Question 20.2: A centrifugal pump delivers 0.3 m³/s against a head of 30 m ......

Given data:

Discharge                                              Q=0.3 \mathrm{~m}^3 / \mathrm{s}

Speed of pump                                 N =1400 rpm

Head                                                  H_m=30 \mathrm{~m}

External diameter of impeller      D_2=0.5 \mathrm{~m}

Width at outlet                                  B_2=0.05 \mathrm{~m}

Manometric efficiency                  \eta_{\text {mano }}=0.80

Tangential velocity of impeller at outlet is given by

U_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 0.5 \times 1400}{60}=36.65 \mathrm{~m} / \mathrm{s}

The velocity of flow at the impeller outlet is

V_{f 2}=\frac{Q}{\pi D_2 B_2}

=\frac{0.3}{\pi \times 0.5 \times 0.05}=3.82 \mathrm{~m} / \mathrm{s}

Manometric efficiency is found from 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                            0.80=\frac{9.81 \times 30}{V_{w 2} \times 36.65}

or                          V_{w 2}=10.04 \mathrm{~m} / \mathrm{s}

The outlet velocity triangle is shown in Fig. 20.8.
From the outlet velocity triangle, we have

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

or          \beta_2=\tan ^{-1} \frac{3.82}{36.65-10.04}=8.17^{\circ}