Prerequisites
Before studying Bernoulli’s theorem, make sure you understand:
- Fluid Pressure — Hydrostatic pressure
- Work-Energy — Energy conservation
- Viscosity — Streamline vs turbulent flow
What is Bernoulli’s Theorem?
Bernoulli’s Theorem is the law of energy conservation applied to fluids in motion.
For an ideal fluid (non-viscous, incompressible) in streamline flow, the total mechanical energy per unit volume remains constant along a streamline.
$$\boxed{P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}}$$Or, dividing by ρg:
$$\boxed{\frac{P}{\rho g} + \frac{v^2}{2g} + h = \text{constant}}$$Interactive Demo: Visualize Bernoulli’s Principle
Explore how pressure, velocity, and height are related in flowing fluids.
where:
- P = pressure (pressure energy per unit volume)
- ½ρv² = kinetic energy per unit volume
- ρgh = potential energy per unit volume
Ever wonder why airplanes fly?
Bernoulli’s principle!
Air flows faster over the curved top of the wing than the flat bottom. According to Bernoulli’s equation:
- Higher velocity (top) → Lower pressure
- Lower velocity (bottom) → Higher pressure
Pressure difference creates upward lift force that keeps the plane airborne!
A Boeing 747 generates about 1.8 million pounds of lift using this principle.
Assumptions for Bernoulli’s Equation
Bernoulli’s theorem applies when:
Fluid is incompressible (ρ = constant)
- True for liquids
- Approximately true for gases at low speeds (v « speed of sound)
Fluid is non-viscous (η = 0)
- No internal friction
- No energy loss to heat
Flow is streamline (laminar, not turbulent)
- Organized, smooth flow
- No eddies or vortices
Flow is steady (velocity at any point doesn’t change with time)
- ∂v/∂t = 0
No energy added or removed along streamline
- No pumps, turbines, or heat exchange
Real fluids have some viscosity and energy losses. Bernoulli’s equation gives approximate results for real fluids, with corrections needed for:
- Viscous drag
- Turbulence
- Compressibility (for gases at high speeds)
Derivation Using Energy Conservation
Consider a fluid element moving along a streamline from point 1 to point 2.
Work done BY pressure forces:
$$W_{pressure} = P_1 A_1 v_1 \Delta t - P_2 A_2 v_2 \Delta t$$Mass of fluid element:
$$\Delta m = \rho A_1 v_1 \Delta t = \rho A_2 v_2 \Delta t$$(continuity equation)
Work-energy theorem:
$$W = \Delta KE + \Delta PE$$ $$P_1 A_1 v_1 \Delta t - P_2 A_2 v_2 \Delta t = \frac{1}{2}\Delta m (v_2^2 - v_1^2) + \Delta m \cdot g(h_2 - h_1)$$Divide by ΔV = A₁v₁Δt = A₂v₂Δt:
$$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) + \rho g(h_2 - h_1)$$Rearranging:
$$\boxed{P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2}$$At any two points on a streamline!
Equation of Continuity
For incompressible fluid in streamline flow:
Mass flow rate is constant at all cross-sections.
$$\boxed{A_1 v_1 = A_2 v_2 = \text{constant}}$$ $$\boxed{Av = \text{constant}}$$where:
- A = cross-sectional area
- v = velocity of fluid
Physical meaning: Where pipe narrows, fluid flows faster!
When you partially cover the end of a garden hose with your thumb:
- Area decreases → Velocity increases (continuity equation)
- Water shoots out faster and goes farther
This is why firefighters use nozzles — to increase water velocity by reducing cross-section!
Water flows through a pipe of diameter 4 cm at 2 m/s. The pipe narrows to 2 cm diameter. Find velocity in narrow section.
Solution:
A₁v₁ = A₂v₂
$$\pi r_1^2 v_1 = \pi r_2^2 v_2$$ $$r_1^2 v_1 = r_2^2 v_2$$ $$(2)^2 \times 2 = (1)^2 \times v_2$$ $$v_2 = \frac{4 \times 2}{1} = \boxed{8 \text{ m/s}}$$Velocity quadruples when radius is halved!
Applications of Bernoulli’s Theorem
1. Torricelli’s Theorem (Efflux Velocity)
Problem: A tank filled with liquid has a small hole at depth h below the surface. Find velocity of liquid emerging from hole.
Solution:
Apply Bernoulli between surface (point 1) and hole (point 2):
At surface:
- P₁ = P₀ (atmospheric)
- v₁ ≈ 0 (large tank)
- height = h
At hole:
- P₂ = P₀ (jet exits to atmosphere)
- v₂ = v (to find)
- height = 0 (reference)
Torricelli’s Law: Velocity of efflux equals velocity of free fall from height h!
This is the same formula as for an object dropped from height h!
The liquid “falls” horizontally through the hole with the same speed it would gain falling vertically through distance h.
If your overhead water tank has a leak at the bottom (10 m below water surface):
$$v = \sqrt{2 \times 10 \times 10} = \sqrt{200} \approx 14 \text{ m/s}$$Water shoots out at 14 m/s (about 50 km/h) — that’s why leaks can spray water quite far!
2. Venturi Meter (Measuring Flow Rate)
Device: Horizontal tube with a constriction (narrow section) used to measure flow rate.
Setup:
- Wide section: area A₁, pressure P₁, velocity v₁
- Narrow section: area A₂, pressure P₂, velocity v₂
- Manometer measures pressure difference
Analysis:
From continuity: $A_1 v_1 = A_2 v_2$
From Bernoulli (horizontal, h₁ = h₂):
$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$ $$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$Using continuity to eliminate v₂:
$$v_1 = A_2 v_2 / A_1$$After algebra:
$$\boxed{v_1 = A_2 \sqrt{\frac{2(P_1 - P_2)}{\rho(A_1^2 - A_2^2)}}}$$Flow rate:
$$\boxed{Q = A_1 v_1}$$In constriction (narrow section):
- Velocity increases (continuity)
- Pressure decreases (Bernoulli)
P₁ > P₂ even though v₁ < v₂!
This seems counterintuitive but is fundamental to Bernoulli’s principle.
3. Atomizer/Sprayer
Principle: Air blown rapidly over a tube creates low pressure, sucking liquid up and spraying it.
How it works:
- Air blows fast over tube opening (high velocity)
- By Bernoulli: high v → low pressure
- Atmospheric pressure in liquid container pushes liquid up
- Liquid emerges and gets atomized by air stream
Examples:
- Perfume sprayers
- Paint sprayers
- Insecticide sprayers
- Carburetors in engines
When you press a perfume sprayer:
- Air rushes rapidly through narrow tube
- Creates low pressure zone at tube exit (Bernoulli)
- Liquid perfume is pushed up by atmospheric pressure from below
- Air stream atomizes liquid into fine mist
No pump needed — Bernoulli’s principle does all the work!
4. Magnus Effect (Spinning Ball)
When a ball spins while moving through air:
- Air velocity is higher on side moving with ball’s motion
- Air velocity is lower on side moving against ball’s motion
- By Bernoulli: pressure difference creates sideways force
Examples:
- Baseball curve ball: Spins to curve the trajectory
- Table tennis spin: Top spin makes ball dip, backspin makes it rise
- Soccer free kicks: “Banana kick” curves around defensive wall
In cricket, bowlers make the ball swing in air using:
- Spin (Magnus effect)
- Seam orientation (creates asymmetric airflow)
- Ball shine (one side rough, one smooth)
Combined with Bernoulli’s principle, the ball can curve dramatically — sometimes more than a meter! This makes it incredibly difficult for batsmen.
5. Chimney and Ventilation
Tall chimneys create better draft (upward flow):
- Wind blows across top of chimney (high velocity)
- Creates low pressure at top (Bernoulli)
- Higher pressure inside causes smoke/air to rise
Height advantage: Taller chimney → greater pressure difference → better draft
6. Aerofoil (Airfoil) Lift
Airplane wing shape:
- Curved top (convex)
- Flat or less curved bottom
Airflow:
- Air travels farther over curved top → flows faster (to meet air from bottom)
- Faster flow on top → lower pressure (Bernoulli)
- Slower flow on bottom → higher pressure
Result: Net upward force (lift)
$$\boxed{F_{lift} = \frac{1}{2}\rho(v_{top}^2 - v_{bottom}^2) \cdot A}$$where A = wing area
Common Mistakes to Avoid
Wrong: Applying Bernoulli to highly viscous fluids like honey.
Correct: Bernoulli assumes non-viscous fluid. For viscous fluids, energy is lost to friction, and Bernoulli’s equation doesn’t hold.
Wrong: Using Bernoulli between any two points in flow.
Correct: Bernoulli equation connects points on the same streamline only. Can’t compare points on different streamlines unless flow is irrotational.
Wrong: In Torricelli’s theorem, P₁ = ρgh, P₂ = 0
Correct: Both surface and hole exit are at atmospheric pressure P₀. These cancel out in the equation.
Wrong: “High speed always creates low pressure”
Correct: High speed means low pressure along a streamline (Bernoulli). But different streamlines can have different total energies. Static pressure may increase with velocity if height or other factors change.
Practice Problems
Level 1: JEE Main Basics
Water flows through a horizontal pipe. At section 1, area = 20 cm², velocity = 2 m/s, pressure = 10⁵ Pa. At section 2, area = 10 cm². Find pressure at section 2. (ρ_water = 1000 kg/m³)
Solution:
From continuity:
$$A_1 v_1 = A_2 v_2$$ $$20 \times 2 = 10 \times v_2$$ $$v_2 = 4 \text{ m/s}$$From Bernoulli (horizontal, h₁ = h₂):
$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$ $$10^5 + \frac{1}{2} \times 1000 \times 4 = P_2 + \frac{1}{2} \times 1000 \times 16$$ $$10^5 + 2000 = P_2 + 8000$$ $$P_2 = 102000 - 8000 = 94000 \text{ Pa}$$P₂ = 9.4 × 10⁴ Pa
Pressure decreased as velocity increased!
A water tank has a hole at depth 5 m below surface. Find speed of water emerging from hole. (g = 10 m/s²)
Solution:
Using Torricelli’s theorem:
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5}$$ $$= \sqrt{100} = \boxed{10 \text{ m/s}}$$Level 2: JEE Main/Advanced
A horizontal pipe has two sections with areas 40 cm² and 10 cm². Water flows at 1 m/s in larger section. Pressure difference between sections is measured as 3000 Pa. Find density of water.
Solution:
Continuity:
$$v_1 = 1 \text{ m/s}, \quad v_2 = \frac{A_1 v_1}{A_2} = \frac{40 \times 1}{10} = 4 \text{ m/s}$$Bernoulli:
$$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$ $$3000 = \frac{1}{2}\rho(16 - 1)$$ $$3000 = \frac{15\rho}{2}$$ $$\rho = \frac{6000}{15} = \boxed{400 \text{ kg/m}^3}$$(This is less than water’s actual density — problem might be for oil or there’s experimental error)
Water flows through a pipe inclined at 30° to horizontal. At lower point (P₁), area = 20 cm², velocity = 2 m/s, pressure = 2 × 10⁵ Pa. At upper point (P₂) 2 m higher, area = 10 cm². Find pressure at P₂.
Solution:
Continuity:
$$v_2 = \frac{A_1 v_1}{A_2} = \frac{20 \times 2}{10} = 4 \text{ m/s}$$Bernoulli:
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$Taking h₁ = 0, h₂ = 2:
$$2 \times 10^5 + \frac{1}{2}(1000)(4) + 0 = P_2 + \frac{1}{2}(1000)(16) + 1000(10)(2)$$ $$200000 + 2000 = P_2 + 8000 + 20000$$ $$P_2 = 202000 - 28000 = \boxed{1.74 \times 10^5 \text{ Pa}}$$Pressure decreased due to both increased velocity and height!
Level 3: JEE Advanced
A cylindrical tank of height 3 m and radius 1 m is filled with water. A small hole of area 1 cm² is made at bottom. Find: (a) Initial velocity of efflux (b) Initial volume flow rate (c) Time to empty the tank (assuming constant velocity — approximation)
Solution:
(a) Initial velocity:
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 3} = \sqrt{60} \approx 7.75 \text{ m/s}$$(b) Volume flow rate:
$$Q = Av = (1 \times 10^{-4}) \times 7.75 = 7.75 \times 10^{-4} \text{ m}^3\text{/s}$$(c) Time to empty:
Volume of tank = πr²h = π(1)²(3) = 3π m³
$$t = \frac{V}{Q} = \frac{3\pi}{7.75 \times 10^{-4}} = \frac{9.42}{7.75 \times 10^{-4}} \approx 12155 \text{ s}$$≈ 3.4 hours
(This is underestimate because velocity decreases as water level drops!)
A large open tank has two small holes in its vertical wall, one at depth h₁ = 1 m and another at h₂ = 4 m from top. If the water streams from both holes hit the ground at the same point, find height of water in tank.
Solution:
Let water surface be at height H from ground.
Hole 1: At depth h₁ = 1 m, height from ground = H - 1
Efflux velocity: $v_1 = \sqrt{2gh_1} = \sqrt{2g}$
Horizontal range (projectile motion):
$$R_1 = v_1 \times t_1 = \sqrt{2g} \times \sqrt{\frac{2(H-1)}{g}} = 2\sqrt{H-1}$$Hole 2: At depth h₂ = 4 m, height from ground = H - 4
Efflux velocity: $v_2 = \sqrt{2gh_2} = \sqrt{8g}$
Range:
$$R_2 = v_2 \times t_2 = \sqrt{8g} \times \sqrt{\frac{2(H-4)}{g}} = 4\sqrt{H-4}$$For same range: R₁ = R₂
$$2\sqrt{H-1} = 4\sqrt{H-4}$$ $$\sqrt{H-1} = 2\sqrt{H-4}$$Square both sides:
$$H - 1 = 4(H - 4)$$ $$H - 1 = 4H - 16$$ $$3H = 15$$ $$\boxed{H = 5 \text{ m}}$$Water height is 5 m from ground.
Summary Table
| Application | Key Formula | Condition |
|---|---|---|
| Bernoulli’s Equation | P + ½ρv² + ρgh = const | Streamline flow, ideal fluid |
| Continuity | Av = constant | Incompressible fluid |
| Torricelli’s Theorem | v = √(2gh) | Horizontal efflux from depth h |
| Venturi Meter | v₁ = A₂√[2ΔP/ρ(A₁²-A₂²)] | Horizontal pipe |
Related Topics
Within Properties of Matter
- Fluid Pressure — Static pressure fundamentals
- Viscosity — Real fluid effects
- Surface Tension — Surface energy considerations
Connected Chapters
- Work-Energy-Power — Energy conservation
- Kinematics — Trajectory of water streams