The problem. The problem. Poiseuille flow of a power law fluid.
A power law fluid flows through a circular pipe with radius $R$ and length $L$. The pipe is inclined by and angle $\chi$, as shown below. The pressures at $z=0,L$ are $p=P_0,P_L$, respectively. Calculate the velocity profile in steady state and the corresponding volumetric flow rate. Exit and entrances losses, and heating due to friction are very small, so that these can be omitted.
 |
| Fig. 1 Problem sketch for the flow of a newtonian fluid through an inclined pipe. |
The solution
According to the sketch in Fig. 1, the geometry of the problem is cylindrical. So, it seems obvious to use cylindrical coordinates for the solution of the problem.
The solution to velocity distribution and flow rate problem
Next, according to the physics of the problem the following conclusions are made,
- the fluid moves only along the axial direction from $z=0$ to $z=L$. This is, the fluid moves along the cylinder axis;
- the fluid does not move radial or angularly. From a transversal view, the fluid does not move left, right or in another radial direction. Also, the fluid does not circle motions;
- considering that the function representing the fluid velocity is $\textbf{v}=(v_r,v\theta, v_z)$ and in agreement with the above statements $v_r=0$, $v_\theta=0$ and $v_z\ne 0$ follow,
- the component $v_z$ does not vanish and may depend on $(r,\theta,z,t)$. In this case, it only depends on $r$. One arrives to this conclusion as follows. The continuity equation assures that the fluid velocity at $z=0$ and $z=L$ is the same, so that it does not depends on $z$. Besides, it is easy to realize that the fluid velocity remains unchanged under angular variations,
- stationary conditions are considered. This means that there is no acceleration and that the derivatives of $\textbf{v}$ are zero or that there is no temporal dependency,
- the pressure difference between $z=0$ and $z=L$ sets the fluid in motion,
- the pressure $p$ is in general a function of $(r, \theta, z,t)$; but in this case it only depends on $z$. One arrives to this conclusion as follows. Since the fluid is not moving radial nor angularly, then there is no dependency of the pressure on $(r,\theta)$,
- the pressure gradient is generated because of the inclination of the pipe. This is, the fluid moves because of the acceleration due to gravity so that its effect should be included in the problem,
- the problem has angular symmetry. This is, none of the variables of the problem as the pressure, the velocity or the stress depend on $\theta$. In other words, it does not matter the assigned value to $\theta$, the system will not change. $\theta=0$ will be used;
- the effect of temperature on the fluid properties is very small and will be discarded. Then, $\rho$ and $\mu$ are constants.
In order to determine the velocity distribution two equations should be considered: one defining the type of fluid and another one that models the fluid motion under any circumstance. These equations are,
- the fluid constitutive equation.- The viscosity of this working fluid is not constant for a given temperature, pressure, and composition. In fact, its viscosity changes as the fluid moves. Then, its constitutive equation is:
$\mathbf{\tau}=-\eta \mathbf{\dot{\gamma}} = -\eta \left[\nabla \mathbf{v} + \left(\nabla \mathbf{v}\right)^T\right]$ Eq. (01)
where $\eta$ is the non-newtonian viscosity and $\mathbf{\dot{\gamma}}$ is the rate of strain tensor. Also, experimental data of $\eta$ versus $\mathbf{\dot{\gamma}}$ are described by a simple power law so that:
From Eq. (02) follows that $m$ and $n$ are free parameters to be determined for a given fluid. The parameter $m$ is called \textit{consistency index} while $n$ is called \textit{flow index}. Finally, the constitutive equation for a power law fluid is:
$\mathbf{\tau}=-m \left[\nabla \mathbf{v} + \left(\nabla \mathbf{v}\right)^T\right]^n$ Eq. (03)
which can be written in extended form in cylindrical coordinates as,
\tau_{rr}\cos^2 \theta-2\tau_{r\theta}\sin \theta \cos \theta+\tau_{\theta \theta}\sin^2 \theta & \tau_{rr}\sin \theta \cos \theta + \tau_{r\theta}\left[\cos^2 \theta-\sin^2 \theta\right]-\tau_{\theta \theta}\sin \theta\cos \theta & \tau_{rz}\cos \theta -\tau_{z\theta}\sin \theta\\
\tau_{rr}\sin \theta \cos \theta+\tau_{r\theta}\left[\cos^2 \theta-\sin^2 \theta\right]-\tau_{\theta \theta}\sin \theta\cos \theta & \tau_{rr}\sin^2 \theta+2\tau_{r\theta}\sin \theta \cos \theta+\tau_{\theta \theta}\cos^2 \theta & \tau_{rz}\sin \theta +\tau_{z\theta}\cos \theta\\
\tau_{rz}\cos \theta -\tau_{z\theta}\sin \theta & \tau_{rz}\sin \theta +\tau_{z\theta}\cos \theta & \tau_{zz}
\end{bmatrix}\nonumber\\
=-m
\begin{bmatrix}
2\dfrac{\partial v_r}{\partial r} & r\dfrac{\partial}{\partial r}\left(\dfrac{v_\theta}{r}\right)+\dfrac{1}{r}\dfrac{\partial v_r}{\partial \theta} & \dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r} \\
r\dfrac{\partial}{\partial r}\left(\dfrac{v_\theta}{r}\right)+\dfrac{1}{r}\dfrac{\partial v_r}{\partial \theta} & 2\left(\dfrac{1}{r}\dfrac{\partial v_\theta}{\partial \theta}+\dfrac{v_r}{r}\right) & \dfrac{1}{r}\dfrac{\partial v_z}{\partial \theta}+\dfrac{\partial v_\theta}{\partial z} \\
\dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r} & \dfrac{1}{r}\dfrac{\partial v_z}{\partial \theta}+\dfrac{\partial v_\theta}{\partial z} & 2\dfrac{\partial v_z}{\partial z}
\end{bmatrix}^n$ Eq. (04)
- the momentum balance equation.- This is a vectorial partial differential equation with components $(r,\theta,z)$ as follows,
$\rho\left(\frac{\partial v_\theta}{\partial t}+v_r\frac{\partial v_\theta}{\partial r}+\frac{v_\theta}{r}\frac{\partial v_\theta}{\partial \theta}+\frac{v_\theta v_r}{r}+v_z\frac{\partial v_\theta}{\partial z}\right)=-\left[\frac{1}{r^2}\frac{\partial \left(r^2\tau_{r\theta}\right)}{\partial r}+\frac{1}{r}\frac{\partial \tau_{\theta \theta}}{\partial \theta}+\frac{\partial \tau_{z\theta}}{\partial z}+\frac{\tau_{\theta r}-\tau_{r\theta}}{r}\right]-\frac{1}{r}\frac{\partial p}{\partial \theta}+\rho g_\theta$ Eq. (06)
$\rho\left(\frac{\partial v_z}{\partial t}+v_r\frac{\partial v_z}{\partial r}+\frac{v_\theta}{r}\frac{\partial v_z}{\partial \theta}+v_z\frac{\partial v_z}{\partial z}\right)=-\left[\frac{1}{r}\frac{\partial \left(r\tau_{rz}\right)}{\partial r}+\frac{1}{r}\frac{\partial \tau_{\theta z}}{\partial \theta}+\frac{\partial \tau_{zz}}{\partial z}\right]-\frac{\partial p}{\partial z}+\rho g_z$ Eq. (07)
The constitutive and the momentum balance equations form a system of differential equations for the stress $\mathbf{\tau}$ and the fluid velocity $\mathbf{v}$. These equations can be simplified by using the above listed observations to the physics of the problem. First, Eq. (04) is simplified to get,
$\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta \theta} & \tau_{z\theta}\\
\tau_{rz} & \tau_{z\theta} & \tau_{zz}
\end{bmatrix}
=-m
\begin{bmatrix}
0 & 0 & \dfrac{\partial v_z}{\partial r} \\
0 & 0 & 0 \\
\dfrac{\partial v_z}{\partial r} & 0 & 0
\end{bmatrix}^n$ Eq. (08)
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta \theta} & \tau_{z\theta}\\
\tau_{rz} & \tau_{z\theta} & \tau_{zz}
\end{bmatrix}
=-m
\begin{bmatrix}
0 & 0 & \dfrac{\partial v_z}{\partial r} \\
0 & 0 & 0 \\
\dfrac{\partial v_z}{\partial r} & 0 & 0
\end{bmatrix}^n$ Eq. (08)
From Eq. (08) it can be seen that all elements in the stress matrix are zero but $\tau_{rz}$. This is comes out after element-to-element comparison between the left hand side and the right hand side matrices. The resulting equation for the shear stress is,
In Eq. (09) $\tau_{rz}$ is in fact negative but because $\dot{\gamma}$ must be positive the sign is placed inside thee parenthesis so that the physical meaning do not change. Next, since $v_z$ depends only on $r$ partial derivatives notation is not needed anymore. The simplification process to obtain Eq. (08) from Eq. (04) is repeated to simplify the components of the momentum balance Eqs. (08-07). The terms in the radial and angular components, Eqs. (05-06), are zero. From the axial component Eq. (07) only the following remains,
In Eq. (10) $g_z$ represents the $z$ component projection of the acceleration due to gravity vector $\mathbf{g}$ parallel to the inclined pipe (see Fig. 2). Then, since $\mathbf{g}$ is a vector pointing downwards $g_z=-g\cos \chi$ is substituted into Eq. (10).
 |
| Fig. 2 Angle sketch. |
At this stage, two different path can be followed to find an ordinary differential equation for $v_z\left(r\right)$. The first one is to combine Eqs. (09-10) to give,
which would not be easily solved, analytically, due to the exponent $n$.
A second path to find $v_z\left(r\right)$ must be followed by solving the ordinary differential equation for $\tau_{rz}$ in Eq. (10). This gives,
which shows an important feature: there is a divergence as $r\rightarrow 0$. In other words, the stress goes to infinity at the center of the pipe which has no physical sense. Then, $C_1=0$.
Next, the resulting $\tau_{rz}$ can be combined with Eq. (09) to have a single equation for $v_z\left(r\right)$ as follows,
$\frac{d v_z\left(r\right)}{d r} = (-1)^{1/n+1} \left[\frac{d p\left(z\right)}{d z}+\rho g \cos \chi\right]^{1/n}\frac{r^{1/n}}{\left(2m\right)^{1/n}}$
$v_z\left(r\right) = (-1)^{1/n+1} \left[\frac{d p\left(z\right)}{d z}+\rho g \cos \chi\right]^{1/n}\frac{r^{1/n+1}}{\left(2m\right)^{1/n}\left( 1/n+1 \right)} + C_2$ Eq. (13)
The constant of integration $C_2$ in $v_z$ Eq. (13) should be determined using a proper boundary condition. For this case it seems clear that the available boundary conditions are,
$v_z=0,\quad\text{at}\quad r= R$ Eq. (15)
However, boundary condition Eq. (14) is redundant since a similar condition was applied to $\tau_{rz}$ at the pipe center. This is good because $C_2$ would have to different values, and that would be a trouble. Evaluation of boundary condition Eq. (15) gives,
$C_2 = -(-1)^{1/n+1}\left[\frac{d p\left(z\right)}{d z}+\rho g \cos \chi\right]^{1/n}\frac{R^{1/n+1}}{\left(2m\right)^{1/n}\left( 1/n+1 \right)}$ Eq. (16)
Next, substitution of $C_2$ from Eq. (16) into $v_z\left(r\right)$ given in Eq. (13) outputs,
$v_z\left(r\right) = (-1)^{1/n+1}\left[\frac{d p\left(z\right)}{d z}+\rho g \cos \chi\right]^{1/n}\frac{r^{1/n+1} - R^{1/n+1}}{\left(2m\right)^{1/n}\left( 1/n+1 \right)}$ Eq. (17)
Next, Eq. (17) can be rearranged as follows:
$v_z\left(r\right) = \left[-\frac{d p\left(z\right)}{d z}-\rho g \cos \chi\right]^{1/n}\frac{ R^{1/n+1}}{\left(2m\right)^{1/n}\left( 1/n+1 \right)} \left[ 1-\left( \frac{r}{R}\right)^{1/n+1} \right]$ Eq. (18)
Equation (18) can already be considered as the velocity distribution across the power law fluid. However, the pressure gradient is to be determined to have a more explicit expression. This is done as follows.
First, two boundary conditions for pressure are to be considered,
$p=P_0$ at $z=0$
$p=P_L$ at $z=L$ Eqs. (19)
since the pipe inclination is related to the height $L$. Next, you have to accept that Eq. (18) is in fact a differential equation for $p$ which can actually be solved easily by separation of variables.
$\dfrac{dp}{dz}= -\left[ \dfrac{v_z\left(r\right) }{\frac{ R^{1/n+1}}{\left(2m\right)^{1/n}\left( 1/n+1 \right)} \left[ 1-\left( \frac{r}{R}\right)^{1/n+1} \right]} \right]^n - \rho g \cos \chi$
$p= -\left[ \dfrac{v_z\left(r\right) }{\frac{ R^{1/n+1}}{\left(2m\right)^{1/n}\left( 1/n+1 \right)} \left[ 1-\left( \frac{r}{R}\right)^{1/n+1} \right]} \right]^n z - \rho g \cos \chi z + C_3$ Eq. (20)
Evaluation of the boundry conditions Eqs. (19) on Eq. (20)should produce the difference $p_0-p_L$,,
$P_0-P_L= \left[ \dfrac{v_z\left(r\right) }{\frac{ R^{1/n+1}}{\left(2m\right)^{1/n}\left( 1/n+1 \right)} \left[ 1-\left( \frac{r}{R}\right)^{1/n+1} \right]} \right]^n L + \rho g \cos \chi L$ Eq. (21)
The real velocity distribution can be isolated from Eq. (21) to be,
The flow rate of a newtonian fluid through an inclined pipe
Finally, the flow rate can be easily found by integration of Eq. (22). The process is very similar to the one followed for thee newtonian fluid case,
$Q = \int_0^{2\pi} \int_0^R v_zr\;dr\;d\theta$
$Q = 2\pi R^2 \int_0^1 v_z \left( \dfrac{r}{R} \right) \;d\left( \dfrac{r}{R} \right)$
$Q = \left[\dfrac{P_0-P_L}{L}-\rho g \cos \chi\right]^{1/n}\dfrac{ 2\pi R^{1/n+3}}{\left(2m\right)^{1/n}\left( 1/n+1 \right)} \left[ \left( \dfrac{r}{R} \right)^2 - \dfrac{1}{1/n+3} \left(\dfrac{r}{R} \right)^{1/n+3} \right]_0^1$
$Q = \left[\dfrac{P_0-P_L}{L}-\rho g \cos \chi\right]^{1/n}\dfrac{ 2\pi R^{1/n+3}}{\left(2m\right)^{1/n}\left( 1/n+3 \right)} $ Eq. (23)
Equation (23) is the flow rate for a power law fluid travelling through an inclined pipe by an angle $\chi$. This is the end of thee problem.
Any question? Write in the comments and I shall try to help.
Other stuff of interest
- The Poiseuille flow of a newtonian fluid through an inclined pipe
- The flow of a plastic Bingham fluid through an inclined pipe
- The upward flow of a newtonian fluid through the annular region of two vertical pipes
- Understanding the mechanics of the Bernoulli equation
- Practical situation. A polymeric fluid flowing through a pipe
- Practical situation. A polyisoprene solution flowing through a pipe
=========
Ildebrando.
No comments:
Post a Comment