Page: 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?
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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 degree 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] Dia. of impeller at, $D_2$ = 40cm

outlet = 0.40 m.

3] Vane angle at outlet = $\phi = 45$ degree

4] Reduction in area =10%

at outlet = 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, No = 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,

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

$= \frac{0.05}{0.02262}$

= 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 $\phi = \frac{vf_2}{u_2 - vw_2}$

$\therefore$ $u_2 - vw_2 = \frac{vf_2}{tan \phi}$

$20.94 - vw_2 = \frac{2.21}{tan 45 degree}$

$20.94 - vw_2 = 2.21 m/s$

$\therefore$ $vw_2 = 20.94 - 2.21$

$\therefore$ $[ vw_2 = 18.73 m/s]$

Step No (3) using equation

$n man = \frac{9 Hm}{vw_2 u_2}$

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

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

$\therefore$ [H_m = 35.98m]

$\therefore$ Total Head generated by pump

= n x Hm

= 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

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

$\therefore$ $n_o = \frac{52.94}{s.p}$

$\therefore$ $s.p = \frac{52.94}{0.80} = 66.175 K.W$

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modified 6 days ago  • written 6 days ago by gravatar for Renu Banswani Renu Banswani10
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