Question 20.21: A centrifugal pump has an impeller of 80 cm in diameter and ......

Given data:

For pump 1,

Diameter of impeller  D_1=80 \mathrm{~cm}=0.8 \mathrm{~m}

Discharge                          Q_1=40 \text { litres } / \mathrm{s}=0.04 \mathrm{~m}^3 / \mathrm{s}

Head                                        H_1=30 \mathrm{~m}

Speed of pump                      N_1=800 \mathrm{~rpm}

For pump 2,

Diameter of impeller                      D_2=40 \mathrm{~cm}=0.4 \mathrm{~m}

Speed of pump                              N_2=1600 \mathrm{~rpm}

Using Eq. (20.31), we get

\left({\frac{Q}{N D^{3}}}\right)_{m}=\left({\frac{Q}{N D^{3}}}\right)_{p}      (20.31)

\left(\frac{Q}{N D^3}\right)_1=\left(\frac{Q}{N D^3}\right)_2

or                          \frac{Q_1}{N_1 D_1^3}=\frac{Q_2}{N_2 D_2^3}

or                        Q_2=Q_1 \times \frac{N_2}{N_1} \times \frac{D_2^3}{D_1^3}=0.04 \times \frac{1600}{800} \times \frac{0.4^3}{0.8^3}=0.01 \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)

\left(\frac{H}{N^2 D^2}\right)_1=\left(\frac{H}{N^2 D^2}\right)_2

or              \frac{H_1}{N_1^2 D_1^2}=\frac{H_2}{N_2^2 D_2^2}

or              H_2=H_1 \times \frac{N_2^2}{N_1^2} \times \frac{D_2^2}{D_1^2}=30 \times \frac{1600^2}{800_2} \times \frac{0.4^2}{0.8^2}=30 \mathrm{~m}

Using Eq. (20.33), we have

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

\left(\frac{P}{N^3 D^5}\right)_1=\left(\frac{P}{N^3 D^5}\right)_2

or                        \frac{P_1}{N_1^3 D_1^5}=\frac{P_2}{N_2^3 D_2^5}

or                        \frac{P_2}{P_1}=\frac{N_2^3}{N_1^3} \times \frac{D_2^5}{D_1^5}=\frac{1600^3}{800^3} \times \frac{0.4^5}{0.8^5}=0.25