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A centrifugal pump is running at 1000 rpm. The outlet vane angle of the impeller is $45^\circ$ and the velocity of flow at the outlet is 2.5 m/s

The discharge through the pump is 200 lit/s when the pump is working against a total head of 20 m. If the manometric efficiency of the pump is 80% determine the outer diameter of the impeller and the width of the impeller at outlet.

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Given :

Speed $N=1000r.p.m$

Outlet vane angle $\phi = 45^\circ$

Velocity of flow at outlet $Vf_2=2.5m/s$

Discharge $Q= 200$ lit/s = $0.2m^2/s$

Head $Hm= 20m$

Manometric efficiency $N_{max}= 80\% = 0.80$

From outlet velocity triangle we have $\tan\phi = \dfrac {Vf_2}{U_2-Vw_2} \\ or \space \space U_2-Vw_2 = \dfrac {Vf_2}{\tan \phi} = \dfrac {2.5}{\tan 45}=2.5\\ \therefore Vw_2 =(U_2-2.5) ----------------------- 1\\ N_{max} = \dfrac {gHm}{Vw_2U_2}\\ 0.80 = \dfrac {9.81\times 20}{Vw_2\times U_2}$

$\therefore Vw_2U_2 = \dfrac {9.81\times 20}{0.80}=245.25 --------- 2$

Substituting the value of $Vw_2$ in equation 1 & 2 we get

$(U_2-2.5)U_2=245.25 \\ U_2^2 -2.5U^2 -245.25=0$

Which is a quadratic equation in $U_2$ ad its solution is

$U_2 = \dfrac {2.5\pm \sqrt{(2.5)^2+4\times 245.25}}2 \\ = \dfrac {2.5\pm \sqrt{6.25+981}}2 \\ = \dfrac {2.5\pm31.42}2 = 16.96 \space or \space -14.46 \\ U_2 = 16.96 \hspace{3cm} (\because \text {-ve value is not possible })$

i) Diameter of impeller $D_2$

using $U_2 = \dfrac {\pi D_2N}{60} \\ 16.96 = \dfrac {\pi D_2 N}{60} = \dfrac {\pi\times D_2\times 1000}{60}\\ D_2 = \dfrac {16.96\times 60}{\pi \times 1000} = 0.324 = 324 mm$

ii) Width of impeller at outlet $B_2$

Discharge $Q= \pi D_2B_2Vf_2\\ 0.2 = \pi \times 0.324\times B_2\times 2.5 \\ B_2 = \dfrac {0.2}{\pi\times 0.34 \times 2.5} = 0.0786 m \\ 78.6 mm$
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