Chapter 9, Problem 103P

Consider steady, incompressible, laminar flow of a Newtonian fluid in an infinitely long round pipe of diameter D or radium R = D/2 inclined at angle a. (Fig- 9-103). There is no applied pressure gradient $\left(\partial P/\partial x=0\right)$ . Instead, the fluid flows down the pipe due to gravity alone. We adopt the coordinate system shown, with x down the axis of the pipe. Derive an expression for the x-component of velocity u as a function of radius r and the other parameters of the problem. Calculate the volume flow rate and average axial velocity through the pipe.
Answer: $\rho \left(\sin a\right)\left(R^2-r^2\right)/4\mu ,\rho g\left(\sin a\right)\pi R^4/8\mu ,\rho g\left(\sin a\right)R^2/8\mu$

FIGURE P9-103

To determine

The expression for average velocity through the pipe.

The expression for volume flow rate through the pipe.

The expression for $x$ component of velocity.

Answer to Problem 103P

The expression for average velocity through the pipe is $\frac{-\pi \rho R^2\sin \alpha }{8\mu }$.

The expression for volume flow rate through the pipe is $\frac{-\pi \rho R^4\sin \alpha }{8\mu }$.

The expression for $x$ component of velocity is $u_x=\frac{\rho \sin \alpha }{4\mu }\left(r^2-R^2\right)$.

Explanation of Solution

Given information:

The diameter of the pipe is $D$, the radius of the

pipe is $R=D/2$ and the pipe is inclined at an angle $\alpha$. The applied pressure gradient is $\partial P/\partial x=0$.

The flow is assumed to be Newtonian flow. The velocity field is assumed to be axial symmetric no swirl $u_{\theta }=0$ and all the derivates with respect to $\theta$ are zero. The flow is assumed to be parallel $u_r=0$.

The no slip condition at the pipe wall implies that when $r=R$ and the velocity is $\overrightarrow{V}=0$.

The axis of the pipe is symmetry when $r=0$.

Write the expression for continuity equation for incompressible fluid.

  $\frac{1}{r}\left(\frac{\partial \left(r{u}_r\right)}{\partial r}\right)+\frac{1}{r}\left(\frac{\partial u_{\theta }}{\partial \theta }\right)+\left(\frac{\partial u_x}{\partial x}\right)=0$  ...... (I)

Here, the change in distance along $r$ direction is $\partial r$, the change along a distance in $\theta$ direction is $\partial \theta$, the change in distance along $x$ direction is $\partial x$, the radial velocity component along $r$ direction is $u_r$, the distance along $r$ direction is $r$, the velocity component along $\theta$ direction is $u_{\theta }$ and the velocity component along $x$ direction is $u_x$.

Write the expression for the velocity component along $x$ direction.

  $u_x=f\left(r\right)$   ....... (II)

Here, the function of the radius is $f\left(r\right)$.

Write the expression for angle $\alpha$.

  $\begin{array}{l}\alpha ={\sin }^{-1}\left(g_x\right)\\ g_x=\sin \alpha \end{array}$   ....... (III)

Write the expression for $x$ component of Navier-Stroke equation.

  $\left[\rho \left(\begin{array}{l}\frac{\partial u_x}{\partial x}+u_r\frac{\partial u_x}{\partial r}\\ +\frac{u_{\theta }}{r}\frac{\partial u_x}{\partial \theta }+u_x\frac{\partial u_x}{\partial x}\end{array}\right)\right]=\left[\begin{array}{l}-\frac{\partial P}{\partial x}+\rho g_x\\ +\mu \left[\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)+\frac{1}{r^2}\left(\frac{{\partial }^2{u}_x}{\partial {\theta }^2}\right)+\frac{{\partial }^2{u}_x}{\partial x^2}\right]\end{array}\right]$..... (IV)

Here, the density is $\rho$, the viscosity is $\mu$ and the change in pressure is $\partial P$.

Write the expression for volume flow rate through pipe.

  $\dot{V}={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{r=0}{\overset{r=R}{\int }}{u}_x\cdot dA}}$   ....... (VI)

Here, the volume flow rate is $\dot{V}$, the change in the area of pipe is $dA$, the upper limit of first integration is $r=R$, the lower limit of first integration is $r=0$, the upper limit of second integration is $\theta =2\pi$ and the lower limit of second integration is $\theta =0$.

Write the expression for change in the area of the pipe.

  $dA=rdrd\theta$   ....... (VII)

Here, the change in the distance along $r$ direction is $dr$ and the change along the distance along $\theta$ direction is $d\theta$.

Write the expression for average velocity through the pipe.

  $V=\frac{\dot{V}}{A}$   ...... (VIII)

Here, the velocity through the pipe is $V$ and the area of the pipe is $A$.

Write the expression for area of pipe.

  $A=\pi R^2$   ....... (IX)

Here, the radius of the pipe is $R$.

Calculation:

Substitute $0$ for $\frac{\partial u_x}{\partial x}$, $0$ for $u_r$, $0$ for $u_{\theta }$ and $0$ for $u_x$ for $\frac{\partial P}{\partial x}$ in Equation (IV).

  $\begin{array}{l}\left[\rho \left(\begin{array}{l}0+\left(0\right)\frac{\partial u_x}{\partial r}\\ +\frac{0}{r}\left(\frac{\partial u_x}{\partial \theta }\right)+u_x\left(0\right)\end{array}\right)\right]=\left[\begin{array}{l}-\left(0\right)+\rho g_x\\ +\mu \left[\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)+\frac{1}{r^2}\left(0\right)+0\right]\end{array}\right]\\ 0=\mu \left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)\right)+\rho g_x\\ \mu \left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)\right)=-\rho g_x\\ \frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)=\frac{-r\rho g_x}{2\mu }\end{array}$   ....... (X)

Integrate Equation (X) with respect to $r$.

  $\begin{array}{l}\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)=\frac{\partial \left(\frac{-r\rho g_x}{2\mu }\right)}{\partial r}\\ r{u}_x=\frac{-\rho g_x}{2\mu }\left(\frac{r^2}{2}\right)+C_1\\ u_x=\frac{-\rho g_x}{2\mu }\left(\frac{r}{2}\right)+\frac{C_1}{r}\end{array}$   ....... (XI)

Here, the constant is $C_1$.

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

  $\begin{array}{l}\frac{\partial u_x}{\partial r}=\frac{-\rho g_x}{2\mu }\left(\frac{0}{2}\right)+\frac{C_1}{0}\\ \frac{\partial u_x}{\partial r}=0+\infty \end{array}$

Double integration of Equation (X) with respect to $r$.

  $\begin{array}{l}\frac{\partial }{\partial r}\left(\frac{\partial u_x}{\partial r}\right)=\frac{\partial }{\partial r}\left(\frac{\partial \left(\frac{-r\rho g_x}{2\mu }\right)}{\partial r}\right)\\ u_x=\frac{-\rho g_x}{2\mu }\left(\frac{r^2}{2}\right)+C_2\\ u_x=\frac{-r^2\rho g_x}{4\mu }+C_2\end{array}$  ....... (XII)

Here, the constant is $C_2$.

Substitute $R$ for $r$ and $0$ for $u_x$ in Equation (XII).

  $\begin{array}{l}0=\frac{-R^2\rho g_x}{4\mu }+C_2\\ C_2=\frac{-R^2\rho g_x}{4\mu }\end{array}$

Substitute $\frac{-R^2\rho g_x}{4\mu }$ for $C_2$ in Equation (XII).

  $\begin{array}{c}{u}_x=\left(\frac{-r^2\rho g_x}{4\mu }\right)+\left(\frac{-R^2\rho g_x}{4\mu }\right)\\ =\frac{-\rho g_x}{4\mu }\left(r^2+R^2\right)\\ =\frac{\rho g_x}{4\mu }\left(r^2-R^2\right)\end{array}$   ....... (XIII)

Substitute $\sin \alpha$ for $g_x$ in Equation (XIII).

  $u_x=\frac{\rho \sin \alpha }{4\mu }\left(r^2-R^2\right)$

Substitute $\frac{\rho g_x}{4\mu }\left(r^2-R^2\right)$ for $u_x$ and $rdrd\theta$ for $dA$ in Equation (V).

  $\begin{array}{c}\dot{V}={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{r=0}{\overset{r=R}{\int }}\frac{\rho g_x}{4\mu }\left(r^2-R^2\right)\cdot rdrd\theta }}\\ ={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}\frac{\rho g_x}{4\mu }\left(\left(\frac{R^{2+2}}{2+2}\right)-R^2\frac{R^{1+1}}{1+1}\right)d\theta }\\ ={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}\frac{\rho g_x}{4\mu }\left(\frac{R^4}{4}-\frac{R^4}{2}\right)d\theta }\\ ={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}\frac{-\rho g_x}{4\mu }\left(\frac{R^4}{4}\right)d\theta }\end{array}$

  $\begin{array}{c}\dot{V}={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}\frac{-\rho g_x}{4\mu }\left(\frac{R^4}{4}\right)d\theta }\\ =\frac{-\rho g_x{R}^4}{16\mu }{\left(\theta \right)}_{\theta =0}^{\theta =2\pi }\\ =\frac{-\rho g_x{R}^4}{16\mu }\left(2\pi \right)\\ =\frac{-\pi \rho g_x{R}^4}{8\mu }\end{array}$  ...... (XIV)

Substitute $\sin \alpha$ for $g_x$ in Equation (XIV).

  $\dot{V}=\frac{-\pi \rho R^4\sin \alpha }{8\mu }$

Substitute $\pi R^2$ for $A$ and $\frac{-\pi \rho g_x{R}^4}{8\mu }$ for $\dot{V}$ in Equation (VII).

  $\begin{array}{c}V=\frac{\frac{-\pi \rho g_x{R}^4}{8\mu }}{\pi R^2}\\ =\frac{-\pi \rho g_x{R}^4}{8\pi \mu R^2}\\ =\frac{-\pi \rho g_x{R}^2}{8\pi \mu }\end{array}$   ....... (XV)

Substitute $\sin \alpha$ for $g_x$ in Equation (XV).

  $V=\frac{-\pi \rho R^2\sin \alpha }{8\mu }$

Conclusion:

The expression for average velocity through the pipe is $\frac{-\pi \rho R^2\sin \alpha }{8\mu }$.

The expression for volume flow rate through the pipe is $\frac{-\pi \rho R^4\sin \alpha }{8\mu }$.

The expression for $x$ component of velocity is $\frac{\rho \sin \alpha }{4\mu }\left(r^2-R^2\right)$.