91Ó°ÊÓ

• The diameter of a large pipe is, ${\mathrm{d}}_{\mathrm{L}}=1.9\mathrm{cm}$.

• The diameter of smaller pipes, ${\mathrm{d}}_{\mathrm{S}}=1.3\mathrm{cm}$.

• The flow rates in the smaller pipes are$26,19\mathrm{and}11\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$\min $}\right.$

An ideal fluid is incompressible and lacks viscosity, and its flow is steady and irrotational. A streamline is a path followed by an individual fluid particle. A tube of flow is a bundle of streamlines. The flow within any tube of flow obeys the equation of continuity:

${\mathbf{R}}_{\mathbf{v}}\mathbf{=}\mathbf{V}\mathbf{=}\mathbf{Constant}$

Where ${\mathbf{R}}_{\mathbf{v}}$is the volume flow rate, $\mathbf{A}$is the cross-sectional area of the tube of flow at any point, and $\mathbf{v}$ is the speed of the fluid at that point.

Find the flow rate in a large pipe having diameter $\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{cm}\mathbf{}$ using the equation of continuity, and to calculate the ratio of speed, assume that both pipes are circular.

Formulae are as follows:

1. The equation of continuity is$\mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2+…+{\mathrm{R}}_{\mathrm{n}}$

2. The equation for water speed is$\mathrm{V}=\frac{\mathrm{R}}{\mathrm{A}}$

3. The equation for area of cross-section is$\mathrm{A}=\frac{{\mathrm{d}}^2}{4}$

Where, $\mathrm{R}$is rate of flow, $\mathrm{A}$is area,$\mathrm{v}$is velocity and d is diameter of pipe.

The equation of continuity can be written as,

$\mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2+…+{\mathrm{R}}_{\mathrm{n}}$

Here, three smaller pipes are given, so $\mathrm{n}=3$.

Hence,

$\mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3$

$=(26+19+11)\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$\min $}\right.$

$=56\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$\min $}\right.$

Hence, the flow rate in the pipe of $1.9\mathrm{cm}$diameter is $56\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$\min $}\right.$.

The equation for water speed can be written as,

$\mathrm{V}=\frac{\mathrm{R}}{\mathrm{A}}$

So, the required ratio can be written as,

$\frac{{\mathrm{V}}_{56}}{{\mathrm{V}}_{26}}=\frac{\frac{{\mathrm{R}}_{56}}{{\mathrm{A}}_{56}}}{{\mathrm{R}}_{26}}=\frac{{\mathrm{R}}_{56}}{{\mathrm{A}}_{56}}×\frac{{\mathrm{A}}_{26}}{{\mathrm{R}}_{26}}$

$=\frac{56}{{\mathrm{Ï€»å}}_{\mathrm{L}}^2}×\frac{\frac{\mathrm{Ï€}}{4}{\mathrm{d}}_{\mathrm{s}}^2}{26}$

$=\frac{56}{1{.9}^2}×\frac{1{.3}^2}{26}$

$≈1.0$

Hence, the ratio of the speed in the $1.9\mathrm{cm}$pipe to that in the pipe carrying $26\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$\min $}\right.$is $1.0$.