Physics problem: viscous flow in a cylindrical pipe

This problem is related to the previous one, where we discussed a setup with a fluid flow between two horizontal plates. The solution method is very similar, with a difference that here we select a cylindrical element and consider forces acting on it.

Problem statement

A liquid is pumped from one container to another through a long pipe of radius $R$ and length $l$. Determine the flow velocity in the pipe as a function of distance to its axis (i.e. to the center of the pipe). Given that the pressure drop between the two ends of the pipe is $\Delta P$ and the dynamic viscosity of the liquid is $\eta$.

Problem statement is taken from Savchenko et al (1981) book and translated by the post author.

Exploiting the symmetry around the axis of the pipe we select a coaxial cylinder of radius $r$ in the flow and investigate forces acting on it.

Figure 1: Flow in pipe of length $l$ and radius $R$.

If we denote the viscosity stress along the cylinder's sides and parallel to its axis as $\tau_x(r)$, then we can write the balance of the viscous and pressure gradient forces acting on it, along the $x$-axis, as:

\begin{align*} & \tau_x(r) \cdot 2\pi r l + \Delta P \cdot \pi r^2 = 0 & \Rightarrow \\ & \tau_x(r) = -\frac{\Delta P r}{2 l} & \end{align*}

Now we can use the expression linking viscous stress and flow velocity gradient through the dynamic viscosity to determine the flow velocity as follows:

\begin{align*} & \tau_x(r) = -\frac{\Delta P r}{2 l} = \eta \frac{dv}{dr} & \Rightarrow \\ & \frac{dv}{dr} = -\frac{\Delta P r}{2 l \eta} & \Rightarrow \\ & \int\limits_{R}^{r} \frac{dv}{dr}\,dr = -\int\limits_{R}^{r}\frac{\Delta P r}{2 l \eta}\,dr & \Rightarrow \\ & v(r) - v(R) = -\frac{\Delta P \left( r^{2} - R^{2} \right) }{4 l \eta} \\ \end{align*}

Now using the non-slip condition at the pipe walls (i.e. $v(R) = 0$), we get the final expression for the flow velocity at a distance $r$ from the pipe axis: \begin{align*} \boxed{ v(r) = -\frac{\Delta P \left( r^{2} - R^{2} \right) }{4 l \eta} }. \end{align*}

Physics problem: viscous flow between two horizontal plates

Problem statement

Determine velocity of fluid in a stationary flow between two fixed horizontal plates as a function of position between the two plates. The distance between the plates is $h$ and the dynamic viscosity of the fluid is $\eta$. The pressure drop in the direction of the flow per unit length is $\frac{dP}{dx}=-\Delta P$. Additionally, compute the fluid discharge per unit of horizontal length perpendicular to the flow.

Problem statement is taken from Savchenko et al (1981) book and translated by the post author.

Solution

Tihs is not a praticularly complicated problem, but I find it important for understanding the momentum exchanges between layers of viscous fluids and it does not require very deep knowledge of the process but allows us to draw some interesting insights.

Let's introduce a coordinate system with $x$-axis along and pointing in the direction of flow, $z$-axis in the vertical direction upwards, and $y$-axis in the horizontal direction transversal to the flow and perpendicular to the plane formed by $x$ and $z$ axes. The origin of the coordinate system $O$ is on the lower plate (see Figure 1).

Figure 1: Vertical cross-section of the viscous flow between to horizontal plates. $y$-axis is perpendicular and points into the figure plane.

Since the flow is stationary and horizontally uniform the flow velocity is only a function of the vertical coordinate $z$: $v(z)$.

We assume a non-slip boundary condition near the horizontal plates: $v(z=0)=v(z=h)=0$. Additionally, we assume that the stress (friction force per unit of contact area between horizontal layers of the fluid, $\tau_x$) due to viscosity decreases towards the plane of symmetry of the flow: $z=h/2$, i.e the layers of the fluid closer to the middle have higher velocities. Therefore, we can determine $v(z)$ for $z \in [0, h/2]$ and then use the symmetry of the flow to find the velocity for $z \in [h/2, h]$.

Let's consider an element of fluid with dimensions $(\Delta x, \Delta y, \Delta z = h - 2z)$ with it's lower face at $z \in [0, h/2]$ and parallel to the axes $x$ and $y$. Assuming that the viscous stress $x$-coordinate $\tau_x(z)$ is negative, we can express the balance of the forces acting on the selected element of fluid along the $x$-axis as follows:

\begin{align*} \tau_x(z) \cdot \Delta x \cdot \Delta y + \tau_x(z + \Delta z) \cdot \Delta x \cdot \Delta y + \Delta P \cdot \Delta x \cdot \Delta y \cdot \Delta z = 0 \end{align*}

It is easy to see that with $\Delta z = h - 2z$ the upper and lower faces of the selected element are symmetric with respect to the middle plane of the flow ($z=h/2$), and are at the same distance from the upper and lower plates respectively (this is by design of our imaginary flow element) which means that $\tau_x(z + \Delta z) = \tau_x(z)$. Then we can simplify the above balance of forces as:

\begin{align*} & 2 \tau_x(z) + \Delta P \cdot \Delta z = 0 & \Rightarrow \\ & 2 \tau_x(z) + \Delta P \cdot (h - 2z) = 0 & \Rightarrow^{\,} \\ & \tau_x(z) = -\frac{\Delta P}{2} (h - 2z) & \end{align*}

The above expression for the stress is valid for $z \in [0, h/2]$.

Let's now determine $v(z)$ for $z \in [0, h/2]$ and then we'll use the symmetry of the flow to extend the solution to the whole range $z \in [0, h]$.

We can now use the relation between the velocity and the stress through the dynamic viscosity $\eta$:

\begin{align*} \tau_x(z) = -\eta\frac{dv}{dz}, \; z \in [0, h/2] \end{align*}

Using previously derived expression for $\tau_x(z)$ and then integrating we get:

\begin{align*} & -\frac{\Delta P}{2} (h - 2z) = -\eta\frac{dv}{dz} & \Rightarrow \\ & v(z) = \frac{\Delta P}{2\eta} \left( hz -z^{2} \right) + C & \end{align*}

where $C$ is the constant of integration.

Then using the non-slip condition at the plates we determine the integration constant as $C = v(z=0)=0 \Rightarrow C=0$ and the final expression for the velocity profile is:

\begin{align*} v(z) = \frac{\Delta P}{2\eta} z (h - z), \; z \in [0, h/2]. \end{align*}

Now for $z \in [h/2, h]$ we can use the flow symmetry to express the flow velocity as:

\begin{align*} \hat{v}(z) = v(h - z) = \frac{\Delta P}{2\eta} (h - z) z, \; z \in [h/2, h]. \end{align*} which, curiously, is the same expression as for $z \in [0, h/2]$, therefore finally the flow velocity profile can be expressed using a single formula:

\begin{align*} \boxed{v(z) = \frac{\Delta P}{2\eta} (h - z) z, \; z \in [0, h]} \end{align*}

In the second part of the problem we are asked to compute the flux density per unit of length in $y$ direction, i.e. $q=\frac{dQ}{dy}$. This is just a simplification for us to use a simpler flux expression ($\int{v\,dz}$) instead of the more general one ($\iint{v\,dy\,dz}$).

Therefore we obtain the flux density by integrating the above profile:

\begin{align*} &q=\int\limits_{0}^{h}v(z)\,dz & \Rightarrow \\ &q=\int\limits_{0}^{h} \frac{\Delta P}{2\eta} (h - z) z \,dz &\Rightarrow \\ &q= \frac{\Delta P}{2\eta} \left(\frac{h^{3}}{2} - \frac{h^{3}}{3} \right) & \Rightarrow \\ &\boxed{q= \frac{\Delta P h^{3}}{12\eta}} & \end{align*}