Question 20.13: A centrifugal pump is used to deliver water and it has an im......

Given data:

External diameter of impeller            D_2=600 \mathrm{~mm}=0.6 \mathrm{~m}

Internal diameter of impeller              D_1=200 \mathrm{~mm}=0.2 \mathrm{~m}

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

Width at inlet                                        B_1=100 \mathrm{~mm}=0.1 \mathrm{~m}

Vane angle at inlet                                \beta_1=15^{\circ}

Vane angle at outlet                              \beta_2=25^{\circ}

Speed of pump                                      N =1500 rpm

Head                                                        H = 30 m

Overall efficiency                                  \eta_o=0.70

(a) Tangential velocity of impeller at inlet is given by

U_1=\frac{\pi D_1 N}{60}=\frac{\pi \times 0.2 \times 1500}{60}=15.71 \mathrm{~m} / \mathrm{s}

Tangential velocity of impeller at outlet is given by

U_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 0.6 \times 1500}{60}=47.12 \mathrm{~m} / \mathrm{s}

The inlet and outlet velocity triangles are shown in Fig. 20.17.

From the inlet velocity triangle, we have

or    V_{f 1}=U_1 \tan \beta_1

=15.71 \tan 15^{\circ}=4.21 \mathrm{~m} / \mathrm{s}

The discharge is given by 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 0.2 \times 0.1 \times 4.21=0.265 \mathrm{~m}^3 / \mathrm{s}

(b) The velocity of flow at the impeller outlet is

V_{f 2}=\frac{Q}{\pi D_2 B_2}=\frac{0.265}{\pi \times 0.6 \times 0.04}=3.51 \mathrm{~m} / \mathrm{s}

From the outlet velocity triangle, we have

V_{w 2}=U_2-V_{f 2} \cot \beta_2

=47.12-3.51 \cot 25^{\circ}=39.59 \mathrm{~m} / \mathrm{s}

The head developed is found to be

H=\frac{V_{w 2} U_2}{g}

=\frac{39.59 \times 47.12}{9.81}=190.16 \mathrm{~m}

(c) The power required to drive the pump is

=\frac{\rho Q g H}{\eta_o}

=\frac{1000 \times 0.265 \times 9.81 \times 190.16}{0.7}=706.21 \times 10^3 \mathrm{~W}=706.21 \mathrm{~kW}

(d) Applying energy equation between inlet and outlet of the impeller, we have

\frac{p_1}{\rho g}+\frac{V_1^2}{2 g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2 g}+z_2-\frac{V_{w 2} U_2}{g}

or            \frac{p_2}{\rho g}-\frac{p_1}{\rho g}=\frac{V_1^2}{2 g}-\frac{V_2^2}{2 g}+\frac{V_{w 2} U_2}{g}         \left[\because z_1=z_2\right]

Now,                V_1=V_{f 1}=4.21 \mathrm{~m} / \mathrm{s}

From the outlet velocity triangle, we have

V_2^2=V_{w2}^2+V_{f2}^2

=3.51^2+39.59^2

or                  V_2=39.75 \mathrm{~m} / \mathrm{s}

The pressure rise is

\frac{p_2}{\rho g}-\frac{p_1}{\rho g}=\frac{V_1^2}{2 g}-\frac{V_2^2}{2 g}+\frac{V_{w 2} U_2}{g}

=\frac{4.21^2}{2 \times 9.81}-\frac{39.75^2}{2 \times 9.81}+\frac{39.59 \times 47.12}{9.81}

=110.53 m