Page: Determine, 1] The diameter of the impeller. 2] The width of the impeller at outlet.
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A centrifugal pump is running at 1000 r.p.m The outlet vane angle of the impeller is 45 degree and velocity of flow at outlet is 2.5m/s The discharge through the pump is 200 liters/s when the pump is working against a total head of 20m If the manometric efficiency of the pump is 80% Determine,

1] The diameter of the impeller.

2] The width of the impeller at outlet.

Solution, Given :

1] Speed, N = 1000 r.p.m

2] Outlet vane angle $\phi = 45$ degree

3] velocity of flow at outlet,

$vf_2 = 2.5m/s$

4] Discharge, Q = 200 lit/s

$= 0.2 m^2/s$

6] Manometric efficiency

n man = 80% = 0.80

From outlet velocity triangles, we have

$tan \phi = \frac{xf_2}{u_2-vw_2}$

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

$u_2 - vw_2 = \frac{2.5}{tan h_5}$

$u_2 - vw_2 = 2.5$

$\therefore$ $vw_2 = (u_2 - 2.5)$ ----(1)

n man $= \frac{9H_m}{vw_2 u_2}$

$0.80 = \frac{9.81 \times 20}{vw_2 u_2}$

$vw_2 u_2 = \frac{9.81 \times 20}{0.80}$

$vw_2 u_2 = 245.25$ -----(2)

substituting the value of $vw_2$ from equation (1) in equation (2), we get,

$\therefore$ $(u_2 - 2.5) u_2 = 245.25$

$u2^2 - 2.5 u_2 - 245.25 = 0$

which is a quadratic equation in $u_2$ and its solution is,

$u_2 = \frac{2.5 + \sqrt{(2.5)^2 + 4 \times 245.25}}{2}$

$= \frac{2.5 + \sqrt{6.25 + 981}}{2}$

$= \frac{2.5 + 31.42}{2}$

$u_2 = 16.96$ or (-14.46)

$\because$ negative value is not possible

[$\therefore$ $u_2 = 16.96]$

Step No (1) Diameter of impeller $(D_2)$

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

$16.96 = \frac{\pi D_2 N}{60}$

$16.96 = \frac{ \pi \times D_2 \times 1000}{60}$

$\therefore$ $D_2 = \frac{16.96 \times 60}{\pi \times 1000}$

$\therefore$ $[ D_2 = 0.324m]$ OR $[ D_2 = 324mm]$

Step No (2) width of impeller at outlet $(B_2)$

$Q = \pi D_2 B_2 Vf_2$

$0.2 = \pi \times 0.324 \times B_2 \times 2.5$

$\therefore$ $B_2 = \frac{0.2}{\pi \times 0.324 \times 2.5}$

$\therefore$ $B_2 = 0.0786m = 78.6mm$

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