Chapter 14, Problem 18SP

Oil flows through a 4.0-cm-i.d. (i.e., inner diameter) pipe at an average speed of 2.5 m/s. Find the flow in m3/s and cm3/s.

To determine

The flow of oil in ${\text{m}}^{\text{3}}\text{/s}$ and ${\text{cm}}^{\text{3}}\text{/s}$ if oil flows through a pipe of $4.0\text{ cm}$ inner diameter at an average speed of $2.5\text{ m/s}$.

Answer to Problem 18SP

Solution:

$3.1\times {10}^{-3}{\text{ m}}^{\text{3}}\text{/s}$ and $3.1\times {10}^3{\text{ cm}}^3\text{/s}$

Explanation of Solution

Given data:

Inner diameter of pipe is $4.0\text{ cm}$.

Average speed of the flow is $2.5\text{ m/s}$.

Formula used:

Write the formula for cross sectional area of circular pipe:

$A=\pi r^2$

Here, $r$ is the radius of the pipe.

Write the expression for radius:

$r=\frac{d}{2}$

Here, $d$ is the diameter of the wire.

Write the expression for discharge rate or flow $\left(J\right)$:

$J=Av$

Here, $A$ is the cross-sectionalarea and $v$ is the average speed of flow.

Explanation:

Recall the expression for radius:

$r=\frac{d}{2}$

Substitute $4.0\text{ cm}$ for $d$

$\begin{array}{c}r=\frac{4.0\text{ cm}}{2}\\ =\left(2.0\text{ cm}\right)\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\\ =2\times {10}^{-2}\text{ m}\end{array}$

Recall the formula for cross sectional area of the pipe:

$A=\pi r^2$

Substitute $2\times {10}^{-2}\text{ m}$ for $r$

$\begin{array}{c}A=\pi {\left(2\times {10}^{-2}\text{ m}\right)}^2\\ =1.25\times {10}^{-3}{\text{ m}}^2\end{array}$

Recall the expression for discharge rate $\left(J\right)$:

$J=Av$

Substitute $1.25\times {10}^{-3}{\text{ m}}^2$ for $A$ and $2.5\text{ m/s}$ for $v$.

$\begin{array}{c}J=\left(1.25\times {10}^{-3}{\text{ m}}^2\right)\left(2.5\text{ m/s}\right)\\ =3.1\times {10}^{-3}{\text{ m}}^3\text{/s}\end{array}$

Convert the flow rate of oil from ${\text{m}}^3\text{/s}$ to ${\text{cm}}^3\text{/s}$.

$\begin{array}{c}3.1\times {10}^{-3}{\text{ m}}^3\text{/s}=3.1\times {10}^{-3}{\text{ m}}^3\text{/s}{\left(\frac{1\text{ cm}}{{10}^{-2}\text{ m}}\right)}^3\\ =3.1\times {10}^3{\text{ cm}}^3\text{/s}\end{array}$

Conclusion:

The flow of oil through the pipein ${\text{m}}^{\text{3}}\text{/s}$ is $3.1\times {10}^{-3}{\text{ m}}^{\text{3}}\text{/s}$ and in ${\text{cm}}^{\text{3}}\text{/s}$ is $3.1\times {10}^3{\text{ cm}}^3\text{/s}$.