Pump specific speed and affinity laws

Specific speed

The specific speed of a centrifugal pump is the speed of a geometrically similar pump that delivers unit discharge (1 m³/s) against unit head (1 m).

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

($N$ in rpm, $Q$ in m³/s, $H_m$ in m.) It is the type number of the pump:

It helps in selecting the pump type for a duty and in predicting performance of similar pumps. (The head $H_m$ is for a single stage in a multistage pump, and $Q$ is for one side for a double suction impeller.)

(a) Specific speed and shaft power

$$N_s=\frac{1450\times\sqrt{0.05}}{25^{0.75}}=\frac{1450\times0.2236}{11.180}=29.0$$

This lies in the radial-flow (centrifugal) range.

$$P_1=\frac{\rho gQH}{\eta_o}=\frac{1000\times9.81\times0.05\times25}{0.72}=17.03\ \text{kW}$$

(b) Speed changed to 1750 rpm

Speed ratio $r=\dfrac{1750}{1450}=1.2069$. For the same pump:

$$Q\propto N,\qquad H\propto N^2,\qquad P\propto N^3$$ $$\begin{aligned} Q_2&=0.05\times1.2069=0.0603\ \text{m}^3/\text{s}\ (60.3\ \text{L/s})\\ H_2&=25\times1.2069^2=36.4\ \text{m}\\ P_2&=17.03\times1.2069^3=29.94\ \text{kW} \end{aligned}$$

Answer: (a) $N_s=29.0$, $P_1=17.03$ kW; (b) $Q_2=0.0603$ m³/s, $H_2=36.4$ m, $P_2=29.94$ kW.