A jumbo jet weighing 350 tonnes hangs in the air, held up by nothing but moving fluid. A pitcher makes a baseball swerve sideways in flight. A perfume atomiser lifts liquid up a tube using only a puff of air, and a partly-blocked artery can quietly balloon into an aneurysm. These wildly different events share one underlying rule: in a smoothly flowing fluid, where the flow speeds up, the pressure drops. That single sentence — made precise — is Bernoulli's theorem, and it falls straight out of Newton's second law once we learn to write it for a fluid.

This page does exactly that. First we apply \mathbf{F} = m\mathbf{a} to a tiny fluid parcel to get Euler's equation for an ideal (frictionless) fluid. Then, by integrating it along a streamline for steady incompressible flow, we squeeze out the famous conserved combination p + \tfrac12\rho v^2 + \rho g z. Finally we read that combination as an energy budget and use it to explain lift, the Venturi throat, and the Pitot tube that tells an aircraft how fast it is going.

Euler's equation: Newton's second law for a fluid parcel

Take a small blob of fluid — a parcel — of volume \delta V and mass \delta m = \rho\,\delta V. Newton's second law says its mass times its acceleration equals the net force on it. The subtle part is the acceleration: because the parcel moves with the flow, the acceleration it actually feels is the material derivative of the velocity, D\mathbf{u}/Dt, not the plain \partial\mathbf{u}/\partial t.

For an ideal (inviscid) fluid — one with no viscosity, so neighbouring parcels exert no sideways drag on each other — only two forces act on the parcel: the pressure of the surrounding fluid squeezing on its faces, and gravity. A pressure that varies across the parcel gives a net push from high pressure toward low, of size -\nabla p per unit volume; gravity contributes \rho\mathbf{g} per unit volume. Dividing Newton's law by \delta V gives Euler's equation:

Read it term by term and it is nothing but m\mathbf{a} = \mathbf{F} written per unit volume: \rho\,D\mathbf{u}/Dt is "mass × acceleration", while -\nabla p (the pressure-gradient force) and \rho\mathbf{g} (weight) are the forces. Euler derived it in 1757; drop the one thing it ignores — viscosity — back in and you get the full \text{Navier–Stokes} equations. So Euler is Navier–Stokes with the friction switched off, and it is the honest starting point for lift, waves, and flow through pipes.

From Euler to Bernoulli: integrating along a streamline

Now specialise to the case that covers an enormous fraction of real flows: steady (\partial\mathbf{u}/\partial t = 0), incompressible (\rho constant), and inviscid. Take Euler's equation and follow one streamline, the curve everywhere tangent to the flow. Using the vector identity (\mathbf{u}\cdot\nabla)\mathbf{u} = \nabla(\tfrac12 v^2) - \mathbf{u}\times(\nabla\times\mathbf{u}) and projecting onto the streamline direction (where the last term drops out, because it is perpendicular to \mathbf{u}), Euler's equation becomes a total derivative along the arc-length s:

If the derivative along the streamline is zero, the quantity in brackets does not change as you move along that streamline — it is a constant of the flow. That is Bernoulli's theorem.

Reading it as energy: pressure, kinetic, potential

Every term in Bernoulli's equation has units of energy per unit volume (\text{J/m}^3 = \text{Pa}). So the theorem is a statement of energy conservation for a parcel of ideal fluid: as it flows along a streamline, its energy budget simply shuffles between three pockets whose total never changes.

• \tfrac12\rho v^2 — the kinetic energy per unit volume, also called the dynamic pressure q. Speed up and this grows.
• \rho g z — the gravitational potential energy per unit volume. Climb and this grows.
• p — the static pressure, the flow-work the surrounding fluid can do; behaving like a stored pressure energy per unit volume.

The consequence is immediate and is the heart of the whole subject: at the same height, if a parcel speeds up, its \tfrac12\rho v^2 pocket fills, so to keep the total fixed its pressure pocket p must empty. Faster flow means lower pressure. Not because speed "causes" low pressure by magic, but because energy is conserved and it has to come from somewhere.

The Venturi: watch the trade happen

Push an incompressible fluid through a tube that narrows to a throat and widens again. Because the same volume per second must pass every cross-section (that is continuity, A_1 v_1 = A_2 v_2), the flow has to go faster where the tube is narrower. Bernoulli then insists the pressure there must be lower. In the figure the streamlines physically crowd together at the throat — a visual signature of faster flow — and that is exactly where a pressure gauge reads its minimum.

The graph below makes the trade-off quantitative. Along the tube, the relative flow speed rises to a peak at the throat while the relative pressure falls to a trough at the very same place — a mirror image. Their sum (kinetic + pressure energy, at constant height) is what stays constant.

Worked examples

Example 1 — pressure drop in a Venturi. Water (\rho = 1000\ \text{kg/m}^3) flows horizontally through a Venturi. In the wide inlet the speed is v_1 = 2\ \text{m/s} at pressure p_1 = 120\ \text{kPa}; at the throat continuity gives v_2 = 6\ \text{m/s}. Since the height is constant, the \rho g z terms cancel and Bernoulli reads

The pressure falls by 16\ \text{kPa} at the throat — enough to suck fuel into a carburettor or lift dye up a side-tube.

Example 2 — a Pitot tube measures airspeed. An aircraft's Pitot tube has one opening facing the flow, where the air is brought to rest (a stagnation point, v = 0, pressure p_0), and a side opening reading the undisturbed static pressure p where the air still moves at the flight speed v. Same height, so Bernoulli gives p_0 = p + \tfrac12\rho v^2, and solving for the speed,

With a stagnation-minus-static difference of p_0 - p = 600\ \text{Pa} in air of density \rho = 1.2\ \text{kg/m}^3, v = \sqrt{2(600)/1.2} = \sqrt{1000} \approx 31.6\ \text{m/s} — about 114\ \text{km/h}. The difference p_0 - p = \tfrac12\rho v^2 is exactly the dynamic pressure.

Example 3 — Torricelli's law (a leaking tank). A tank of water has a small hole a depth h below the free surface. Take a streamline from the calm top surface (pressure atmospheric, speed \approx 0) down to the jet at the hole (also at atmospheric pressure, since it exits into the air). The pressure terms cancel and Bernoulli reduces to \tfrac12\rho v^2 = \rho g h, so

For h = 5\ \text{m} and g = 10\ \text{m/s}^2, v = \sqrt{100} = 10\ \text{m/s}. Strikingly, this is the very speed the water would reach if it simply fell the height h — the pressure energy at depth has been perfectly cashed in for kinetic energy.