Fluid Mechanics and Hydraulic Machinery Miscellaneous
 A sprinkler shown in the figure rotates about its hinge point in a horizontal plane due to water flow discharged through its two exit nozzles.
The total flow rate Q through the sprinkler is 1 litre/sec and the crosssectional area of each exit nozzle is 1 cm². Assuming equal flow rate through both arms and a frictionless hinge, the steady state angular speed of rotation (rad/s) of the sprinkler is _______ (correct to two decimal places).

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Correct Option: A
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 A 60 mm diameter water jet strikes a plate containing a hole of 40 mm diameter as shown in the figure. Part of the jet passes through the hole horizontally, and the remaining is deflected vertically. The density of water is 1000 kg/m^{3}. If velocities are as indicated in the figure, the magnitude of horizontal force (in N) required to hold the plate is _________.

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Correct Option: A
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 An ideal water jet with volume flow rate of 0.05 m^{3}/s strikes a flat plate placed normal to its path and exerts a force of 1000 N. Considering the density of water as 1000 kg/m^{3}, the diameter (in mm) of the water jet is ______.

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Correct Option: A
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 Water, having a density of 1000 kg/m^{3}, issues from a nozzle with a velocity of 10 m/s and the jet strikes a bucket mounted on a Pelton wheel. The wheel rotates at 10 rad/s. The mean diameter of the wheel is 1 m. The jet is split into two equal streams by the bucket, such that each stream is deflected by 120° as shown in the figure. Friction in the bucket may be neglected. Magnitude of the torque exerted by the water on the wheel, per unit mass flow rate of the incoming jet, is

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Correct Option: D
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 An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ____ kg/s (round off to two decimal places).

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D = 2 x 50 = 100 mm
N = 3000 rpm.
Power, P = m.u²u = πDN = 90 ⇒ x = π × 100 × 3000 60 60
= 15707.963 mm/sec.
= 15.7079 m/sec.
⇒ 2000 = m(15.7079)²
⇒ m8.105 kg/s ≈ 8.11 kg/sCorrect Option: A
D = 2 x 50 = 100 mm
N = 3000 rpm.
Power, P = m.u²u = πDN = 90 ⇒ x = π × 100 × 3000 60 60
= 15707.963 mm/sec.
= 15.7079 m/sec.
⇒ 2000 = m(15.7079)²
⇒ m8.105 kg/s ≈ 8.11 kg/s