Question 18.19: A 75 mm diameter jet of water having a velocity of 20 m/s st......

Given data

Diameter of jet              d = 75 mm = 0.075 m

Velocity of jet                 V = 20 m/s

Velocity of plate            u = 5 m/s 

Discharge at A              Q_2=\frac{Q}{4}

Area of jet is                a=\frac{\pi}{4} d^2=\frac{\pi}{4}(0.075)^2=0.00442 \mathrm{~m}^2

Let the plate is inclined at an angle θ in the direction of liquid jet.
From Eq. (18.8), we get

Q_2=\frac{Q}{2}(1-\cos \theta)

or                                  \frac{Q}{4}=\frac{Q}{2}(1-\cos \theta)

or                              1-\cos \theta=\frac{1}{2}

or                            \cos \theta=1-\frac{1}{2}=\frac{1}{2}

or                            \theta=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}

The component of force in the direction of jet is obtained by using Eq. (18.18) as

F_x=\rho a(V-u)^2 \sin ^2 \theta

=1000 \times 0.00442 \times(20-5)^2 \sin ^2 60^{\circ}=745.875 \mathrm{~N}

The component of force normal to the direction of jet is given by Eq. (18.4) as

F_{y}=\rho a V^{2}\sin\theta\cos\theta          (18.4)

F_y=\rho a(V-u)^2 \sin \theta \cos \theta

=1000 \times 0.00442 \times(20-5)^2 \sin 60^{\circ} \times \cos 60^{\circ}=430.63 \mathrm{~N}