Laminar flow in an inclined circular pipe. Poiseuille parallel flow

 The problem. Poiseuille flow of a newtonian fluid.

A newtonian 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 velocitty 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 of an inclined pipe

The solution to velocity distribution and flow rate problem

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. 

Assumptions and comments on the physics of the flow of a newtonian fluid through an inclined pipe at low speed

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.

Constitutive equation and momentum balance equation for a newtonian fluid travelling through an incined pipe

In order to determine the velocity distribution it is necessary to consider two equations. One defining the kid of fluid you are dealing with and another one modeling the fluid motions under any condition. These are,

• the fluid constitutive equation.- Since the fluid is newtonian its constitutive equation is,

$\mathbf{\tau}=-\mu \left[\nabla \mathbf{v}+\left(\nabla \mathbf{v}\right)^2\right]$        Eq. (1)

        which written in extended manner in cylindrical coordinates becomes,

$\begin{bmatrix}
\tau_{rr}\cos^2 \theta-2\tau_{r\theta}\sin \theta \cos \theta+\tau_{\theta \theta}\sin^2 \theta & \tau_{rr}\sin \theta \cos \theta +\tau_{r\theta}\left[\cos^2 \theta -\sin^2 \theta\right]-\tau_{\theta \theta}\sin \theta \cos \theta & \tau_{rz}\cos \theta -\tau_{z\theta}\sin \theta\\
\tau_{rr}\sin \theta \cos \theta+\tau_{r\theta}\left[\cos^2 \theta -\sin^2 \theta\right]-\tau_{\theta \theta}\sin \theta \cos \theta & \tau_{rr}\sin^2 \theta+2\tau_{r\theta}\sin \theta \cos \theta+\tau_{\theta \theta} \cos^2 \theta & \tau_{rz}\sin \theta+\tau_{z\theta}\cos \theta\\
\tau_{rz}\cos \theta-\tau_{z\theta}\sin \theta & \tau_{rz}\sin \theta+\tau_{z\theta}\cos \theta & \tau_{zz}
\end{bmatrix}$

$=-\mu\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}$        Eq. (2)

• the momentum balance equation.- This is a vectorial equation whose components in $(r, \theta, z)$ are as follows,

$ \rho\left(\frac{\partial v_r}{\partial t} + v_r\frac{\partial v_r}{\partial r}+\frac{v_{\theta}}{r}\frac{\partial v_r}{\partial \theta}-\frac{v_{\theta}^2}{r}+v_z\frac{\partial v_r}{\partial z}\right)=-\left[\frac{1}{r}\frac{\partial \left(r\tau_{rr}\right)}{\partial r}+\frac{1}{r}\frac{\partial \tau_{\theta r}}{\partial \theta}+\frac{\partial \tau_{rz}}{\partial z}-\frac{\tau_{\theta \theta}}{r}\right] - \frac{\partial p}{\partial r}+\rho g_r$        Eq. (3)

$\rho\left(\frac{\partial v_{\theta}}{\partial t} + v_r\frac{\partial v_{\theta}}{\partial r}+\frac{v_{\theta}}{r}\frac{\partial v_{\theta}}{\partial \theta}+\frac{v_{\theta}v_{r}}{r}+v_z\frac{\partial v_{\theta}}{\partial z}\right)=-\left[\frac{1}{r^2}\frac{\partial \left(r^2\tau_{r\theta}\right)}{\partial r}+\frac{1}{r}\frac{\partial \tau_{\theta \theta}}{\partial \theta}+\frac{\partial \tau_{z\theta}}{\partial z}+\frac{\tau_{\theta r}-\tau_{r\theta}}{r}\right] - \frac{1}{r}\frac{\partial p}{\partial \theta}+\rho g_{\theta}$        Eq. (4)

$\rho\left(\frac{\partial v_{z}}{\partial t} + v_r\frac{\partial v_{z}}{\partial r}+\frac{v_{\theta}}{r}\frac{\partial v_{z}}{\partial \theta}+v_z\frac{\partial v_{z}}{\partial z}\right)=-\left[\frac{1}{r}\frac{\partial \left(r\tau_{rz}\right)}{\partial r}+\frac{1}{r}\frac{\partial \tau_{\theta z}}{\partial \theta}+\frac{\partial \tau_{zz}}{\partial z}\right] - \frac{\partial p}{\partial z}+\rho g_z$        Eq. (5)

The constitutive and momentum equations form a system of equations for the stress $\mathbf{\tau}$ and the velocity of the fluid $\mathbf{v}$. These should be simplified by using the above conclusions. First, Eq. (2) is simplified to obtain,

$\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{\theta r} & \tau_{\theta \theta} & \tau_{\theta z}\\
\tau_{zr} & \tau_{z\theta} & \tau_{zz}
\end{bmatrix}=-\mu
\begin{bmatrix}
0 & 0 & \dfrac{\partial v_z}{\partial r}\\
0 & 0 & 0\\
\dfrac{\partial v_z}{\partial r} & 0 & 0
\end{bmatrix}$        Eq. (6)

From Eq. (6) it can be seen that all components in the stress matrix are zero except $\tau_{rz}$. This becomes obvious after comparison of each element of the stress matrix with its analogous at the other side of the equal sign. The result is,

$\tau_{rz}=-\mu \frac{d v_z(r)}{d r}$        Eq. (7)

Since $v_z$ only depends on a single variable it can not have partial derivatives. The same process of simplification followed to obtain Eq. (6) from Eq. (2) is used again to simplify the components of the momentum balance equation given in Eqs. (3)-(5). From the radial and angular components given in Eqs. (3)-(4) all terms vanish. From the axial component given in Eq. (5) only the following terms remain,

$0=-\frac{1}{r}\frac{d\left(r\tau_{rr}\right)}{dr}-\frac{dp(z)}{dz}+\rho g_z$        Eq. (8)

In Eq. (8) the coefficient $g_z$ represents the $z$ component of the gravity vector wich can be calculated by using the following arrangement obtained from the main schematics shown at the beginning of this post. It follows that $g_z = \mathbf{g} \cos \chi$.

Fig. 2 Schematics to represent the $g_z$ component of the acceleration due to gravity $\mathbf{g}$

On the other hand, Eqs. (7) and (8) may be combined. The result is,

$0=\frac{\mu}{r}\frac{d}{dr}\left(r\frac{dv_z(r)}{dr}\right)-\frac{dp(z)}{dz}+\rho g_z$        Eq. (9)

After simplifying Eq. (9) and reordering it is obtained,

$r^2\frac{d^2v_z(r)}{dr^2}+r\frac{dv_z(r)}{dr}=\frac{r^2}{\mu}\left[\frac{dp(z)}{dz}-\rho g \cos \chi\right]$        Eq. (10)

Equation (10) is in fact an Euler ordinary differential equation with a polynomial inhomogeneity that can be easily solved. Its solution is,

$v_z(r)=C_1+C_2\ln (r)+\frac{r^2}{4\mu}\left[\frac{dp(z)}{dz}-\rho g \cos \chi\right]$        Eq. (11)

The integration constants in solution Eq. (11) are calculated from a set of boundary conditions. One of these demands that the fluid velocity at the wall should be zero due to the no-slip condition. This is, the viscous fluid adheres to the wall and in that point $r = R$ its velocity is the same of the wall, zero. In other words,

$v_z(r=R)=0$        Eq. (12)

The other boundary condition is a little more complex. Since there are not to much points to evaluate the velocity $v_z$ in $r$ the center of the pipe $r = 0$, is considered. Because a priori one does not know the value of the velocity at $r = 0$ it is reasonable to ask that in that position the fluid velocity may take a finite or quantifiable value. Mathematically, this is written as,

$v_z(r=0)=\text{finite value}$        Eq. (13)

Evaluating the boundary condition Eq. (12) it is obtained,

$C_1=-C_2\ln (R)-\frac{R^2}{4\mu}\left[\frac{dp(z)}{dz}-\rho g \cos \chi\right]$        Eq. (14)

Next, from evaluation of boundary condition Eq. (13) at $r=0$ it follows that the velocity is infinite, which is physically impossible because the fluid velocity the pipe never will be infinite, in fact the fluid moves slowly. Besides, the boundary condition estates that $v_z$ has a finite or quantifiable value at $r = 0$; then $C_2$ must be zero. Therefore, the fluid velocity becomes,

$v_z(r)=\frac{R^2}{4\mu}\left[\frac{dp(z)}{dz}-\rho g \cos \chi\right]\left[\left(\frac{r}{R}\right)^2-1\right]$        Eq. (15)

In Eq. (15) pressure is still unknown. In fact, the velocity Eq. (15) is a first order ordinary differential equation for $p(z)$ and can be solved by separation of variables. Its solution is,

$p(z)=\left(\frac{4\mu}{R^2}\frac{v_z(r)}{\left(\frac{r}{R}\right)^2-1}+\rho g \cos \chi\right)z+C_3$        Eq. (16)

The constant of integration $C_3$ can be calculated with help of the pressure boundary conditions given for this problem,

$p(z=0)=P_0$        Eq. (17)

$p(z=L)=P_L$        Eq. (18)

After evaluation of the boundary conditions Eqs. (17)-(18) two expressions having $C_3$ are obtained. Combining these two last expressions, it is obtained,

$P_0-P_L=-\left(\frac{4\mu}{R^2}\frac{v_z(r)}{\left(\frac{r}{R}\right)^2-1}+\rho g \cos \chi\right)L$        Eq. (18)

Combining Eqs. (16) and (19) follows,

$\frac{dp(z)}{dz}=\frac{4\mu}{R^2}\frac{v_z(r)}{\left(\frac{r}{R}\right)^2-1}+\rho g \cos \chi=-\frac{P_0-P_L}{L}$        Eq. (20)

Finally, the velocity distribution for a fluid flowing through an inclined pipe is,

$v_z(r)=-\frac{R^2}{4\mu}\left[\frac{P_0-P_L}{L}+\rho g \cos \chi\right]\left[\left(\frac{r}{R}\right)^2-1\right]$        Eq. (21)

The flow rate of a newtonian fluid through an inclined pipe

The volumetric flow through the pipe is calculated with the integral of area for the velocity. In cylindrical coordinates this is,

$Q=\int_{0}^{2\pi} \int_{0}^{R} v_z(r)drd\theta$

$Q=-\frac{\pi R^2}{2\mu}\left[\frac{P_0-P_L}{L}+\rho g \cos \chi\right] \int_{0}^{R} \left[\left(\frac{r}{R}\right)^2-1\right]dr$

$Q=-\frac{\pi R^4}{8\mu}\left[\frac{P_0-P_L}{L}+\rho g \cos \chi\right]$        Eq. (22)

Equation (22) represents the volumetric flow of a fluid flowing through an inclined pipe. This is the end of the solution of the problem.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• The Poiseuille flow of a power law 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

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Ildebrando.