Chapter 6, Problem 23P

Water enters a 7-cm-diameter pipe steadily with a uniform velocity of 2 mi’s and exits with the turbulent flow velocity distribution given by $u=u_{\max }{\left(1-r/R\right)}^{1/7}$ . If the pressure drop along the pipe is 10 kPa, determine the drag force exerted on the pipe by water flow.

To determine

The drag force exerted on the pipe by water flow.

Answer to Problem 23P

The drag force exerted on the pipe by water flow is $38.19\text{N}$.

Explanation of Solution

Given information:

The diameter of the pipe is $7\text{cm}$, the velocity of the water is $2\text{m}/\text{s}$ and the pressure drop along the pipe is $10\text{kPa}$.

Expression of flow area of the pipe,
$A=\frac{\pi d^2}{4}$...... (I)
Here, the diameter of the pipe is $d$.

Expression of turbulent flow velocity distribution,
$u=U_{\text{max}}{\left(1-\frac{r}{R}\right)}^{1/7}$...... (II)
Here, the maximum velocity in the pipe is $U_{\text{max}}$, the distance from circumference to the centre of the pipe is $r$ and the radius of the pipe is $R$.

Expression of $x$ component of momentum equation,
$-\rho U_{\text{avg}}^{2}A+\rho {\displaystyle \underset{0}{\overset{R}{\int }}{u}^2}dA=\left(p_1-p_2\right)A-F_D$...... (III)
Here,the mean velocity is $U_{\text{avg}}$, the density of the water is $\rho$, the drag force is $F_D$, the pressure at inlet is $p_1$, the pressure at outlet is $p_2$ and the elemental area of the pipe is $dA$.

Substitute $U_{\text{max}}{\left(1-\frac{r}{R}\right)}^{1/7}$ for $u$ in Equation (III).

$-\rho U_{\text{avg}}^{2}A+\rho {\displaystyle \underset{0}{\overset{R}{\int }}{U}_{\text{max}}^2{\left(1-\frac{r}{R}\right)}^{2/7}}dA=\left(p_1-p_2\right)A-F_D$....... (IV)

Expression of elemental area of the pipe,
$dA=2\pi rdr$

Expression of maximum velocity for turbulent flow,
$U_{\max }=1.2255U_{\text{avg}}$

Substitute $1.2255U_{\text{avg}}$ for $U_{\max }$, $2\pi rdr$ for $dA$ in Equation (IV).

$\begin{array}{l}-\rho U_{\text{avg}}^{2}A+\rho {\displaystyle \underset{0}{\overset{R}{\int }}{\left(1.2255U_{\text{avg}}\right)}^2{\left(1-\frac{r}{R}\right)}^{2/7}}2\pi rdr=\left(p_1-p_2\right)A-F_D\\ -\rho U_{\text{avg}}^{2}A+3\rho U_{\text{avg}}^2\pi {\displaystyle \underset{0}{\overset{R}{\int }}{\left(1-\frac{r}{R}\right)}^{2/7}rdr}=\left(p_1-p_2\right)A-F_D\end{array}$....... (V)
Consider $B$ for ${\displaystyle \underset{0}{\overset{R}{\int }}{\left(1-\frac{r}{R}\right)}^{2/7}rdr}$

$B={\displaystyle \underset{0}{\overset{R}{\int }}{\left(1-\frac{r}{R}\right)}^{2/7}rdr}$...... (VI)
Substitute $B$ for ${\displaystyle \underset{0}{\overset{R}{\int }}{\left(1-\frac{r}{R}\right)}^{2/7}rdr}$ in Equation (V).

$-\rho U_{\text{avg}}^{2}A+3\rho U_{\text{avg}}^2\pi Brdr=\left(p_1-p_2\right)A-F_D$...... (VII)
Consider $x$ for $\left(1-\frac{r}{R}\right)$

$\begin{array}{l}\left(1-\frac{r}{R}\right)=x\\ \frac{r}{R}=1-x\\ r=R\left(1-x\right)\end{array}$...... (VI)
Differentiate Equation (VI) with respect to $x$ on both sides.

$\begin{array}{l}\frac{d}{dx}\left(r\right)=\frac{d}{dx}\left\{R\left(1-x\right)\right\}\\ \frac{dr}{dx}=R\left\{\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)\right\}\\ dr=R\left(-1\right)dx\\ dr=-Rdx\end{array}$

Substitute $0$ for $r$ in Equation (VI).

$\begin{array}{l}0=R\left(1-x\right)\\ 0=\left(1-x\right)\\ x=1\end{array}$

Substitute $R$ for $r$ in Equation (VI).

$\begin{array}{l}R=R\left(1-x\right)\\ 1=\left(1-x\right)\\ x=0\end{array}$

Substitute $x$ for $\left(1-\frac{r}{R}\right)$, $R\left(1-x\right)$ for $r$ lower limit as $1$ and upper limit as $0$ and $-Rdx$ for $dr$ in Equation (VI).

$\begin{array}{c}B={\displaystyle \underset{1}{\overset{0}{\int }}{x}^{2/7}R\left(1-x\right)\left(-Rdx\right)}\\ =-R^2{\displaystyle \underset{1}{\overset{0}{\int }}\left(x^{2/7}-x^{9/7}\right)dx}\\ =-R^2{\left[\frac{x^{\left(2/7+1\right)}}{9/7}-\frac{x^{\left(9/7+1\right)}}{16/7}\right]}_1^0\\ =-R^2\left[\frac{0^{\left(2/7+1\right)}}{9/7}-\frac{0^{\left(9/7+1\right)}}{16/7}-\frac{1}{9/7}+\frac{1}{16/7}\right]\end{array}$

$B=\frac{49}{144}{R}^2$

Substitute $-\frac{49}{144}{R}^2$ for $B$ in Equation (VII).

$\begin{array}{l}-\rho U_{\text{avg}}^{2}A+3\rho U_{\text{avg}}^2\pi \left[\frac{49}{144}{R}^2\right]=\left(p_1-p_2\right)A-F_D\\ -\rho U_{\text{avg}}^{2}A+\frac{147}{144}\pi \rho U_{\text{avg}}^2{R}^2=\left(p_1-p_2\right)A-F_D\\ F_D=\left(p_1-p_2\right)A+\rho U_{\text{avg}}^{2}A-3.207\rho U_{\text{avg}}^2{R}^2\end{array}$...... (VIII)

Expression of pressure drop,
$\Delta p=p_1-p_2$

Substitute $\Delta p$ for $p_1-p_2$ in Equation (VIII).

$F_D=\Delta pA+\rho U_{\text{avg}}^{2}A-3.207\rho U_{\text{avg}}^2{R}^2$...... (IX)

Expression of radius of pipe
$R=d/2$...... (X)

Calculation:

Substitute $7\text{cm}$ for $d$ in Equation (I).

$\begin{array}{c}A=\frac{\pi {\left(7\text{cm}\right)}^2}{4}\\ =\frac{\pi }{4}{\left\{\left(7\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right\}}^2\\ =0.785\times \left(0.0049{\text{m}}^{\text{2}}\right)\\ =0.00385{\text{m}}^{\text{2}}\end{array}$

Substitute $7\text{cm}$ for $d$ in Equation (IX).

$\begin{array}{c}R=\frac{7\text{cm}}{2}\\ =\left(\frac{7\text{cm}}{2}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.035\text{m}\end{array}$

Substitute $10\text{kPa}$ for $\Delta p$, $0.00385{\text{m}}^{\text{2}}$ for $A$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2\text{m}/\text{s}$ for $U_{\text{avg}}$ and $0.035\text{m}$ for $R$ in Equation (IX).

$\begin{array}{c}{F}_D=\left[\begin{array}{l}\left(10\text{kPa}\right)\left(0.00385{\text{m}}^{\text{2}}\right)+\left(1000\text{kg}/{\text{m}}^{\text{3}}\right){\left(2\text{m}/\text{s}\right)}^2\left(0.00385{\text{m}}^{\text{2}}\right)\\ -3.207\left(1000\text{kg}/{\text{m}}^{\text{3}}\right){\left(2\text{m}/\text{s}\right)}^2{\left(0.035\text{m}\right)}^2\end{array}\right]\\ =\left[\begin{array}{l}\left(10\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{1\text{Pa}}\right)\left(0.00385{\text{m}}^{\text{2}}\right)\\ +\left(1000\text{kg}/{\text{m}}^{\text{3}}\right){\left(2\text{m}/\text{s}\right)}^2\left(0.00385{\text{m}}^{\text{2}}\right)\\ -3.207\left(1000\text{kg}/{\text{m}}^{\text{3}}\right){\left(2\text{m}/\text{s}\right)}^2{\left(0.035\text{m}\right)}^2\end{array}\right]\\ =\left(38.5\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)+\left(15.4\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)-\left(15.71\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(38.19\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\end{array}$

$F_D=38.19\text{N}$

Conclusion:

The drag force exerted on the pipe by water flow is $38.19\text{N}$.