written 6.0 years ago by
teamques10
★ 67k

•
modified 6.0 years ago

Given:
$D_2 = 0.5m D_1 = 0.25m, B_2 = 0.05m \\
N = 1200rpm, H_m = 48m, V_f = V_{f1} = V_{f2} = 3m/s\\
\phi = 40^\circ$
Solution:
$u_1 = \frac{\pi D_1N}{60} = \frac{\pi \times 0.25 \times1200}{60} = 15.7m/s$
$u_2 = \frac{\pi D_2N}{60} = \frac{\pi \times 0.5 \times 1200}{60} = 31.4m/s$
$Q = \pi D_2B_2V_{f2}\\
= \pi \times 0.5 \times 0.05\times3\\
= 0.2355m^3/s$
Vane angle at inlet,
$\tan\theta =\frac{V_{f1}}{u_1} = \frac{3}{15.7} = 0.1911$
$\theta = \tan^{1}(0.1911) = 10.81^\circ$
Work done by impeller on water per sec,
$W.D = \frac{W}{g}\times V_{w2}.u_2= \frac{\rho gQ}{g}\times V_{w2}.u_2$
$= \frac{1000\times 9.81 \times 0.2355}{9.81}\times V_{w2} \times 31.4$
$\tan\phi = \frac{V_{f2}}{u_2  V_{w2}} = \frac{3}{31.4  V_{w2}}$
$\tan40 = \frac{3}{31.4  V_{w2}}$
$V_{w2} = 27.82 m/s$
$WD = \frac{1000 \times 9.81 \times 0.2355}{9.81} \times27.82\times31.4$
$ = 205720.55\hspace{0.05cm} Nm/s$
Manometric efficiency: $\eta_{mano} = \frac{gH_m}{V_{w2}.u_2}\\
=\frac{9.81\times 48}{27.82 \times 31.4}\\
=0.539\approx53.9\%$