and the vanes are set back at the outlet at an angle of 45$^{\circ}$ to the periphery. The area of the flow is constant from inlet to outlet of the impeller and is 0.6m$^{3}$

Given

$D_{2}$=0.5m

N=600 rpm

Q=8000 lit/min=$\frac{8000}{1000}\times \frac{1}{60}m^{3}/s$

$\frac{2}{15}m^{3}/s$

Hm=8.5

At inlet whirel velocity Vw1=0

i.e $\alpha=90^{\circ}$

inner diameter $D_{1}$=0.25m

Q=45$^{\circ}$

Area of flow Af1=AF2=0.6$m^{2}$

u1=$\frac{\pi\times D_{1}N}{60}$

$\frac{\pi\times 0..25\times 600}{60}$=7.854 m/s

u2=$\frac{\pi\times p2\times N}{60}$

$\frac{\pi\times 0.5\times 600}{60}$= 15.708 m/s

Q=$Af\times Vf1$

$\frac{2}{15}=0.6\times Vf1$

Vf1=0.222 m/s=vf2

i) Vane angle at inlet $\Theta$ …