Question 1.6: Consider an axial flow pump, which has rotor diameter of 32 ......
Consider an axial flow pump, which has rotor diameter of 32 cm that discharges liquid water at the rate of 2.5 m³/min while running at 1450 rpm. The corresponding energy input is 120 J/kg, and the total efficiency is 78%. If a second geometrically similar pump with diameter of 22 cm operates at 2900 rpm, what are its (1) flow rate, (2) change in total pressure, and (3) input power?
Using the geometric and dynamic similarity equations,
\frac{Q_1} {N_1D^2_1} = \frac{Q_2}{ N_2D_2^2}
Therefore,
Q_2 = \frac{Q_1N_2D_2^2}{ N_1D^2_1}=\frac{(2.5)(2900)(0.22)^2}{(1450)(0.32)^2} = 2.363 m³/min
As the head coefficient is constant,
W_2 = \frac{W_1N_2^2D_2^2}{ N_1^2D^2_1}=\frac{(120)(2900)^2(0.22)^2}{(1450)^2(0.32)^2} = 226.88 J/kg
The change in total pressure is:
ΔP = W_2η_{tt}ρ = (226.88)(0.78)(1000) N/m²
= (226.88)(0.78)(1000) 10^{-5} = 1.77 bar
Input power is given by
P = \dot{m}W_2 = \frac{(1000)(2.363)(0.22688)}{60} = 8.94 kW