Question 18.18: A 12 cm diameter jet of water with a velocity of 15 m/s stri......

Given data:

Diameter of jet                        d = 12 cm = 0.12 m

Velocity of jet                          V = 15 m/s

Velocity of plate                    u = 5 m/s

Area of jet is                            a=\frac{\pi}{4} d^2=\frac{\pi}{4}(0.12)^2=0.0113 \mathrm{~m}^2

(a) The force exerted by the jet on the plate is given by Eq. (18.15) as

F_{x}=\rho a(V-u)^{2} (18.15)

F=\rho a(V-u)^2

=1000 \times 0.0113 \times(15-5)^2=1130 \mathrm{~N}

(b) The rate of work done by the jet is

=F \times u=1130 \times 5=5650 \mathrm{~Nm} / \mathrm{s}=5650 \mathrm{~W}

(c) The input power of jet is

\text { = Kinetic enrgy of jet/s }

=\frac{1}{2} m V^2=\frac{1}{2} \rho a V \times V^2=\frac{1}{2} \rho a V^3

=\frac{1}{2} \times 1000 \times 0.0113 \times 15^2=19068.75 \mathrm{~W}

The output power of jet is

= Work done/s = 5650 N-m/s = 5650 W

The efficiency of the jet is

\eta=\frac{\text { Output power }}{\text { Input power }}=\frac{5650}{19068.75}=0.2963 \text { or } 29.63 \%