Chapter 13, Problem 58P

(a)

To determine

The speed of flow of water in the pipe of radius $0.25$ m.

Answer to Problem 58P

  ${\upsilon }_2=4.9$ m/s

Explanation of Solution

Given:

Length of conical section of pipe $\left(L\right)$ through which the water flows is $1.0$ m

Radius of cylindrical pipe on the left side of the conical section of pipe $\left(r_1\right)$ is $0.45$ m

Radius of cylindrical pipe on the right side of the conical section of pipe $\left(r_2\right)$ is $0.25$ m

Speed of water that flows in to the $0.45$ m radius pipe $\left({\upsilon }_1\right)$ is $1.50$ m/s

Formula used:

FIGURE: 1

In the figure1, $d_1$ is the diameter of cylindrical pipe on the left side of the conical section of pipe,

  $d_2$ is the diameter of cylindrical pipe on the right side of the conical section of pipe,

  ${\upsilon }_1$ is the speed of water that flows in to the $0.45$ m radius pipe,

  ${\upsilon }_2$ is the speed of water in the $0.25$ m radius pipe,

  $\upsilon$ is the speed of flow of water at a position $x$ in the conical section.

Let $A_1$ and $A_2$ be the area of cross section of cylindrical pipe on the left and right side of the conical section of pipe respectively.

Now applying the continuity equation to the flow of water in the two sections of cylindrical pipes,

  ${\upsilon }_1{A}_1={\upsilon }_2{A}_2$

  $\Rightarrow {\upsilon }_2=\frac{{\upsilon }_1{A}_1}{A_2}$$\left(1\right)$

Area of cross section of cylindrical pipes is,

  $A=\pi r^2=\pi {\left(\frac{d}{2}\right)}^2=\frac{1}{4}\pi d^2$

Applying this to the equation gives,

  ${\upsilon }_2={\upsilon }_1\frac{\frac{1}{4}\pi d_1^2}{\frac{1}{4}\pi d_2^2}={\upsilon }_1\frac{d_1^2}{d_2^2}={\upsilon }_1{\left(\frac{d_1}{d_2}\right)}^2$$\left(2\right)$

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  ${\upsilon }_2=\left(\text{1}\text{.50 m/s}\right){\left(\frac{\text{0}\text{.90 m}}{\text{0}\text{.50 m}}\right)}^{\text{2}}=4.9\text{m/s}$

Conclusion:

The speed of flow of water in the pipe of radius $0.25$ m is $4.9$ m/s.

(b)

To determine

The speed of flow of water at a position $x$ in the conical section, if $x$ is the distance measured from the left-hand end of the conical section of the pipe.

Answer to Problem 58P

  $\upsilon \left(x\right)=\frac{\text{0}{\text{.304 m}}^{\text{3}}\text{/s}}{{\left(\text{0}\text{.45 m-0}\text{.20 }x\right)}^{\text{2}}}$

Explanation of Solution

Given:

Length of conical section of pipe $\left(L\right)$ through which the water flows is $1.0$ m

Radius of cylindrical pipe on the left side of the conical section of pipe $\left(r_1\right)$ is $0.45$ m

Radius of cylindrical pipe on the right side of the conical section of pipe $\left(r_2\right)$ is $0.25$ m

Speed of water that flows in to the $0.45$ m radius pipe $\left({\upsilon }_1\right)$ is $1.50$ m/s

Formula used:

Now applying the continuity equation to the flow of water in the conical section of pipe,

  $\upsilon \left(0\right)A\left(0\right)=\upsilon \left(x\right)A\left(x\right)$

  $\Rightarrow \upsilon \left(x\right)=\frac{\upsilon \left(0\right)A\left(0\right)}{A\left(x\right)}$$\left(3\right)$

The upper half of the conical section of pipe can be redrawn as shown in the figure 2.

FIGURE: 2

Now using the coordinates shown on the line in the figure 2, one can write the expression,

  $\frac{r\left(x\right)-0.45\text{m}}{x-0}=\frac{\text{0}\text{.25 m - 0}\text{.45 m}}{L-0}$

  $\begin{array}{l}\Rightarrow \frac{r\left(x\right)-0.45\text{m}}{x}=\frac{\text{0}\text{.25 m -0}\text{.45 m}}{L}\\ \Rightarrow r\left(x\right)-0.45\text{m}=\frac{\left(\text{0}\text{.25 m - 0}\text{.45 m}\right)}{L}x\end{array}$

  $\Rightarrow r\left(x\right)=0.45\text{m}-\frac{\text{0}\text{.20 m}}{L}x$$\left(4\right)$

Where, $L$ is the length of the conical section

The expression for variation of cross-sectional area of conical section with $x$ is given by,

  $A\left(x\right)=\pi {\left[r\left(x\right)\right]}^2$

Substituting for $r\left(x\right)$ from equation $\left(4\right)$ ,

  $A\left(x\right)=\pi {\left[\text{0}\text{.45 m-}\frac{\text{0}\text{.20 m}}{\text{L}}x\right]}^2$

Substituting for $A\left(x\right)$ in equation $\left(3\right)$ ,

  $\Rightarrow \upsilon \left(x\right)=\frac{\upsilon \left(0\right)A\left(0\right)}{\pi {\left[\text{0}\text{.45 m-}\frac{\text{0}\text{.20 m}}{\text{L}}x\right]}^2}$

Since area of cross section $A\left(0\right)=\frac{1}{4}\pi d_1^2$ , above expression becomes

  $\upsilon \left(x\right)=\frac{\upsilon \left(0\right)\frac{1}{4}\pi d_1^2}{\pi {\left[0.45\text{m}-\frac{\text{0}\text{.20}\text{m}}{L}x\right]}^2}$

  $\upsilon \left(x\right)=\frac{\upsilon \left(0\right)d_1^2}{4{\left[\text{0}\text{.45}\text{m-}\frac{\text{0}\text{.20}\text{m}}{\text{L}}x\right]}^2}$$\left(4\right)$

Calculation:

Substituting the numerical values in equation $\left(4\right)$ ,

  $\begin{array}{l}\upsilon \left(x\right)=\frac{\left(\text{1}\text{.5}\text{m/s}\right){\left(\text{0}\text{.90 m}\right)}^2}{4{\left[0.45\text{m}-\frac{\text{0}\text{.20}\text{m}}{\text{1}\text{.0}\text{m}}x\right]}^2}\\ \upsilon \left(x\right)=\frac{\text{0}\text{.304}{\text{m}}^{\text{3}}\text{/s}}{{\left(0.45\text{m}-0.20x\right)}^2}\end{array}$

Conclusion:

The speed of flow of water at a position $x$ in the conical section, if $x$ is the distance measured from the left-hand end of the conical section of the pipe is $\upsilon \left(x\right)=\frac{\text{0}{\text{.304 m}}^{\text{3}}\text{/s}}{{\left(\text{0}\text{.45 m-0}\text{.20 }x\right)}^{\text{2}}}$ .