velocity and velocity of flow is constant. Assume number of frictional classes. find speed of wheel and vane angle at exit

Given:-

H=30m

$D_{1}$=1.2 m

$D_{2}$=0.6 m

Guide blade angle $\alpha=15^{\circ}$

inlet vane angle $\Theta =90^{\circ}$

i.e $Vr_{1}=vf_{1} and Vw_{1}$=0

$vw_{2}$=0

$vf_{1}=vf_{2}=vf$

speed of whee, Nand vane angle $\phi$ at exit

$u_{1}vw_{2}=v_{1}cos\alpha=v_{1}cos/5$

$vf_{1}=v_{1}sin \alpha=vf_{2}=v_{2}$

$vf_{1}=vf_{2}=v_{2}=v_{1}sin/5$

H=$\frac{Vw_{1.u_{1}}}{g}+\frac{V_{2}^{2}}{2g}$

30=$\frac{V_{1}cos 15\times V_{1}cos15}{9.81}+\frac{(v_{1}sin15)^{2}}{2\times 9.81}$

$\Theta=\tan^{-1}\frac{Vf_{1}}{BD}=tan^{-1}(\frac{2.15}{0.093})=87.52^{\circ}.....$

$\Theta=tan^{-1}\frac{vf_{2}}{u_{2}}=tan^{-1}(\frac{2.15}{6.05})=19.564^{\circ}...ans$

$v_{2}=vf_{2}$=2.15m/s

ii) Head at inlet of turbine H:

H=$\frac{vw_{1}}{9}+\frac{v^{2}_{2}}{29}=\frac{12.193\times 12.1}{9.81}+\frac{(2.15)^{2}}{2\times 9.81}$= 15.275m

iii) Power developed P:

P=$\frac{mVw_{1}-u_{1}}{1000}K/ N=\frac{1857.5\times 12.193\times 12.1}{1000}$=274.05 kW

iv)Hydralic efficiency $n_{h}$

$h_{h}:\frac{vw_{1}u_{1}}{9^{m}}=\frac{12.193\times 12.1}{9.81\times15.275}$=0.9846 or 98.46% ans

v) Inlet and outlet diagrams are shown in fig

$30\times 9.81=0.933v_{1}^{2}+0.0335v_{1}^{2}$

$v_{1}$=17.45 m/s

$u_{1}=v_{1}cos \alpha=17.45 cos 15$

=16.86 m/s

$u_{1}=\frac{\pi D_{1}N}{60}$

16.86=$\frac{\pi \times 1.2\times N}{60}$

$Vf_{1}=vf_{2}=v_{1}sin\alpha$

=17.45 sin 15

4.516m/s

$u_{2}=\frac{\pi D_{2}N}{60}$

$\frac{\pi\times 0.6\times 268.34}{60}$

=8.43 m/s

$\phi=tan^{-1}(\frac{vf_{2}}{u_{2}})$

$tan^{-1}(\frac{4.516}{843})$

=28.18$^{\circ}$