Bernoulli's Equation: Derivation and Limits

1. Setting and assumptions

Four assumptions, all essential: the flow is steady, incompressible, inviscid between the two points considered, and the points lie on one streamline. Each assumption removes a term from the full Navier–Stokes equation; what remains integrates exactly.

2. Derivation — Euler's equation along a streamline

Write the streamwise momentum equation for an element of length ds along a streamline, with s the arc coordinate and z(s) its elevation. Forces per unit volume: pressure gradient −∂p/∂s and the streamwise component of gravity −ρg dz/ds. With steady flow the acceleration is purely convective, v dv/ds:

Every term is a perfect derivative in s. Integrate along the streamline with ρ constant:

Each term is an energy per unit volume: pressure work capacity, kinetic energy, potential energy. Dividing by ρg gives the equivalent head form used by hydraulic engineers, with each term in metres.

3. Reading it physically

Where a streamline's velocity rises, its pressure must fall (same elevation) — the mechanism behind venturi meters, aerofoil lift, atomizers and the curve of a spinning ball. The trade is not mystical: to accelerate fluid, a net pressure force must push it, so pressure ahead must be lower.

4. Where it fails

Across a pump or turbine (energy added or removed), through long pipes (viscous head loss — extend with the Hagen–Poiseuille or Darcy–Weisbach terms), in boundary layers (viscosity dominant), and in gases beyond roughly Mach 0.3 (compressibility). A computed P₂ below absolute zero signals cavitation — the equation announcing its own breakdown. Explore with the Bernoulli calculator.

5. Extended worked examples

Example 1 — venturi meter. Water passes from D₁ = 10 cm (v₁ = 2 m/s, by continuity from Q) to a D₂ = 5 cm throat, so v₂ = 8 m/s. Same elevation: Δp = ½ρ(v₂² − v₁²) = 0.5 × 998 × (64 − 4) = 29 940 Pa ≈ 29.9 kPa. Measure that Δp with a manometer and invert the formula: you have a flow meter with no moving parts — the venturi, still standard in water works a century after its invention.

Example 2 — pitot tube (how aircraft know their speed). A pitot tube brings oncoming air to rest, converting dynamic pressure to a measurable difference: q = ½ρv². An airliner's sensor reads q = 500 Pa at low altitude (ρ = 1.225): v = √(2q/ρ) = √(816) = 28.6 m/s. At cruise the same principle holds with the local ρ — which is why blocked pitot tubes (ice, insects) have caused crashes: the aircraft literally loses this equation.

Example 3 — Torricelli's jet. A tank drains through a small hole 5 m below the surface. Streamline from the quiet surface (p_atm, v ≈ 0, z = 5) to the jet (p_atm, v, z = 0): ½v² = gz → v = √(2×9.81×5) = 9.90 m/s — the speed of free fall from the same height, because the fluid is doing exactly that, collectively.

Example 4 — cavitation check in a pump inlet. Water at 20 °C (vapour pressure 2.34 kPa) approaches a pump at v₁ = 1 m/s, p₁ = 101.3 kPa abs. The impeller eye accelerates it to 15 m/s at the same level: p₂ = p₁ + ½ρ(v₁² − v₂²) = 101 300 + 499(1 − 225) = 101 300 − 111 800 = −10 500 Pa. A negative absolute pressure is impossible: long before reaching it the water flashes to vapour at 2.34 kPa — cavitation bubbles form, then collapse violently against the blades. Bernoulli did not fail; it predicted its own limit, and pump engineers run this exact calculation as NPSH (net positive suction head).

6. Common misconceptions

"Faster flow causes low pressure." Backwards causality. A pressure gradient causes acceleration (Newton); Bernoulli records the energy consequence. The popular equal-transit-time explanation of wing lift — air "must" race over the longer top surface — is simply false (experiments show top-surface air arrives earlier); lift is correctly the pressure field required to turn the flow, with Bernoulli as accountant, not agent.

"Bernoulli applies between any two points." Only along a streamline (or throughout irrotational flow). Comparing a point inside a jet with a point in the still room air across the shear layer is illegitimate — the constant differs streamline to streamline once rotation or losses intervene.

"Add pump pressure into the same equation." A machine adds or removes mechanical energy; the balance needs an explicit work term (the extended Bernoulli/energy equation), otherwise the bookkeeping is broken.

"It's exact for water because water is incompressible." Incompressibility is one assumption of four. In a long pipeline the inviscid assumption fails first — friction converts the "constant" steadily into heat, and the head line slopes downhill at exactly the Darcy–Weisbach rate.

7. Where this shows up

Flow measurement is Bernoulli's kingdom: venturis, orifice plates and pitot-statics together meter most of the world's water, gas and flight speeds. Medicine uses the same algebra: Doppler echocardiography converts a jet velocity across a narrowed heart valve into a pressure drop via the simplified Δp ≈ 4v² (mmHg, m/s) — a bedside Bernoulli. Atomizers, carburettors, chimney draft, the curve of a football free kick, prairie-dog burrow ventilation (wind over a raised mound sucks air through the tunnel) — each is the pressure-velocity trade in costume. Even the roof lifted off in a storm follows: fast air above, still air inside, net upward Δp = ½ρv² ≈ 0.5×1.225×40² ≈ 980 Pa — a tonne of lift per 10 m².

8. Practice problems

P1. Wind at 25 m/s blows across a chimney top; inside air is still. Pressure deficit driving the draft? Solution: Δp = ½×1.225×625 = 383 Pa.

P2. Water in a horizontal pipe: p₁ = 200 kPa, v₁ = 3 m/s; downstream v₂ = 9 m/s. Find p₂. Solution: p₂ = 200 000 + 499(9 − 81) = 200 000 − 35 928 = 164.1 kPa.

P3. A fire hose nozzle points straight up; water leaves at 20 m/s. Ideal maximum height? Solution: h = v²/2g = 400/19.62 = 20.4 m — air drag will steal several metres in practice; name the assumption that broke.

P4. An aortic-stenosis jet measures 4 m/s on Doppler. Estimate the valve pressure gradient. Solution: Δp ≈ 4v² = 64 mmHg (≈ 8.5 kPa) — clinically severe; the cardiologist's Bernoulli.

9. Going deeper

For unsteady irrotational flow the derivation survives with one extra term: introducing the velocity potential u = ∇φ, the momentum equation integrates to ∂φ/∂t + p/ρ + ½|u|² + gz = f(t) — the unsteady Bernoulli equation, the tool for sloshing tanks, water hammer onset and wave theory. For compressible gas flow, replace p/ρ by the enthalpy integral ∫dp/ρ and the same streamline integration produces the compressible Bernoulli/Saint-Venant relations, the gateway to nozzle gas dynamics. Both extensions preserve the lesson of these notes: Bernoulli is not a formula but a first integral of momentum, and every added physical effect either respects that integral or leaves a named, calculable footprint on it.

10. Historical context

Daniel Bernoulli's Hydrodynamica (1738) contains the physics — he measured flowing blood pressure with open standpipes and understood the pressure–velocity trade — but the clean equation is Euler's (1752), who first wrote the momentum equations the theorem integrates. The family drama is real: Daniel's father Johann published a competing Hydraulica, backdated to 1732, attempting to claim priority over his own son. The instruments came fast: Pitot's tube (1732) predates the theory's final form; Venturi's converging-diverging meter followed in 1797; and by the twentieth century the equation was aviation's daily bread. The "head" language of hydraulic engineers — pressure head, velocity head, elevation head, each in metres — is the equation divided by ρg, and the hydraulic and energy grade lines drawn on every pipeline profile are Bernoulli's constant made visible, sloping downward at exactly the rate friction steals it.

11. Another way to see it: work–energy on a stream tube

Follow a slug of fluid occupying a thin stream tube between stations 1 and 2. In time dt, pressure p₁ pushes the rear face through a distance v₁dt, doing work p₁A₁v₁dt; the front face does negative work −p₂A₂v₂dt against p₂. By continuity A₁v₁ = A₂v₂ = Q, so net pressure work = (p₁ − p₂)Q dt. The work–energy theorem equates this to the change in kinetic plus gravitational energy of the slug, which (steady flow) is the difference between what exits and what enters: ρQ dt[½v₂² − ½v₁² + g(z₂ − z₁)]. Divide by Q dt:

— Bernoulli as bookkeeping of work done by pressure. This derivation exposes the physical meaning of each term (p is flow work per unit volume, available energy, not merely a force), makes the inviscid assumption transparent (friction would siphon work into heat), and extends naturally: insert a pump adding w_s per unit volume or a loss term h_L and you have the engineering energy equation used to size every pumping station on earth.

12. Frequently asked questions

Why does the shower curtain attack me? Partly Bernoulli (fast water-driven air sheet at lower pressure), partly a buoyant hot-air chimney, partly a driven vortex — controlled studies implicate the vortex most. A good reminder that "Bernoulli explains everything with moving air" is a reflex to resist.

A ping-pong ball hovers in a hair-dryer jet — Bernoulli? The stable trapping is genuinely pressure-field physics: the ball sitting off-centre sees faster flow on the jet side, lower pressure, and is pushed back in (with the Coandă effect helping the jet wrap the ball). The support against gravity is mostly direct momentum flux. Layered physics, honest decomposition.

So how do planes really fly? The wing turns air downward; the reaction is lift (momentum), and the associated pressure field — low above, high below — is consistent with Bernoulli along streamlines (energy). Momentum and energy accounts of one phenomenon, both correct; only the equal-transit-time story is wrong.

Why quote pressures in "metres of head"? Divide the equation by ρg and every term is a length — convenient when the driving quantity is a height (reservoirs, water towers) and when drawing grade lines on terrain profiles. A pump rated "40 m head" adds 40ρg ≈ 392 kPa to water regardless of flow-loop details.

Can Bernoulli hold in a rotating frame? Yes, per streamline with a centrifugal potential term added — the working form for turbomachinery passages and tropical-cyclone balance models alike.

13. Further practice

P5. A siphon crests 1.5 m above the reservoir surface and discharges 3 m below it. Find exit speed and crest pressure (water, ideal). Solution: exit v = √(2g×3) = 7.67 m/s; at the crest, p = p_atm − ρg(1.5) − ½ρv² = 101.3 − 14.7 − 29.4 = 57.2 kPa abs — comfortably above vapour pressure, so this siphon runs; raise the crest past ~9 m and it cavitates dead.

P6. A venturi with area ratio 2.5 reads Δp = 12 kPa on water. Throat and inlet velocities? Solution: v_t = 2.5v_i; Δp = ½ρ(v_t² − v_i²) = ½×998×5.25v_i² → v_i = √(24 000/998/5.25/0.5×?) — cleanly: v_i² = 2Δp/(ρ(2.5²−1)) = 24 000/(998×5.25) = 4.58 → v_i = 2.14 m/s, v_t = 5.35 m/s.

P7. Estimate the pressure rise at a building's windward face in a 20 m/s storm and the force on a 2 m × 3 m window. Solution: stagnation q = ½×1.225×400 = 245 Pa; F ≈ 245×6 = 1 470 N — a person's weight leaning on the glass, from the humblest term of the equation.

14. Worked exam problem

Problem. Design check for a courtyard fountain: the jet must reach 6 m. The 2 cm nozzle sits at ground level. Find (a) the required exit velocity, (b) the flow rate, (c) the gauge pressure needed in the supply line just before the nozzle (line velocity negligible).

Solution. (a) Torricelli inverted: v = √(2gh) = √(2×9.81×6) = 10.85 m/s. (b) Q = vA = 10.85×π(0.01)² = 3.41×10⁻³ m³/s ≈ 3.4 L/s. (c) Streamline from the slow, pressurised line to the free jet: p_gauge = ½ρv² = 0.5×998×117.7 = 58.7 kPa — equivalently 6 m of head, which is the answer you could have written immediately: the pressure head converts wholly to elevation head at the jet's peak. Real fountains add ~20–30% for nozzle and pipe losses — the inviscid assumption's known bill.

15. Key takeaways

Bernoulli is the first integral of momentum along a streamline for steady, inviscid, incompressible flow: p + ½ρv² + ρgz constant — pressure, kinetic and potential energy per volume trading places. Check all four assumptions before use; add explicit work and loss terms for machines and friction; never compare across streamlines casually; and read "fast means low pressure" as accounting, not causation.

16. Where to go next

Its parent momentum balance is Navier–Stokes; its partner constraint is continuity; the friction that erodes its constant is quantified from the Hagen–Poiseuille analysis onward. The Bernoulli calculator solves two-station problems with every step visible, and the formula page holds the compact reference.