Question

The cross-sectional area of a pipe is 25 square metres. What is the volumetric flow rate of the water if the velocity of the water is 10m/s?

Answer 1

Hint: Assume that the cross-sectional radius of the pipe is r. Use the fact that the area of the cross-section of the pipe or radius r is equal to $\pi {{r}^{2}}$. Hence find the radius of the pipe. Use the fact that the volume of the cylinder of radius r and height h is given by $V=\pi {{r}^{2}}h$. Hence determine the volumetric flow of water through the pipe.

Complete step-by-step answer:
Let the cross-sectional radius of the pipe be r.

We know that the area of a circle of radius r is given by $\pi {{r}^{2}}$
Since the cross-section of the pipe is circular, we have
$A=\pi {{r}^{2}}$, where A is the area of the cross-section of the pipe.
Since the cross-sectional area of the pipe is 25 square metres, we have
$\pi {{r}^{2}}=25$
Dividing both sides by $\pi $, we get
${{r}^{2}}=\dfrac{25}{\pi }$
Hence, we have
$r=\dfrac{5}{\sqrt{\pi }}$
Length of the column of the water flowing through the pipe in 1 second = 10m(Since the flow rate is 10m/s)
We know that the volume of the cylinder of radius r and height h is given by $\pi {{r}^{2}}h$.
Hence, we have
The volume of the column of the water flowing through the pipe in 1 second $=\pi {{\left( \dfrac{5}{\sqrt{\pi }} \right)}^{2}}\times 10=250{{m}^{3}}$
Hence, we have
The volumetric flow rate of water $=250{{m}^{3}}{{s}^{-1}}$

Note: Alternative Solution:
We have
The volume of a column of water of height h is given by
\[V=\pi {{r}^{2}}h=ar\left( \text{cross-section} \right)\times h\]
Hence, we have
The volumetric flow rate $=ar\left( \text{cross-section} \right)\times \text{velocity of water}$
Hence, we have
The volumetric flow rate $=25\times 10{{m}^{3}}{{s}^{-1}}=250{{m}^{3}}{{s}^{-1}}$, which is the same as obtained above.