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Apply Continuity First, Then Bernoulli

Moving fluids obey two relations. Continuity says the flow rate is the same everywhere in a pipe, and Bernoulli's equation says that where a fluid speeds up its pressure drops. Together they explain spray pumps, wings, roofs in the wind and water leaving a tank.

NSEJS · Properties of Matter and Fluid Mechanics · 9 linked questions · 100% coverage

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Continuity and Bernoulli

Continuity: A1 v1 = A2 v2 for an incompressible fluid, so the speed doubles when the cross-section halves;

in a tube whose area falls from 80 cm² to 40 cm², a speed of 3 m/s becomes 6 m/s.

For a spray pump with a tube of radius R feeding n fine holes of radius r, π R²

v = n π r² v', so the ejection speed is v' = v R² / (n r²)

Bernoulli: P + (1/2) ρ v² + ρ g h is constant along a streamline, so higher speed means lower pressure at the same height.

Flow out of a tank

Speed of efflux from a hole at depth h below the surface is v = sqrt(2 g h) (Torricelli), so a hole in the bottom of a 5 m tank gives v = 10 m/s and a flow of A v = 10⁻⁴ x 10 = 10⁻³ m³ per second for a 1 cm² hole. The jet then falls as a projectile. For a hole at depth y in a tank of water height h resting on the ground, the hole is a height (h - y) above the ground, so the range is x = 2 sqrt(y (h - y)), which is greatest at y = h/2 with x_max = h. If the tank stands on a platform, the hole is higher above the ground, so write the fall height as (platform height + h - y), recompute the range, and maximise that expression instead of reusing the half-height result.

Wings, roofs and sprayers

The pressure difference across a surface with air speeds v_upper and v_lower

ΔP = (1/2) ρ (v_upper² - v_lower²), and the force is ΔP x area

For a wing with speeds v below and sqrt(5) v above,

ΔP = (1/2) ρ (5 v² - v²) = 2 ρ v², so the dynamic lift is 2 ρ v² A

Wind blowing at 40 m/s over a roof of area 250 m² with atmospheric pressure inside gives a net upward force (1/2) x 1.2 x 40² x 250 = 2.4 x 10⁵ N. A scent sprayer works on the same principle: fast air across the top of a tube lowers the pressure there, and atmospheric pressure pushes the liquid up. A falling stream from a tap narrows because its speed increases while the flow rate stays fixed.

OlympIQ exam lens

✓Continuity: A v is constant along a pipe; a smaller cross-section means a higher speed
✓Bernoulli: P + (1/2) ρ v² + ρ g h is constant along a streamline
✓Efflux speed from a hole at depth h is sqrt(2 g h); flow rate = A sqrt(2 g h)
✓For a tank on level ground the jet range from a hole at depth y is 2 sqrt(y (h - y)), largest at half the liquid height; on a platform the fall height is larger, so recompute and maximise
✓Lift or net force = (1/2) ρ (v_upper² - v_lower²) x area; sprayers and wings use the same idea

OlympIQ trap

Do not apply Bernoulli's equation across two points that are not on the same streamline of a steady flow, and do not forget the continuity step. Find the speeds from A v = constant first, then use the pressure formula.

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