Question

A centrifugal pump running at 500 rpm and at its maximum efficiency is delivering a head of 30 m at a flow rate of 60 L/min. If the rpm is changed to 1000, then the head ‘H’ in meter and flow rate ‘Q’ in L/min at maximum efficiency is estimated to be
1. H = 60, Q = 120
2. H = 120, Q = 120
3. H = 60, Q = 480
4. H = 120, Q = 30

Answer 1

Correct Answer - Option 2 : H = 120, Q = 120

Concept:

For constant diameter, N α √H

\(\frac{{{Q_1}}}{{D_1^3{N_1}}} = \frac{{{Q_2}}}{{D_2^3{N_2}}}\)

Calculation:

Given:

H1 = 30 m, Q1 = 60 liters/min, N₁ = 500 rpm, N₂ = 1000 rpm, H₂ = ? and Q₂ = ?

\(⇒ \frac{{{H_2}}}{{{H_1}}}={\left( {\frac{{{N_2}}}{{{N_1}}}} \right)^2} ⇒ \frac{{{H_2}}}{{30}} = {\left( {\frac{{1000}}{{500}}} \right)^2}\)

⇒ H₂ = 120 m

\(\frac{{{Q_1}}}{{D_1^3{N_1}}} = \frac{{{Q_2}}}{{D_2^3{N_2}}}\)

\(⇒ {Q_2} = \left( {\frac{{1000}}{{500}}} \right) \times 60= 120~lit/min\)