The Bernoulli Equation
Integrate Euler's momentum equation along a streamline, and a constant of the motion falls out.
Every plumber, pilot, and windsurfer relies on one equation: where a fluid speeds up, its pressure drops. The relation is so useful it has its own name — Bernoulli — and it is everywhere: atomizers, carburetors, roof lift in a storm, the curve of a baseball. The surprise is where it comes from. Not from energy bookkeeping, but from the same momentum balance you just wrote: Euler's equation, integrated along a streamline.
Integrating Euler's equation for an ideal fluid
Take the differential relation from Euler's equation — dp/ρ + V dV + g dz = 0 — and integrate it along a streamline between two points. Three restrictions make the integral easy and exact: the flow is steady (nothing changes with time), the fluid is incompressible (constant ρ, so 1/ρ comes outside the pressure integral), and the fluid is inviscid (Euler's equation applies in the first place). Under those three conditions the integration gives a single constant of the motion along the streamline.
The result, per unit mass, is the form most physics courses reach first. Engineers more often divide through by g, turning every term into a length in metres — a 'head'. Heads are wonderfully concrete: a velocity head of 0.5 m literally means the kinetic energy of the flow is equivalent to a 0.5 m column of the fluid, and you can read each term straight off a manometer. Both forms are the same equation; pick whichever your problem wants.
Bernoulli is powerful precisely because it is restrictive. Check all five before you use it:
- Steady — the flow does not change with time at any point.
- Incompressible — density is constant (an excellent approximation for liquids, and for gases below roughly Mach 0.3).
- Inviscid (frictionless) — negligible viscous effects, so no energy is dissipated to internal energy. (This is the assumption real flows violate most often.)
- Along a single streamline — and, for the constant to be the same everywhere, the flow must additionally be irrotational.
- No shaft work and no heat transfer between the two points — no pump, turbine, or heater doing work on the fluid.
When friction or a pump is present, Bernoulli in this form is simply the wrong tool — and that is exactly what Lessons 4 and 5 are for.
Here is a subtlety worth pausing on. For an ideal fluid (inviscid, adiabatic, no shaft work), applying the First Law of Thermodynamics to a control volume happens to produce the same algebraic result as integrating Euler's momentum equation. That is reassuring — physics must be consistent — but it is a consistency check, not a derivation, and it holds only in the frictionless limit.
The moment you add friction, the two roads diverge. The momentum road (Euler → Bernoulli) cannot see dissipation at all; it simply stops being valid. The thermodynamic road keeps going, because the First Law explicitly tracks internal energy and heat — so friction shows up naturally as 'head loss.' That is why the general Energy Equation in Lesson 4 is a genuinely different and more powerful object, not merely 'Bernoulli with losses tacked on.' Keep the threads separate and the whole module will click.
- Apply Bernoulli between 1 and 2 along the streamline, with z₁ = z₂ (horizontal): p₁/ρ + V₁²/2 = p₂/ρ + V₂²/2.
- Solve for p₂: p₂ = p₁ + (ρ/2)(V₁² − V₂²).
- Substitute: p₂ = 300{,}000 + (1000/2)((3)² − (12)²) = 300{,}000 + 500(9 − 144).
- Compute: 500 × (−135) = −67{,}500, so p₂ = 300{,}000 − 67{,}500 = 232{,}500 Pa.
- Check the sign: speed rose from 3 to 12 m/s, so pressure must fall — and it does, by 67.5 kPa.
- p₂ = p₁ + (ρ/2)(V₁² − V₂²) = 250{,}000 + 500(16 − 64) = 250{,}000 − 24{,}000 = 226{,}000 Pa.
- p₂ = 226 kPa.
- Pressure head: p/ρg = 200{,}000/(1000 × 9.81) = 20.39 m.
- Velocity head: V²/2g = (3)²/(2 × 9.81) = 9/19.62 = 0.46 m.
- Total head H = 20.39 + 0.46 + 1.0 = 21.85 m.
Check your understanding
- Bernoulli's equation is Euler's momentum equation integrated along a streamline for steady, incompressible, inviscid flow.
- In head form: p/ρg + V²/2g + z = H — pressure head + velocity head + elevation head = total head, conserved for an ideal fluid.
- It requires steady flow, incompressibility, no viscosity, no shaft work, and no heat transfer — and applies along a streamline.
- For an ideal fluid the thermodynamic First Law gives the same result, but only as a consistency check; the two derivations diverge once friction appears.
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