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Mechanical Properties of Fluids - Streamline Flow and Equation of Continuity

Grade 11CBSEPhysics

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

🔑Concepts

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Steady Flow (Streamline Flow): A flow in which the velocity of the fluid particles reaching a particular point remains constant with time. Every particle passing through that point follows the same path as the preceding particle.

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Streamline: A curve whose tangent at any point indicates the direction of the fluid velocity v⃗\vec{v} at that point. In steady flow, streamlines do not intersect.

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Turbulent Flow: When the velocity of fluid exceeds a certain critical value (Critical Velocity vcv_c), the flow becomes irregular and complex, characterized by eddies and whirlpools.

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Equation of Continuity: It is based on the Law of Conservation of Mass. For an incompressible, non-viscous fluid in a steady flow, the product of the cross-sectional area and the fluid velocity is constant throughout the pipe.

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Conservation of Mass: If the fluid is incompressible (density ρ\rho is constant), the mass of fluid entering a pipe section per unit time must equal the mass of fluid leaving it per unit time.

📐Formulae

A1v1=A2v2A_1 v_1 = A_2 v_2

Av=constantAv = \text{constant}

Volume flow rate (Discharge) Q=dVdt=Av\text{Volume flow rate (Discharge)} \, Q = \frac{dV}{dt} = A v

Mass flow rate dmdt=ρAv\text{Mass flow rate} \, \frac{dm}{dt} = \rho A v

v1=A2A1v2v_1 = \frac{A_2}{A_1} v_2

💡Examples

Problem 1:

Water flows through a horizontal pipe of varying cross-section. At a point where the radius is r1=4 cmr_1 = 4 \text{ cm}, the velocity of flow is v1=2 m/sv_1 = 2 \text{ m/s}. Calculate the velocity of flow v2v_2 at a point where the radius is r2=2 cmr_2 = 2 \text{ cm}.

Solution:

Given: r1=4 cmr_1 = 4 \text{ cm}, r2=2 cmr_2 = 2 \text{ cm}, v1=2 m/sv_1 = 2 \text{ m/s}. From the equation of continuity: A1v1=A2v2A_1 v_1 = A_2 v_2 Since A=πr2A = \pi r^2, we have: πr12v1=πr22v2\pi r_1^2 v_1 = \pi r_2^2 v_2 v2=v1(r1r2)2v_2 = v_1 \left( \frac{r_1}{r_2} \right)^2 Substituting the values: v2=2×(42)2v_2 = 2 \times \left( \frac{4}{2} \right)^2 v2=2×22=2×4=8 m/sv_2 = 2 \times 2^2 = 2 \times 4 = 8 \text{ m/s}

Explanation:

According to the equation of continuity, as the cross-sectional area of the pipe decreases, the velocity of the fluid must increase to maintain a constant volume flow rate. Here, the radius is halved, so the area becomes one-fourth, causing the velocity to quadruple.

Problem 2:

A pipe has a diameter of 0.02 m0.02 \text{ m} at the inlet and 0.01 m0.01 \text{ m} at the outlet. If water enters at a speed of 1.5 m/s1.5 \text{ m/s}, find the volume of water discharged per second.

Solution:

The volume flow rate QQ is given by A×vA \times v. Radius at inlet r=0.022=0.01 mr = \frac{0.02}{2} = 0.01 \text{ m}. Area at inlet A=πr2=π(0.01)2=10−4π m2A = \pi r^2 = \pi (0.01)^2 = 10^{-4}\pi \text{ m}^2. Q=Av=(10−4π)×1.5Q = A v = (10^{-4}\pi) \times 1.5 Q=1.5π×10−4 m3/sQ = 1.5\pi \times 10^{-4} \text{ m}^3/\text{s} Using π≈3.14\pi \approx 3.14: Q≈4.71×10−4 m3/sQ \approx 4.71 \times 10^{-4} \text{ m}^3/\text{s}

Explanation:

The volume flow rate is the product of the area of cross-section and the velocity at any point. Since the fluid is incompressible, this value remains constant throughout the length of the pipe.