Question 9.3.1: Find the main dimensions of a propeller = pump which is requ......

Given: Q = 210  l/s = 0.210  m^3/s,  H = 2.75  m,  K_u = 2.1, \text{ and } K_f = 0.45

Using the expression for the specific speed of a pump, we get

N=N_s\frac{H^{3/4}}{\sqrt{Q} }= 200 \times \frac{2.75^{3/4}}{\sqrt{0.21} }= 935 \mathrm{rpm}

Induction motor operating at 960 rpm is available in the market. Therefore, we choose N = 960.

Peripheral speed u=K_u\times \sqrt{2gh} =2.1\times \sqrt{2\times 9.81 \times 2.75} =3.305  m/s

But u = πDN/60 → D = 0.305 m

Velocity of flow =V_f=k_f\times \sqrt{2gH} =0.45 \times \sqrt{2\times 9.81\times 2.75} =3.305  m/s

Area of flow =\frac{\pi}{4}\left(D^2-d^2\right) =\frac{\pi}{4} D^2\left(1 – n^2\right)

where n = d/D and d is the diameter of the hub. Using the above determined values and writing in the continuity equation, we get

Q=\frac{\pi}{4}D^2\left(1-n^2\right)\times V_f\rightarrow n=0.35 \text{ and } d=nD=107  mm