Properties of Solids and Liquids

For steady, incompressible, non-viscous flow along a streamline:

P + ½ρv² + ρgh = constant

That is: pressure energy + kinetic energy per volume + potential energy per volume is conserved. Direct consequence of energy conservation for fluids.

4 assumptions (violating any one breaks Bernoulli):

1. Steady flow — velocity at each point is time-independent.
2. Incompressible — ρ constant. Reasonable for liquids; questionable for gases above Mach 0.3.
3. Non-viscous — no internal friction.
4. Along a streamline — particles in a streamtube. Different streamlines can have different Bernoulli constants.

Continuity equation (mass conservation): for incompressible flow, A₁v₁ = A₂v₂.

Smaller cross-section → faster flow. This is why water shoots out of a hose nozzle.

Applications:

1. Venturi meter. Narrowing in a pipe increases v, decreases P. Pressure difference measures flow rate.
ΔP = ½ρ(v₂² − v₁²)

2. Aerofoil lift. Air over a curved wing surface travels faster (longer path) → lower pressure → lift force upward.

3. Torricelli's theorem. Speed of efflux from a hole at depth h in an open tank:
v = √(2gh) — same as a freely-falling body from height h.

4. Pitot tube. Measures airspeed of aircraft. Compares total (stagnation) pressure with static pressure.

5. Carburetors and atomizers. Fast air at constricted neck reduces pressure → sucks fuel/perfume into the airstream.

Worked example. Water flows through a horizontal pipe of cross-section 4 cm² at speed 3 m/s. The pipe narrows to 1 cm². Find the speed and pressure drop in the narrow section.

By continuity: A₁v₁ = A₂v₂ → 4 × 3 = 1 × v₂ → v₂ = 12 m/s.

Bernoulli (horizontal, so ρgh terms cancel): P₁ + ½ρv₁² = P₂ + ½ρv₂².
ΔP = P₁ − P₂ = ½ρ(v₂² − v₁²) = ½ × 1000 × (144 − 9) = 67,500 Pa = 67.5 kPa.

That's why high-pressure water lines avoid sudden narrowings — both for energy loss and pipe stress.