# 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.

**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**

- The forces exerted by the jet at the plate
- Work is done by the jet on the plate per second
- Power of jet in kW
- 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 % **