Given:
velocity of jet $v_1$ = 35 m/s
Velocity of vane $u_1 = u_2$ = 20 m/s
Angle of jet at inlet$ \alpha = 30^0$
Angle made by the jet at outlet with the direction of motion of vanes = $120^0$
$\text{Angle B} = 180^0 - 120^0 = 60^0$
Angle of vanes tips
From inlet velocity triangle
$V_{w1}= V_1 cos \alpha = 35 cos 30^0 = 30.31 m/s$
$V_{f1}= V_1 sin \alpha = 35 sin 30^0 = 17.50 m/s$
$u_1 = 11 m/s$
$tan \theta = \frac{V_{f1}}{V_{w1}- u_1} = \frac{17.50}{30.31 - 20} = 1.697$
$\therefore \theta = tan^{-1} (1.697) = 60^0$
By sine rule,
$\frac{V_{r1}}{sin 90^0} = \frac{V_{f1}}{sin \theta} or \frac{V_{r1}}{1} = \frac{17.50}{sin 60^0}$
$V_{r1} = \frac{17.50}{0.866} = 20.25 m/s$
$\text{Now} V_{r1} = V_{f1} = 20.25m/s$
From outlet velocity triangle by sine rule
$\frac{V_{r2}}{sin 120^0} = \frac{u_2}{sin (60^0 - \phi)} or \frac{20.25}{0.866} = \frac{20}{sin (60^0 - \phi)}$
$sin(60^0 - \phi) = \frac{20 \times 0.866}{20.25} = 0.855$
$(60^0 - \phi) = sin (0.855)$
$60^0 - \phi = 58.75^0$
$\phi = 60^0 - 58.75 = 1.25^0$
Work done per unit weight of water entering $\frac{1}{g}(V_{w1} + V_{w2}) \times u_1$
$V_{w2} = V_{r2} cos \phi - u_2 = 20.25 cos 1.25^0 - 20.0 = 0.24 m/s$
$\therefore \text{work done/unit weight} = \frac{1}{9.81}[30.31 + 0.24] \times 20 = 62.28 Nm/N$
$\text{Efficiency} = \frac{\text{work done per kg}}{\text{Energy supplied per kg}} = \frac{62.28}{\frac{V_1^2}{2g}}$
$\frac{62.28 \times 2 \times 9.81}{35 \times 35} = 99.74 %$