The diameter of an impeller of a centrifugal pump at outlet are 60 cm and at inlet 30cm respectively. The velocity of flow at outlet is 20 m/s and the vanes are set back at an angle of 45 degree at the outlet. determine the minimum starting speed of the pump, if the manometric efficiency is 70%

**Solution Given:**

Diameter at inlet, $D_1$ = 30 cm = 0.30 m.

Diameter at outlet, $D_2$ = 60 cm = 0.60 m.

Velocity of flow at outlet, $vf_2$ = 2.0 m/s.

Vane angle at outlet, $\phi$ = 45 degree.

manometric effiency, $n_{man}$ = 70% = 0.70.

Let the minimum starting speed = N

Step No (1)

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

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

$u_2 - v_{w2} = \frac{2.0}{tan 45}$

$\therefore$ $v_{w2} = u_2 - 20$

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

$u_2 = \frac{\pi \times 0.60 \times N}{60}$

$u_2 = 0.03141 N$

$\therefore$ $v_{w2} = (0.03141 N - 20)$

Using the equation for minimum starting speed

Step No (2)

$N = \frac{120 \times n_{man} \times vw_2 \times F_2}{\pi (D_2 - D1^2]}$

$\frac{120 \times 0.70 \times (0.03141 N - 2.0 ) \times 0.6)}{\pi [0.6^2 - 0.3^2]}$

$= \frac{50.4 \times (0.03141 N - 2.0)}{\pi [0.36 - 0.09]}$

N = 59.417 [0.0314 N - 2.0]

N = 1.866 N - 118.834

$\therefore$ 1.866N - N = 118.834

$\therefore$ $N = \frac{118.834}{0.866}$

[ N = 137.22 r.p.m]