Bernoulli’s Principle and Real Applications

Bernoulli’s principle states that along a streamline in the steady flow of an ideal fluid, the sum p + ½ρv2 + ρgh is constant, so where the speed of the fluid increases, its pressure decreases. In brief: the equation is conservation of energy per unit volume: pressure energy, kinetic energy and gravitational potential energy. It explains the lift on a wing (faster flow over the top gives a lower pressure), the Venturi effect, the speed of water leaving a hole in a tank (Torricelli’s law, v = √(2gh)) and the Pitot tube used to measure airspeed. It does not apply to viscous or turbulent flows without corrections.

Aircraft, perfume atomizers, chimneys and sailboats rely on Bernoulli’s principle. This guide states and interprets the equation, applies it to the Venturi meter, Torricelli’s law, the Pitot tube and lift, discusses the limits and common misconceptions, compares the applications in a table, works through numerical examples and corrects common mistakes.

What does Bernoulli’s equation say?

For an incompressible, non-viscous fluid in steady flow, the work done by the pressure forces on a small volume of fluid moving along a streamline changes its kinetic and potential energy. This gives, between points 1 and 2 on the same streamline:

p1 + ½ρv12 + ρgh1 = p2 + ½ρv22 + ρgh2.

Each term has the units of pressure (Pa) or energy per unit volume (J/m3). The term ½ρv2 is the dynamic pressure. At constant height, an increase in speed means a decrease in static pressure: p + ½ρv2 = constant. In combination with the continuity equation, a narrower pipe has a higher speed and lower pressure. The principle holds for fluids with negligible viscosity, in steady flow, for incompressible fluids (for gases at speeds below about 30 percent of the speed of sound), along a single streamline, with no energy added or removed by pumps or turbines.

What are the main applications?

Venturi tube with pressure gauge tubes Δp 16.9 kPa p1 p2 1.5 m/s, 4.0 cm 6.0 m/s, 2.0 cm
A Venturi tube with the water levels in two open vertical tubes (schematic). In the narrow throat the water moves faster (6.0 m/s instead of 1.5 m/s), so the pressure is lower by ½ρ(v₂² − v₁²) = 16.9 kPa and the liquid in the throat tube stands lower.

How does it explain lift?

An airfoil is shaped and angled so that the air flows faster over the curved upper surface than under the lower one. By Bernoulli’s principle, the pressure on top is lower than the pressure below, and the pressure difference over the wing area gives an upward force. A complete explanation also includes Newton’s third law: the wing deflects the air downward (downwash), and the air pushes the wing up. Both descriptions are consistent. The “equal transit time” explanation, which says that air must travel the longer upper path in the same time as the lower, is incorrect: air over the top actually arrives at the trailing edge earlier. Lift increases with the square of the speed and with the angle of attack, up to the stall angle, when the flow separates from the wing.

Applications compared

Applications of Bernoulli’s principle
Application Where the speed is high Effect
Venturi tubeIn the narrow throatPressure drops, measures flow
Airplane wingOver the upper surfaceLower pressure above, lift
AtomizerAcross the top of a tubeLiquid is drawn up
Pitot tube—Stagnation pressure shows airspeed

Worked example: Venturi, tank and wing

  1. Venturi. Water flows at 1.5 m/s in a pipe of diameter 4.0 cm and at 6.0 m/s in the 2.0 cm throat (from the continuity equation). Δp = ½ρ(v22 − v12) = 0.5 × 1,000 × (36 − 2.25) = 16,875 Pa = 16.9 kPa lower in the throat.
  2. Torricelli. A hole is 2.0 m below the surface of the water in a tank. v = √(2gh) = √(2 × 9.8 × 2.0) = 6.26 m/s. The jet from a hole at the side lands at a horizontal distance determined by the height of the hole above the ground: for a hole 1.0 m above the ground, the fall time is √(2 × 1.0 ÷ 9.8) = 0.45 s, and the range is 6.26 × 0.45 = 2.8 m.
  3. Wing. Air (ρ = 1.2 kg/m3) flows at 70 m/s over the top and 60 m/s under the bottom of a wing of area 16 m2. Δp = ½ × 1.2 × (702 − 602) = 0.6 × 1,300 = 780 Pa. Lift = Δp × A = 780 × 16 = 12,480 N, enough to support a mass of 1,270 kg.
  4. Pitot tube. A Pitot tube on an aircraft measures a pressure difference of 600 Pa between the stagnation and static ports. v = √(2Δp ÷ ρ) = √(2 × 600 ÷ 1.2) = 31.6 m/s (114 km/h).
  5. Pressure in a rising pipe. Water flows at constant speed through a pipe that rises by 3.0 m. By Bernoulli, with the same speed at both points, p2 = p1 − ρgΔh = p1 − 1,000 × 9.8 × 3.0 = p1 − 29.4 kPa.

Who discovered the principle?

Daniel Bernoulli published Hydrodynamica in 1738, relating pressure and speed. Leonhard Euler derived the equation in its modern form in 1752. Giovanni Venturi investigated the flow in a constricted pipe in 1797. Henri Pitot invented the tube in 1732 to measure the flow in the Seine. The Wright brothers’ wind tunnel experiments (1901) and Ludwig Prandtl’s boundary layer theory (1904) developed the understanding of lift.

Which mistakes are common?

Mistake 1: Applying the equation across streamlines or to rotational flow without caution

It holds along a streamline (or throughout irrotational flow).

Mistake 2: Using the equal transit time explanation for lift

It is not correct. The air over the top is faster for other reasons, linked to the circulation around the wing.

Mistake 3: Saying that the higher pressure causes high speed in a narrow section

The pressure falls in the narrow section, and the pressure difference accelerates the fluid into it.

Mistake 4: Ignoring viscosity and compressibility

Real flows lose energy to viscosity, and at high Mach numbers density changes matter.