A three stage centrifugal pump has impellers 40 cm in diameter and 2 cm wide at outlet. The vanes are waved back at the outlet at $45^o$ and reduce the circumferential area by 10%. The manometric efficiency is 90% and the overall efficiency is 80% Determine the head generated by the pump when running at 1000 r.p.m delivering 50 liters per second what should be the shaft horse power?

**Solution Given:**

1] Number of stages, n = 3

2] Diagram of impeller at outlet, $D_2$ = 40cm = 0.40 m.

3] Vane angle at outlet = $\phi = 45^0$

4] Reduction in area at outlet =10% = 0.1

5] Width at outlet, $B_2$ = 2cm = 0.02 m

$\therefore$ Area of flow at outlet = $0.9 \times \pi D_2 \times B_2$

= $0.9 \times \pi \times 0.4 \times 0.02$

$= 0.02262 m^2$

6] Manometric efficiency, $n_{man}$ = 90% = 0.90

7] Overall efficiency, $n_o$ = 80% = 0.80

8] Speed, N = 1000 r.p.m

9] Discharge, Q = 50 lit/s = 0.05 $m^3/s$

**To find :**

1] Head generated by the pump.

2] shaft power.

Step No (1) Velocity of flow at outlet,

$vf_2 = \frac{Discharge}{\text{Area of flow}}$

$= \frac{0.05}{0.02262}$

= 2.21 m/s

Step No (2) Tangential velocity of impeller,

$u_2 = \frac{ \pi D_2 N}{60}$

$= \frac{\pi \times 0.4 \times 1000}{60}$

= 20.94 m/s

From velocity triangle at outlet,

tan $\phi = \frac{vf_2}{u_2 - v_{w2}}$

$\therefore$ $u_2 - v_{w2} = \frac{vf_2}{tan \phi}$

$20.94 - v_{w2}= \frac{2.21}{tan 45 degree}$

$20.94 - v_{w2}= 2.21 m/s$

$\therefore$ $v_{w2} = 20.94 - 2.21$

$\therefore$ $[ v_{w2} = 18.73 m/s]$

Step No (3) using equation

$n_{man} = \frac{9 H_m}{v_{w2} u_2}$

$\therefore$ $0.90 = \frac{9.81 \times Hm}{18.73 \times 20.94}$

$\therefore$ $H_m = \frac{0.90 \times 18.73 \times 20.94}{9.81}$

$\therefore$ [H_m = 35.98m]

$\therefore$ Total Head generated by pump

= n x $H_m$

= 3 x 35.98

= 107.94 m

Step No (4) Power output of the pump.

$= \frac{pg \times Q \times total head}{1000}$

$= \frac{1000 \times 9.81 \times 107.94 \times 0.05}{1000}$

= 52.94 KW

$\therefore$ $n_o = \frac{\text{power output of pump}}{\text{power input to the pump}}$

$\therefore$ $n_o = \frac{52.94}{S.P}$

$\therefore$ S.P = $\frac{52.94}{0.80}$ = 66.175 K.W