Question 18.27: A 50 mm diameter water jet having a velocity of 20 m/s impin......
A 50 mm diameter water jet having a velocity of 20 m/s impinges on a curved vane which is moving in the same direction as that of the jet with a velocity of 5 m/s. The jet leaves the vane at an angle of 150° with the direction of motion of vane. Find neglecting friction (a) the force exerted by the jet in the direction of motion of the vane, (b) the power developed by the jet and (c) the efficiency of the vane.
Given data:
Diameter of jet d = 50 mm = 0.05 m
Velocity of jet V_1=20 \mathrm{~m} / \mathrm{s}
Velocity of vane u = 5 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 = 150°
\therefore \beta_2=180^{\circ}-150^{\circ}=30^{\circ}
Area of jet is a=\frac{\pi}{4} d^2=\frac{\pi}{4}(0.05)^2=0.00196 \mathrm{~m}^2
The inlet and outlet velocity triangles are shown in Fig. 18.22.
Here u_1=u_2=u=5 \mathrm{~m} / \mathrm{s}
V_{r 1}=V_1-u_1=20-5=15 \mathrm{~m} / \mathrm{s}
V_{r 2}=V_{r 1}=15 \mathrm{~m} / \mathrm{s} \text { ( } \because \text { friction neglected) }
Again from the outlet velocity triangle, we get
V_{w 2}=V_{r 2} \cos \beta_2-u_2
=15 \cos 30^{\circ}-5=12.32 \mathrm{~m} / \mathrm{s}
(a) The force exerted by the jet in the direction of motion of the vane is given by Eq. (18.1) as
F_{x}=\rho a V^{2} (18.1)
F_x=\rho a V_{r 1}\left(V_{w 1}-V_{w 2}\right)
=1000 \times 0.00196 \times 15 \times[20-(-12.32)]=950.2 \mathrm{~N}
(b) The work done by the jet per second is
=F_x \times u=950.2 \times 5=4751 \mathrm{~W}
(c) The efficiency of the vane is
\eta=\frac{\text { Work done by the jet per second }}{\text { Kinetic energy of supplied jet per second }}
=\frac{F_x \times u}{\frac{1}{2} m V^2}=\frac{F_x \times u}{\frac{1}{2} \rho a V \times V^2}
=\frac{2 \times 950.2 \times 5}{1000 \times 0.00196 \times 20 \times 20^2}=0.606 \text { or } 60.6 \%
