Given

External Dia $= D_1 = 0.9m $

Internal Dia $= D_2= 0.45 $

Speed $N=200 rpm$

Width at inlet $ B_1 = 200mm =0.2 m$

Velocity of flow $vf_1 =vf_2 =1.8 m/s $

Guide blade angle $\alpha =10^\circ $

Discharge at outlet = Radial

$\therefore \space\space\space\space B=90^\circ $ and $Vw_2=0$

Tangential velocity of wheel at inlet and outlet are :-

$U_1 =\dfrac {\pi D_1N}{60}=\dfrac {\pi\times 0.9\times 200}{60}=9.424 m/s \\ U_2 =\dfrac {\pi D_2 N}{60} =\dfrac {\pi\times0.45\times 200}{60}= 4.712 m/s $

ii. Absolute velocity of water at inlet of the runner i.e. $V_1 $

From inlet velocity triangle

$V_1\sin \alpha = Vf_1 \\ V_1= \dfrac {Vf_1}{\sin\alpha} = \dfrac {1.8}{\sin 10^\circ}=10.365 m/s $

ii. Velocity of whirl at inlet i.e $Vw_1\\ Vw_1 = V_1\cos \alpha = 10.365 \times \cos 10^\circ \\ =10.207 m/s $

iii. Relative Velocity at inlet i.e $Vr_1$

$Vr_1 =\sqrt{Vr_1^2+ (Vw_1 - U_1)^2}\\ =\sqrt{1.8^2+(10.207-9.424)^2} \\ \sqrt{3.24+0.613}\\ =1.963m/s $

iv. The runner blade angle means the angle $\theta $ and $ \phi$

$\tan \theta = \dfrac {Vf_1}{(Vw_1-U_1)} =\dfrac {1.8}{(10.207-9.424)}=2.298 \\ \theta = \tan^{-1} 2.298 = 66.48^\circ \space or \space 66^\circ 29'$

From outlet velocity triangle we have

$\tan \theta = \dfrac {Vf_2}{U_2} =\dfrac {1.8}{4.712}= \tan 20.9^\circ \\ \phi = 20.9^\circ \space or \space 20^\circ 54.4'$

v. Width of runner at outlet i.e $B_2$

$\pi D_1 B_1 Vf_1 = \pi D_2B_2Vf_2 \space or \space D_1B_1=D_2B_2 \\ (\because \pi Vf_1 =\pi Vf_2 \space as \space Vf_1 =Vf_2) \\ B_2 =\dfrac {D_1B_1}{D_2} =\dfrac {0.90\times 0.20}{0.45} = 0.40 m=400 mm $

vi. Mass of water flowing through the runner per second

$Q=\pi D_1B_1Vf_1= \pi \times 0.8\times 0.20 \times 1.8 \\ =1.0178 m^3/s$