses dominate and Bernoulli alone will mislead you.
  • Same streamline, no machines between. The two points compared must lie on one streamline, with no pump or turbine adding or removing energy in between.

Air resistance is the everyday reminder of that third point. Drag on a falling object is a viscous, turbulent affair, which is why terminal velocity needs its own treatment rather than a quick Bernoulli argument.

A practical habit worth stealing from engineers: apply Bernoulli first for the big picture, then ask which assumption your system bends, and correct for it. Real pipe design adds “head-loss” terms to the equation for exactly this reason.

How Bernoulli’s Principle Relates to Other Physics

Bernoulli’s equation is not new physics — it is familiar physics wearing fluid clothing. Spotting the family resemblance makes it far easier to remember.

The ½ρv² term is the kinetic energy formula ½mv² divided by volume, since density is just mass per volume. The ρgh term is gravitational potential energy mgh on the same per-volume diet. And pressure enters the ledger through the work one parcel of fluid does on the next.

Two neighbouring ideas deserve separating. Hydrostatic pressure in a still liquid (increasing by ρgh with depth) is Bernoulli with the speed terms switched off. Buoyancy comes from pressure differences with depth, not with speed — Archimedes and Bernoulli answer different questions.

Even projectile motion makes a cameo. The instant a jet of water leaves a hole in a tank, Bernoulli’s job is done and gravity’s begins, arcing the stream like any thrown ball. Torricelli’s theorem — exit speed v = √(2gh) — is nothing more than Bernoulli applied between the tank’s calm surface and the hole.

h open surface, atmospheric pressure small hole Exit speed: v = √(2gh)

Torricelli’s theorem is Bernoulli between two points — the still surface and the jet at the hole. Depth alone sets the exit speed.

Worked Problems

Grab a calculator and work along — it is the fastest way to own this topic. Use ρ = 1000 kg/m³ for water and g = 9.81 m/s², and keep pressures in pascals until the final line.

Problem 1
Water flows through a horizontal pipe at 2.0 m/s where the pressure is 150 kPa. Downstream, the pipe narrows and the speed rises to 5.0 m/s. What is the pressure in the narrow section? Take the density of water as 1000 kg/m³.
Show Solution
Solution: Step 1: The pipe is horizontal, so h₁ = h₂ and the height terms cancel. Bernoulli gives P₂ = P₁ + ½ρ(v₁² − v₂²). Step 2: Substitute with units: P₂ = 150 000 Pa + ½ × 1000 kg/m³ × (2.0² − 5.0²) m²/s² = 150 000 Pa + 500 × (4.0 − 25) Pa. Step 3: P₂ = 150 000 Pa − 10 500 Pa = 139 500 Pa. Answer: P₂ ≈ 1.40 × 10⁵ Pa (about 140 kPa) — lower than upstream, exactly as the principle predicts.
Problem 2
A water pipe of diameter 8.0 cm narrows to 4.0 cm. The speed in the wide section is 1.5 m/s and the pressure there is 200 kPa. The pipe is horizontal. Find the speed and the pressure in the narrow section.
Show Solution
Solution: Step 1: Continuity first: A₁v₁ = A₂v₂, and area scales with diameter squared, so v₂ = v₁ × (d₁/d₂)² = 1.5 m/s × (8.0/4.0)² = 1.5 × 4 = 6.0 m/s. Step 2: Bernoulli (horizontal): P₂ = P₁ + ½ρ(v₁² − v₂²) = 200 000 Pa + 500 kg/m³·(1.5² − 6.0²) m²/s². Step 3: P₂ = 200 000 Pa + 500 × (2.25 − 36) Pa = 200 000 − 16 875 = 183 125 Pa. Answer: v₂ = 6.0 m/s and P₂ ≈ 183 kPa.
Problem 3
An open storage tank holds water. A small hole forms in its side 3.2 m below the water surface. How fast does water leave the hole? (This is Torricelli's theorem.)
Show Solution
Solution: Step 1: Apply Bernoulli between the surface (point 1) and the hole (point 2). Both are open to the air, so P₁ = P₂ = atmospheric, and those terms cancel. The tank is wide, so v₁ ≈ 0. Step 2: What remains is ρgh₁ = ½ρv₂² + ρgh₂, so v₂ = √(2g(h₁ − h₂)) = √(2gh). Step 3: v₂ = √(2 × 9.81 m/s² × 3.2 m) = √62.8 m²/s² = 7.92 m/s. Answer: v ≈ 7.9 m/s — the same speed the water would reach falling freely through 3.2 m.
Problem 4
Water at 250 kPa moves at 4.0 m/s through a vertical pipe of constant diameter at ground level. Find the pressure at a point 5.0 m higher up the same pipe.
Show Solution
Solution: Step 1: Constant diameter means continuity forces v₂ = v₁, so the speed terms cancel and Bernoulli reduces to P₂ = P₁ − ρg(h₂ − h₁). Step 2: Substitute: P₂ = 250 000 Pa − 1000 kg/m³ × 9.81 m/s² × 5.0 m. Step 3: P₂ = 250 000 Pa − 49 050 Pa = 200 950 Pa. Answer: P₂ ≈ 201 kPa. Height alone costs this flow about 49 kPa.
Problem 5
A horizontal Venturi meter carrying water has an inlet area of 10 cm² and a throat area of 5.0 cm². The gauge reads a pressure drop of 6.0 kPa between inlet and throat. Find the throat speed and the volume flow rate.
Show Solution
Solution: Step 1: Continuity: v₁ = (A₂/A₁)v₂ = 0.5 v₂. Bernoulli: ΔP = ½ρ(v₂² − v₁²). Step 2: Substitute v₁: 6000 Pa = ½ × 1000 × (v₂² − 0.25v₂²) = 375 v₂², so v₂² = 16 m²/s². Step 3: v₂ = 4.0 m/s. Flow rate Q = A₂v₂ = 5.0 × 10⁻⁴ m² × 4.0 m/s = 2.0 × 10⁻³ m³/s. Answer: v₂ = 4.0 m/s and Q = 2.0 L/s. This is exactly how the meter converts a pressure reading into a flow rate.
Problem 6
In a simplified model, air streams over the top of a wing at 72 m/s and under it at 63 m/s. Taking air density as 1.225 kg/m³ and wing area as 24 m², estimate the pressure difference and the lift force.
Show Solution
Solution: Step 1: Same height above and below (to a good approximation), so ΔP = ½ρ(v_top² − v_bottom²). Step 2: ΔP = ½ × 1.225 kg/m³ × (72² − 63²) m²/s² = 0.6125 × (5184 − 3969) Pa = 0.6125 × 1215 Pa = 744 Pa. Step 3: Lift F = ΔP × A = 744.2 Pa × 24 m² = 17 860 N. Answer: ΔP ≈ 744 Pa and F ≈ 1.79 × 10⁴ N — enough to hold up about 1.8 tonnes. (Real lift analysis integrates pressure over the whole flow field; this classic estimate shows the scale.)
Problem 7
Water enters a building at ground level through a 6.0 cm diameter pipe at 1.2 m/s and 400 kPa. It emerges from a 3.0 cm diameter tap on a floor 9.0 m higher. Find the exit speed and the exit pressure.
Show Solution
Solution: Step 1: Continuity: v₂ = v₁ × (d₁/d₂)² = 1.2 m/s × (6.0/3.0)² = 4.8 m/s. Step 2: Full Bernoulli: P₂ = P₁ + ½ρ(v₁² − v₂²) − ρg(h₂ − h₁). Step 3: P₂ = 400 000 Pa + 500 × (1.44 − 23.04) Pa − 1000 × 9.81 × 9.0 Pa = 400 000 − 10 800 − 88 290 = 300 910 Pa. Answer: v₂ = 4.8 m/s and P₂ ≈ 301 kPa. Both the climb and the speed-up take their share of the pressure.

Frequently Asked Questions

What does Bernoulli's principle state in simple terms?
In a smoothly flowing fluid, where the flow is faster the pressure is lower, and where the flow is slower the pressure is higher, comparing points at the same height. It works because a moving fluid trades pressure energy for kinetic energy, so the total — static pressure plus ½ρv² plus ρgh — stays constant along a streamline.
Why does pressure decrease when a fluid speeds up?
Because the pressure difference is what causes the speeding up. A fluid parcel accelerates only when pushed from a higher-pressure region towards a lower-pressure one, so fast flow is always found on the low-pressure end of that push. The kinetic energy gained is paid for exactly by the drop in pressure energy — nothing is lost, only converted.
Does Bernoulli's principle explain how aeroplanes fly?
Partly, and only when used correctly. The pressure above a lifting wing really is lower than below it, and that difference produces lift, consistent with Bernoulli’s equation. But the popular equal-transit-time reason for the faster upper airflow is false. A complete explanation also needs the wing’s angle of attack and the downward turning of the airflow.
What are the conditions for using Bernoulli's equation?
The flow should be steady, effectively incompressible and low in friction, and the two points compared must lie on the same streamline with no pump or turbine between them. Liquids meet the incompressibility condition easily; air does too below roughly 100 m/s. Strong turbulence or viscous flow in long narrow pipes breaks the equation.
What is the difference between Bernoulli's principle and the Bernoulli equation?
The principle is the qualitative statement: faster flow means lower pressure at the same height. The equation is its quantitative form, P + ½ρv² + ρgh = constant, which lets you calculate exact pressures and speeds. In everyday use the names are swapped freely, with “equation” preferred once actual numbers are involved.
Who discovered Bernoulli's principle and when?
The Swiss mathematician and physicist Daniel Bernoulli published the principle in his 1738 book Hydrodynamica, where he applied energy conservation to flowing fluids. The familiar modern form of the equation was written down soon afterwards by Leonhard Euler, Bernoulli’s colleague, but the principle rightly carries Bernoulli’s name.
Does Bernoulli's principle work for gases like air?
Yes, provided the gas behaves as if incompressible, which holds when flow speeds stay below about a third of the speed of sound — roughly 100 m/s in air. That comfortably covers breezes, ventilation, atomisers and light-aircraft speeds. For fast jets and supersonic flight, density changes matter and compressible-flow relations replace the simple equation.
P

Written by PhysicsFundamentals Editorial Team

Articles on PhysicsFundamentalsinfo.com are researched, written, and fact-checked by our editorial team. Every piece is reviewed for accuracy before publishing, with formulas and worked examples checked against standard physics references.

View All Authors →