Chapter 12, Problem 59P

To determine

The kinetic energy correction factor for this flow.

Explanation of Solution

Given:

The velocity profile for turbulent flow in a circular pipe $u\left(r\right)$ is $u_{\max }{\left(1-\frac{r}{R}\right)}^{1/n}$

The value of $\left(n\right)$  is 9.

Calculation:

Draw velocity profile for turbulent flow in a circular pipe as in Figure 1.

Show the area of circular pipe $\left(A\right)$ using the formula;

  $A=\pi r^2$

Differentiate the above equation with respect to r.

  $\begin{array}{c}\frac{dA}{dr}=2\pi r\\ dA=2\pi rdr\end{array}$

Determine the correction factor $\left(\alpha \right)$  using the relation;

  $\begin{array}{c}\alpha =\frac{1}{A}{\displaystyle \underset{A}{\int }{\left(\frac{u\left(r\right)}{V_{avg}}\right)}^3}dA\\ =\frac{1}{A{V}_{avg}{}^3}{\displaystyle \underset{A}{\int }\left(u{\left(r\right)}^3\right)}dA\end{array}$

  $\begin{array}{c}\alpha =\frac{1}{A}{\displaystyle \underset{A}{\int }{\left(\frac{u\left(r\right)}{V_{avg}}\right)}^3}dA\\ =\frac{1}{A{V}_{avg}{}^3}{\displaystyle \underset{A}{\int }\left(u{\left(r\right)}^3\right)}dA\end{array}$

Substitute $\pi R^2$ for A and $u_{\max }{\left(1-\frac{r}{R}\right)}^{1/n}$ for $u\left(r\right)$.

  $\begin{array}{c}\alpha =\frac{1}{\pi R^2{V}_{avg}{}^3}{\displaystyle \underset{A}{\int }{\left(u_{\max }{\left(1-\frac{r}{R}\right)}^{1/n}\right)}^3}dA\\ =\frac{u_{\max }{}^3}{\pi R^2{V}_{avg}{}^3}{\displaystyle \underset{r=0}{\overset{R}{\int }}\left({\left(1-\frac{r}{R}\right)}^{3/n}\right)}dA\end{array}$

Substitute $2\pi rdr$ for $dA$.

  $\alpha =\frac{u_{\max }{}^3}{\pi R^2{V}_{avg}{}^3}{\displaystyle \underset{r=0}{\overset{R}{\int }}\left({\left(1-\frac{r}{R}\right)}^{3/n}\right)}2\pi rdr$

  $\alpha =\frac{2u_{\max }{}^3}{R^2{V}_{avg}{}^3}{\displaystyle \underset{r=0}{\overset{R}{\int }}\left({\left(1-\frac{r}{R}\right)}^{3/n}\right)}rdr$        (1)

Calculate the average velocity $\left(V_{avg}\right)$ using the relation;

  $\begin{array}{c}{V}_{avg}=\frac{1}{A}{\displaystyle \underset{A}{\int }u\left(r\right)dA}\\ =\frac{1}{\pi R^2}{\displaystyle \underset{r=0}{\overset{R}{\int }}\left(u_{\max }{\left(1-\frac{r}{R}\right)}^{1/n}\right)2\pi rdr}\end{array}$

  $V_{avg}=\frac{2u_{\max }}{R^2}{\displaystyle \underset{r=0}{\overset{R}{\int }}{\left(1-\frac{r}{R}\right)}^{1/n}rdr}$        (2)

Using the integral table formula.

  ${{\displaystyle \int \left(a+bx\right)}}^{n}xdx=\frac{{\left(ax+bx\right)}^{n+2}}{b^2\left(n+2\right)}-\frac{a{\left(a+bx\right)}^{n+1}}{b^2\left(n+1\right)}$

Calculate the value of ${\displaystyle \underset{r=0}{\overset{R}{\int }}u\left(r\right)rdr}$ using integral table formula;

  $\begin{array}{c}{\displaystyle \underset{r=0}{\overset{R}{\int }}u\left(r\right)rdr}={\displaystyle \underset{r=0}{\overset{R}{\int }}{\left(1-\frac{r}{R}\right)}^{\frac{1}{n}}rdr}\\ ={\left[\frac{{\left(1-\frac{r}{R}\right)}^{\frac{1}{n}+2}}{\frac{1}{R^2}\left(\frac{1}{n}+2\right)}-\frac{{\left(1-\frac{r}{R}\right)}^{\frac{1}{n}+1}}{\frac{1}{R^2}\left(\frac{1}{n}+1\right)}\right]}_{r=0}^R\\ =\frac{n^2{R}^2}{\left(n+1\right)\left(2n+1\right)}\end{array}$

Substitute $\frac{n^2{R}^2}{\left(n+1\right)\left(2n+1\right)}$ for ${\displaystyle \underset{r=0}{\overset{R}{\int }}{\left(1-\frac{r}{R}\right)}^{1/n}rdr}$ in the equation (2).

  $\begin{array}{c}{V}_{avg}=\frac{2u_{\max }}{R^2}\frac{n^2{R}^2}{\left(n+1\right)\left(2n+1\right)}\\ =\frac{2u_{\max }{n}^2}{\left(n+1\right)\left(2n+1\right)}\end{array}$

Substitute 9 for n.

  $\begin{array}{c}{V}_{avg}=\frac{2u_{\max }{\left(9\right)}^2}{\left(9+1\right)\left(2\left(9\right)+1\right)}\\ =0.8167u_{\max }\end{array}$

Calculate the value of ${\displaystyle \underset{r=0}{\overset{R}{\int }}u{\left(r\right)}^{3}rdr}$ using integral table formula;

  $\begin{array}{c}{\displaystyle \underset{r=0}{\overset{R}{\int }}u{\left(r\right)}^{3}rdr}={\displaystyle \underset{r=0}{\overset{R}{\int }}{\left(1-\frac{r}{R}\right)}^{\frac{3}{n}}rdr}\\ ={\left[\frac{{\left(1-\frac{r}{R}\right)}^{\frac{3}{n}+2}}{\frac{1}{R^2}\left(\frac{3}{n}+2\right)}-\frac{{\left(1-\frac{r}{R}\right)}^{\frac{3}{n}+1}}{\frac{1}{R^2}\left(\frac{3}{n}+1\right)}\right]}_{r=0}^R\\ =\frac{n^2{R}^2}{\left(n+3\right)\left(2n+3\right)}\end{array}$

Substitute $\frac{n^2{R}^2}{\left(n+3\right)\left(2n+3\right)}$ for ${\displaystyle \underset{r=0}{\overset{R}{\int }}\left({\left(1-\frac{r}{R}\right)}^{3/n}\right)}rdr$ in the equation (1).

  $\begin{array}{c}\alpha =\frac{2u_{\max }{}^3}{R^2{V}_{avg}{}^3}\frac{n^2{R}^2}{\left(n+3\right)\left(2n+3\right)}\\ =\frac{2u_{\max }{}^3}{R^2}{\left(\frac{2n^2{R}^2}{\left(n+1\right)\left(2n+1\right)}\right)}^{-3}\frac{n^2{R}^2}{\left(n+3\right)\left(2n+3\right)}\\ =\frac{{\left(n+1\right)}^3{\left(2n+1\right)}^3}{4n^4\left(n+3\right)\left(2n+3\right)}\\ =\frac{{\left(9+1\right)}^3{\left(2\left(9\right)+1\right)}^3}{4{\left(9\right)}^4\left(9+3\right)\left(2\left(9\right)+3\right)}\\ =1.04\end{array}$

Thus, the kinetic energy correction factor for this flow is $\underset{\_}{\text{1}\text{.04}}$.