The velocity of flow through the impeller is constant and equal to 2.5 m/s. The vanes are set back at an angle of 400 at outlet. If the outer diameter of the impeller is 500 mm and width at outlet is 50 mm, determine: i) Vane angle at inlet, ii) Work done by impeller on water per second and iii) Manometric efficiency.

Given:

$N = 1000 \hspace{0.05cm} rpm, H_m = 40\hspace{0.05cm}m, V_f = V_{f1} = V_{f2} =2.5\hspace{0.05cm}m/s\\ \phi = 40^\circ, D_2 = 500\hspace{0.05cm}mm = 0.50\hspace{0.05cm}m\\ D_1 = \frac{D_2}{2} = 0.25\hspace{0.05cm}m, B_2 = 0.05\hspace{0.05cm}m$

Solution:

$u_1 = \frac{\pi D_1N}{60} = \frac{\pi \hspace{0.05cm}\times\hspace{0.05cm} 0.25\hspace{0.05cm} \times\hspace{0.05cm} 1000}{60} = 13.09\hspace{0.05cm}m/s$

$u_2 = \frac{\pi D_2N}{60} = \frac{\pi \hspace{0.05cm}\times\hspace{0.05cm} 0.5\hspace{0.05cm} \times\hspace{0.05cm} 1000}{60} = 26.18\hspace{0.05cm}m/s$

$Q = \pi D_2B_2V_{f2}\\ \hspace{0.2cm}=\pi \times 0.5 \hspace{0.05cm}\times\hspace{0.05cm}0.05\hspace{0.05cm}\times\hspace{0.05cm}0.25 = 0.1963\hspace{0.05cm}m^3/s$

Vane angle at inlet : $\tan\theta = \frac{V_{f1}}{u_1} = \frac{2.5}{13.09}\\ \hspace{0.5cm}\theta = \tan^{-1}(\frac{2.5}{13.09}) = 10.81^\circ$

Work done by impeller on water per sec

$W.D = \frac{w}{g} \times V_{w2}.u_2 = \frac{\rho g Q}{g} \times V_{w2}.u_2\\ \hspace{1cm}= \frac{1000 \times 9.81 \times 0.1963}{9.81} \times V_{w2}.u_2$

$\tan\phi = \frac{V_{f2}}{u_2 - V_{w2}} =\frac{2.5}{26.18 - V_{w2}}$

$tan\hspace{0.1cm}40 = \frac{2.5}{26.18 - V_{w2}}$

$V_{w2} = 23.2\hspace{0.05cm}m/s$

W.D = $\frac{1000\hspace{0.05cm}\times\hspace{0.05cm}9.81\hspace{0.05cm}\times\hspace{0.05cm}0.1963}{9.81} \times\hspace{0.05cm}23.2\hspace{0.05cm}\times\hspace{0.05cm}26.18 = 119227.9\hspace{0.05cm}Nm/s$

Manometric efficiency:

$\eta_{mano} = \frac{gH_m}{V_{w2}.u_2}\\ \hspace{0.8cm}=\frac{9.81\hspace{0.05cm}\times\hspace{0.05cm}40}{23.2\hspace{0.05cm}\times\hspace{0.05cm}26.18}\\ \hspace{0.8cm}=0.646 = 64.6\%$