What is Bernoulli's Principle?
Bernoulli's Principle states that for an ideal (non-viscous, incompressible) fluid in steady flow, the total mechanical energy per unit volume remains constant along a streamline. This means that where fluid speed increases, pressure decreases, and vice versa. It is derived from the law of conservation of energy applied to fluid flow.
Core Fluid Mechanics Formulas
| Symbol | Quantity | Unit | Notes |
|---|
| P | Static pressure | Pa (N/m²) | Force per unit area by fluid |
| ρ | Fluid density | kg/m³ | Water = 1000; Air = 1.225 |
| v | Flow velocity | m/s | Magnitude of fluid velocity |
| h | Height above datum | m | Potential energy term |
| g | Gravitational acceleration | m/s² | 9.8 m/s² on Earth |
| Q | Volume flow rate | m³/s | Q = Av (constant for incompressible flow) |
| A | Cross-sectional area | m² | Larger area → slower flow |
| ΔP | Pressure difference | Pa | Measured by manometer in Venturi meter |
Key Concepts & Real-World Applications
The Three Energy Terms
Bernoulli's equation is fundamentally an energy conservation equation for fluid flow. Each term has units of pressure (Pa = J/m³):
Aircraft Wing (Aerofoil)
Air flows faster over the curved upper surface of a wing than the flat lower surface. By Bernoulli, higher velocity means lower pressure on top. The pressure difference creates lift force upward.
Venturi Meter
A pipe narrows at the throat (smaller A). By continuity, v increases; by Bernoulli, P decreases. The pressure difference ΔP is measured by a manometer and used to calculate flow rate Q.
Atomizer / Spray Bottle
Fast-moving air across the top of a tube creates low pressure. Liquid rises up the tube and is carried away as a spray. Used in perfume bottles, carburettors, and paint sprayers.
Bunsen Burner
Gas flows rapidly through a narrow nozzle, creating low pressure. This draws air through the side holes by Bernoulli suction. The air-gas mixture creates a clean, hot flame.
Magnus Effect (Spinning Ball)
A spinning ball drags air with it. On one side, ball and air move together (faster); on the other, they oppose (slower). The pressure difference causes lateral deflection — used in swing bowling and free kicks.
Blood Flow in Arteries
At an arterial constriction (stenosis), blood velocity increases and pressure drops. The reduced pressure can cause arterial walls to collapse inward, a real medical concern. Bernoulli's principle helps model this.
Assumptions and Limitations
Bernoulli's equation is valid under specific conditions: (1) Steady flow (velocity at any point doesn't change with time). (2) Incompressible fluid (constant density; valid for liquids and low-speed gases). (3) Non-viscous fluid (no friction losses). (4) Flow along a streamline. Real fluids (water, oil, air at high speed) deviate from ideal Bernoulli behaviour due to viscosity, turbulence, and compressibility.
Solved Examples (NCERT / JEE Level)
Example 01 — Pressure Drop in a Constricted Pipe
Water flows through a horizontal pipe. At point 1 (radius 10 cm), the velocity is 2 m/s and pressure is 2 × 105 Pa. Find the pressure at point 2 where the radius narrows to 5 cm. (ρ = 1000 kg/m³)
By continuity: A₁v₁ = A₂v₂
A₁ = π(0.10)² = 0.03142 m²
A₂ = π(0.05)² = 0.007854 m²
v₂ = A₁v₁ / A₂ = 0.03142 × 2 / 0.007854 = 8 m/s
Bernoulli (horizontal, h₁ = h₂ = 0):
P₂ = P₁ + ½ρ(v₁² − v₂²)
P₂ = 2 × 10&sup5; + ½ × 1000 × (4 − 64)
P₂ = 200000 + 500 × (−60)
P₂ = 200000 − 30000
P₂ = 1.70 × 10&sup5; Pa | Pressure drops by 30 kPa where velocity quadruples
Example 02 — Torricelli's Theorem
A large tank holds water with a water level 3.2 m above a small hole at the bottom. Calculate the speed of water emerging from the hole and the time to fall 1 m below the hole. Take g = 10 m/s².
Torricelli's theorem: v = sqrt(2gh)
v = sqrt(2 × 10 × 3.2)
v = sqrt(64)
v = 8 m/s
The water emerges horizontally. Time to fall 1 m:
1 = ½ × 10 × t²
t² = 0.2 → t = 0.447 s
Horizontal distance = v × t = 8 × 0.447 = 3.58 m
Speed of efflux = 8 m/s | Horizontal throw distance = 3.58 m in 0.447 s
Example 03 — Venturi Meter Flow Rate
A Venturi meter has inlet diameter 8 cm and throat diameter 4 cm. The pressure difference is 5000 Pa. Find the volume flow rate for water (ρ = 1000 kg/m³).
A₁ = π(0.04)² = 5.027 × 10−3 m²
A₂ = π(0.02)² = 1.257 × 10−3 m²
Q = A₁A₂ × sqrt(2ΔP / (ρ(A₁² − A₂²)))
A₁² = 2.527 × 10−5; A₂² = 1.580 × 10−6
A₁² − A₂² = 2.369 × 10−5
2ΔP / (ρ × ΔA²) = 10000 / (1000 × 2.369 × 10−5) = 421.9
Q = 5.027 × 10−3 × 1.257 × 10−3 × sqrt(421.9)
Q = 6.319 × 10−6 × 20.54 = 1.298 × 10−4 m³/s
Q = 1.30 × 10−4 m³/s = 0.130 L/s = 7.8 L/min
Frequently Asked Questions (FAQs)
Why does pressure decrease when fluid velocity increases?
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This is a consequence of energy conservation. The total energy per unit volume of a fluid (P + ½ρv² + ρgh) must remain constant along a streamline. When velocity (kinetic energy) increases, pressure (pressure potential energy) must decrease to keep the total constant. A familiar analogy: a garden hose — when you partially block the opening (reducing area), water speeds up and you can feel the reduced pressure near the constriction.
What is the continuity equation and why does it matter in Bernoulli problems?
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The continuity equation A₁v₁ = A₂v₂ comes from the conservation of mass. For an incompressible fluid in a pipe, the volume flow rate Q (m³/s) is constant. A narrower pipe (smaller A) forces higher velocity v. The continuity equation is almost always used together with Bernoulli's equation in JEE problems — continuity gives v₂ in terms of v₁, which is then substituted into Bernoulli to find pressures.
What is Torricelli's theorem and how is it derived?
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Torricelli's theorem gives the speed of efflux (outflow) from a hole in a tank: v = sqrt(2gh), where h is the height of the fluid surface above the hole. It is derived by applying Bernoulli's equation between the tank surface (point 1: P = P₀, v ≈ 0, height h) and the hole (point 2: P = P₀, v = efflux speed, height 0). The atmospheric pressures cancel, giving ½ρv² = ρgh, so v = sqrt(2gh). This is analogous to free-fall — the fluid exits at the same speed as an object dropped from height h.
How does a Venturi meter measure flow rate?
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A Venturi meter is inserted into a pipe and has a gradually narrowing throat section. By the continuity equation, fluid speeds up at the throat. By Bernoulli's equation, pressure drops. A manometer measures the pressure difference ΔP between the inlet and throat. Since both the geometry (A₁, A₂) and fluid density (ρ) are known, the flow rate Q can be calculated precisely from Q = A₁A₂ × sqrt(2ΔP / (ρ(A₁² − A₂²))). Venturi meters are widely used in water treatment plants, oil pipelines, and HVAC systems.
Can Bernoulli's equation be applied to air? Does it explain how airplanes fly?
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Yes, Bernoulli's equation applies to air at low speeds (below ~0.3 Mach, where compressibility effects are negligible). The aerofoil (wing) is shaped so air must travel a longer path over the curved upper surface, causing it to speed up relative to the lower surface. By Bernoulli, the faster air on top has lower pressure, creating a net upward lift force. However, the full explanation of aircraft lift also involves the Kutta condition and circulation theory (angle of attack), which go beyond simple Bernoulli in modern aerodynamics.