Page: Minimum speed for starting a centrifugal pump.

If the pressure rise in the impeller is more than or equal to manometric head (Hm) the centrifugal pump will start delivering water, otherwise, the pump will not discharge any water, though the impeller is rotating. when impeller is rotating, the water in contact with the impeller is also rotating. this is the case of forced vortex.

$= \frac{w^2 r2^2}{2g} - \frac{w^2 r1^2}{2g}$

$Wr_2$ = Tangential velocity of impeller at outlet = $u_2$

$Wr_2$ = Tangential velocity of impeller at inlet = $u_1$

$\therefore$ Head due to pressure.

$= \frac{u2^2}{2g} - \frac{u1^2}{2g}$

The flow of water will commence only if,

1] Head sue to pressure rise in impeller > $H_m$

i.e. $(\frac{u2^2}{2g} - \frac{u1^2}{2g} \gt H_m)$

2] For minimum speed, we must have,

$( \frac{u2^2}{2g} - \frac{u1^2}{2g} = H_m)$ -------(1)

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

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

Substituting the value of Hm in equation (1),

$\frac{u2^2}{2g} - \frac{u1^2}{2g} = n_{man} \times \frac{v_{w2} u_w}{g}$ --------(2)

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

$u_1 = \frac{\pi D_1 N}{60}$

Substituting values of $u_1$ and $u_2$, in equation (2)

$\frac{1}{2g} (\frac{ \pi D_2 N}{60})^2 - \frac{1}{2g} (\frac{\pi D_1 N}{60})^2 = n_{man} \times \frac{v_{w2} \times \pi D_2 N}{g \times 60}$

Divide by $\frac{\pi N}{9 \times 60}$, we will get,

$\frac{\pi N D_2^2}{120} - \frac{\pi N D_1^2}{120} = n_{man} \times v_{w2} \times D_2$

$\frac{\pi N}{120} [D_2^2 - D_1^2] = n_{man} \times v_{w2} \times D_2$

$\therefore$ $[ N = \frac{120 \times n_{man} \times v_{w2} \times D_2}{\pi [D_2^2 - D_1^2]}]$

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modified 2 days ago by gravatar for Abhishek Tiwari Abhishek Tiwari50 written 9 weeks ago by gravatar for RB RB ♦♦ 100
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