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Properties of Bulk Matter - Bernoulli's Principle

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Bernoulli's Principle states that for an incompressible, non-viscous fluid undergoing streamline flow, the sum of the pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant throughout the flow: P+12ρv2+ρgh=constantP + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}.

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The principle is a direct consequence of the Law of Conservation of Energy applied to moving fluids.

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The Equation of Continuity (A1v1=A2v2A_1 v_1 = A_2 v_2) is a prerequisite, stating that for an incompressible fluid, the volume flow rate is constant; thus, velocity vv increases where the cross-sectional area AA decreases.

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Torricelli's Law: A special application of Bernoulli's principle that determines the speed of efflux of a liquid from an orifice at depth hh as v=2ghv = \sqrt{2gh}.

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Venturi Meter: A device used to measure the flow speed of an incompressible fluid in a pipe, based on the pressure difference created by varying the cross-sectional area.

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Magnus Effect: The phenomenon where a spinning cylinder or ball curves away from its principal path due to the pressure difference created by air velocity variations on opposite sides.

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Dynamic Lift: The upward force on an airplane wing (aerofoil) occurs because the air velocity is higher over the top surface than the bottom, leading to lower pressure on top according to P+12ρv2=constantP + \frac{1}{2}\rho v^2 = \text{constant}.

📐Formulae

P+12ρv2+ρgh=constantP + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}

A1v1=A2v2A_1 v_1 = A_2 v_2

v=2ghv = \sqrt{2gh}

Pρg+v22g+h=constant\frac{P}{\rho g} + \frac{v^2}{2g} + h = \text{constant}

P1−P2=12ρ(v22−v12)P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)

💡Examples

Problem 1:

Water flows through a horizontal pipe of varying cross-section. At a point where the pressure is 4×104 Pa4 \times 10^4 \text{ Pa}, the velocity is 2 m/s2 \text{ m/s}. Calculate the pressure at another point where the velocity is 5 m/s5 \text{ m/s}. (Density of water ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3)

Solution:

Given: P1=4×104 PaP_1 = 4 \times 10^4 \text{ Pa}, v1=2 m/sv_1 = 2 \text{ m/s}, v2=5 m/sv_2 = 5 \text{ m/s}, ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3. Since the pipe is horizontal, h1=h2h_1 = h_2. Using Bernoulli's equation: P1+12ρv12=P2+12ρv22P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 4×104+12(1000)(22)=P2+12(1000)(52)4 \times 10^4 + \frac{1}{2}(1000)(2^2) = P_2 + \frac{1}{2}(1000)(5^2) 40000+2000=P2+1250040000 + 2000 = P_2 + 12500 P2=42000−12500=29500 PaP_2 = 42000 - 12500 = 29500 \text{ Pa} P2=2.95×104 PaP_2 = 2.95 \times 10^4 \text{ Pa}.

Explanation:

Since the pipe is horizontal, the potential energy term ρgh\rho gh cancels out. As the velocity increases from 2 m/s2 \text{ m/s} to 5 m/s5 \text{ m/s}, the kinetic energy increases, causing the pressure to decrease to maintain the energy balance.

Problem 2:

A tank filled with water has a small hole at a depth of 20 m20 \text{ m} below the free surface. What is the velocity of efflux of water from the hole? (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Using Torricelli's Law: v=2ghv = \sqrt{2gh} v=2×10×20v = \sqrt{2 \times 10 \times 20} v=400v = \sqrt{400} v=20 m/sv = 20 \text{ m/s}.

Explanation:

According to Bernoulli's principle, the potential energy of the water at the surface is converted into kinetic energy at the orifice. The velocity depends only on the depth hh and is independent of the liquid's density.

Bernoulli's Principle - Revision Notes & Key Formulas | ICSE Class 11 Physics