Chapter 9, Problem 5SP

(a)

To determine

The cross-sectional area of the wide and narrow portions of the pipe.

Answer to Problem 5SP

The cross-sectional area of the wide and narrow portions of the pipe are $50.26{\text{ cm}}^2$ and $19.6{\text{ cm}}^2$ respectively.

Explanation of Solution

Given info: The diameter of the wider portion of the pipe is $8\text{ cm}$. The diameter of the narrower portion of the pipe is $5\text{ cm}$. The velocity of water in the wider portion of the pipe is $1.5\text{m}/\text{s}$.

Write the expression to find the area of the wider portion of the pipe.

$\begin{array}{c}{A}_w=\pi r_w^2\\ =\pi {\left(\frac{d_w}{2}\right)}^2\end{array}$

Here,

$d_w$ is the diameter of the narrow pipe

$r_w$ is the radius of the narrow pipe

Substitute $8\text{ cm}$ for $d_w$ in the above equation.

$\begin{array}{c}{A}_w=\pi {\left(\frac{8\text{ cm}}{2}\right)}^2\\ =50.26{\text{ cm}}^2\end{array}$

Write the expression to find the area of the narrow pipe.

$\begin{array}{c}{A}_n=\pi r_n^2\\ =\pi {\left(\frac{d_n}{2}\right)}^2\end{array}$

Here,

$d_n$ is the diameter of the narrow pipe

$r_n$ is the radius of the narrow pipe

Substitute $5\text{ cm}$ for $d_n$ in the above equation.

$\begin{array}{c}{A}_n=\pi {\left(\frac{5\text{ cm}}{2}\right)}^2\\ =19.6{\text{ cm}}^2\end{array}$

Conclusion:

Therefore, the cross-sectional area of the wide and narrow portions of the pipe are $50.26{\text{ cm}}^2$ and $19.6{\text{ cm}}^2$ respectively.

(b)

To determine

The speed of water in narrow portion of the pipe.

Answer to Problem 5SP

The speed of water in narrow portion of the pipe is $3.85\text{m}/\text{s}$.

Explanation of Solution

Write the expression to find relation between velocity and cross-sectional area.

$v_n{A}_n=v_w{A}_w$

Here,

$v_n$ is the velocity of the narrower section of pipe

$A_n$ is the cross-sectional area of the narrower section of pipe

$v_w$ is the velocity of the wider section of pipe

$A_w$ is the cross-sectional area of the wider section of pipe

Substitute $1.5\text{m}/\text{s}$ for $v_w$, $50.26{\text{ cm}}^2$ for $A_w$, $19.6{\text{ cm}}^2$ for $A_n$ and rearrange the above equation to find the speed of water in the narrowed portion of the pipe.

$\begin{array}{c}{v}_n=\frac{v_w{A}_w}{A_n}\\ =\frac{\left(1.5\text{m}/\text{s}\right)\left(50.26{\text{ cm}}^2\right)}{\left(19.6{\text{ cm}}^2\right)}\\ =3.85\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of water in narrow portion of the pipe is $3.85\text{m}/\text{s}$.

(c)

To determine

Whether the pressure in the narrow portion of the pipe is greater than, less than, or equal to the pressure in the wider portion.

Answer to Problem 5SP

The pressure in the narrow portion is less than the pressure in the wider portion of the pipe.

Explanation of Solution

The expression for the continuity equation is $v_n{A}_n=v_w{A}_w$. The expression for Bernoulli’s equation for a horizontal streamline is given by $P_w-P_n=\frac{1}{2}\rho \left(v_n^2-v_w^2\right)$, here, $P_n$ is the pressure in the narrow portion of the pipe, $P_w$ is the pressure in the wide portion of the pipe and $\rho$ is the density of water

The velocity of the narrow portion of the pipe is greater than the velocity of the wide portion of the pipe, $v_n>v_w$. Thus according the Bernoulli’s equation, the pressure in the narrow portion will be less than the pressure in the wider portion of the pipe.

Conclusion:

Therefore, the pressure in the narrow portion is less than the pressure in the wider portion of the pipe.