Question

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

Answer 1

Hint: Volume flux of water in a pipe is defined as the rate of volume of water flowing through the pipe in unit time. For a pipe, volume flux is equal to the product of area of cross-section of the pipe and velocity of water, at a particular point in the pipe.
Formula used:
$1)\dfrac{dV}{dt}=Av$
$2)A=\pi {{r}^{2}}$

Complete answer:
From the question, we can understand that water is allowed to flow through a pipe of certain radius. Water is flowing at the rate of $15{{m}^{3}}{{\min }^{-1}}$. Velocity of water is given as $3m{{s}^{-1}}$, at a particular point in the pipe. We are required to find the radius of the pipe at this particular point.
We know that volume flux of water in a pipe is defined as the rate of volume of water flowing through the pipe with respect to time. Volume flux is represented by
$\dfrac{dV}{dt}$
where
$dV$ is the change in volume of water
$dt$ is the change in time
Let this be equation 1.
According to the question provided, the value of volume flux is given by
 $\dfrac{dV}{dt}=15{{m}^{3}}{{\min }^{-1}}=\dfrac{15}{60}{{m}^{3}}{{s}^{-1}}$
Now, let us consider a point in the pipe where the value of velocity of water is given by
$v=3m{{s}^{-1}}$
where
we have assumed that $v$ is the velocity of water at that particular point.
Let this be equation 2.
For water flowing in a pipe, we know that volume flux is equal to the product of area of cross-section of pipe and velocity of water at a particular point. Mathematically, volume flux can be represented as
$\dfrac{dV}{dt}=Av$
where
$A$ is the area of cross-section of pipe at a particular point
$v$ is the velocity of water at that particular point
Let this be equation 3.
Using equation 3, we can calculate the area of cross-section of the pipe, used in our question.
Therefore,
$\dfrac{dV}{dt}=Av\Rightarrow \dfrac{15}{60}{{m}^{3}}{{s}^{-1}}=A\times 3m{{s}^{-1}}\Rightarrow A=\dfrac{15}{180}{{m}^{2}}$
Let this be equation 4.
Now, we know that area of cross-section of a pipe is equal to the area of a circle, and is given by
$A=\pi {{r}^{2}}$
where
$A$ is the area of cross-section of pipe
$r$ is the radius of cross-section of pipe
Let this be equation 5.
Substituting the value of $A$ from equation 4, equation 5 becomes
$A=\pi {{r}^{2}}\Rightarrow \dfrac{15}{180}=\pi {{r}^{2}}\Rightarrow {{r}^{2}}=\dfrac{15}{180\pi }=\dfrac{15}{180\times 3.14}=\dfrac{1}{12\times 3.14}=0.02653\Rightarrow r=\sqrt{0.02653}=0.I62m$
Therefore, the radius of pipe where water velocity is $3m{{s}^{-1}}$, is equal to $0.162m$.

Note:
Students need to be thorough with conversion formulas as well as the value of $\pi $. Conversion formula used here is given below.
$\begin{align}
  & 1\min =60s \\
 & 15{{m}^{3}}{{\min }^{-1}}=15{{m}^{3}}{{(60s)}^{-1}}=\dfrac{15}{60}{{m}^{3}}{{s}^{-1}} \\
\end{align}$
It is to be noted that converting parameters to the SI system of units is the easiest as well as the safest method to solve such problems. Also, the value of $\pi $ is taken as $3.14$, in such types of questions.