Question 18.29: An orifice of diameter 50 mm fitted in the vertical side of ......

Given data:
Diameter of orifice                                             d = 50 mm = 0.05 m

Head over the center of orifice                         H = 4 m

Coefficient of velocity                                          C_ν=0.96

Area of orifice is                                                  a=\frac{\pi}{4} d^2=\frac{\pi}{4}(0.05)^2=0.00196 \mathrm{~m}^2

The velocity of jet is given by Eq. (18.34) as

V=C_ν \sqrt{2 g H}

=0.96 \times \sqrt{2 \times 9.81 \times 4}=8.5 \mathrm{~m} / \mathrm{s}

(a) The force exerted by the jet of water on the tank when the tank is stationary is obtained from Eq. (18.35) as

F=\rho a V(V-0)=\rho a V^{2}      (18.35)

F=\rho a V^2=1000 \times 0.00196 \times 8.5^2=141.61 \mathrm{~N}

(b) The velocity of tank u = 2 m/s
The force exerted by the jet of water on the tank when the tank is moving is given by Eq. (18.36) as

F=\rho a(V+u)[(V+u)-u]=\rho a V(V+u)    (18.36)

F=\rho a V(V+u)=1000 \times 0.00196 \times 8.5 \times(8.5+2)=174.93 \mathrm{~N}