Question 20.1: A centrifugal pump running at 1200 rpm delivers water. The d......

Given data:

Speed of pump                                               N =1200 rpm

Internal diameter of impeller                      D_1=100 \mathrm{~mm}=0.1 \mathrm{~m}

External diameter of impeller                    D_2=300 \mathrm{~mm}=0.3 \mathrm{~m}

Width at inlet                                                B_1=50 \mathrm{~mm}=0.05 \mathrm{~m}

Width at outlet                                            B_2=20 \mathrm{~mm}=0.02 \mathrm{~m}

Velocity of flow at inlet                                V_{f 1}=2.2 \mathrm{~m} / \mathrm{s}

Manometric efficiency                                \eta_{\text {mano }}=0.75

The discharge is found from Eq. (20.3) as

{Q}=\pi D_{1}B_{1}V_{f1}=\pi D_{2}B_{2}V_{f2}            (20.3)

Q=\pi D_1 B_1 V_{f 1}

=\pi \times 0.10 \times 0.05 \times 2.2=0.0345 \mathrm{~m}^3 / \mathrm{s}

The velocity of flow at outlet is found to be

V_{f 2}=\frac{Q}{\pi D_2 B_2}=\frac{0.0345157}{\pi \times 0.3 \times 0.02}=1.83 \mathrm{~m} / \mathrm{s}

Tangential velocity of impeller at outlet is given by

U_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 0.3 \times 1200}{60}=18.85 \mathrm{~m} / \mathrm{s}

The outlet velocity triangle is shown in Fig. 20.7.
From the outlet velocity triangle, we get

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

or                    \tan 30^{\circ}=\frac{1.83}{18.85-V_{w 2}}

or              V_{w 2}=18.85-\frac{1.83}{\tan 30^{\circ}}=18.85-3.17=15.68 \mathrm{~m} / \mathrm{s}

Manometric efficiency is given by Eq. (20.10) as

\eta_{m a n 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_m}{V_{w 2} U_2}

or                      0.75=\frac{9.81 \times H_m}{15.68 \times 18.85}

or                      H_m=\frac{0.75 \times 15.68 \times 18.85}{9.81}=22.597 \mathrm{~m}