Bernoulli Equation
The Bernoulli equation is energy conservation for flowing fluids: where the speed rises, the static pressure drops.
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Formula
p + \frac{1}{2} \rho v^2 + \rho g h = \text{const.}Variables & units – Bernoulli Equation
| Symbol | Meaning | Unit |
|---|---|---|
| p | Static pressure in the fluid | Pa |
| ρ | Density of the fluid | kg/m³ |
| v | Flow speed | m/s |
| g | Gravitational acceleration (9.81 m/s²) | m/s² |
| h | Height above the reference level | m |
Derivation & background – Bernoulli Equation
Daniel Bernoulli published the equation in 1738 in his "Hydrodynamica". It is the energy-density balance along a streamline: static pressure, dynamic pressure ½ρv² and elevation pressure ρgh add up to a constant. Valid for steady, frictionless, incompressible flow. In a pipe constriction v rises (continuity equation A₁v₁ = A₂v₂), so p falls there, the Venturi effect. The special case of outflow from a tank yields the Torricelli formula v = √(2gh).
Exam blueprint
Validity range
Holds along a streamline for steady, frictionless, incompressible flow. With strong viscosity, turbulence or compressible gases (high speeds) it is only an approximation.
Derivation steps
Energy conservation for a flowing volume element: pressure work converts into kinetic and potential energy.
- 1The pressure forces do the work (p₁ − p₂)·V on the volume element.
- 2Equating with ΔE_kin + ΔE_pot and dividing by V yields p + ½ρv² + ρgh = const.
Rearrangements
Speed from pressure difference
Horizontal flow; the measuring principle of the Pitot tube.
Efflux speed (Torricelli)
Special case: open tank, hole at depth h.
Dynamic pressure
The part that appears as pressure when the flow is brought to rest.
Task variant
Water stands 5 m above a small hole in a tank. At what speed does it flow out?
Torricelli: v = √(2gh) = √(2 × 9.81 × 5) = √98.1 ≈ 9.9 m/s.
Wind hits a wall at 30 m/s (ρ_air = 1.2 kg/m³). What is the dynamic pressure?
q = ½ρv² = 0.5 × 1.2 × 900 = 540 Pa, about one two-hundredth of atmospheric pressure.
Common mistakes
Confusing static pressure p and total pressure.
p is only the static part; the sum of p, dynamic and elevation pressure is constant.
Forgetting the continuity equation.
Speeds follow from A₁·v₁ = A₂·v₂; only then use Bernoulli for the pressures.
Applying the equation to strongly viscous flow (narrow long pipes).
Friction losses dominate there; real pressures fall below the Bernoulli values.
Exam context
- Typical: Venturi tube with continuity plus Bernoulli, Torricelli efflux and qualitative explanations (atomiser, aerodynamic lift) via the pressure drop at high speed.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Fluid dynamics
Connects pressure and density with energy conservation of moving fluids.
Worked example
Water (ρ = 1000 kg/m³) flows horizontally from v₁ = 2 m/s to v₂ = 8 m/s into a constriction: p₂ = p₁ + ½ρ(v₁² − v₂²) = 200 kPa + 0.5 × 1000 × (4 − 64) Pa = 170 kPa.
Applications
Airflow around wings, Venturi nozzles and carburettors, Pitot tube (aircraft speed measurement), water jet pump, atomisers
Quanta exam set
Curated exam set for "Bernoulli Equation":
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Which formula describes Bernoulli Equation?
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How do you rearrange p + ½ρv² + ρgh = const. for Speed from pressure difference?
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Which common mistake happens with Bernoulli Equation?
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Frequently asked questions about Bernoulli Equation
What does the Bernoulli equation say intuitively?+
It is energy conservation for flowing liquids and gases. Three energy densities share a fixed budget: the static pressure p (what a sensor drifting along would measure), the dynamic pressure ½ρv² (motion share) and the elevation pressure ρgh (height share). Their sum is constant along a streamline. This yields the famous core statement: where the flow speeds up, for instance in a constriction, the static pressure must drop, and vice versa. This seems paradoxical at first; many expect higher pressure in the narrow section. In fact it is precisely the pressure difference that accelerates the fluid into the constriction, pushing it from high towards low pressure.
How do you solve a typical Bernoulli problem?+
Choose two points on the same streamline and write the equation for both: p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂. Cancel what is equal, for horizontal flow the height terms. Missing speeds come from the continuity equation A₁·v₁ = A₂·v₂. Example: water at p₁ = 200 kPa flows at 2 m/s into a pipe that narrows so that v₂ = 8 m/s. Then p₂ = p₁ + ½ρ(v₁² − v₂²) = 200,000 + 500 × (4 − 64) = 170,000 Pa = 170 kPa. The pressure drops by 30 kPa although nothing is pumped, purely due to the acceleration into the constriction.
What is the Torricelli formula and how does it follow from Bernoulli?+
Torricelli describes the efflux speed from an open tank: v = √(2gh). It follows as a special case: compare the calm water surface (point 1) with the hole at depth h (point 2). Atmospheric pressure acts at both places, so the pressure terms cancel; for a large tank the surface sinks negligibly slowly, so v₁ ≈ 0. What remains is ρgh = ½ρv², solved as v = √(2gh). At h = 5 m the water exits at √(2 × 9.81 × 5) ≈ 9.9 m/s, exactly as fast as if it had fallen freely from 5 m. That is no coincidence: both calculations are the same energy conservation.
Does Bernoulli explain why aeroplanes fly?+
Partly, and the popular short version is often told wrongly. What is correct: air flows faster over the curved upper surface of the wing, so by Bernoulli the static pressure there is lower than below, and the pressure difference carries the aircraft. What is wrong is the widespread claim that two neighbouring air parcels must arrive at the trailing edge simultaneously; demonstrably they do not. The deeper cause of the faster flow lies in the circulation around the profile and the angle of attack: the wing deflects air downwards, and by Newton third law the reaction force acts upwards. Bernoulli and the momentum view are two consistent perspectives on the same physics, not competitors.
What are the limits of the Bernoulli equation?+
The equation assumes four idealisations: steady flow (no changes in time), absence of friction, incompressibility and validity along a streamline. In narrow, long pipes viscosity dominates and the real pressure drop exceeds the calculated one (Hagen-Poiseuille law). For gases Bernoulli only works while the density change stays small, as a rule of thumb up to about 0.3 times the speed of sound (roughly 100 m/s in air); beyond that compressible flow theory is needed. In turbulent eddies and across streamlines the equation likewise does not apply directly. For exams this means: Bernoulli for short, smooth flow paths with water or slow air, not for capillaries or supersonic flow.
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How do you calculate with Bernoulli Equation?
Here is how to work through a typical Bernoulli Equation (p + ½ρv² + ρgh = const.) task step by step:
- 1
Task
Water stands 5 m above a small hole in a tank. At what speed does it flow out?
Solution path
Torricelli: v = √(2gh) = √(2 × 9.81 × 5) = √98.1 ≈ 9.9 m/s.
- 2
Task
Wind hits a wall at 30 m/s (ρ_air = 1.2 kg/m³). What is the dynamic pressure?
Solution path
q = ½ρv² = 0.5 × 1.2 × 900 = 540 Pa, about one two-hundredth of atmospheric pressure.