Fluid mechanics and hydraulics miscellaneous
- A venturi meter, having a diameter of 7.5 cm at the throat and 15 cm at the enlarged end, is installed in a horizontal pipeline of 15 cm diameter. The pipe carries an incompressible fluid at a steady rate of 30 litres per second. The difference of pressure head measured in terms of the moving fluid in between the enlarged and the throat of the venturimeter is observed to be 2.45 m. Taking the acceleration due to gravity as 9.81 m/s², the coefficient of discharge of the venturimeter (correct up to two places of decimal) is __________.
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0.93 to 0.96
Q = Cd A1A2 √2gh √A²1 - A²2 Cd = Q A1A2 √2gh √A²1 - A²2 = 30 × 10-3 π/4 × (0.15)² × (0.075) √ × 2 × 9.81 × 2.45 √(0.15)² - (0.075)²
= 0.95Correct Option: D
0.93 to 0.96
Q = Cd A1A2 √2gh √A²1 - A²2 Cd = Q A1A2 √2gh √A²1 - A²2 = 30 × 10-3 π/4 × (0.15)² × (0.075) √ × 2 × 9.81 × 2.45 √(0.15)² - (0.075)²
= 0.95
- An incompressible homogeneous fluid is flowing steadily in a variable diameter pipe having the large and small diameters as 15 cm and 5 cm, respectively. If the velocity at a section at the 15 cm diameter portion of the pipe is 2.5 m/s, the velocity of the fluid (in m/s) at a section falling in 5 cm portion of the pipe is ___________.
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22 to 23
π × (15)² × 2.5 = π × 5² × V 4 4 ⇒ V = 2.5 × 
15² 
25
= 22.5 m/sCorrect Option: C
22 to 23
π × (15)² × 2.5 = π × 5² × V 4 4 ⇒ V = 2.5 × 15² 25
= 22.5 m/s
Direction: A rectangular open channel needs to be designed to carry a flow of 2.0 m³/s under uniform flow conditions. The Manning’s roughness coefficient is 0.018. The channel should be such that the flow depth is equal to half the width, and the Froude number is equal to 0.5.
- The bed slope of the channel to be provided is
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R = A = B.y = 2y.y = y P B + 2y 2y + 2y 2 Froude’s number = V = 0.5 √gy
Q = VA = 0.5√gy × 2y²
∴ y = 0.836 mV = 1 R2/3S1/2(Manning’s equation) n 0.5√gy = 1 × y 2/3 S1/2 0.018 2
∴ S = 0.0021.Correct Option: B
R = A = B.y = 2y.y = y P B + 2y 2y + 2y 2 Froude’s number = V = 0.5 √gy
Q = VA = 0.5√gy × 2y²
∴ y = 0.836 mV = 1 R2/3S1/2(Manning’s equation) n 0.5√gy = 1 × y 2/3 S1/2 0.018 2
∴ S = 0.0021.
Direction: An upward flow of oil (mass density 800 kg/m³, dynamic viscosity 0.8 kg/m–s) takes place under laminar conditions in an inclined pipe of 0.1 m diameter as shown in the figure. The pressures at section 1 and 2 are measured p1 = 435 kN/m² and p² = 200 kN/m². 
- If the flow is reversed, keeping the same discharge, and the pressure at section 1 is maintained as 435 kN/m², the pressure at section 2 is equal to
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P2 + sin 45° = P1 + 32 μVL r r PgD² P2 = 
435 × 10³ + 32 × 0.8 × 16.2 × 5 - 5 
× 800 × 9.81 = 614.48kN/m² 800 × 9.81 800 × 9.81 × (0.1)² V2
= 614.48 kN/m².Correct Option: D
P2 + sin 45° = P1 + 32 μVL r r PgD² P2 = 435 × 10³ + 32 × 0.8 × 16.2 × 5 - 5 × 800 × 9.81 = 614.48kN/m² 800 × 9.81 800 × 9.81 × (0.1)² V2
= 614.48 kN/m².
- The discharge in the pipe is equal to
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Loss of head, hf = flv² 2gD f = 64 = 64μ Re ρVrD ∴ hf = 64μ × 1 × V² = 32Lvμ ρVD 2gD PgD² = 32 × 5 × V × 0.8 0.8 × 9.81 × (0.1)² × 10³
Applying Bernoulli’s equation and substituting values,P1 = P2 + 5 . sinθ + hf r r
we get, V = 16.2 m/sQ = π × d² = 0.127 m³/s 4 Correct Option: B
Loss of head, hf = flv² 2gD f = 64 = 64μ Re ρVrD ∴ hf = 64μ × 1 × V² = 32Lvμ ρVD 2gD PgD² = 32 × 5 × V × 0.8 0.8 × 9.81 × (0.1)² × 10³
Applying Bernoulli’s equation and substituting values,P1 = P2 + 5 . sinθ + hf r r
we get, V = 16.2 m/sQ = π × d² = 0.127 m³/s 4
