Page: Determine. 1] vane angle at inlet. 2] work done (per second) 3] manometric efficiency.
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A centrifugal pump having outer diameter equal to two times the inner diameter and remaining at 1000 r.p.m works against a total head of 40m. 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 40 degree at outlet. if the outer diameter of the impeller is 500 mm and width at outlet is 50mm, Determine.

1] vane angle at inlet.

2] work done (per second)

3] manometric efficiency.

Solution: Given :

1] speed N = 1000 r.p.m

3] velocity of flow,

$vf_1$ = 2.5 m/s

$vf_2$ = 2.5 m/s

4] vane angle at $\phi$ = 40 degree outlet

5] Outer diameter of impeller, D2 = 500mm = 0.50m

6] Inner dia, $D_1 = \frac{D_2}{2} = \frac{0.50}{2}$

= 0.25m

7] width at outlet, B2 = 50mm

= 0.05 m

Tangential velocity of impeller at inlet and outlet,

$u_1 = \frac{\pi D_1 N}{60}$

$\frac{\pi \times 0.25 \times 1000}{60}$

u1 = 13.09 m/s

$u_2 = \frac{\pi D_2 N}{60}$

$= \frac{\pi \times 0.50 \times 1000}{60}$

u2 = 26.18 m/s

$\therefore$ Discharge $Q = \pi D_2 B_2 \times vf_2$

$= \pi \times 0.50 \times 0.05 \times 2.5$

$Q = 0.1963 m^3/s$

step no (1) vane angle at inlet ( $\theta$ )

$\theta = \frac{vf_1}{u_1} = \frac{2.5}{13.09} = 0.191$

$\theta = tan^-1 (0.191)$

[ $\theta = 10.81$ degree ]

Step no (2) work done

$= \frac{w}{g} \times vw_2 \times u_2$

$\frac{p \times g \times Q}{g} \times vw_2 \times u_2$

$= \frac{1000 \times 9.81 \times 0.1963}{9.81} \times vw_2 \times 26.18$ ---(1)

But from outlet velocity triangle, we get,

$tan \phi = \frac{vf_2}{u_2 - vw_2}$

$tan \phi = \frac{2.5}{(26.18 - vw_2)}$

$26.18 - vw_2 = \frac{2.5}{tan 40 degree}$

vw2 = 26.18 - 2.979

[ vw2 = 23.2 m/s ]

substitute the value in equation (1)

$= \frac{1000 \times 9.81 \times 0.1963 \times 23.2 \times 26.18}{9.81}$

= 119227.9 Nm/s

Step no (3) manometric efficiency ( n man)

$n man = \frac{g Hm}{vw_2 u_2}$

$= \frac{9.81 \times 40}{23.2 \times 26.18}$

= 0.646

[ n man = 64.6 % ]

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