Virtual laboratory/Physics/Hydrostatics/Bernoulli's principle — pressure in a flow

Bernoulli's principle — pressure in a flow

Narrow the pipe and watch what happens to the fluid. From the continuity equation the flow speed rises at the narrow point, and from Bernoulli's principle p+½ρv²=const a rising speed means a drop in pressure. This is how an aerofoil and an atomiser work.

Bernoulli's principle — pressure in a flow
Physics · Hydrodynamics · Bernoulli's equation (FIZ.HYDRO.PP.02)
Pipe cross-section at the constriction SS2.6 cm²
In a pipe constriction (smaller cross-section, faster flow) the fluid pressure:
cross-section S
2.6 cm²
speed v
1.92 m/s
static press. p
8.2 kPa
dynamic press. ½ρv²
1.8 kPa
🎯 Challenge: hit the speed
streak: 0

I'll give you a target speed in the constriction — tune the pipe cross-section to reach it. S·v = const.

What just happened?

You narrowed the pipe, the fluid stream sped up and the static pressure dropped. It is not magic — it is a direct consequence of energy conservation for a flowing fluid.

Continuity equation (incompressible fluid):

S⋅v=constS \cdot v = \text{const}

— the same amount of fluid passes through every pipe cross-section in the same time. A smaller cross-section SS ⇒ a higher speed vv. That is why the streamlines bunch up in the constriction.

Bernoulli's equation (horizontal pipe):

p+12ρv2=constp + \tfrac{1}{2}\rho v^2 = \text{const}

— the sum of the static pressure pp and the dynamic pressure 12ρv2\tfrac12\rho v^2 is constant. As the speed rises, the "dynamic" term takes more and more, so the static pressure pp must drop. ρ\rho — the fluid density.

What does this mean in practice?

Fast flow = low pressure. This is how a carburettor and an atomiser work (the Venturi effect), how an aircraft gains lift, and why two boats sailing close together are "sucked" toward each other.

Why you need this for the exam:

In pipe-constriction problems, first use S⋅v=constS\cdot v=\text{const} to find the speed in the constriction, then p+12ρv2=constp+\tfrac12\rho v^2=\text{const} to compute the pressure. Trick: where it is tighter and faster, the pressure is lower.

Remember: faster-flowing fluid exerts lower pressure — p + ½ρv² = const. In the constriction the speed rises and the pressure drops.

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Theory & formulas

Continuity equation

For an incompressible fluid the product of cross-sectional area and speed is constant: S v=constS\,v = \text{const} In a constriction (smaller SS) the speed vv must rise.

Bernoulli's principle

The sum of the static and dynamic pressure along a streamline is constant (at one level): p+12ρv2=constp + \tfrac{1}{2}\rho v^2 = \text{const} Since vv rises in the constriction, the static pressure pp must drop.

Applications

Aerofoil lift, the carburettor, the atomiser, the Venturi tube, a "curved" ball.

Notation: SS — pipe cross-sectional area, vv — flow speed, pp — static pressure, ρ\rho — fluid density, 12ρv2\tfrac{1}{2}\rho v^2 — dynamic pressure.

Frequently asked questions

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Why does the pressure drop in a constriction?

Because the sum of the static and dynamic pressure is constant (p+½ρv²=const); as the speed v rises, the static pressure p must drop.

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Why does the speed rise in a constriction?

From the continuity equation S·v=const — a smaller cross-section S forces a higher speed v so that the flow is conserved.

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Where does Bernoulli's principle apply?

Aerofoil lift, the carburettor, the atomiser, the Venturi tube, a "curved" ball.

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