Question 18.5: A jet of water 75 mm diameter strikes a stationary flat plat......

Given data:
Diameter of jet                            d = 75 mm = 0.075 m

Angle of inclination                    θ = 30°

Force exerted in the direction of flow    F_x=1620 \mathrm{~N}

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

(a) The force exerted on the plate in the direction of flow is given by Eq. (18.5) as

Q_{1}-Q_{2}-Q\cos\theta=0  (18.5)

F_x=\rho a V^2 \sin ^2 \theta

or                                1620=1000 \times 0.00442 \times V^2 \times \sin ^2 30^{\circ}

or                              V^2=1466.06

or                            V=38.29 \mathrm{~m} / \mathrm{s}

(b) The force normal to the plate is given by Eq. (18.6) as

Q=Q_{1}+Q_{2}          (18.6)

F_n=\rho a V^2 \sin \theta

=1000 \times 0.00442 \times 38.29^2 \times \sin ^2 30^{\circ}=3240 \mathrm{~N}

(c) The volume flow rate of water is

Q=a V=0.00442 \times 38.29=0.1692 \mathrm{~m}^3 / \mathrm{s}