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 '$N_s$'

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 manometric head ($H_m$) as

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

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

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

Substituting the values of D in equation (2),

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

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

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

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

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

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

where K is a constant of proportionality.

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

N becomes = $N_s$

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,

$Q = N_s^2 \frac{H_m^{3/2}}{N^2}$

$N_s^2 = \frac{N^2Q}{H_m^{3/2}}$

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

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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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