Wednesday - June 5, 2024
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The inclined plate moving of jet

Force on the inclined plate moving in the direction of the jet

Let a jet of water strikes an inclined plate, which is moving with a uniform velocity in the direction of the jet as shown in Fig.

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Let

v = absolute velocity of a jet of water,

u = velocity from the plate into the direction of the jet,

a = Cross-sectional area of jet, and

θ = Angle between jet and plate.

  • The relative velocity of a jet of water = (v – u)
  • The velocity with which jet strikes = (v – u)
  • Mass of water striking per second = ρ x a x (v – u)

If the plate is smooth and energy loss due to the jet’s impact is assumed zero, the jet of water will leave the inclined plate with a velocity equal to (v- u).

The force exerted by the jet of water on the plate in the direction normal to the plate is given as

Fn = Mass striking per second x [Initial velocity in the normal direction with which jet strikes – Final velocity]

= ρa (v – u) [(v – u) sinθ – 0]

= ρa (v – u)^2 sinθ 

  • This normal force Fn is resolved into two components namely Fx and Fy in the direction of the jet and perpendicular to the direction of the jet respectively.

Fx = Fn sinθ

= ρa (v – u)^2 sin^2θ

Fy = Fn cosθ

= ρa (v – u)^2 sinθ cosθ 

Work was done per second by the jet on the plate 

= Fx x Distance per second in the direction of x

= Fx x u

= ρa(v – u)^2 sin^2θ x u

= ρa (v – u)^2 u sin^2θ N m/s. 

Ex. A jet of water of diameter 10 cm strikes a flat plate normally with a velocity of 15 m/s. Each plate is moving at a velocity of 6 m/s in some direction of the jet and also away from the jet.

Find

  1. The forces exerted by the jet at the plate
  2. Work is done by the jet on the plate per second
  3. Power of jet in kW
  4. Efficiency of jet

Given

The diameter of the jet,

d = 10 cm = 0.1 m,

Area,

a = (π/4)d^2

= (π/4)(0.1)^2

= 0.007854 m^2

The velocity of the jet,

v = 15 m/s

The velocity of the plate,

u = 6 m/s.

Solution

This force exerted by the jet to a moving flat vertical plate is given by,

Fx = ρa (v – u)^2

= 1000 x 0.007854 x (15 – 6)^2

= 636.17 N

Work done per second by the jet

= Fx x u

= 636.17 x 6

= 3817.02 Nm/s 

Power of jet

= (work is done per second)/1000

= 3817.02/1000

= 3.817 kW 

The efficiency of the jet (η)

= Output of jet per second/Input of jet per second

Where,

the output of jet per second  = work done by jet per second

= 3817.02 Nm/s and

the input of jet per second = Kinetic energy of the jet per second

= (1/2) m v2 /s

= (1/2) (m/s) v2

= (1/2) (ρav) v2

= (1/2) ρav^3

= (1/2) 1000 x 0.007854 x 15^3

= 13253.6 Nm/s

The efficiency of the jet (η) 

= 3817.02/13253.6

= 0.288

= 28.8 %