Draw the triangles of velocities at inlet and outlet and find

1)The angle of vanes tips so that water enters and leaves without shock

2)The work done per unit weight of water entering the vanes

3) The efficiency

Given

Jet veloicty $v_{2}$=35m/s

vane velocity, 4=20m/s

$\alpha=30^{\circ},\beta=180-120=60^{\circ}$

1] Vane angle at inlet $\Theta$ and exit , $\phi$

from inlet velocity $\triangle ACD$

$V_{w}=Ac=v_{1}cos\alpha$=30.31 m/s

$vs_{1}=CD=V_{1}sin alpha$

35 sin 30=17.50m/s

BC=$Vw_{1}-\mu=30.31$-20=10.31m/s

$v_{r_{1}}BD=\sqrt{(BC)^{2}+(D)^{2}}$

$\sqrt{(10.31)^{2}+(17.8)^{2}}$

20.31 m/s

$\Theta=tan^{-1}(\frac{CD}{BC})$

$\tan^{-1}(\frac{17.5}{10.31})$=59.5

Consider outlet velocity $\triangle EGH$ and neglecting blade friction i,e

$Vr_{2}=vr_{1}$=20.31 m/s

Applying sine rule to $\triangle EFG$

$\frac{Vr_{2}}{sin(180-\beta)}=\frac{4}{sin(B-\phi)}$

$\frac{20.31}{sin(180-60)}=\frac{20}{sin(60-\Phi)}$

sin$(60-\phi)$=0.8528

$(60-\phi)=58.52^{\circ}$

$\phi=1.48^{\circ}$

FH==$Vw_{2}$=EH-EF

$vr_{2}cos\phi$-4

20.31cos1.48-20

=0.303 m/s

ii) work done /N weight of water,W

w=$\frac{1}{2}(Vw_{1}+Vw_{2})u=\frac{1}{9.81}\times (30.31+0.303)$

=62.41 Nm/N weight of water

iii) Hydraullic efficiency

nh=$\frac{Work done/kg.w}{k.E supplied (w^{2}/2)}$

$\frac{2(Vw_{1}+Vw_{2})}{v^{2}}$

$\frac{2\times(30.31+0.303)\times 20}{35^{2}}$

=0.9996

=99.96