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 $\Theta$ from inlet velocity $\triangle$ABc

$\Theta=tan^{-1}(\frac{vf1}{u1})$

$tan^{-1}(\frac{02222}{7.854})$

=1.62$^{\circ}$

ii) Manometric efficiency of pump

Consider exit velocity $\triangle$EGF

EH=EF-HF

$Vw_{2}=15.708-\frac{0.222}{tan 45}$=15.486 m/s

$n_{mano}=\frac{gHm}{Vw_{2}-u_{2}}=\frac{9.81\times 8.5}{15.486\times 15.708}$=0.3248=34.28%

iii) Minimum starting speed of pumo N1:

$N_{1}=\frac{60}{\pi}\frac{\sqrt{2gHm}}{\sqrt{D^{2_{2}}-p_{1}^{2}}}$

$\frac{60}{\pi}\times \frac{\sqrt{2\times 9.81\times 8.5}}{\sqrt{(0.3)^{2}-(0.25)^{2}}}$

=569.6 rpm