# 11.2.1 Overview of fluid dynamics

This section will make extensive use of vector calculus concepts and notation: if you are a bit rusty on that, check section 4.1.4 first.

The governing equations for fluid flow involve two rather different ingredients. First, we will use universal laws of physics: Newton’s law and the conservation of mass. The second ingredient is more empirical, and the details could vary depend on which fluid we are interested in, and what kind of flow regime we wish to study. This ingredient is a constitutive law, relating the internal forces (described via a stress tensor) to the fluid motion (described by a rate of strain tensor, provided our fluid can be approximated by a classical “Newtonian fluid” with no “memory”). If our fluid is compressible, we also need a thermodynamic relation of some kind to relate density to pressure.

Suppose that our fluid has velocity $\underline{u}(\underline{r},t)$, density $\rho(\underline{r},t)$ and pressure $p(\underline{r},t)$. First, we will look at the requirement for conservation of mass. Consider the small brick-shaped element of volume sketched in Fig. 1, with one corner at position $\underline{r}=(x,y,z)$ and sides of length $\delta x$, $\delta y$ and $\delta z$ aligned with the $x,y,z$ axes of a Cartesian coordinate system. This element has volume $\delta V= \delta x \delta y \delta z$.

The rate of change of mass inside the volume is approximately

$$\dfrac{\partial \rho}{\partial t} \delta V \tag{1}$$

where the derivative of density is evaluated at the centre of the element. Now we need to look at the rate at which mass is flowing into or out of the volume, through the various faces. For the two vertical faces perpendicular to the $x$ axis, the net mass flow into the volume is

$$\left[\rho(x,y,z,t) u_1(x,y,z,t) – \rho (x + \delta x,y,z,t) u_1(x + \delta x,y,z,t) \right] \delta y \delta z$$

$$\approx -\delta V \dfrac{\partial (\rho u_1)}{\partial x} \tag{2}$$

where $(u_1,u_2,u_3)$ are the components of $\underline{u}$ so that $u_1$ is the velocity perpendicular to the faces we are thinking about. The other two pairs of faces contribute similar terms, so the net mass flow into the volume is

$$-\delta V \dfrac{\partial (\rho u_1)}{\partial x} -\delta V \dfrac{\partial (\rho u_2)}{\partial y} -\delta V \dfrac{\partial (\rho u_3)}{\partial z} = -\delta V \nabla \cdot (\rho \underline{u}) . \tag{3}$$

Equating this to the rate of increase of mass inside the volume from (1), we obtain the equation

$$\dfrac{\partial \rho}{\partial t} = – \nabla \cdot (\rho \underline{u}) . \tag{4}$$

Now we can use a standard vector identity

$$\nabla (\rho \underline{u}) = \underline{u} \cdot \nabla \rho + \rho \nabla \cdot \underline{u}\tag{5}$$

to rewrite equation (4) in the form

$$\dfrac{\partial \rho}{\partial t} + \underline{u} \cdot \nabla \rho = – \rho \nabla \cdot \underline{u}\tag{6}$$

or

$$\dfrac{D \rho}{D t} = – \rho \nabla \cdot \underline{u}\tag{7}$$

where $\frac{D\rho}{Dt} = \frac{\partial \rho}{\partial t} + \underline{u} \cdot \nabla \rho$ is the rate of change of density following the flow, in other words the rate of change of density of a fluid particle.

Equation (7) immediately tells us that an incompressible flow, for which $\frac{D\rho}{Dt} =0$, must satisfy

$$\nabla \cdot \underline{u} = 0 \mathrm{~~~~(incompressible~flow)} . \tag{8}$$

Equation (7) also gives us an example of a physical interpretation of the divergence: $\nabla \cdot \underline{u}$ must be the rate of increase of volume of a fluid element.

Fluid flow can also give us a physical example of the curl of a vector field. To see this, we will look at the local flow field in the immediate neighbourhood of a chosen point, which we can define to be $\underline{r}=0$. The first two terms of a Taylor expansion of the velocity field then give

$$u_i \approx u_i^{(0)} + \sum_j{c_{ij} r_j} \mathrm{~~~for~~}i=1,2,3 \tag{9}$$

where it is convenient to work in terms of components of the vectors for this particular analysis, so $\underline{u} = (u_1,u_2,u_3)$ and $\underline{r}=(r_1,r_2,r_3)$ (which are just another name for $x,y,z$). In equation (9), $u_i^{(0)}$ are the components of the velocity at the reference position $\underline{r}=0$, and the $3 \times 3$ matrix (or more technically the second-rank tensor)

$$c_{ij} = \dfrac{\partial u_i}{\partial r_j} \mathrm{~~~for~~}i,j=1,2,3 . \tag{10}$$

Now we separate $c_{ij}$ into the sum of its symmetric part and its antisymmetric part:

$$c_{ij} = \frac{1}{2}\left[c_{ij} + c_{ji} \right]+\frac{1}{2}\left[c_{ij} – c_{ji} \right] = e_{ij} + \omega_{ij} . \tag{11}$$

The antisymmetric term has a simple interpretation. By writing everything out in full and then comparing, it can be easily verified that

$$\sum_j{\omega_{ij} r_j}=\frac{1}{2}[\underline{\omega} \times \underline{r}]_i \mathrm{~~~for~~}i=1,2,3 \tag{12}$$

where the vector

$$\underline{\omega} = \nabla \times \underline{u} .\tag{13}$$

This vector is called the vorticity. Equation (12) describes rigid rotation, at angular velocity $\underline{\omega}/2$.

The effect of the symmetric term $e_{ij}$ (known technically as the rate of strain tensor) can also be described quite easily. This is a symmetric $3 \times 3$ matrix, so by a standard result from linear algebra it has three mutually perpendicular eigenvectors, and if we rotate our coordinate axes to those three directions, the matrix is diagonalised. This means that the associated term in equation (9) describes a flow which would turn a sphere into an ellipsoid, with axes aligned with those three eigenvectors.

We can summarise all this with a geometric description. If you were to inject a blob of dye to make a small sphere around the chosen reference point, the subsequent flow can be described as the sum of three components. The blob will rotate with an angular velocity $(1/2)\nabla \times \underline{u}$, its volume will grow at rate $\nabla \cdot \underline{u}$, and the remaining part of the flow will cause the sphere to distort into an ellipsoid.

We are nearly ready to apply Newton’s law to our flowing fluid, and deduce the governing equation. But first we need to relate the rate of strain tensor $e_{ij}$ to the internal forces, which are described by another $3 \times 3$ matrix, the stress tensor $\sigma_{ij}$. The definition of this stress tensor is based on a hypothetical measurement. Suppose we imagine a small element of area somewhere within the flow, described by a vector area $\underline{dA}$. If we could somehow embed a force-measuring sensor in this little piece of area, it would find that there was a force acting across the two faces. That force could in principle be in any direction: for example the effect of pressure in the fluid would give a force normal to the surface, parallel to the vector $\underline{dA}$, whereas a shearing flow in the presence of viscosity would give a shear force, acting parallel to the surface and thus perpendicular to $\underline{dA}$. If this force has components $(p_1,p_2,p_3)$ per unit area, the stress tensor is defined so that

$$p_i=\sum_j{\sigma_{ij} dA_j}\mathrm{~~~for~~}i=1,2,3 . \tag{14}$$

If the fluid is at rest, the only internal force acting is pressure. At any given point, the pressure always acts normal to any surface element $\underline{dA}$, and the value of that force is the same, whatever the orientation of the surface. This means that the stress tensor takes a very simple form:

$$e_{ij}=-p \delta_{ij} \tag{15}$$

where $\delta_{ij}$ is the mathematician’s standard name for the identity matrix:

$$\delta_{ij} = \left\lbrace \begin{matrix} {1 \mathrm{~~if~~} i=j}\\{0 \mathrm{~~if~~} i\ne j}\end{matrix} \right. \tag{16}$$

When the fluid is flowing, we can generalise this result and define the pressure by

$$p=-\dfrac{1}{3} \sum_{i=1}^3{e_{ii}} \tag{17}$$

Now we can think about Newton’s law. The acceleration of a fluid particle is given by $\frac{D\underline{u}}{Dt}$: obviously we need the rate of change of velocity following the particle. To combine this with what we know about the internal forces, it is clearest to apply Newton’s law to a region of the flow field, rather than to a small element. So suppose we have a volume $V$ within the flow, with a closed surface $S$ forming its outside “skin”. The result, for the $i\mathrm{th}$ component, is

$$\int{\int{\int_V{\rho \dfrac{Du_i}{Dt} dV}}}=\int{\int{\int_V{\rho f_i dV}}}+\int{\int_S{\sum_j{\sigma_{ij} dA_j }}} \tag{18}$$

where $f_i$ is the $i\mathrm{th}$ component of any external force (such as gravity) that may be acting throughout the body of our fluid. It is expressed here as a force per unit mass, so that if it is in fact the gravitational force we would have $\underline{f}=\underline{g}$ in terms of the usual gravitational acceleration $\underline{g}$. Now we can apply a version of the divergence theorem to the final term, to convert it from a surface integral to a volume integral:

$$\int{\int{\int_V{\rho \dfrac{Du_i}{Dt} dV}}}=\int{\int{\int_V{\rho f_i dV}}}+\int{\int{\int_V{\sum_j{\dfrac{\partial \sigma_{ij}}{\partial r_j} dV }}}} . \tag{19}$$

Now we have a sum of three volume integrals equal to zero, but the volume over which we have integrated is entirely arbitrary. The only way this result could be true for every choice of volume $V$ would be if

$$\rho \dfrac{Du_i}{Dt}=\rho f_i+\sum_j{\dfrac{\partial \sigma_{ij}}{\partial r_j}}\mathrm{~~~for~~}i=1,2,3 \tag{20}$$

everywhere. This is the governing equation we want, but it still remains to express $\sigma_{ij}$ in terms of $e_{ij}$, which is turn depends on $\underline{u}$.

From here on I will only present the simplest case, when the flow is incompressible so that equation (8) is satisfied. We want a relation between $e_{ij}$ and $\sigma_{ij}$. If we make the assumptions (i) that this relation is linear; and (ii) that the physical properties of the fluid are isotropic (in other words without any preferred directions), then a standard result from tensor algebra tells us what the form of the relationship must be:

$$\sigma_{ij} = -p \delta_{ij} + 2 \mu e_{ij} \tag{21}$$

where $\mu$ is a constant, the viscosity. Substituting and tidying up leads to the equation

$$\rho \dfrac{D \underline{u}}{Dt}= -\nabla p \mathrm{~} – \rho \underline{f}+\mu \nabla^2 \underline{u}. \tag{22}$$

This is the Navier-Stokes equation for the case of incompressible flow. Notice that the only nonlinearity here is the term $\underline{u} \cdot \nabla \underline{u}$ hidden within the $D/Dt$ term.

The next interesting thing we can deduce is called Bernoulli’s principle. Again, I will just present the very simplest version of this, which is enough to give the flavour. Suppose we have a flow which is steady, incompressible and inviscid ($\mu=0$), with density the same everywhere. Our Navier-Stokes equation then reads

$$\rho \underline{u} \cdot \nabla \underline{u}= -\nabla p \mathrm{~} – \rho \underline{f} . \tag{23}$$

If the body force is gravitational, then it can be written as the gradient of the gravitational potential $gz$ where $g$ is the usual gravitational acceleration and $z$ is the vertical coordinate, measured upwards. Now note that

$$\underline{u} \cdot \nabla \underline{u} = \frac{1}{2} \nabla (\underline{u} \cdot \underline{u}) . \tag{24}$$

Every term in the equation is now a gradient, and it can be rearranged into

$$\nabla \left(\dfrac{q^2}{2}+\dfrac{p}{\rho}+gz\right) = 0 \tag{25}$$

where $q^2=\underline{u} \cdot \underline{u}$ is the square of the flow speed. It follows that

$$\dfrac{q^2}{2}+\dfrac{p}{\rho}+gz\ = \mathrm{~constant.} \tag{26}$$

We can see explanations of more than one phenomenon in this simple equation. First, it tells us about the increase in static pressure with depth, for example under water. If the water is not flowing, so $q=0$, it tells that $p=K \rho-\rho g z$. Recalling that we have defined $z$ to be positive upwards, if a diver goes down into the sea they will encounter pressure rising with depth, just as we expect. This is called the hydrostatic pressure, equal to $\rho g |z|$. More directly relevant to our interest in wind instruments, if the gravity term is insignificant the equation tells us that as flow speed $q$ increases, so the pressure must drop by an amount proportional to $q^2$. This will be significant when we look at the excitation of reed and brass instruments.

For completeness, it is useful to summarise (without derivations) some extensions of Bernoulli’s principle with less restrictive assumptions. If the fluid is not incompressible, the next simplest assumption is that it is barotropic: this means that the density is an instantaneous function of the pressure, $\rho(p)$. In that case, we define the enthalpy

$$I(p)=\int{\dfrac{dp’}{\rho(p’)}} \tag{27}$$

and then Bernoulli’s principle for steady irrotational flow (i.e. $\nabla \times \underline{u}=0$) says that

$$\dfrac{q^2}{2}+I(p)+gz = \mathrm{~constant.} \tag{28}$$

Next, if we have steady flow with vorticity, we define the quantity

$$H=\dfrac{q^2}{2}+I(p)+gz , \tag{29}$$

and then Bernoulli’s principle states that $\nabla H$ is perpendicular to both $\underline{u}$ and $\underline{\omega} = \nabla \times \underline{u}$, in other words $H$ is constant along both streamlines and vortex lines.

Finally, we can say a little about vorticity and how it evolves. If we take the curl of equation (22) and then make use of some vector identities, we can manipulate it into the form

$$\dfrac{D \underline{\omega}}{Dt}=\underline{\omega} \cdot \nabla\underline{u} + \nu \nabla^2 \underline{\omega} \tag{30}$$

where $\nu=\mu/\rho$ is called the “kinematic viscosity”. This equation tells us about two rather different aspects of vorticity dynamics. First, in the inviscid case ($\nu=0$), it reveals a powerful and unexpected result. If we think about a small “thread” of material, on a line element $\underline{\delta l}$, the analysis leading to equation (9) tells us that this material element evolves according to

$$\dfrac{D \underline{\delta l}}{Dt}=\underline{\delta l} \cdot \nabla\underline{u} . \tag{31}$$

This is exactly the same equation that governs the evolution of vorticity $\underline{\omega}$ in an inviscid flow field. So if we choose a material line element that lies along a vortex line, we see that the vorticity must evolve with the flow. As it is often stated, “vortex lines are frozen in the fluid”. So, for example, if the material line gets stretched by the flow, the vorticity will also increase in the same proportion. This is part of the mechanism by which tropical storms and tornadoes can “spin up”: the rising of hot air stretches the vortex lines, creating faster rotation.

This result tells us something important. For any flow started from rest, there must initially be no vorticity anywhere. If the fluid is inviscid, it appears that this must remain true in the subsequent flow. Indeed, there is a result called the Kelvin circulation theorem for an inviscid fluid. If you consider a closed material curve $C$, the integral of the velocity round $C$, called the circulation $K$, is constant in time:

$$K=\oint_C{\underline{u} \cdot \underline{dl}} . \tag{32}$$

So how does circulation or vorticity ever arise in a flow? We can see one route for this to happen if we restore the viscosity term of equation (30), and note that the equation then has the character of a diffusion equation. If the first term on the right-hand side were omitted, it would have exactly the same form as the equation governing heat diffusion in a solid body, for example.

So viscosity causes vorticity to diffuse through the fluid. An important example of this is illustrated in Fig. 2. As explained in section 11.2, when fluid flows over a fixed surface the no-slip boundary condition on the surface leads to the formation of a boundary layer, with a flow profile similar to the one shown in the figure. The figure shows a small material element in the boundary layer. It is subjected to shearing flow, which automatically means that the element has some net rotation — in other words, it has non-zero vorticity. So boundary layers on solid surfaces generate vorticity, and equation (30) indicates that vorticity may then diffuse out into the flow pattern.

There is another, more vigorous, way that non-zero circulation can appear in a flow. It still involves boundary layers, but rather than diffusion, the process involves vortex shedding. It will be relevant to several aspects of wind instruments, but I will introduce the idea via a different application: how do aeroplanes fly? I will illustrate, with a series of hand-drawn sketches of streamline patterns (which should not be regarded as more than qualitative).

Figure 3 shows a schematic airfoil cross-section. When the aircraft first starts to taxi down the runway, the flow pattern in the initial moments may look a little like this sketch. There are two stagnation points, marked by red rings: the streamlines joined to these points are the ones that separate flow that passes above the wing from flow that passes below it.

But there is a big problem with this flow. The trailing edge of the wing has a sharp edge, marked by the green ring. The flow speed close to this sharp corner will get very high. Bernoulli then tells us that the pressure must get very low. If this was a propellor blade and the fluid was water, this would be the mechanism of cavitation: when the pressure drops low enough to cancel the hydrostatic pressure, bubbles can be seen streaming from the propellor blades.

But for the aircraft wing, something different happens, which is indicated very schematically in Fig. 4. The boundary layer separates from the surface and forms a separation bubble. This bubble then separates entirely from the wing, and becomes an independent vortex with a circulation direction that is anticlockwise in the diagram. In order to preserve the total circulation, and thus satisfy Kelvin’s theorem, there is equal and opposite net circulation around the wing.

The result is as shown (schematically again) in Fig. 5. The vortex which was shed is “left behind on the airfield”. The wing then has just the right amount of clockwise circulation (indicated by the red arrow) that the rear stagnation point is shifted exactly to the sharp trailing edge, as sketched. This eliminates the need for very fast local flow, because (as the name suggests) the flow speed goes to zero at a stagnation point. The effect of the clockwise circulation round the wing is that the air flowing over the top surface is speeded up, while the air flowing over the bottom surface is slowed down. Bernoulli then tells us that the pressure is lower on the top surface than on the bottom surface, and it is this difference of pressures that lifts the aircraft off the ground.

Any flow interacting with a sharp corner is likely to lead to vortex shedding, and this is relevant to many problems, including some involving musical instruments. Examples might be air flow in and out of tone-holes, or flow in and out of the f-holes of a violin near the Helmholtz resonance frequency. More immediately important for this chapter, the excitation mechanism of flute-like instruments involves these ingredients, as we will see in section 11.6.