Question 20.20: A centrifugal pump rotating at 1800 rpm delivers 0.4 m³/s at......

Given data:

Speed of pump                                      N_1=1800 \mathrm{~rpm}

Discharge                                                Q_1=0.4 \mathrm{~m}^3 / \mathrm{s}

Head                                                        H_1=16 \mathrm{~m}

Overall efficiency                                \eta_o=0.70

The power input to the pump is given by

P_1=\frac{\rho Q_1 g H_1}{\eta_o}

=\frac{1000 \times 0.4 \times 9.81 \times 16}{0.7}=89.69 \times 10^3 \mathrm{~W}=89.69 \mathrm{~kW}

The specific speed of the pump is given by Eq. (20.30) as

N_{S P}={\frac{N{\sqrt{Q}}}{(H)^{3/4}}}        (20.30)

N_s=\frac{N_1 \sqrt{Q_1}}{H_1^{3 / 4}}

=\frac{1800 \sqrt{0.4}}{(16)^{3 / 4}}=142.3 \text { units }

Let Q_2, H_2 \text { and } P_2 be the discharge, head and power required for homologous condition, respectively. Using Eq. (20.31), we have

\left(\frac{Q}{N D^3}\right)_m=\left(\frac{Q}{N D^3}\right)_p     (20.31)

\frac{Q_1}{N_1}=\frac{Q_2}{N_2}                     \left[\because D_1=D_2\right]

or                                  Q_2=Q_1 \frac{N_2}{N_1}=0.4 \times \frac{900}{1800}=0.2 \mathrm{~m}^3 / \mathrm{s}

Using Eq. (20.32), we obtain

\left({\frac{g H}{N^{2}D^{2}}}\right)_{m}=\left({\frac{g H}{N^{2}D^{2}}}\right)_{p}            (20.32)

\frac{H_1}{N_1^2}=\frac{H_2}{N_2^2}

or                              H_2=H_1 \frac{N_2^2}{N_1^2}=16 \times \frac{900^2}{1800^2}=4 \mathrm{~m}

Using Eq. (20.33), one can write

\left({\frac{P}{\rho N^{3}D^{5}}}\right)_{m}=\left({\frac{P}{\rho N^{3}D^{5}}}\right)_{p}    (20.33)

\frac{P_1}{N_1^3}=\frac{P_2}{N_2^3}

or                  P_2=P_1 \frac{N_2^3}{N_1^3}=89.69 \times \frac{900^3}{1800^3}=11.21 \mathrm{~kW}