Question 20.4: A centrifugal pump running at 900 rpm is working against a h......

Given data:

Speed of pump                                        N = 900 rpm

Head                                                            H_m=16 \mathrm{~m}

External diameter of impeller              D_2=360 \mathrm{~mm}=0.36 \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_{\text {mano }}=0.85

Tangential velocity of impeller at outlet is given by

U_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 0.36 \times 900}{60}=16.96 \mathrm{~m} / \mathrm{s}

From Eq. (20.10), manometric efficiency can be written 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                          0.8=\frac{9.81 \times 16}{V_{w 2} \times 16.96}

or                            V_{w 2}=\frac{9.81 \times 16}{0.8 \times 16.96}=11.57 \mathrm{~m} / \mathrm{s}

The outlet velocity triangle is shown in Fig. 20.10.
From the outlet velocity triangle, we get

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

or          \tan 30^{\circ}=\frac{V_{f 2}}{16.96-11.57}=\frac{V_{f 2}}{5.39}

or              V_{f 2}=5.39 \times \tan 30^{\circ}=3.11 \mathrm{~m} / \mathrm{s}

Discharge is found from Eq. (20.3) as

Q=\pi D_2 B_2 V_{f 2}

=\pi \times 0.36 \times 0.04 \times 3.11=0.1407 \mathrm{~m}^3 / \mathrm{s}