Two homologous pumps A and B are to run at the same speed. Pump Ahas an impeller diameter of 50 cm and discharge 0.4 m³/s under the head of 50 m. What will be the diameter of pump B, if its discharge is 0.2 m³/s? (Assuming 2^(1/3)= 1.26)

Concept:

Head coefficient

\(\frac{{H}}{{{D^2}N^2}} = C \)

Discharge coefficient

\(\frac{Q}{{{D^3}N}} = C \)

Calculation:

Given:

For Pump A, D_A = 50 cm, Q _A = 0.4 m³/s, H_A = 50 m

For Pump B, Q_B = 0.2 m³/s, D_B = ?

Also, N_A = N_B

We know that

 \(\frac{Q}{{{D^3}N}} = C \)

\({\left( {\frac{{{D_B}}}{{{D_A}}}} \right)^3} = \frac{{{Q_B}}}{{{Q_A}}} \)

\(\left( {\frac{{{D_B}}}{{50}}} \right) = {\left( {\frac{{0.2}}{{0.4}}} \right)^{\frac{1}{3}}}\)

D_B = 39.68 cm

Unit Quantities: (First Preference)

\({Q_u} = \frac{Q}{{\sqrt H }}\;\;\;\;\;\;\;\;\;\;\;{N_u} = \frac{N}{{\sqrt H }}\;\;\;\;\;\;\;\;\;\;\;{P_u} = \frac{P}{{{H^{\frac{3}{2}}}}}\)

Model Studies: (Second Preference)

\(\frac{{H}}{{{D^2}N^2}} = C \;\;\;\;\;\;\;\;\;\;\frac{Q}{{{D^3}N}} = C\;\;\;\;\;\;\;\;\;\;\;\frac{P}{{{D^5}{N^3}}} = C\)

Specific Speed: (Third Preference)

\({N_s} = \frac{{N\sqrt P }}{{{H^{\frac{5}{4}}}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;For\;Turbines\)

\({N_s} = \frac{{N\sqrt Q }}{{{H^{\frac{3}{4}}}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;For\;Pumps\)

 where ‘P’ in kW, ‘N’ in rpm