Given:- H=12m, hub dia $D_{b}=0.35\times runner \ dia D$

N=100 rpm $\phi=15^{\circ}$

flow ratio $k_{f}=\frac{vf_{1}}{\sqrt{2gH}}$=0.6

$vf_{1}=0.6\sqrt{2gH}$

=0.6$\times\sqrt{2\times 9.81\times 12}$=9.206 m/s

$vf_{2}=vf_{1}$=9.206 m/s

$vm_{2}$=0 since discharge is areial

From Outlet velocity $\triangle EFG$

tan$\phi =\frac{vf_{2}}{u_{2}}$

tan15=$\frac{9.206}{u_{2}}$

$u_{2}$=34.357 m/s

But $u_{1}=u_{2}=u$=34.357 m/s

i) Diameter of runner $D_{o}$

$u_{1}=\frac{\pi DoN}{60}$

34.357=$\frac{\pi\times D_{o}\times 100}{60}$

$D_{o}$=6.562 m

ii) Diameter if boss $D_{b}$:

$D_{b}$=0.35

$D_{b}=0.35\times 6.562$

=2.297m

iii) Discharge through runner Q:

Q=$\frac{\pi}{4}(D^{2}_{o}-D^{2}_{o})$

=$\frac{\pi}{4}(6.6562^{2}-2.297^{2})\times 9.206$

=273.19 $m^{3}/s$