14.1 Introduction

In this section, we will derive the fluid mechanics of the circular flow tube. Flow tubes as control volumes were introduced in section 10.1.4. They represent a very simple type of control volume and can be used whenever a fluid flow is parallel. The circular flow tube is the simplest case of a flow tube. It exploits the fact that a cylindrical coordinate, although being a three-dimensional object, can be treated mathematically as a two-dimensional problem if the mechanics are derived in cylindrical coordinates (see section 7). In general, the cylindrical coordinate system uses three axes: ${\overrightarrow{e}}_r$ (radius), ${\overrightarrow{e}}_{\phi }$ (rotation around ${\overrightarrow{e}}_z$, azimuthal angle), and ${\overrightarrow{e}}_z$. However, in cylindrical flow tubes, changes in the azimuthal axis can be neglected assuming a symmetric flow profile. This reduces the number of variables to two, making the problem significantly simpler mathematically. In addition, we consider the flow to be parallel; therefore there is no flow along the radial or azimuthal direction.

The principle geometrical relations on the circular flow tube are shown in Fig. 14.1. In the following, we will derive the laws of conservation for mass (continuity equation), momentum (Navier-Stokes equation), and energy. The application of the thermodynamic equations of state as discussed in section 12.8 applies to the flow tube as well and is therefore not treated separately. The flow tube is characterized by its radius r and the length s.

14.2 Conservation of Mass: The Continuity Equation

The conservation of mass is given by Eq. 10.8 as

$\begin{array}{l}\dot{m}=\rho \dot{V}=\rho Av\hfill \\ \stackrel{!}{=}\text{const}\hfill \end{array}$

If we consider the entrance and the exit of the flow tube we find that

$\begin{array}{l}\dot{m}=\rho A_1{v}_1=\rho A_2{v}_2\hfill \\ \stackrel{!}{=}\text{const}\hfill \end{array}$

14.3 Conservation of Momentum: The Navier-Stokes Equation

The second equation we will derive is the conservation of momentum, i.e., the Navier-Stokes equation. As stated, the Navier-Stokes is effectively Newton’s second law of motion (see Eq. 11.1), which is given by

$\begin{array}{ll}{\displaystyle \sum _{i}d{\overrightarrow{F}}_i}\hfill & =dma\hfill \\ \hfill & =\left(\frac{\partial v}{\partial t}+\left(v\cdot \overrightarrow{\nabla }\right)v\right)\rho dV\hfill \\ \hfill & =\left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial s}\right)\rho 2\pi rdrds\hfill \end{array}$

where we used the acceleration in the Lagrangian frame of reference (see Eq. 10.3) as our chosen coordinate systems moves with the flow tube. Please note that we also simplified $\left(v\cdot \overrightarrow{\nabla }\right)v$ to $v\frac{\partial v}{\partial s}$ because there is no flow in the direction of the azimuthal and also no flow along the radial direction.

The forces we need to consider are given by the pressure forces at the entrance and the exit of the flow tube, the weight of the flow tube, and the friction forces at the outside and the inside of the flow tube.

14.4 Euler Equation

Inserting Eq. 14.3, Eq. 14.5, and Eq. 14.6 into Eq. 14.2 results in the Euler equation. As stated, this equation is a simplified version of the Navier-Stokes equation for incompressible fluids in the circular flow tube.

$\begin{array}{ll}\left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial s}\right)\rho 2\pi rdrds\hfill & =-\frac{\partial p}{\partial s}ds2\pi rdr-g\rho 2\pi rdrds\cos \phi +\mathcal{T}2\pi drds+\frac{\partial \mathcal{T}}{\partial r}2\pi rdrds\hfill \\ \rho \left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial s}\right)\hfill & =-\frac{\partial p}{\partial s}-g\rho \cos \phi +\frac{\mathcal{T}}{r}+\frac{\partial \mathcal{T}}{\partial r}\hfill \\ \hfill & =-\frac{\partial p}{\partial s}-g\rho \frac{dz}{ds}+\frac{\mathcal{T}}{r}+\frac{\partial \mathcal{T}}{\partial r}\hfill \\ \frac{\partial v}{\partial t}+v\frac{\partial v}{\partial s}\hfill & =-\frac{1}{\rho }\frac{\partial p}{\partial s}-g\frac{dz}{ds}+\frac{1}{\rho }\left(\frac{\mathcal{T}}{r}+\frac{\partial \mathcal{T}}{\partial r}\right)\hfill \end{array}$

Here we have used Eq. 14.4. The Euler equation describes the different force contributions of the flow field as

• inertia forces: $\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial s}$

• pressure forces: $\frac{1}{\rho }\frac{\partial p}{\partial s}$

• gravity forces: $g\rho \frac{dz}{ds}$

• friction (viscous) forces: $\frac{\mathcal{T}}{r}+\frac{\partial \mathcal{T}}{\partial r}$

The Euler equation lends itself well to putting the individual force contributions into relation and thus allows estimating which effects dominate and which ones may be neglected. In many applications, dropping some terms from this equation will make calculating the flow field significantly simpler. The two most important dimensionless numbers used with the Euler equation are

• the Froude number (see section 9.9.11), which puts in relation inertia and gravity forces

• the Reynolds number (see section 9.9.8), which puts in relation inertia and viscous forces

14.5 Bernoulli Equation

If we consider only stationary fluid flows (in which case, all derivatives with respect to time are 0) and ignore effects of the fluid friction (in which case, the viscous force terms are set to 0), the Euler equation (Eq. 14.7) can be simplified further. In this case, we find

$v\frac{dv}{ds}=-\frac{1}{\rho }\frac{dp}{ds}-g\frac{dz}{ds}$

where we can use the following trick to simplify the left-hand side of the equation (see section 3.3.2 for a description of this technique)

$v\frac{dv}{ds}=\frac{d}{ds}\left(\frac{v^2}{2}\right)$

in which case we find Eq. 14.9 to be

$d\left(\frac{v^2}{2}\right)=-\frac{1}{\rho }dp-gdz$

which can be integrated. Usually it is integrated between two points 1 and 2 of the flow tube that can correspond, e.g., to the inlet and the outlet of a microfluidic channel, respectively. The integration yields the Bernoulli¹equation.

$\begin{array}{ll}\frac{v_2^2}{2}-\frac{v_1^2}{2}\hfill & =-\frac{1}{\rho }\left(p_2-p_1\right)-g\left(z_2-z_1\right)\hfill \\ \frac{v_2^2}{2}+\frac{p_2}{\mathrm{p}}+g{z}_2\hfill & =\frac{v_1^2}{2}+\frac{p_1}{\rho }+g{z}_1\hfill \end{array}$

Please note that during this integration, we assumed the density ρ to be a constant; otherwise, we would have to express ρ as a function of the pressure p before performing the integration. Eq. 14.11 is therefore the Bernoulli equation for incompressible fluids. In order to account for compressible fluids, i.e., gases, we can use Eq. 6.4 to account for changing values of the density with changing pressure.

The Bernoulli equation is one of the most important equations in fluid mechanics because it allows the easy and fast derivation of important field variables in many applications. For example, if the flowrate is given for a system, the Bernoulli equation allows deriving the pressure at any given point in the fluidic system. The Bernoulli equation is a special case of the Euler equation, namely the case where fluid friction is ignored.

14.6 Conservation of Energy

We will now quickly derive the last conservation equation for the flow tube, which is the conservation of energy. The change of energy from point 1 to point 2 (e.g., the inlet or the outlet, respectively) is given by

$\Delta W_{1\to 2}={\dot{m}}_2\left(h_2+\frac{v_2^2}{2}+g{h}_2\right)=-{\dot{m}}_1\left(h_1+\frac{v_1^2}{2}+g{h}_1\right)\stackrel{!}{=}W+Q$

which is a simplified form of the first law of thermodynamics (see Eq. 6.32). Because the total mass flow is constant, i.e., ${\dot{m}}_2={\dot{m}}_1$, Eq. 14.12 can be divided by ${\dot{m}}_1$, resulting in

$\left(h_2+\frac{v_2^2}{2}+g{h}_2\right)-\left(h_1+\frac{v_1^2}{2}+g{h}_1\right)=\dot{w}+\dot{q}$

Eq. 14.13 is the conservation of energy for the flow tube. If no work or heat is brought into the system, $\dot{w}=\dot{q}=0$, and Eq. 14.13 is identical to the Bernoulli equation (see Eq. 14.11).

14.7 Deriving the Euler Equation by a Coordinate System Transformation

Laplace Operator in Cylindrical Coordinates. As stated at the beginning to this section, we could have made our lives a little easier by simply applying a coordinate system transformation to the Navier-Stokes equation we derived for the Cartesian coordinate system (see Eq. 11.40). We already introduced coordinate system transformations in section 7.3 and derived the transformed version of the Navier-Stokes equation in cylindrical coordinates (see Eq. 13.2). We can further simplify this equation by noting that all derivatives with respect to ϕ (no change in the azimuthal direction) and z (fully developed flow profile) are zero. Furthermore, there is only one relevant velocity component, which is the velocity along the axis. In cylindrical coordinates, this axis is usually the z-axis (see section 7). We therefore simplify Eq. 13.2 to

$\begin{array}{ll}\rho \left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial z}\right)\hfill & =k_z-\frac{\partial p}{\partial z}+\frac{\eta }{r}\frac{\partial }{\partial r}\left(r\frac{\partial rv}{\partial r}\right)\hfill \\ \hfill & =k_z-\frac{\partial p}{\partial z}+\frac{\eta }{r}\left(\frac{\partial v}{\partial r}+r\frac{{\partial }^{2}v}{\partial r^2}\right)\hfill \\ \hfill & =k_z-\frac{\partial p}{\partial z}+\eta \left(\frac{1}{r}\frac{\partial v}{\partial r}+\frac{{\partial }^{2}v}{\partial r^2}\right)\hfill \end{array}$

The only relevant volume force is gravity given by

$k_z=-\frac{m}{V}g\frac{dz}{ds}=-\rho g\frac{dz}{ds}$

which allows us to rewrite Eq. 14.14 to

$\begin{array}{ll}\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial s}\hfill & =\frac{1}{\rho }\frac{\partial p}{\partial s}-g\frac{dz}{ds}+\frac{\eta }{\rho }\left(\frac{1}{r}\frac{\partial \mathrm{v}}{\partial \mathrm{r}}\frac{{\partial }^{2}v}{d{r}^2}\right)\hfill \\ \hfill & =\frac{1}{\rho }\frac{\partial p}{\partial s}-g\frac{dz}{ds}+v\left(\frac{1}{r}\frac{\partial v}{dr}+\frac{{\partial }^{2}v}{d{r}^2}\right)\hfill \end{array}$

where we replaced the independent variable z with the path s along the cylinder axis. Please note that, even though this independent variable can be used for the flow inside the tube, gravity must be calculated by using z because it refers to the global coordinate system. As you can see, Eq. 14.15 and Eq. 14.8 are identical. As stated, the coordinate system transformation of the Navier-Stokes equation to cylindrical coordinates yields the Euler equation.

14.8 Summary

In this section, we introduced the concept of flow tubes, which are convenient control volumes that lend themselves well to describing constraint flow volumes, e.g., pipes and tubes. We have derived the fundamental equations, i.e., the continuity, the Navier-Stokes, as well as the energy equation for this control volume. We also derived the Euler and the Bernoulli equations, which are special forms of the Navier-Stokes equation in cylindrical coordinates. Obviously, we could have simply copied the converted forms of the fundamental equations in cylindrical coordinates and gotten the same equations. However, as we have seen, it is not difficult to derive these equations “from scratch.”