Question 18.26: A jet of water 80 mm diameter impinges on a curved vane whic......

Given data:  
Diameter of jet                       d = 80 mm = 0.08 m

Rate of flow                             Q = 80 \text{ litres/s} = 0.08  m^3/s

Velocity of vane                      u = 8 m/s

Angle made by jet with the direction of motion of vane at entry      \alpha_1=0^{\circ}
Angle made by leaving jet with the direction of motion of vane = 60°

\therefore \quad \alpha_2=180^{\circ}-60^{\circ}=120^{\circ}

Area of jet is                  a=\frac{\pi}{4} d^2=\frac{\pi}{4}(0.08)^2=0.005 \mathrm{~m}^2

The velocity of jet is              V_1=\frac{Q}{a}=\frac{0.08}{0.005}=16 \mathrm{~m} / \mathrm{s}

The inlet and outlet velocity triangles are shown in Fig. 18.21.

u_1=u_2=u=8 \mathrm{~m} / \mathrm{s}

V_{r 1}=V_1-u_1=16-8=8 \mathrm{~m} / \mathrm{s}

Here              V_{r 2}=V_{r 1}=8 \mathrm{~m} / \mathrm{s}             \text { ( } \because \text { friction neglected) }

Again from the outlet velocity triangle, we get

V_{w 2}=u_2-V_{r 2} \cos \beta_2

=8-8 \cos 60^{\circ}=4 \mathrm{~m} / \mathrm{s}

(a) The force exerted by the jet in the direction of motion of the vane is given by Eq. (18.24) as

F_{x}=\rho a V_{r1}(V_{w1}-V_{w2})    (18.24)

F_x=\rho Q\left(V_{w 1}-V_{w 2}\right)

=1000 \times 0.08 \times(16-4)=960 \mathrm{~N}

The work done by the jet per second is

=F_x \times u=960 \times 8=7680 \mathrm{~W}