Bernoulli's Equation

Take the fluid lying between cross-sections 1 and 2 of a narrow stream tube and follow it for a short time Δt. At the back, the fluid behind pushes it forwards with force P₁A₁ through a distance v₁Δt, doing work P₁A₁v₁Δt = P₁ΔV. At the front it advances against P₂A₂ through v₂Δt, so the work done on it there is −P₂ΔV; continuity for an incompressible fluid guarantees the same ΔV at both ends. Because the flow is steady, everything between the two ends looks the same afterwards as before. The only net change is that a mass ρΔV has left section 1, with speed v₁ and height y₁, and appeared at section 2 with v₂ and y₂. Work–energy then reads (P₁ − P₂)ΔV = ½ρΔV(v₂² − v₁²) + ρΔVg(y₂ − y₁). Divide by ΔV and sort the subscripts: P₁ + ½ρv₁² + ρgy₁ = P₂ + ½ρv₂² + ρgy₂. No new physics was needed, only W = ΔK + ΔU with no friction term.

A pascal is a newton per square metre, and N m⁻² = J m⁻³, so all three terms are energy densities and can be compared directly. For water at 3.0 m s⁻¹ the kinetic term is ½ × 1000 × 3.0² = 4.5 kPa; lifting water through 1.0 m costs ρgy = 1000 × 9.81 × 1.0 = 9.8 kPa; the atmosphere supplies 101 kPa. In household plumbing, then, the speed term is a modest correction while a few metres of height matter a great deal. For air at 30 m s⁻¹, ½ρv² = ½ × 1.20 × 30² = 540 Pa, about 0.5% of atmospheric pressure, which is why air at such speeds barely compresses and the constant-density model still serves. Two bookkeeping rules follow. Pressure may be absolute or gauge, provided both points use the same kind, because the atmospheric offset cancels. Height may be measured from any datum, for the same reason. The term ½ρv² is often called the dynamic pressure, and P the static pressure.

When y₁ = y₂ the height terms cancel and the equation becomes a two-way trade between P and ½ρv². The speeds are not free: the continuity equation of the previous topic fixes v₂ = v₁A₁/A₂. Put the two together and P₁ − P₂ = ½ρv₁²[(A₁/A₂)² − 1]. Water that approaches at 1.2 m s⁻¹ a throat of one fifth the area passes through it at 6.0 m s⁻¹, and the pressure falls by ½ × 1000 × 1.2² × (25 − 1) = 1.7 × 10⁴ Pa. A drop that size is easy to read on a manometer, which turns the constriction into a flow meter, the Venturi meter. The direction of the effect is dictated by Newton's second law. A parcel entering the throat is accelerating forwards, so the net force on it points forwards, so the pressure behind it must exceed the pressure ahead. Where the pipe widens again the parcel decelerates and the pressure recovers; in an ideal fluid it recovers completely.

Open a small hole a depth h below the surface of a wide tank. Choose point 1 on the surface and point 2 in the emerging jet. Both touch the atmosphere, so P₁ = P₂; the surface is far larger than the hole, so v₁ ≈ 0. What is left is ρgh = ½ρv², giving Torricelli's result v = √(2gh), the speed of a body that has fallen freely through h. For h = 0.80 m, v = √(2 × 9.81 × 0.80) = 4.0 m s⁻¹, and the density has cancelled, so petrol would leave at the same speed. A siphon uses all three terms at once. With its outlet 1.2 m below the tank surface the water leaves at √(2 × 9.81 × 1.2) = 4.9 m s⁻¹, so in a tube of constant bore ½ρv² = ρg × 1.2 m = 11.8 kPa all along it. At a crest 0.50 m above the surface ρgy = 4.9 kPa, and the pressure is atmospheric less both terms: 101.3 − 4.9 − 11.8 = 84.6 kPa, below atmospheric. It is the atmosphere that pushes the water over the crest, which is why no water siphon can climb much more than 10 m.

Hold a blunt tube facing into a flow. The streamline that meets the centre of its mouth ends at a stagnation point, where the fluid has been brought to rest. With v = 0 there and no height change, Bernoulli gives the stagnation pressure P₀ = P + ½ρv²: the static pressure of the undisturbed stream plus the whole of its kinetic energy density. A Pitot-static probe senses P₀ at its nose and P through side holes that the flow slides past undisturbed; the difference is ½ρv², so v = √(2(P₀ − P)/ρ). A light aircraft in air of density 1.20 kg m⁻³ whose probe reads P₀ − P = 735 Pa is moving at √(2 × 735/1.20) = √1225 = 35.0 m s⁻¹ relative to the air. The same idea explains why the pressure on the upstream face of a bridge pier, or on a palm held out of a car window, exceeds the ambient pressure: it is the one place where the flow's kinetic energy has been handed back as pressure.

The derivation leaned on four things, and each can fail. Steady flow: the pattern must not change with time, so a valve being opened or a sloshing tank is excluded, although slow draining can be treated as steady at each instant. Incompressible: ρ is constant, safe for liquids and for gases slower than roughly 0.3 of the speed of sound, about 100 m s⁻¹ in air. Non-viscous: no mechanical energy becomes thermal energy. Water in a long level pipe of constant bore loses pressure steadily along its length although v and y do not change; Bernoulli predicts no loss at all, so that loss is pure viscosity, the subject of the next topic. One streamline, with no pump, fan or turbine between the two points: the constant belongs to a streamline, and different streamlines share one value only when they all begin in the same still reservoir or the same uniform stream, as they do ahead of a Pitot probe. Finally, distrust any answer in which the absolute pressure falls below the liquid's vapour pressure, 2.3 kPa for water at 20 °C: the liquid cavitates instead.