Question

Water flows through a horizontal pipe of varying area of cross section at the rate of 15 cubic metre per minute. Find the radius of pipe when water velocity is 3 $m{s^{ - 1}}$.

Answer 1

Convert the rate of flow of water in SI units, then use the relation between volume of water flowing per second R, area of cross section of the pipe A, and the velocity of fluid (here, water) flowing inside the pipe, v.

Complete Step by Step Answer:
We are given the rate of flow of water, R = 15 ${m^3}/\min$.
Converting into SI units we get,
$R = \dfrac{{15}}{{60}} = 0.25{m^3}/\sec$
As we know, $R = A$ where R=volume of water flow per second, A=area of cross section and v=velocity of water. Using the relation,
$\Rightarrow A = \dfrac{R}{v}$
$\Rightarrow A = \dfrac{{0.25{m^3}/\sec }}{{3m/\sec }}$
$\Rightarrow A = \dfrac{1}{{12}}{m^2}$
We get, the area of the cross section where the velocity of water is 3 $m{s^{ - 1}}$ will be $\dfrac{1}{{12}}{m^2}$.
Now we use the relation $A = \pi {r^2}$ , where A is the area of the cross section of pipe, and r is the radius of the pipe.
Rearranging the above equation and substituting the values,
$\Rightarrow r = \sqrt {\dfrac{A}{\pi }}$
$\Rightarrow r = \sqrt {\dfrac{1}{{12 \times \pi }}}$
$\Rightarrow r = 0.162m$
$\Rightarrow r = 16.2cm$

Note: Remember to always change all given values into SI Units, prior to solving the question.