91Ó°ÊÓ

• The number of showerhead openings is,$n=20$
• The radius of each circular opening is,$\mathrm{r}=1.0\mathrm{mm}\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)=1×{10}^{-3}\mathrm{m}$
• The speed of water in the pipe,$v_2=3.0\mathrm{m}/\mathrm{s}$
• The radius of the pipe is,$r_2=0.80\mathrm{cm}\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)=1×{10}^{-3}\mathrm{m}$

In mathematics, a "transport equation" or "continuity equation" explains the movement of a given quantity. A preserved quantity is especially simple and effective, but it may be extended to apply to any large amount.

The continuity equation can be expressed as,

$A_1{v}_1=A_2{v}_2$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(1)

Where A and v are the cross-sectional area and velocity of the fluid in a particular pipe.

The area of the circular opening is evaluated by,

$$\begin{gathered} A_1=\mathrm{π}×{\left(1×{10}^{-3}\mathrm{m}\right)}^2 \\ =3.14×{10}^{-6}{\mathrm{m}}^2 \end{gathered}$$

The area of the pipe can be evaluated by,

$$\begin{gathered} A_2=\mathrm{π}×{\left(8×{10}^{-3}\mathrm{m}\right)}^2 \\ =2.01×{10}^{-4}{\mathrm{m}}^2 \end{gathered}$$

To evaluate speed at shower opening using equation (1),

$$\begin{gathered} v_1=\left(\frac{\left(2.01×{10}^{-4}{\mathrm{m}}^2\right)\left(3.0\mathrm{m}/\mathrm{s}\right)}{3.14×{10}^{-6}{\mathrm{m}}^2}\right) \\ =192.1\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the showerhead is 192.1m/s