The impeller is 0.5m in diameter and 3 cm wide at outer and runs at a speed of 1200 rpm. The exit blade angle is 20$^{\circ}$.if the manometer efficiency of the pump is 85%. Determine the discharge and the pressure at the suction and delivery ends at the pump

Given

Static lift $H_{3}$=h_{s}+h_{d}=40m

Suction lift $h_{3}$=3m

$d_{3}=d_{d}$=35cm=0.35 m

h$f_{3}$=2m,hfd=6m

impeller outer diameter

$O_{2}$=0.5m

width $B_{2}$=3 cm=0.03m

N=1200 rpm

Exit blade angle

$\phi=20^{\circ}$

$n_{mano}$=1200 rpm

Exit blade angle

$\phi=20^{\circ}$

1) Discharge Q

$u_{2}=\frac{\pi\times D_{2}N}{60}=\frac{\pi\times 0.5\times 1200}{60}$=31.42 m/s

$n_{mano}=\frac{Vw_{s}-u2}{gH_{m}}$

Hm=$\frac{Vws=u2}{gHm}$

Hm=Hs+$hf_{3}+hfd+\frac{{Vd}^{2}}{2g}$

on neglecting $\frac{Vd^{2}}{2}$ since it is negligible the manometer head

$Hm=Hg+hfs+hfd$

=40+2+6

=48m

0.85=$\frac{vw_{2}\times 31.42}{9.81\times 48}$

Vw2=12.74m/s

We have

$Vw2=u2-HF$

$u2-\frac{Vf2}{tan\phi}$

12.74=81.42-$\frac{Vf2}{tan 20}$

Vf2=6.8m/s

Q=$\pi D_{2}B_{2}Vf_{2}$

=$\pi \times 0.5\times 0.03\times 6.8$

=0.3204m^{3}/s

2 Pressure Head at suction $H_{3}$

Velocity in suction and delivery pipes

$V_{Q}=V_{d}=\frac{Q}{A}=\frac{0.3201}{{\frac{\pi}{4}}(0.35)^{2}}$=3.33m/s

Suction head H3=h3+hfs+$\frac{Vs}{2g}$

=3+2+$\frac{(3.33)^{2}}{2\times 9.81}$

=5.565m

1. Delivery head Hd:

Hd=Hm-Hs-$\frac{{Vd}^{2}}{2}$

=48-5.565-$\frac{(3.33)^{2}}{2\times 9.81}$

=41.87m