Laminar flow through an annular region. Newtonian fluid

The problem of the flow of a newtonian fluid through an annular region

A newtonian fluid flows axially through the annular region between two vertical concentric circular pipes, both of length $L$. The outer pipe has radii $R$ and inner pipe has radii $\kappa R$ where $0>\kappa<1$. 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 upward flow of a newtonian fluid through the annular region between two cylinders.

The solution for the velocity distribution

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 through an annular region

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 parallel to the cylinder axis;
• the fluid moves upwards through the annular region between the two concentric pipes;
• 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 externally and not shown in Fig. 1. However, since the fluid moves upwards the acceleration due to gravity has the opposite effect;
• 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.- Since the working fluid is newtonian the corresponding equation is

$\mathbf{\tau}=-\mu \left[\nabla \mathbf{v} + \left(\nabla \mathbf{v}\right)^T\right]$        Eq. (01)

which can be written in extended form in cylindrical coordinates as,

$\begin{bmatrix}
\tau_{rr}\cos^2 \theta-2\tau_{r\theta}\sin \theta \cos \theta+\tau_{\theta \theta}\sin^2 \theta & \tau_{rr}\sin \theta \cos \theta + \tau_{r\theta}\left[\cos^2 \theta-\sin^2 \theta\right]-\tau_{\theta \theta}\sin \theta\cos \theta & \tau_{rz}\cos \theta -\tau_{z\theta}\sin \theta\\
\tau_{rr}\sin \theta \cos \theta+\tau_{r\theta}\left[\cos^2 \theta-\sin^2 \theta\right]-\tau_{\theta \theta}\sin \theta\cos \theta & \tau_{rr}\sin^2 \theta+2\tau_{r\theta}\sin \theta \cos \theta+\tau_{\theta \theta}\cos^2 \theta & \tau_{rz}\sin \theta +\tau_{z\theta}\cos \theta\\
\tau_{rz}\cos \theta -\tau_{z\theta}\sin \theta & \tau_{rz}\sin \theta +\tau_{z\theta}\cos \theta & \tau_{zz}
\end{bmatrix}\nonumber\\
=-\mu
\begin{bmatrix}
2\dfrac{\partial v_r}{\partial r} & r\dfrac{\partial}{\partial r}\left(\dfrac{v_\theta}{r}\right)+\dfrac{1}{r}\dfrac{\partial v_r}{\partial \theta} & \dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r} \\
r\dfrac{\partial}{\partial r}\left(\dfrac{v_\theta}{r}\right)+\dfrac{1}{r}\dfrac{\partial v_r}{\partial \theta} & 2\left(\dfrac{1}{r}\dfrac{\partial v_\theta}{\partial \theta}+\dfrac{v_r}{r}\right) & \dfrac{1}{r}\dfrac{\partial v_z}{\partial \theta}+\dfrac{\partial v_\theta}{\partial z} \\
\dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r} & \dfrac{1}{r}\dfrac{\partial v_z}{\partial \theta}+\dfrac{\partial v_\theta}{\partial z} & 2\dfrac{\partial v_z}{\partial z}
\end{bmatrix}$        Eq. (02)

• the momentum balance equation.- This is a vectorial partial differential equation with components $(r,\theta,z)$ 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. (03)

$\rho\left(\frac{\partial v_\theta}{\partial t}+v_r\frac{\partial v_\theta}{\partial r}+\frac{v_\theta}{r}\frac{\partial v_\theta}{\partial \theta}+\frac{v_\theta v_r}{r}+v_z\frac{\partial v_\theta}{\partial z}\right)=-\left[\frac{1}{r^2}\frac{\partial \left(r^2\tau_{r\theta}\right)}{\partial r}+\frac{1}{r}\frac{\partial \tau_{\theta \theta}}{\partial \theta}+\frac{\partial \tau_{z\theta}}{\partial z}+\frac{\tau_{\theta r}-\tau_{r\theta}}{r}\right]-\frac{1}{r}\frac{\partial p}{\partial \theta}+\rho g_\theta$        Eq. (04)

$\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. (05)

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. (02) is simplified to get,

$\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta \theta} & \tau_{z\theta}\\
\tau_{rz} & \tau_{z\theta} & \tau_{zz}
\end{bmatrix}
=-\mu
\begin{bmatrix}
0 & 0 & \dfrac{\partial v_z}{\partial r} \\
0 & 0 & 0 \\
\dfrac{\partial v_z}{\partial r} & 0 & 0
\end{bmatrix}$        Eq. (06)

From Eq. (06) 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,

$\tau_{rz}=-\mu \frac{d v_z\left(r\right)}{d r}$        Eq. (07)

Since $v_z$ depends only on $r$ partial derivatives notation is not needed anymore. The simplification process to obtain Eq. (06) from Eq. (02) is repeated to simplify the components of the momentum balance Eqs. (03-05). The terms in the radial and angular components, Eqs. (03-04), are zero. From the axial component Eq. (05) only the following remains,

$0=-\frac{1}{r}\frac{d \left(r\tau_{rz}\right)}{d r}-\frac{d p\left(z\right)}{d z}+\rho g_z$        Eq. (08)

In Eq. (08) $g_z$ represents the $z$ component of the acceleration due to gravity vector $\mathbf{g}$. As mentioned early the fluid travels through a vertical annular region so that $g_z=-g$ since the vector points downwards. On the other hand, Eqs. (07-08) can be combined to obtain,

$0=\frac{\mu}{r}\frac{d}{d r}\left(r\frac{d v_z\left(r\right)}{d r}\right)-\frac{d p\left(z\right)}{d z}-\rho g$        Eq. (09)

Equation (09) can be simplified and rewritten as,

$r^2\frac{d^2 v_z\left(r\right)}{dr^2}+r\frac{d v_z\left(r\right)}{dr}=\frac{r^{2}}{\mu}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]$        Eq. (10)

Equations (09-10) are the same. The only difference is that the version shown in Eq. (10) allows us to see that this is in fact an Euler equation type with a polynomial inhomogeneity. Solution to Eq. (10) is straight,

$v_z\left(r\right) = C_1+C_2 \ln \lvert r \rvert + \frac{r^2}{4\mu}\left[\frac{d p\left(z\right)}{d z}+\rho g\right] $        Eq. (11)

The constants of integration in $v_z$ Eq. (11) should be determined using proper boundary conditions. For this case it seems clear that the corresponding boundary conditions for $v_z$ should be given at the inner and outer pipe walls as follows,

$v_z=0,\quad\text{at}\quad r=\kappa R$
$v_z=0,\quad\text{at}\quad r= R$        Eq. (12)

Boundary conditions Eqs. (12) correspond to the adherence condition which states that the fluid velocity vanishes at the walls because of the friction and viscous nature of the fluid; but it also implies that between the walls the fluid velocity is not zero and may reach an extrema. Notice that for this case demanding a finite value for $v_z$ at the mid distance between the two pipe walls is useless since there is no divergence at $r=\left(R-\kappa R\right)/2$ as in the case of the fluid traveling through a circular pipe.

Evaluation of boundary conditions Eqs. (12) gives,

$C_1+C_2 \ln \lvert \kappa R \rvert + \frac{\kappa^2 R^2}{4\mu}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]=0$        Eq. (13)

$C_1+C_2 \ln \lvert R \rvert + \frac{R^2}{4\mu}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]=0$        Eq. (14)

Equations (13-14) are set of simultaneous algebraic equations for $C_1$ and $C_2$. This is,

$C_1=\frac{R^2}{4\mu \ln \lvert 1/\kappa \rvert}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]\left[\left(1-\kappa^2\right) \ln \lvert R \rvert - \ln \lvert 1/\kappa \rvert \right]$

$C_2=-\frac{R^2\left(1-\kappa^2\right)}{4\mu \ln \lvert 1/\kappa \rvert}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]$        Eq. (15)

The constants of integration Eqs. (15) are now used to rewrite $v_z$ given in Eq. (11). Simplification could be difficult for novices so that some steps are shown to get a grip on the algebra,

$v_z\left(r\right)= \frac{R^2}{4\mu\ln \lvert 1/\kappa \rvert}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]\left[\left(1-\kappa^2\right) \ln \lvert R \rvert - \ln \lvert 1/\kappa \rvert \right] - \frac{R^2\left(1-\kappa^2\right)\ln \lvert r \rvert}{4\mu \ln \lvert 1/\kappa \rvert}\left[\frac{d p\left(z\right)}{d z}+\rho g\right] +\frac{r^2}{4\mu}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]$

$v_z\left(r\right)=\frac{R^2}{4\mu}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]\left[\frac{\ln \lvert R \rvert-\kappa^2\ln \lvert R \rvert-\ln \lvert r \rvert+\kappa^2\ln \lvert r \rvert }{\ln \lvert 1/\kappa \rvert}-1+\frac{r^2}{R^2}\right]$

and after simplification of the logarithms $v_z$ can be expressed as,

$v_z\left(r\right)=\frac{R^2}{4\mu}\left[\frac{d p\left(z\right)}{d z}+\rho g\right]\left[\frac{1-\kappa^2}{\ln \lvert 1/\kappa \rvert}\ln \left| \frac{R}{r} \right| - 1 + \left(\frac{r}{R}\right)^2\right]$        Eq. (16)

Although Eq. (16) represents the fluid velocity distribution it is still implicit because the pressure gradient has not been determined yet. One should notice that Eq. (16) is first order ordinary differential equation for $p$ and it can be easily solved by separation of variables. The result is,

$p\left(z\right)= \left\lbrace \frac{4\mu v_z\left(r\right)}{R^2}\left[\frac{1-\kappa^2}{\ln \lvert 1/\kappa \rvert}\ln \left| \frac{R}{r} \right| - 1 + \left(\frac{r}{R}\right)^2\right]^{-1}-\rho g \right\rbrace z+C_3 $        Eq. (17)

$C_3$ can be found with help of the corresponding boundary condition for $p(z)$. There is in fact two boundary conditions for this variable,

$p=P_0,\quad\text{at}\quad z=0$
$p=P_L,\quad\text{at}\quad z=L$        Eq. (18)

Despite two available equations for $C_3$ obtained after the evaluation of Eqs. (18) advantage can be taken from this situation. Combination of the two last expressions provides,

$P_0-P_L=-\left\lbrace \frac{4\mu v_z\left(r\right)}{R^2}\left[\frac{1-\kappa^2}{\ln \lvert 1/\kappa \rvert}\ln \left| \frac{R}{r} \right| - 1 + \left(\frac{r}{R}\right)^2\right]^{-1}-\rho g \right\rbrace L$        Eq. (19)

so that,

$p\left(z\right)=-\left(P_0-P_L\right)\frac{z}{L}+C_3$

$\frac{d p\left(z\right)}{d z}=-\frac{\left(P_0-P_L\right)}{L}$        Eq. (20)

Further substitution of Eq. (20) into Eq. (16) gives,

$v_z\left(r\right)=\frac{R^2}{4\mu}\left[\frac{\left(P_0-P_L\right)}{L}-\rho g\right]\left[1 - \left(\frac{r}{R}\right)^2 - \frac{1-\kappa^2}{\ln \lvert 1/\kappa \rvert}\ln \left| \frac{R}{r} \right| \right] $        Eq. (21)

which is an explicit expression for the velocity profile in terms of measurable parameters. Next, the volumetric flow rate can be calculated by using $v_z$ as given in Eq. (21). 

The flow rate of a newtonian fluid through annular region

Then, $v_z$ is integrated over the cross-section area of the annular region as follows,

$Q=\int_{0}^{2\pi}\int_{\kappa R}^{R} v_zrdrd\theta, \quad\text{introduce change of variable}\quad w=\frac{r}{R}$

$Q=\left.\frac{\pi R^2}{2\mu}\left[\frac{\left(P_0-P_L\right)}{L}-\rho g\right]\left[ \frac{w^2}{2}-\frac{w^4}{4}-\frac{1-\kappa^2}{\ln \lvert 1/\kappa \rvert}\left(\frac{w^2}{4} +\frac{w^2}{2}\ln \left| \frac{1}{w} \right|\right) \right]\right|_{r=\kappa R}^{r=R}$

$\text{Return to the original variable with }\quad w=\frac{r}{R}\quad \text{and evaluate}$

$Q=\frac{\pi R^2}{2\mu}\left[\frac{\left(P_0-P_L\right)}{L}-\rho g\right]\left[ \frac{1}{4}+\frac{\kappa^4}{4}-\frac{\kappa^2}{2} + \frac{1-\kappa^2}{\ln \lvert 1/\kappa \rvert}\left( \frac{\kappa^2}{4}-\frac{1}{4}+\frac{\kappa^2}{2}\ln \left| \frac{1}{\kappa} \right| \right) \right]$

$Q=\frac{\pi R^2}{8\mu}\left[\frac{\left(P_0-P_L\right)}{L}-\rho g\right] \left[ 1-\kappa^4- \frac{\left(1-\kappa^2\right)^2}{\ln \lvert 1/\kappa \rvert}\right]$Eq. (22)

Equation (22) represents the flow rate for a newtonian fluid through a annular region. This is the end of the problem solution.

Watch the following YouTube video for a more dynamic explanation of the solution.

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

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