Question 20.22: A 1/5^th scale model of a centrifugal pump absorbs 15 kW whe......

Given data:

Power absorbed by model            P_m=15 \mathrm{~kW}

Head of model                                H_m=5 \mathrm{~m}

Speed of model                              N_m=2000 \mathrm{~rpm}

Head of prototype                          H_p=30 \mathrm{~m}

Let D_m \text { and } D_p be the diameters of the impeller of the model and prototype respectively. Then \frac{D_m}{D_p}=\frac{1}{5} 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)_m=\left(\frac{H}{N^2 D^2}\right)_p

or                        \left(\frac{\sqrt{H}}{N D}\right)_m=\left(\frac{\sqrt{H}}{N D}\right)_p

or                      \frac{\sqrt{H_m}}{N_m D_m}=\frac{\sqrt{H_p}}{N_p D_p}

or                    N_p=N_m \times \frac{D_m}{D_p} \times \frac{\sqrt{H_p}}{\sqrt{H_m}}=2000 \times \frac{1}{5} \times \frac{\sqrt{30}}{\sqrt{5}}=979.79 \mathrm{~rpm}

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)_m=\left(\frac{P}{N^3 D^5}\right)_p

or                      \frac{P_m}{N_m^3 D_m^5}=\frac{P_p}{N_p^3 D_p^5}

or                      P_p=P_m \times \frac{N_p^3}{N_m^3} \times \frac{D_p^5}{D_m^5}=15 \times \frac{979.79^3}{2000^3} \times 5^5=5511.25 \mathrm{~kW}

Using Eq. (20.31), we get

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

or                        \frac{Q_m}{N_m D_m^3}=\frac{Q_p}{N_p D_p^3}

or                        \frac{Q_p}{Q_m}=\frac{N_p}{N_m} \times \frac{D_p^3}{D_m^3}=\frac{979.79}{2000} \times 5^3=61.237