Question

A six-stage centrifugal pump delivers water at the rate of 150 L.s\(^{-1}\) against a net pressure rise of 5003 kN.m\(^{-2}\). The pump impeller rotates at 1450 rpm. If the acceleration due to gravity is 9.81 m.s\(^{-2}\), the specific speed of the pump is _________ (Answer in integer).

Hint:

To calculate the specific speed of a centrifugal pump, first convert flow rate and pressure rise to the appropriate units (m\(^3\).s\(^{-1}\), meters) before applying the formula.

Answer 1

Step 1: Convert the flow rate from L.s\(^{-1}\) to m\(^3\).s\(^{-1}\). The flow rate is given as 150 L.s\(^{-1}\). We convert this to cubic meters per second: \[ Q = 150 \, {L.s}^{-1} = 150 \times 10^{-3} \, {m}^3.{s}^{-1} = 0.15 \, {m}^3.{s}^{-1}. \] Step 2: Convert the pressure rise to head. The pressure rise is given as 5003 kN.m\(^{-2}\). We convert this to head using the formula: \[ H = \frac{P}{\rho g}, \] Where: - \(P\) is the pressure rise (5003 kN.m\(^{-2}\)), - \(\rho\) is the density of water (\(1000 \, {kg/m}^3\)), - \(g\) is the acceleration due to gravity (\(9.81 \, {m/s}^2\)). Substituting the values: \[ H = \frac{5003 \times 10^3}{1000 \times 9.81} = \frac{5003 \times 10^3}{9810} \approx 510 \, {m}. \] Step 3: Calculate the specific speed. Now that we have \(N = 1450\) rpm, \(Q = 0.15\) m\(^3\).s\(^{-1}\), and \(H = 510\) m, we can calculate the specific speed using the formula: \[ N_s = \frac{1450 \sqrt{0.15}}{510^{3/4}} \approx \frac{1450 \times 0.3873}{510^{0.75}} \approx \frac{561.5}{167.7} \approx 631. \] Thus, the specific speed of the pump is 631.