The jet males an angle of $30^{\circ}$ to the direction of motion of vanes when entering and leaves of an angle 120$^\circ$. Draw the velocity triangles and inlet at outlet and determine

1) The inlet and outlet vane angles so that water enters and leaves without shocks

ii) The work done per N of water entering the Vanes

iii) The efficiency of the system

Inlet and outlet Velocity diagram are

$v_{1}$=40m/s u=20m/s

$\alpha=30^{\circ}= \ \ \ \beta=180-20=60^{\circ}$

i) Inlet Vane angle $\Theta$and Outlet Vane angle $\phi$

Consider inlet velocity $\triangle ACD$

AD=$Vw_{1}=V_{1}cos\alpha$

40Cos 30=34.64m/s

CD=$V_{f}=V_{1}sin\alpha=40 sin30$

=20m/s

BD=AD-AB

=34.64-10=14.64m/s

$Vr_{1}=\sqrt{(BD)^{2}+(CD)^{2}}$

=\sqrt{(14.64)^{2}+(20)^{2}}=24.78m/s

Vr_{2}=Vr_{1}=24.78 m/s

(On neglecting blace friction)

$\Theta=tan^{-1}(\frac{Vf_{1}}{BD})=tan^{-1}(\frac{20}{14.64})$

$\Theta=53.79^{\circ}$

Consider outlet $\triangle$ EFG and apply sine rule.

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

$\frac{24.78}{sin(180-60)}=\frac{20}{sin(20-\phi)}$

sin(60-$\phi)$=0.699

$\phi$=15.65

$Vw_{2}=fH=Vr_{2}Cos\phi-mu$=

=24.78 Cos 15.65-20

=3.86m/s

ii)Workdone/N of water, W

W==$\frac{1}{9}(Vw_{1}+Vw_{2})4$

$\frac{1}{9.81}\times(34.64+3.86)\times20$

=78.491 Nm/N

iii) Efficiency of system i,e Hydraullie of efficiency

$\frac{Workdone/kg}{K.E supplied/kg}$

=$\frac{Vw_{1}+Vw_{2}}{V^{2}_{2}/2}$

=$\frac{2\times (34.64+3.86)20}{(40^{2})}$

=0.9625

=96.25%