Question 20.14: The following data refer to a centrifugal pump which is desi......

Given data:

Speed of pump                                N =1000 rpm

Head                                                  H = 70 m

Internal diameter of impeller        D_1=300 \mathrm{~mm}=0.3 \mathrm{~m}

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

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

Width at outlet                                  B_2=50 \mathrm{~mm}=0.05 \mathrm{~m}

Vane angle at outlet                          \beta_2=30^{\circ}

Discharge                                            Q=400 \text { litres } / \mathrm{s}=0.4 \mathrm{~m}^3 / \mathrm{s}

Overall efficiency                                \eta_o=0.70

The inlet and outlet velocity triangles are shown in Fig. 20.18.

(a) The velocity of flow at inlet is found to be

V_{f 1}=\frac{Q}{\pi D_1 B_1}=\frac{0.4}{\pi \times 0.3 \times 0.1}=4.244 \mathrm{~m} / \mathrm{s}

Tangential velocity of impeller at inlet is given by

U_1=\frac{\pi D_1 N}{60}=\frac{\pi \times 0.3 \times 1000}{60}=15.708 \mathrm{~m} / \mathrm{s}

From the inlet velocity triangle, we have

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

or          \beta_1=\tan ^{-1}\left(\frac{4.244}{15.708}\right)=15.12^{\circ}

(b) The velocity of flow at outlet is found to be

V_{f 2}=\frac{Q}{\pi D_2 B_2}=\frac{0.4}{\pi \times 0.6 \times 0.05}=4.244 \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 1000}{60}=31.412 \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 30^{\circ}=\frac{4.244}{31.412-V_{w 2}}

or                  V_{w 2}=31.412-\frac{4.244}{\tan 30^{\circ}}=31.412-7.351=24.061 \mathrm{~m} / \mathrm{s}

Theoretical head developed is given by Eq. (20.4) as

H_m=\frac{V_{w 2} U_2}{g}=\frac{24.061 \times 31.412}{9.81}=77.04 \mathrm{~m}

(c) The manometric efficiency is given by

\eta_{\text {mano }}=\frac{H}{H_m}=\frac{70}{77.04}=0.9086 \text { or } 90.86 \%

(d) The power required to drive the pump is

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

=\frac{1000 \times 0.4 \times 9.81 \times 77.04}{0.7}=431.86 \times 10^3 \mathrm{~W}=431.86 \mathrm{~kW}

The corresponding mechanical efficiency is found from Eq. (20.14) as

\eta_{o}=\eta_{ma n o}\times\eta_{m}          (20.14)

\eta_m=\frac{\eta_o}{\eta_{\text {mano }}}=\frac{0.7}{0.9086}=0.7704 \text { or } 77.04 \%