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The impeller of a centrifugal pump having external and internal diameters 500 mm and 250 mm respectively, width at outlet 50 mm and running at 1200 r.p.m. works against a head of 48 m.

The velocity of flow through the impeller is constant and equal to 3 m/s. the vans are set back at an angle of 400 at outlet.Determine: 1. Inlet vane angle, 2. Work done by the impeller on water per second and 3. manometric efficiency

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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:

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$u_1 = \frac{\pi D_1N}{60} = \frac{\pi \times 0.25 \times1200}{60} = 15.7m/s$

$u_2 = \frac{\pi D_2N}{60} = \frac{\pi \times 0.5 \times 1200}{60} = 31.4m/s$

$Q = \pi D_2B_2V_{f2}\\ = \pi \times 0.5 \times 0.05\times3\\ = 0.2355m^3/s$

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: $\eta_{mano} = \frac{gH_m}{V_{w2}.u_2}\\ =\frac{9.81\times 48}{27.82 \times 31.4}\\ =0.539\approx53.9\%$

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