Mechanical
Bernoulli's Principle
Where the fluid speeds up, the pressure drops
Bernoulli's equation p + ½ρv² + ρgh = constant relates pressure, velocity, and height along a streamline of an inviscid incompressible flow. It's the engineering shorthand for energy conservation in fluids: pressure work plus kinetic energy plus gravitational potential per unit volume stays constant. It explains how a Pitot tube reads aircraft airspeed, why a curveball curves, why an airfoil generates lift, and why a Venturi nozzle drops pressure as flow accelerates. It also has clean limits — viscous, compressible, and unsteady flows need corrected forms.
- Equationp + ½ρv² + ρgh = const
- Dynamic pressureq = ½ρv²
- Stagnation pressurep_total = p + q
- ValiditySteady, inviscid, M < 0.3
- DiscoveredDaniel Bernoulli, 1738
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What the equation says
For steady, inviscid, incompressible flow along a streamline, Bernoulli's equation states that
p + ½ρv² + ρgh = constant
The three terms have units of energy per unit volume (Pa = J/m³):
- p — static pressure, the force per unit area the fluid exerts on a wall parallel to flow.
- ½ρv² — dynamic pressure, the kinetic energy density of the moving fluid.
- ρgh — hydrostatic head, the gravitational potential energy density.
The sum stays constant along a streamline. So if velocity v increases between two points, pressure p must decrease (assuming negligible height change). It's energy conservation in fluid form: kinetic energy can only grow at the expense of pressure work or gravitational drop.
Worked example: Pitot tube airspeed
A Pitot tube has two openings on a streamlined probe:
- A stagnation port facing into the flow. Air decelerates to zero velocity here; pressure rises to the stagnation (total) pressure p_total.
- A static port on the side of the probe, parallel to the flow. It reads the undisturbed static pressure p_static.
Apply Bernoulli between the freestream and the stagnation point (same streamline, same height):
p_static + ½ρv² = p_total + 0
v = √(2(p_total − p_static) / ρ)
Numerical example. An aircraft cruises at altitude where ρ = 0.71 kg/m³ (about 6,000 m). The Pitot–static pressure differential reads 4,800 Pa. Then:
v = √(2 × 4,800 / 0.71) = √(13,521) = 116.3 m/s ≈ 226 knots
This is the indicated airspeed (IAS). To convert to true airspeed, the avionics correct for density variation with altitude — but the underlying physics is just Bernoulli applied along one streamline.
Aircraft motion ─────► freestream: p_∞, v_∞
┌──┐
Pitot probe ────► │∞ │ ────► stagnation port: p = p_total, v = 0
│∞ │ ____ static port: p = p_static
└──┘
│
└─► differential pressure transducer
reads p_total − p_static = ½ρv²
Venturi effect
When a pipe narrows from area A_1 to A_2, conservation of mass requires v_2 = v_1 · (A_1/A_2). Bernoulli then says the pressure drops:
p_1 + ½ρv_1² = p_2 + ½ρv_2²
p_1 − p_2 = ½ρ(v_2² − v_1²) = ½ρv_1²·((A_1/A_2)² − 1)
For water (ρ = 1000 kg/m³) flowing at v_1 = 2 m/s through a pipe that narrows from 50 mm to 25 mm diameter (A_1/A_2 = 4):
v_2 = 2 × 4 = 8 m/s
p_1 − p_2 = ½ × 1000 × (64 − 4) = 30,000 Pa = 30 kPa
The Venturi effect powers carburetors (the venturi at the throat creates a pressure drop that pulls fuel into the airstream), atomizers, water aspirators, and a class of flowmeters used in pipelines and HVAC ducts.
Forms of Bernoulli
| Form | Validity | Equation | Application |
|---|---|---|---|
| Incompressible (canonical) | M < 0.3, ρ constant | p + ½ρv² + ρgh = const | Water flow, low-speed air |
| Compressible (subsonic) | M < 1, isentropic | (γ/(γ−1))·(p/ρ) + ½v² = const | Subsonic compressible flow, jet engines |
| Stagnation form | Steady inviscid | p_total = p + ½ρv² | Pitot tube, total/static measurement |
| Total head form (engineering) | Steady inviscid + losses | H_1 = H_2 + h_loss (h = p/ρg + v²/2g + z) | Pipe flow with friction (Hazen–Williams) |
| Unsteady form | Time-varying flow | p + ½ρv² + ρgh + ρ·∫(∂φ/∂t)ds = const | Water hammer, oscillating flow |
| Viscous-corrected (Darcy–Weisbach) | Pipe flow with friction | Add (f·L/D)·(½ρv²) loss term | Real pipe networks |
Bernoulli and airfoil lift
The popular explanation of wing lift goes: the upper surface is curved, so air over the top has further to travel, so it goes faster, so by Bernoulli the pressure is lower, so the wing is sucked up. The intuition is right; the "equal transit time" reasoning is wrong. There's no physical principle saying air over the top must arrive at the trailing edge at the same time as air under — and measurements show the upper-surface air actually arrives first.
The accurate picture: a wing at positive angle of attack creates a net circulation of air around itself (Kutta–Joukowski theorem). The circulation speeds up flow over the top and slows it underneath. Bernoulli then correctly predicts the pressure differential that produces lift. So Bernoulli is necessary but not sufficient — you also need the right velocity field, which comes from the wing's geometry and the no-slip condition at the trailing edge.
Lift coefficient C_L for a typical light aircraft is about 0.3 in cruise. Lift L = ½ρv²·S·C_L, where S is the wing area. A Cessna 172 (S = 16.2 m², ρ = 1.0 kg/m³ at altitude, v = 60 m/s) generates L = 0.5 × 1.0 × 3600 × 16.2 × 0.3 = 8,748 N — about 890 kg of lift, balancing a fully-loaded 1,100-kg airplane (the rest comes from C_L closer to 0.4 in real cruise).
The Magnus effect
A spinning ball moving through air creates an asymmetric flow: on the side rotating with the airstream, the air goes faster; on the opposite side, it goes slower. Bernoulli pressure differential pushes the ball perpendicular to both its motion and its spin axis.
For a baseball (D = 73 mm, m = 145 g) thrown at 35 m/s with backspin of 1,800 rpm (188 rad/s):
Surface speed at top of ball = 0.0365 × 188 = 6.86 m/s relative to ball
Effective velocity asymmetry: top moves at 35 + 6.86 = 41.9 m/s, bottom at 35 − 6.86 = 28.1 m/s
Estimated Magnus force ≈ ½ρv²·A·C_L (with C_L ~ 0.2 for backspin) ≈ 0.5 N
Trajectory deflection over 18 m flight: roughly 30 cm vertical lift
Soccer balls (Roberto Carlos's famous 1997 free-kick), tennis topspin shots, and golf-ball trajectories all rely on Magnus. The cm-level deflection is exactly what Bernoulli predicts when you account for the ball's spin-induced velocity asymmetry.
Where Bernoulli breaks
Viscous flow. In a long pipe, friction at the walls dissipates fluid energy as heat. Bernoulli without losses overpredicts downstream pressure. The Darcy–Weisbach equation adds a friction-loss term: Δp_loss = (f·L/D)·½ρv², where f is the Moody friction factor (0.02–0.03 for typical pipes). For a 100-m pipeline carrying water at 2 m/s, this loss is roughly 5–10 kPa per metre — easily 100% of what Bernoulli alone would predict.
Compressible flow. Above Mach 0.3, density varies significantly with pressure and the constant-ρ assumption fails. The compressible form (γ/(γ−1))·(p/ρ) + ½v² = const replaces ρgh and tracks enthalpy along the streamline. For supersonic flow, shocks dissipate energy entirely outside Bernoulli's framework.
Unsteady flow. Water hammer in a closing valve generates pressure spikes from inertial deceleration of the fluid column. Steady-flow Bernoulli predicts none of this. The unsteady form adds ρ·∫(∂φ/∂t)ds along the streamline; for water hammer, the result is the Joukowsky pressure rise Δp = ρ·c·Δv, where c is the speed of sound in the pipe — easily 10× the dynamic pressure for a sudden valve closure.
Across streamlines. Bernoulli relates two points on the same streamline. Across streamlines (e.g., from inside a vortex core to outside), pressure varies according to centripetal balance, not Bernoulli. Confusing the two leads to wrong predictions in tornado modeling and vortex generators.
Real-world applications
- Aircraft Pitot–static system. Reads dynamic pressure from p_total − p_static; combined with temperature gives indicated airspeed, calibrated airspeed, and (with altitude correction) true airspeed. The same probe drives the altimeter (using p_static alone) and rate-of-climb indicator.
- Carburetor. A venturi in the air intake creates suction that pulls fuel from a float bowl through a metering jet. Throttle controls the venturi area and thus the fuel-air ratio. Now mostly replaced by fuel injection, but every small piston aircraft and 1970s car had one.
- Spray atomizer / paint gun. Compressed air at high velocity creates low pressure that draws liquid up a feed tube, then breaks it into droplets. The same physics powers perfume sprayers and laboratory aspirators.
- Bunsen burner air valve. The gas jet from the orifice creates a low-pressure region that sucks in primary air through adjustable holes. Closing the holes makes the flame yellow and sooty (incomplete combustion); opening them makes it blue and hot.
- Sailing close-hauled. A sail trimmed at an angle to the wind acts like a vertical airfoil. The high-velocity flow on the leeward side generates suction; sailing close-hauled uses this Bernoulli lift to make headway against the wind.
- Roof loss in hurricanes. High wind across a roof creates low static pressure above; if interior pressure is higher (windows closed, gusts trapped inside), the differential lifts the roof off. Code now requires hurricane-zone roofs to be tied down for uplift loads computed from Bernoulli.
Common misapplications
- Capillary action via Bernoulli. Water rising in a thin tube is driven by surface tension, not pressure-velocity tradeoffs. Bernoulli predicts no rise in still water. Don't use it where surface tension dominates.
- Pipe flow without friction loss. Engineering pipe-flow problems must include head loss; pure Bernoulli gives nonsense for pipes longer than a few diameters. Always add the Darcy–Weisbach (or Hazen–Williams) friction term.
- Compressible flow at high subsonic Mach. Above M = 0.3, the constant-ρ assumption fails. Aircraft drag, jet-engine intake design, and gas pipelines all need the compressible-flow form (or full computational fluid dynamics).
- Pressure across streamlines. A swirl in a coffee cup has lower pressure at the centre than the rim — but Bernoulli applied across streamlines gives the wrong sign because it doesn't include the centripetal term. Use a momentum balance for cross-streamline pressure variation.
- Equal-transit-time wing argument. The popular textbook explanation that air over the top "must arrive at the same time" is empirically wrong and physically unmotivated. Lift requires circulation; Bernoulli explains only the pressure-velocity link, not the velocity field itself.
- Static pressure during transients. Pumping rates ramping up, valves closing, surge tanks emptying — all unsteady. Steady-Bernoulli numbers can be off by 10× during a transient. Use the unsteady form or a numerical model.
Frequently asked questions
Where does Bernoulli's principle not apply?
It fails for viscous flow (where friction dissipates energy along the streamline), compressible flow at Mach > 0.3 (where density varies with pressure), unsteady flow (where d/dt terms can't be ignored), and flow across streamlines (the equation only relates points on the same streamline). Misapplying Bernoulli to capillary tube flow or supersonic nozzles gives wrong answers; you need the Navier–Stokes equations or a compressible-flow form.
Does Bernoulli explain airfoil lift?
Partly. The flow over a wing's upper surface goes faster, lowering pressure; the flow under goes slower, raising pressure; the difference creates lift. But the popular "equal-transit-time" argument is wrong — air over the top doesn't have to arrive at the trailing edge at the same time as air under. Better: lift comes from circulation around the wing, with Bernoulli explaining the pressure-velocity relationship along each streamline.
How does a Pitot tube measure airspeed?
A Pitot tube has two openings: one facing into the flow (stagnation port, where v = 0 and pressure = p_total) and one perpendicular to the flow (static port, reading p_static). The difference is dynamic pressure: p_total − p_static = ½ρv². So v = √(2(p_total − p_static)/ρ). Aircraft and Formula 1 cars both use this principle for their primary speed measurement.
What causes the Magnus effect on a spinning ball?
A spinning ball drags surrounding air, speeding up flow on the side rotating with the airstream and slowing it on the side rotating against. Bernoulli says faster flow has lower pressure — so the ball experiences a sideways force perpendicular to both its motion and its spin axis. A baseball with backspin curves upward by 30+ cm over 18 m; soccer balls and tennis balls show similar deflections.
Why does flow speed up in a narrow pipe section?
Conservation of mass — the continuity equation A_1·v_1 = A_2·v_2. If the cross-section halves, velocity doubles. Bernoulli then explains that the pressure drops to compensate: ½ρv² goes up, so p must go down. This is the Venturi effect, used in carburetors, atomizers, and flow measurement nozzles.
Is Bernoulli's principle the same as conservation of energy?
It's a special case. The equation p + ½ρv² + ρgh = constant is the energy equation for an inviscid, incompressible, steady flow along a streamline — pressure work plus kinetic energy plus gravitational potential per unit volume stays constant. With viscous losses, you add a "head loss" term; with compressibility, you replace the pressure term with enthalpy; with unsteadiness, you add a time derivative.