Water flows through a horizontal pipe at a speed ' $\mathrm{V}$ '. Internal diameter of the pipe is '…
Water flows through a horizontal pipe at a speed ' $\mathrm{V}$ '. Internal diameter of the pipe is ' $\mathrm{d}$ '. If the water is coming out at a speed ' $V_1$ ' then the diameter of the nozzle is
$d \sqrt{\frac{V_1}{V}}$
$d \sqrt{\frac{V}{V_1}}$
$\frac{d V}{V_1}$
$\frac{\mathrm{V}_1}{\mathrm{dV}}$
Solution
From equation of continuity,
$\begin{aligned}
& \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\
& \text { Given: } \mathrm{V}_1=\mathrm{V} \text { and } \mathrm{V}_2=\mathrm{V}_1 \\
& \Rightarrow \mathrm{A}_1 \mathrm{~V}=\mathrm{A}_2 \mathrm{~V}_1 \\
& \frac{\pi \mathrm{d}^2}{4} \mathrm{~V}=\frac{\pi \mathrm{d}_{\mathrm{n}}^2 \mathrm{n}}{4} \mathrm{~V}_1 \quad \ldots\left(\because \mathrm{A}=\frac{\pi \mathrm{d}^2}{4}\right)
\end{aligned}$
where $d_n$ is the diameter of the nozzle.
$\begin{aligned}
& \therefore \quad \mathrm{d}_{\mathrm{n}}^2=\mathrm{d}^2 \frac{\mathrm{V}}{\mathrm{V}_1} \\
& \therefore \quad \mathrm{d}_{\mathrm{n}}=\mathrm{d} \sqrt{\frac{\mathrm{V}}{\mathrm{V}_1}}
\end{aligned}$
: