Chapter 4, Problem 4.1P

(a)

Interpretation Introduction

Interpretation:

The average velocity of the given liquid flowing in the pipe is to be calculated.

Concept Introduction:

The formula to calculate the average velocity of a stream which is flowing through a pipe of cross-sectional area S is:

  $\overline{V}=\frac{1}{S}{\displaystyle \underset{S}{\int }udS}$ ...... (1)

Here, u is the local velocity of the stream in the differential cross-sectional area dS .

Answer to Problem 4.1P

The average velocity of the given liquid flowing in the pipe is $\overline{V}=0.808\text{m}/\text{s}$ .

Explanation of Solution

Given information:

In a $75\text{ mm}$ pipe, a liquid is flowing at a steady state. The variation of the local velocity of the liquid with the distance from the pipe axis is tabulated as:

  

The cross-sectional area of the pipe is defined as:

  $S=\pi r_w^2$ ...... (2)

Here, $r_w$ is the radius of the pipe. Differentiate equation (2) as shown below:

  $dS=2\pi rdr$ ...... (3)

Here, $r$ is the radial distance from the pipe axis.

Use equation (2) and (3) in equation (1) and simplify further as

  $\begin{array}{c}\overline{V}=\frac{1}{S}{\displaystyle \underset{S}{\int }udS}\\ =\frac{1}{\pi r_w^2}{\displaystyle \underset{0}{\overset{r_w}{\int }}u\left(2\pi rdr\right)}\\ =\frac{2}{r_w^2}{\displaystyle \underset{0}{\overset{r_w}{\int }}urdr}\end{array}$

The table for $u*r$ is:

  

Now, plot the graph of $u*r$ versus $r$ using the excel tool for graphing for the given data as:

  

The area under this curve is defined as ${\displaystyle \underset{0}{\overset{r_w}{\int }}urdr}$ . Thus,

  ${\displaystyle \underset{0}{\overset{r_w}{\int }}urdr}=567.8\times {10}^{-6}{\text{m}}^3/\text{s}$

Now, use this value to calculate the average velocity as:

  $\begin{array}{c}\overline{V}=\frac{2}{r_w^2}{\displaystyle \underset{0}{\overset{r_w}{\int }}urdr}\\ =\frac{2}{{\left(37.5\times {10}^{-3}\text{ m}\right)}^2}\times \left(567.8\times {10}^{-6}{\text{m}}^3/\text{s}\right)\\ =0.808\text{m}/\text{s}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The correction factor for the kinetic energy of the given liquid flowing in the pipe is to be calculated.

Concept Introduction:

The formula to calculate the correction factor $\left(\alpha \right)$ for the kinetic energy of a stream which is flowing through a pipe of cross-sectional area S is:

  $\alpha =\frac{{\displaystyle \underset{S}{\int }{u}^{3}dS}}{{\overline{V}}^{3}S}$ ...... (4)

Here, u is the local velocity of the stream in the differential cross-sectional area dS and $\overline{V}$ is the average velocity of the stream.

Answer to Problem 4.1P

The correction factor for the kinetic energy of the given liquid flowing in the pipe is $1.188$ .

Explanation of Solution

Given information:

In a $75\text{ mm}$ pipe, a liquid is flowing at a steady state. The variation of the local velocity of the liquid with the distance from the pipe axis is tabulated as

  

The cross-sectional area of the pipe is defined as:

  $S=\pi r_w^2$ ...... (2)

Here, $r_w$ is the radius of the pipe. Differentiate equation (2) as shown below:

  $dS=2\pi rdr$ ...... (3)

Here, $r$ is the radial distance from the pipe axis.

Use equation (2) and (3) in equation (4) and simplify further as:

  $\begin{array}{c}\alpha =\frac{{\displaystyle \underset{S}{\int }{u}^{3}dS}}{{\overline{V}}^{3}S}\\ =\frac{{\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^3\left(2\pi rdr\right)}}{{\overline{V}}^3\pi r_w^2}\\ =\frac{2}{{\overline{V}}^3{r}_w^2}{\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^{3}rdr}\end{array}$

Use the value of $\overline{V}$ as calculated in part (a) as:

  $\overline{V}=0.808\text{m}/\text{s}$

The table for $u^{3}r$ is:

  

Now, plot the graph of $u^{3}r$ versus $r$ using the excel tool for graphing for the given data as:

  

The area under this curve is defined as ${\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^{3}rdr}$ . Thus,

  ${\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^{3}rdr}=0.000441{\text{m}}^5/{\text{s}}^3$

Now, use this value to calculate the correction factor $\left(\alpha \right)$ for the kinetic energy as:

  $\begin{array}{c}\alpha =\frac{2}{{\overline{V}}^3{r}_w^2}{\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^{3}rdr}\\ =\frac{2}{{\left(0.808\text{m}/\text{s}\right)}^3{\left(37.5\times {10}^{-3}\text{ m}\right)}^2}\times \left(0.000441{\text{m}}^5/{\text{s}}^3\right)\\ =1.188\end{array}$

(c)

Interpretation Introduction

Interpretation:

The correction factor for the momentum of the given liquid flowing in the pipe is to be calculated.

Concept Introduction:

The formula to calculate the correction factor $\left(\beta \right)$ for the momentum of a stream which is flowing through a pipe of cross-sectional area S is:

  $\beta =\frac{1}{S}{\displaystyle \underset{S}{\int }{\left(\frac{u}{\overline{V}}\right)}^{2}dS}$ ...... (5)

Here, u is the local velocity of the stream in the differential cross-sectional area dS and $\overline{V}$ is the average velocity of the stream.

Answer to Problem 4.1P

The correction factor for the momentum of the given liquid flowing in the pipe is $0.546$ .

Explanation of Solution

Given information:

In a $75\text{ mm}$ pipe, a liquid is flowing at steady state. The variation of the local velocity of the liquid with the distance from the pipe axis is tabulated as:

  

The cross-sectional area of the pipe is defined as:

  $S=\pi r_w^2$ ...... (2)

Here, $r_w$ is the radius of the pipe. Differentiate equation (2) as shown below:

  $dS=2\pi rdr$ ...... (3)

Here, $r$ is the radial distance from the pipe axis.

Use equation (2) and (3) in equation (5) and simplify further as:

  $\begin{array}{c}\beta =\frac{1}{S}{\displaystyle \underset{S}{\int }{\left(\frac{u}{\overline{V}}\right)}^{2}dS}\\ =\frac{1}{\pi r_w^2}{\displaystyle \underset{0}{\overset{r_w}{\int }}{\left(\frac{u}{\overline{V}}\right)}^2\left(2\pi rdr\right)}\\ =\frac{2}{{\overline{V}}^2{r}_w^2}{\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^{2}rdr}\end{array}$

Use the value of $\overline{V}$ as calculated in part (a) as:

  $\overline{V}=0.808\text{m}/\text{s}$

The table for $u^{2}r$ is:

  

Now, plot the graph of $u^{2}r$ versus $r$ using the excel tool for graphing for the given data as:

  

The area under this curve is defined as ${\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^{2}rdr}$ . Thus,

  ${\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^{2}rdr}=0.000501{\text{m}}^4/{\text{s}}^3$

Now, use this value to calculate the correction factor $\left(\alpha \right)$ for the kinetic energy as:

  $\begin{array}{c}\beta =\frac{2}{{\overline{V}}^2{r}_w^2}{\displaystyle \underset{0}{\overset{r_w}{\int }}{u}^{2}rdr}\\ =\frac{2}{{\left(0.808\text{m}/\text{s}\right)}^2{\left(37.5\times {10}^{-3}\text{ m}\right)}^2}\times \left(0.000501{\text{m}}^4/{\text{s}}^3\right)\\ =0.546\end{array}$