# Force Exerted by Jet an Inclined Plate

## Force Exerted by a Jet an Inclined Flat Plate.

- Permit a jet of water, proceeding out from the nozzle, strikes a tilted flat plate as shown in Fig.

**Let**

θ = Angle inside between the jet and plate,

A = Area of a cross-section of the jet.

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Fig. Jet striking stationary sloping plate

- If the plate is smooth and if it is assumed that there is no loss of energy due to impact of the jet, then the jet will move over the plate after striking with a velocity equal to initial velocity i.e., with a velocity v.
- Let us find the force exerted by the jet on the plate in the direction normal to the plate.
- Let this force is represented by Fn

Then

Fn = Mass of jet striking per sec. x [ Initial velocity of the jet before striking in the way from n - The final velocity of jet after striking in the way from n]

= ρav [v sinθ - 0]

**= ρav^2 sinθ**

- This force can be determined into two components, one in the direction of the jet also other perpendiculars to the direction of flow.

Fx = Parts of Fn in the direction of flow.

= Fn cos (90° - θ)

**= Fn sinθ**

= ρav2 sinθ x sinθ

**( ∵ Fn = ρav^2 sinθ )****= ρav^2 sin^2θ**

And, Fy = Parts of Fn perpendicular to flow.

= Fn sin(90° - θ)

= Fn cosθ

**= ρav^2 sinθ cosθ**

**Q. An jet of water of diameter 75 mm moving by a velocity of 25 m/s strikes a fixed plate in such a direction that the angle within the jet and plate is 60°. Determine the force exerted by the jet toward the plate**

- Toward the direction normal at the plate and
- Into the direction from the jet.

**Given**

A diameter of a jet, d = 75 mm = 0.075 m

Area, a = (π/4)d2

= (0.075)^2

= 0.004417 m^2

Velocity of jet, v = 25 m/s

The angle between jet and plate θ = 60°

**Solution**

- This force exerted by the jet of water in the direction normal to the plate is given as

Fn= ρav^2 sinθ

= 1000 x 0.004417 x 25^2 x sin(60)

= 2390.7 N

=2.39 kN

- The force in the direction of the jet is given as

Fx = ρav^2 sin^2θ

= 1000 x 0.004417 x 25^2 x sin^2(60)

= 2070.4 N

= 2.070 kN

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