Question 18.3: Water flowing at the rate of 0.03 m³/s strikes a flat plate ......

Given data:

Volume flow rate of water                    Q=0.03 \mathrm{~m}^3 / \mathrm{s}

Force exerted on the plate                  F_n=520 \mathrm{~N}

The force exerted on the plate is given by Eq. (18.1) as

F_{x}=\rho a V^{2}        (18.1)

F_n=\rho a V^2=\rho V Q                         [\because a V=Q]

or                                                520=1000 \times V \times 0.03

or                                                V=17.33 \mathrm{~m} / \mathrm{s}

Let d be the diameter of the stream of water. The diameter of the stream of water is found to be

\frac{\pi}{4} d^2 V=Q

or                                                  \frac{\pi}{4} d^2 \times 17.33=0.03

or                                            d=0.047 \mathrm{~m}=47 \mathrm{~mm}