Question 20.19: A centrifugal pump running at 900 rpm delivers 100 litres/s ......

Given data:

Speed of pump                                 N = 900 rpm

Discharge                                        Q=100 \text { litres } / \mathrm{s}=0.1 \mathrm{~m}^3 / \mathrm{s}

External diameter of impeller      D_2=250 \mathrm{~mm}=0.25 \mathrm{~m}

Width at outlet                                B_2=40 \mathrm{~mm}=0.04 \mathrm{~m}

Vane angle at outlet                        \beta_2=30^{\circ}

Manometric efficiency                    \eta_{\operatorname{mano}}=0.90

Tangential velocity of impeller at outlet is given by

U_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 0.25 \times 900}{60}=11.78 \mathrm{~m} / \mathrm{s}

The velocity of flow at outlet is found to be

V_{f 2}=\frac{Q}{\pi D_2 B_2}=\frac{0.1}{\pi \times 0.25 \times 0.04}=3.183 \mathrm{~m} / \mathrm{s}

The outlet velocity triangle is shown in Fig. 20.22.

From the outlet velocity triangle, we get

\tan \beta_2=\frac{V_{f 2}}{U_2-V_{w 2}}

or                  \tan 45^{\circ}=\frac{3.183}{11.78-V_{w 2}}

or            V_{w 2}=11.78-\frac{3.183}{\tan 45^{\circ}}=11.78-3.183=8.597 \mathrm{~m} / \mathrm{s}

From Eq. (20.10), manometric efficiency can be written as

\eta_{m a r o}={\frac{H_{m}}{\frac{V_{w2}U_{2}}{g} }}={\frac{g H_{m}}{V_{w2}U_{2}}}        (20.10)

\eta_{\text {mano }}=\frac{g H}{V_{w 2} U_2}

or                      0.9=\frac{9.81 \times H}{8.597 \times 11.78}

or                    H=\frac{0.9 \times 8.597 \times 11.78}{9.81}=9.29 \mathrm{~m}

The specific speed of the pump is given by Eq. (20.30) as

N_s=\frac{N \sqrt{Q}}{H^{3 / 4}}

=\frac{900 \sqrt{0.1}}{(9.29)^{3 / 4}}=53.48 \text { unit }