Page: Specific speed of a centrifugal pump (Ns)
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"The speed of a geometrically similar pump which would deliver one wbic meter of liquid per second against a head of one meter" is known as specific head.

It is denoted by 'Ns'

Expression for specific speed of a pump:

Discharge is given by 'Q'

$\therefore$ Q = Area x Velocity of flow.

$Q = \pi D \times B \times vf$ -----(1)

where, D = Dia of impeller of the pump.

B = width of the impeller.

we know that [ B $\alpha$ D]

$\therefore$ From equation (1), we have [ Q $\alpha D^2 \times vf$] ------(2)

we also know that tangential velocity is given by,

$u = \frac{\pi DN}{60} \alpha DN$ ------(3)

Now the tangential velocity (u) and velocity of flow (vf) are related to the mano metric head (Hm) as

$u \alpha vf \alpha \sqrt Hm$ -----(4)

substituting the value of (u) in equation (3), we get,

[$ \sqrt Hm \alpha DN$] or [ $ D \alpha \frac{ \sqrt {Hm}}{N}] $

substituting the values of D in equation (2),

$Q \alpha \frac{Hm}{N^2} \times vf$

$Q \alpha \frac{Hm}{N^2} \times vf$

$\alpha \frac{Hm}{N^2} \times \sqrt Hm$

[$\because vf \alpha \sqrt Hm$]

$\alpha \frac{Hm^3/2}{N^2}$

$\therefore Q = K \frac{Hm^3/2}{N^2}$ -----(5)

where K is a constant of proportionality.

If Hm = 1m, Q = 1 $m^3/s$

N becomes = Ns

substituting these values in equation (5), we get,

$1 = K \frac{1^3/2}{N^2}$

$= \frac{k}{Ns^2}$

$\therefore$ [$K = Ns^2$]

substituting the values of K in equation (5), we get,

$Q = Ns^2 \frac{Hm^3/2}{N^2}$

$Ns^2 = \frac{N^2Q}{Hm^3/2}$

$\therefore$ $[ Ns = \frac{N \sqrt Q}{Hm^3/4}]$

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written 6 days ago by gravatar for Renu Banswani Renu Banswani10
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