Question 20.3: The following data refer to a centrifugal pump: The diameter......

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

External diameter of impeller    D_2=200 \mathrm{~mm}=0.2 \mathrm{~m}

Speed of pump                              N =1000 rpm

Vane angle at inlet                             \beta_1=30^{\circ}

Vane angle at outlet                            \beta_2=40^{\circ}

Tangential velocity of impeller at inlet is given by

U_1=\frac{\pi D_1 N}{60}=\frac{\pi \times 0.1 \times 1000}{60}=5.236 \mathrm{~m} / \mathrm{s}

Tangential velocity of impeller at outlet is given by

U_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 0.2 \times 1000}{60}=10.472 \mathrm{~m} / \mathrm{s}

The inlet and outlet velocity triangles are shown in Fig. 20.9.
From the inlet velocity triangle, we have

\tan \beta_1=\frac{V_{f 1}}{U_1}

or            V_{f 1}=U_1 \tan \beta_1=5.236 \times \tan 30^{\circ}=3.023 \mathrm{~m} / \mathrm{s}

Since the velocity of flow is constant, the velocity of

flow at outlet is

V_{f 2}=V_{f 1}=3.023 \mathrm{~m} / \mathrm{s}

From the outlet velocity triangle, we get

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

or                    \tan 40^{\circ}=\frac{3.023}{10.472-V_{w 2}}

or                    10.472-V_{w 2}=\frac{3.023}{\tan 40^{\circ}}=3.603

or                      V_{w 2}=10.472-3.603=6.869 \mathrm{~m} / \mathrm{s}

The work done by impeller per kg of water is given by Eq. (20.4) as

H={\frac{V_{w2}U_{2}}{g}}          (20.4)

=\frac{V_{w 2} U_2}{g}=\frac{6.869 \times 10.472}{9.81}=7.332 \mathrm{~N}-\mathrm{m} / \mathrm{N}