written 6.0 years ago by
teamques10
★ 67k

•
modified 6.0 years ago

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