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Mechanical Properties of Fluids - Bernoulli's Principle

Grade 11CBSEPhysics

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

🔑Concepts

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Ideal Fluid: Bernoulli's principle applies to an ideal fluid, which is incompressible (constant density ρ\rho) and non-viscous (no internal friction).

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Streamline Flow: It refers to a steady flow where every particle of the liquid passing through a point travels along the same path as the preceding particles.

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Equation of Continuity: For an incompressible fluid in steady flow, the mass flow rate is constant: A1v1=A2v2A_1v_1 = A_2v_2. This implies velocity vv is inversely proportional to the cross-sectional area AA.

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Bernoulli's Theorem: It states that for a streamline flow of an ideal fluid, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline.

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Conservation of Energy: Bernoulli's principle is essentially a restatement of the Law of Conservation of Energy for flowing fluids.

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Torricelli's Law: The speed of efflux of a liquid through an orifice at a depth hh below the free surface is equal to the speed a body would acquire in falling freely through the same height hh.

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Venturi-meter: A device based on Bernoulli's principle used to measure the flow speed of an incompressible fluid in a pipe.

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Dynamic Lift: The force that acts on a body, such as an airplane wing (airfoil) or a spinning ball (Magnus Effect), due to its motion through a fluid, caused by pressure differences.

📐Formulae

A1v1=A2v2A_1v_1 = A_2v_2

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

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2

v=2ghv = \sqrt{2gh}

v1=A22(P1−P2)ρ(A12−A22)v_1 = A_2 \sqrt{\frac{2(P_1 - P_2)}{\rho(A_1^2 - A_2^2)}}

💡Examples

Problem 1:

Water is flowing through a horizontal pipe of non-uniform cross-section. At a point where the velocity of water is 2 m/s2\text{ m/s}, the pressure is 4×104 Pa4 \times 10^4\text{ Pa}. Calculate the pressure at another point where the velocity of water is 4 m/s4\text{ m/s}. (Density of water ρ=1000 kg/m3\rho = 1000\text{ kg/m}^3)

Solution:

For a horizontal pipe, 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. Given: P1=4×104 PaP_1 = 4 \times 10^4\text{ Pa}, v1=2 m/sv_1 = 2\text{ m/s}, v2=4 m/sv_2 = 4\text{ m/s}, ρ=1000 kg/m3\rho = 1000\text{ kg/m}^3. P2=P1+12ρ(v12−v22)P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) P2=40000+12(1000)(22−42)P_2 = 40000 + \frac{1}{2}(1000)(2^2 - 4^2) P2=40000+500(4−16)P_2 = 40000 + 500(4 - 16) P2=40000−6000=34000 PaP_2 = 40000 - 6000 = 34000\text{ Pa} or 3.4×104 Pa3.4 \times 10^4\text{ Pa}.

Explanation:

Since the pipe is horizontal, the potential energy terms cancel out. As the velocity increases at the second point, the pressure must decrease to satisfy the conservation of energy.

Problem 2:

A tank filled with water has a small hole at a depth of 5 m5\text{ m} from the top surface. If the tank is open to the atmosphere, find the speed with which water emerges from the hole. (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

Using Torricelli's Law: v=2ghv = \sqrt{2gh}. Given: h=5 mh = 5\text{ m}, g=10 m/s2g = 10\text{ m/s}^2. v=2×10×5v = \sqrt{2 \times 10 \times 5} v=100=10 m/sv = \sqrt{100} = 10\text{ m/s}.

Explanation:

The speed of efflux depends only on the depth of the hole below the free surface and is independent of the density of the liquid.

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