Fluid mechanics and hydraulics miscellaneous
- A triangular gate with a base width of 2 m and a height of 1.5 m lies in a vertical plane. The top vertex of the gate is 1.5 m below the surface of a tank which contains oil of specific gravity 0.8. Considering the density of water and acceleration due to gravity to be 1000 kg/m³ and 9.81 m/s² respectively, the hydrostatic force (in kN) exerted by the oil on the gate is ___________
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29.43
Force on gate = 1 × 1.5 × 2 × G.yw 
1.5 + 2 × 1.5 
2 3
= 0.8 × 9810 × 2.5 × 1.5
= 29.43 kNCorrect Option: C
29.43
Force on gate = 1 × 1.5 × 2 × G.yw 1.5 + 2 × 1.5 2 3
= 0.8 × 9810 × 2.5 × 1.5
= 29.43 kN
- A nozzle is so shaped that the average flow velocity changes linearly from 1.5 m/s at the beginning to 15 m/s at its end in a distance of 0.375 m. The magnitude of the convective acceleration (in m/s²) at the end of the nozzle is _________
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54
Convective acceleration = a ∂u + v ∂u + ω∂u ∂x ∂y ∂δ = 1.5 × (15 - 1.5) = 54 m/s² 0.3375 Correct Option: B
54
Convective acceleration = a ∂u + v ∂u + ω∂u ∂x ∂y ∂δ = 1.5 × (15 - 1.5) = 54 m/s² 0.3375
- A short reach of a 2 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000. It carries a discharge of 4 m³/s and its Manning’s roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach, the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance, dy/dx, at this section is ___(use g = 10 m/s²)
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0.0032
S0 = 1 10000
θ = 4 m³/s, n = 0.01, y = 0.5m
Rate of change of water depth, dy/dxQ = 1 AR2/3Sƒ1/2 n 4 = 1 × (2 × 0.5) × 2 × 0.5 2/3 × Sƒ1/2 0.01 2 + 1
∴ Sf = 6.92 × 10–3Fr = V = Q √gy A √gy = 4 = 1.79 (2 × 0.5) × √10 × 5 ∴ dy = (1/10000) - 6.92 × 10-3 dx 1 - (1.79)²
= 3.2 × 10-3 = 0.0032Correct Option: A
0.0032
S0 = 1 10000
θ = 4 m³/s, n = 0.01, y = 0.5m
Rate of change of water depth, dy/dxQ = 1 AR2/3Sƒ1/2 n 4 = 1 × (2 × 0.5) × 2 × 0.5 2/3 × Sƒ1/2 0.01 2 + 1
∴ Sf = 6.92 × 10–3Fr = V = Q √gy A √gy = 4 = 1.79 (2 × 0.5) × √10 × 5 ∴ dy = (1/10000) - 6.92 × 10-3 dx 1 - (1.79)²
= 3.2 × 10-3 = 0.0032
- Two reservoirs are connected through a 930 m long, 0.3 m diameter pipe, which has a gate valve. The pipe entrance is sharp (loss coefficient= 0.5) and the valve is half-open (loss coefficient = 5.5). The head difference between the two reservoirs is 20 m. Assume the friction factor for the pipe as 0.03 and g =10 m/s². The discharge in the pipe accounting for all minor and major losses is _________ m³/s.
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0.1413
Total loss = difference in head = 20m.Entry loss = 0.5v² 2g Loss due to value = 5.5v² 2g Exit loss = v² 2g Friction loss = ƒLv² 2gd ∴ 0.5v² + 5.5v² + v² + ƒLv² = 20 2g 2g 2g 2gd v² 0.5 + 5.5 + 1 + 0.03 × 930 = 20 2g 0.3 ⇒ v² × 100 = 20 ⇒ v = 2 m/s 2g θ π × d² × v 4 π × (0.3)² × 2 = 0.1413 m³/s 4 Correct Option: C
0.1413
Total loss = difference in head = 20m.Entry loss = 0.5v² 2g Loss due to value = 5.5v² 2g Exit loss = v² 2g Friction loss = ƒLv² 2gd ∴ 0.5v² + 5.5v² + v² + ƒLv² = 20 2g 2g 2g 2gd v² 0.5 + 5.5 + 1 + 0.03 × 930 = 20 2g 0.3 ⇒ v² × 100 = 20 ⇒ v = 2 m/s 2g θ π × d² × v 4 π × (0.3)² × 2 = 0.1413 m³/s 4
- A horizontal nozzle of 30 mm diameter discharges a steady jet of water into the atmosphere at a rate of 15 litres per second. The diameter of inlet of the nozzle is 100 mm. The jet impinges normal to a flat stationary plate held close to the nozzle end. Neglecting air friction and considering the density of water as 1000 kg/m³, the force exerted by the jet (in N) on the plate is _______
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31.8 to 31.9
Force = raV²= ra Q ² = 100 × 15² × 10-6 × 1 = 31.8 N a (π/4) × 30² 10-6
= 31.8 NCorrect Option: B
31.8 to 31.9
Force = raV²= ra Q ² = 100 × 15² × 10-6 × 1 = 31.8 N a (π/4) × 30² 10-6
= 31.8 N
