Chapter 2 Mathematical Modelling of Fluid Flow

Consider a thin layer of material sandwiched between two plane walls, as shown in Figure 2.1. Under the action of the applied traction, $\sigma$, each layer of the material slides proportional to the distance from the lower wall, which is held fixed. The displacement is given by $\xi =\gamma (t)y$. The shear in this scenario is $d\xi /dy=\gamma (t)$. At long times, $\gamma (t)$ approaches a constant for a solid. But for a fluid, $\gamma (t)$ keeps increasing at a constant rate, i.e., $\gamma (t)=\dot{\gamma }t$, where $\dot{\gamma }$ is the rate of shear.

There are other definitions of a fluid that, while appropriate to be used at more elementary treatments of fluid dynamics, are inadequate for the purposes of deriving mathematical models for describing their behaviour. One such definition is that a fluid is a substance that assumes the shape of its container. This definition is inadequate for our purpose because there is no mention of gravity, the driving force that makes some of those substances, namely liquids, conform to the shapes of their containers. But even for gases, this definition does not furnish the defining element instrumental in deriving the governing equations for fluid flow – the internal stress experienced by elements of the fluid and the resulting deformation.

While our definition addresses some of the shortcomings of the more elementary definitions, it has left some terms undefined. For instance, what is a continuum substance? What is a shear stress? How does one quantify continual deformation? Also note that the definition does not constrain the magnitudes of the shear stress or that of the deformation, nor does it specify any relation between the two.

Based on this guiding definition, we derive mathematical representations of the laws of conservation of mass, momentum and angular momentum in §2.8, §2.9 and §2.10, respectively. The concepts of deformation rate and stress will emerge in the process. A fluid is any substance that obeys the assumptions that go into the derivations of these mathematical representations. On the one hand, such an exercise rigorously constrains the materials to which the governing equations could be applied. And on the other hand it broadens the scope to include materials that are not commonly classified as fluids but whose mechanics are governed by the equations we derive because they satisfy the underlying assumptions. Understanding the significance of the assumptions that form the basis of the derviation of the governing equations is the objective of this chapter.

Matter on earth is made up of atoms and molecules, each obeying classical or quantum laws of physics. When dealing with a large collection of molecules, the granularity of molecular composition somehow is lost and/or becomes irrelevant. Properties of matter can then be considered to vary smoothly with location within the material. The emergence of thermodynamics is an example of this. Similarly, when describing motion of macroscopic matter on length scales much larger than molecular scales, it becomes possible to consider the material as a infinitely divisible continuum. The theory of mechanics that arises out of such a consideration is called continuum mechanics.

The continuum approach is accurate when the mean free path $\mathrm{\ell }$, which is the length a molecule of the medium travels between collisions with other molecules, is much smaller than the dimensions of the flow $L$, so that $\mathrm{\ell }/L\ll 1$ (the continuum limit takes $\mathrm{\ell }/L\to 0$). Note that $\mathrm{\ell }/L$ is a ‘dimensionless number or parameter’, whose value does not depend on the system of units used. When $\mathrm{\ell }/L\approx 1$ or $\mathrm{\ell }\gg L$, microscopic methods (molecular dynamics for liquids and kinetic theory for gases) are required.

It is worthwhile to ponder briefly about cases where continuum hypothesis fails. Here are two situations:

1. 1.

   Flow of water through a carbon nanotube: Nanotubes are modern materials consisting of a tube of diameter as small as 0.4 nm. Figure 2.2 shows the use of nanotubes made out of carbon for desalinating water. The basic idea is that while water can flow through the tube, the impurities do not. In this case, the water near the centre of the tube flows faster compared to that near the tube walls, where friction impedes the flow. Thus, the water speed changes over a distance compared to the diameter of the tube, so $L\approx 0.4$ nm. The molecular mean-free path is about 0.25 nm. In this case, $\mathrm{\ell }/L\approx 0.25/0.4=0.625$, which is not small. Therefore, the continuum hypothesis is likely to fail.

   

   

2. 2.

   Motion of the International Space Station (ISS): The ISS is about 79 m $\times$ 109 m, say approximately $L=100$ m across and orbits the earth at an altitude of 400 km. The atmosphere there is so rarified that the mean free path of gas molecules is $\mathrm{\ell }\approx 20$ km. The ratio $\mathrm{\ell }/L=20\times {10}^3/100=200$ is clearly not small, so the surrounding gas cannot be treated as a continuum.

In summary, the continuum approximation of the fluid may fail because (i) the length-scale over which the flow varies is too small, or (ii) the mean free path of the fluid molecules is too large. Both these cases are adequately encapsulated by the ratio $\mathrm{\ell }/L$.

In day-to-day situation, e.g. flow of water through ordinary pipes, continuum approximation does apply. In that case, every point in space is assigned physical properties such as density, pressure, velocity, stress, etc. Continuous functions of space and time, called fields, are used to represent these physical variables.

There are five main properties that will be important for us in this module.

1. 1.

   Density ($\rho$): The mass of the fluid per unit volume. Air has a density of about 1.2 kg/m${}^3$ and water of 1000 kg/m${}^3$. Physically, the density of a fluid is determined by its thermodynamic state, and could thus change with pressure and temperature, for instance. If the conditions are such that the density of the fluid does not change appreciably in the flow then such a flow is termed “incompressible”. Under ordinary circumstances, flows of air and water are incompressible. For flow of air around a jet plane, the flow is compressible. And it is certainly so for flow around a supersonic plane.

2. 2.

   Pressure ($p$): The force a fluid exerts on a wall per unit area. At sea level, air exerts a pressure of about ${10}^5$ N/m${}^2$. In SI system, N/m${}^2$ = Pa, a Pascal. Other common units of pressure, especially in atmospheric sciences, are the bar, 1 bar = ${10}^5$ Pa, and the atmosphere, 1 atm = 101325 Pa. Pressure is also measured in terms of a height of a fluid column, where the fluid is commonly but not necessarily either water or mercury. Normal human blood pressure is purported to range from 80 to 120 mm of mercury.

3. 3.

   Dynamic Viscosity ($\mu$): Also known as Newtonian viscosity. It represents the resistance of a fluid to shear. The viscosity is the ratio of the traction applied to the shear rate generated in steady state. In the terminology used in Figure 2.1, $\mu =\sigma /\dot{\gamma }$. Dynamic viscosity has the SI units of Ns/m${}^2$ or, equivalently, Pa s. An “ideal” fluid or an “inviscid” fluid is one with zero viscosity.

4. 4.

   Kinematic viscosity ($\nu$): The kinematic viscosity is the dynamic viscosity divided by fluid density, i.e. $\nu =\mu /\rho$. In the SI system, it has units of m${}^2$/s (note the absence of kg in the units, hence the epithet kinematic).

5. 5.

   Surface tension ($\sigma$): The surface of a liquid behaves like a stretched membrane. The tension in this “membrane”, measured as force per unit length, is called surface tension. Surface tension has SI units of N/m.

The conditions termed as incompressible and inviscid or ideal are idealizations of real conditions. They are never exactly realized.

Watch the video on ‘Lagrangian and Eulerian descriptions of fluid flow’ narrated by Prof. John Lumley of Cornell University from the National Committee for Fluid Mechanics. YouTube link: www.youtube.com/watch?v=mdN8OOkx2ko, duration: 27 mins.

The objective in developing the continuum theory of fluids is to decompose the fluid into an infinite number of infinitesimal fluid elements and impose Newton’s laws of motion on each one of them. To do so, it is necessary to label each infinitesimal element of a fluid. There are two common approaches historically employed for doing so. These approaches are called the Eulerian and the Lagrangian descriptions of fluid flow.

Watch the video on ‘Flow Visualisation’ narrated by Prof. Stephen J. Kline from Stanford University. YouTube link: www.youtube.com/watch?v=nuQyKGuXJOs, duration: 31 mins.

The most common way to visualise a flow, theoretically (once we have $𝒖$) or experimentally, is by drawing a set of curves. Four types of curves are commonly used in fluid dynamics. Some are purely theoretical concepts, while others are used for experimental convenience.

To visualise a flow field experimentally, one usually places illuminated passive tracers (e.g. dye, bubbles or beads that don’t alter the flow) into the flow (e.g. Figure 2.7). The various curves are then realized by the choice of the initial release of the tracer particles and the visualization technique.

1. 1.

   Particle paths or pathlines: The paths of fluid elements (‘particle paths’) $F_i(t)$, which are simply the locus of points traced by a Lagrangian fluid element. Equation (2.1) may be used to theoretically determine these from an Eulerian description of the velocity field. Typically a small amount of illuminated passive tracers are introduced in the flow, and the resulting images superimposed in post-processing to reveal the pathlines for a flow. Long exposures, if possible, also yield pathlines.

2. 2.

   Streaklines: The instantaneous locus of all the Lagrangian fluid elements that have passed through or will pass through a fixed point in space is a ‘streakline’. It is easier to visualize streaklines experimentally, which can be done by introducing a steady streak of a passive tracer in the fluid (hence the name). Theoretically, streakline at time $t$ passing through a point ${𝑿}_0$ may be obtained by solving the equation

   $𝑭(𝑿,s)={𝑿}_0,t_i\le s\le t_f,$

   (2.9)

   for $𝑿(s)$, where $t_i$ and $t_f$ are the initial and final times of interest, and then plotting $𝑭(𝑿(s),t)$ for fixed $t$ while varying $s$ as a parameter.

3. 3.

   Timelines: A ‘time line’ is the subsequent shape of a curve made of Lagrangian fluid elements. It is more convenient to visualize a timeline experimentally than it is to calculate it theoretically. For the experimental visualization, a passive tracer is released at a time instance in the shape of a curve. The subsequent evolution of the curve naturally visualizes the timeline. To construct the timeline from an Eulerian velocity field, we start with a curve parameterized as ${𝑿}_0(s)$ by parameter $s$. The subsequent shape of the curve at time $t$ is $𝑭({𝑿}_0(s),t)$, which is the timeline at time $t$.

4. 4.

   Streamlines: A ‘streamline’ is a curve which is everywhere tangent to the velocity vector of an Eulerian flow field. Streamlines are mathematical constructions, and, therefore, are easier to construct theoretically. Parameterized by an independent parameter $s$, the streamlines $(x,y,z)=(x(s),y(s),z(s))$ at time $t$ are obtained by solving

   $${\frac{\partial x_i(s)}{\partial s}|}_{\text{fixed }t}=u_i.$$

   (2.10)

   To visualize streamlines in an experiment, it is necessary to measure the Eulerian velocity field. This measurement can be accomplished using techniques such as particle image velocimetry (PIV).

   Streamlines, although theoretical, have deep physical significance, especially for incompressible flows. Because the volume of a Lagrangian fluid element is conserved for an incompressible fluid, the only way a fluid element can move is by displacing the one ahead of it. At the same instance, the one ahead in turn displaces the one ahead of it, and so on. An incompressible fluid can flow only by forming such chains of instantaneous displacement of fluid elements. A streamline is nothing but one such chain.

   Because of the aforementioned interpretation, streamlines cannot terminate in an incompressible fluid. They must extend all the way to infinity or form closed curves.

In general, the particle paths differ from the streamlines, which differ from streaklines. The following example illustrates that.

The pathlines are obtained from solving

$$\frac{\partial F_1}{\partial t}=u_0,\frac{\partial F_2}{\partial t}=kt\frac{\partial F_3}{\partial t}=0$$

for fixed initial position $(F_1,F_2,F_3)=(X,Y,Z)$ at $t=0$. Therefore,

$$(F_1,F_2,F_3)=(u_{0}t+X,\frac{1}{2}k{t}^2+Y,Z)$$

which, notably, could be used to transform from Eulerian to Lagrangian coordinates (by expressing $(x,y,z)$ in terms of $(X,Y,Z))$. We can eliminate $t$ to find the particle paths

$$y=\frac{k}{2}{\left(\frac{x-X}{u_0}\right)}^2+Y$$

which are parabolas with their minimum at $(X,Y)$ and $k/2u_0^2$ as the coefficient of the quadratic term (see figure 2.8 for the pathline through the origin).

The streamlines come from

$$\frac{\partial x}{\partial s}=u_0,\frac{\partial y}{\partial s}=kt,\frac{\partial z}{\partial s}=0$$

for a fixed $t$. Therefore

$$(x,y,z)=(u_{0}s+a,kts+b,c)$$

for $a,b,c$ constant. Eliminating $s$ we find that

$$y=\frac{kt}{u_0}x+(b-kta/u_0)$$

showing that the streamlines are straight lines whose slope will increase in time. The result is figure 2.8.

The streaklines are also easy to calculate. Upon solving equation (2.9), we get for $𝑿(s)$

$$X(s)=X_0-us,Y(s)=Y_0-\frac{1}{2}k{s}^2,Z(s)=Z_0.$$

The equation for the curve for the streakline is $𝑭(𝑿(s),t)$, which is

$$x=X_0-us+ut,y=Y_0-\frac{1}{2}k{s}^2+\frac{1}{2}k{t}^2,z=Z_0.$$

These are to be plotted for fixed $t$ while varying $s$ as a parameter, which means that they are inverted parabolas through $(X_0+ut,Y_0+k{t}^2/2,Z_0)$. The one through the origin is plotted in figure 2.8.

The study of the deformation of continuous media (including solid and fluid dynamics) is a course in its own right, so we will only consider the concepts we will require to formulate the equations of fluid mechanics. In particular, we will attempt to understand how flow deforms fluid elements; as it is this (rate of) deformation that is related to the internal stress (and hence forces) required for formulating Newtons second law for the fluid.

The velocity can be decomposed into fundamental components which specify the kinematics of the flow (the geometry of the motion), i.e. how small fluid elements are deformed by the flow. To do so, consider the flow at an infinitesimal distance $\delta x_j$ from a reference point $x_j$, see figure 2.9.

The Taylor expansion of $u_i(x_j+\delta x_j)$, keeping only the terms up to the first power in $\delta x_j$ inclusive, gives

$$u_i(x_j+\delta x_j,t)=u_i(x_j,t)+\frac{\partial u_i}{\partial x_j}\delta x_j=\underset{\text{Rigid Body Motions}}{\underbrace{\underset{\text{Translation}}{\underbrace{u_i(x_j)}}+\underset{\text{Rotation}}{\underbrace{r_{ij}\delta x_j}}}}+\underset{\text{Shearing and Extension}}{\underbrace{e_{ij}\delta x_j}}$$

(2.11)

where we have decomposed ${\partial }_{x_j}{u}_i$ into an anti-symmetric ($r_{ij}=-r_{ji}$) rate of rotation tensor

$$r_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i}\right)\text{or in vector notation}𝒓=\frac{1}{2}\left(\mathbf{\nabla }𝒖-\mathbf{\nabla }{𝒖}^T\right),$$

and the symmetric ($e_{ij}=e_{ji}$) rate of strain tensor

$$e_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)\text{or in vector notation}𝒆=\frac{1}{2}\left(\mathbf{\nabla }𝒖+\mathbf{\nabla }{𝒖}^T\right).$$

Note that $\mathbf{\nabla }𝒖=𝒆+𝒓$.

Each of the terms contributing to the velocity

$$u_i(x_j+\delta x_j,t)=u_i^T+u_i^R+u_i^S,\text{where}{u}_i^T=u_i(x_j),u_i^R=r_{ij}\delta x_j,u_i^S=e_{ij}\delta x_j$$

(2.12)

can now be analysed. The first term in (2.11) is due to rigid body translation $u_i^T=u_i(x_j,t)$. If $r_{ij}=e_{ij}=0$, then all fluid elements have the same velocity.

Due to the antisymmetry of $r_{ij}$, there are only 3 independent components (all diagonal elements zero) and we can rewrite the term $u_i^R=r_{ij}\delta x_j$ as a cross product $u_i^R={ϵ}_{ijk}{\mathrm{\Omega }}_j\delta x_k$ or ${𝐮}^R=𝛀\times \delta 𝐱$ (see figure 2.11), where $𝛀=(r_{32},r_{13},r_{21})$ is the rate of rotation vector (you can check this). Thus we see that this term is associated with rigid body rotation about the reference point with rotational rate $𝛀$.

Notably, the rate of rotation vector $𝛀$ is equal to half the vorticity vector

$$𝝎=\nabla \times 𝐮=2𝛀,\text{so that}{𝐮}^R=\frac{1}{2}𝝎\times \delta 𝐱$$

(2.13)

which is an important fluid mechanical quantity we will repeatedly encounter - now we can recognise it as a measure of the local rate of rotation in the fluid. In Cartesian coordinates it is given by

For a two-dimensional flow $𝐮=(u(x,y,t),v(x,y,t),0)$, the vorticity has only one component $(0,0,\omega )$ where $\omega =\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}$, with a rotation axis in the $z$-direction.

As the first two terms in (2.11) are associated with rigid body motion, it is the third term $u_i^S=e_{ij}\delta x_j$ which represents the straining motions which distinguish the flow through the relative motion of adjacent fluid elements. The rate of strain tensor $e_{ij}$ can be further decomposed into (a) diagonal elements which are associated with extensional motion and (b) off-diagonal elements give the shear rate of strain.

Consider how the components of $e_{ij}$ deform an infinitesimal rectangular fluid element of size $\delta x\times \delta y$ (figure 2.12) in a two-dimensional flow $𝐮=(u(x,y,t),v(x,y,t),0)$. For simplicity, we will remove translation of the fluid element by considering flow relative to the point $A$, so that a Taylor expansion gives the flow components at adjacent points to $A$ shown in figure 2.12.

From figure (2.13) we can see that extensional rate of strain acts to increase the lengths of the sides. For example, taking the line $AB$, in a time $\delta t$ the point B moves so that the new length of $AB$ at time $t+\delta t$ is

$$\delta x(t+\delta t)=\delta x(t)+\underset{\text{speed}}{\underbrace{\frac{\partial u}{\partial x}\delta x}}\underset{\text{time}}{\underbrace{\delta t}}.$$

(2.14)

Therefore, the (infinitesimal) strain (extension/original length)

$$\delta S_x=(\delta x(t+\delta t)-\delta x(t))/\delta x(t)\stackrel{(\mathit{\text{<a href="\#E14" title="2.14 ‣ 2.6 Deformation of infinitesimal fluid elements ‣ Chapter 2 Mathematical Modelling of Fluid Flow ‣ MA3D1 – Fluid Dynamics and MA4L0 – Advanced Topics in Fluids Lecture notes" class="ltx\_ref ltx\_font\_italic" style="font-size:70\%;"><span class="ltx\_text ltx\_ref\_tag">2.14</span></a>}})}{=}\frac{\partial u}{\partial x}\delta t\text{so the rate of strain is}\frac{\partial S_x}{\partial t}=\frac{\partial u}{\partial x}=e_{11}$$

as expected.

It can be shown that the extensional components change the volume $V$ of the fluid element as

$$\frac{1}{V}\frac{DV}{Dt}=e_{kk}\text{symbolically}\frac{1}{V}\frac{DV}{Dt}=\nabla \cdot 𝐮$$

(2.15)

so that for an incompressible fluid, where the volume of the fluid element does not change in time ($D_{t}V=0$), we have $e_{kk}=e_{11}+e_{22}+e_{33}=0$. In other words, extension of a fluid element in one direction must lead to contraction in one of the other directions.

The shear rate of strain acts to drive the element from its rectangular shape. This can be quantified by measuring the angle between the sides. It can be shown that (see Examples Sheet 2) the angle $\theta$ at $CAB$ in figure 2.12 evolves according to

$$-\frac{\partial \theta }{\partial t}=2e_{12}.$$

So we have seen that it is the vorticity and rate of strain that give us all of the information we require to understand how fluid elements are locally deformed, with the former telling us about the rotation of the flow and the latter about the relative motion of fluid elements (which is related to viscous stress).

Let’s consider all the different contributions to (2.12).

• •

  The translational term is just $u_i^T=(\beta y,0,0)$.

• •

  The vorticity will only have a component in the $z$-direction $𝝎=(0,0,\omega )$ which is

  $$\omega =\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=-\beta .$$

  So there is rotation in the flow, despite the streamlines and particle paths being straight. This is because the velocity at the point $C$ is greater than that at $A$ so the line $CA$ rotates clockwise (figure 2.14). The rate of rotation will be $\omega /2=-\beta /2$.

  Therefore, the contribution to (2.12) is

  $$u_i^R=\frac{1}{2}{ϵ}_{ijk}{\omega }_j\delta x_k=\frac{1}{2}(\beta dy,-\beta dx,0)$$

• •

  The only non-zero entries in the rate of strain tensor will be due to ${\partial }_{y}u=\beta$. Therefore, extensional strains are all zero (ensuring incompressibility) and the only non-zero entries are $e_{12}=e_{21}=\beta /2$. Consequently, the contribution to (2.12) is

  $$u_i^S=\frac{1}{2}(\beta dy,\beta dx,0).$$

• •

  Therefore, even in this simple flow there are rigid body motions of translation and rotation as well as deformation.

In continuum mechanics, it is customary to refer to laws of nature as conservation laws. An obvious example is the law of conservation of mass. As the fluid flows, mass can neither be created nor could it disappear. A less obvious one is Newton’s second law of motion. While momentum is not “conserved”, the law constrains the rate of change of momentum to be equal to force. Thus, by considering forces to be “sources” of momentum, one may still use the language of “conservation laws” to describe Newton’s second law. Examples of quantities to develop conservation laws for include mass, (linear) momentum, angular momentum (or moment of momentum), internal energy, total energy, and entropy. We will limit ourselves to mass, momentum and angular momentum.

Three main concepts need to be introduced to develop the conservation law for an abstract quantity, which we will call $B$. These are the volumetric density of $B$, the volumetric rate of generation (also known as the source) of $B$, and the rate of exchange of $B$ between neighbouring fluid volumes.

A concrete example of such a term is the action of fluid pressure to generate force, or rate of change of momentum, on a volume $\mathrm{\Omega }$. Due to the pressure, $h^B=-p\widehat{𝒏}$ because the net force due to pressure is

$${\int }_{\partial \mathrm{\Omega }}-p\widehat{𝒏}dA.$$

Generally, the strength of such a surface source term is proportional to area of the bounding surface, rather than volume as for the volumetric source term. In general, the contribution to the rate of change of $B$ due to the surface source is written as

$${\int }_{\partial \mathrm{\Omega }}{h}^B𝑑A.$$

The tensorial rank of $h^B$ and of $b$ are equal. Clearly, $h^B$ must also be a field, i.e. $h^B(𝒙,t)$. One important way $h^B$ differs from both $b$ and $q^B$ is that in addition to being a field, $h^B$ also has dependency on the orientation of the surface $\partial \mathrm{\Omega }$. In other words, if a point $𝑿$ is on the boundary of two different volumes, $h^B$ can have different values depending on the unit outward normal to their bounding surfaces. This is represented through an explicit dependence on the unit normal as $h_B(𝒙,t;\widehat{𝒏})$.

The conservation law for $B$ may now be written as

${\displaystyle \frac{D}{Dt}}{\displaystyle {\int }_{\mathrm{\Omega }}}b𝑑\mathrm{\Omega }={\displaystyle {\int }_{\mathrm{\Omega }}}{q}^B𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\partial \mathrm{\Omega }}}{h}^B𝑑A.$

(2.16)

Now we use the (classical) physical law that mass can neither be created or destroyed. This implies, when $b=\rho$, then $q^B=T_j^B=0$, and (2.22-2.24) becomes

	${\displaystyle \frac{D}{Dt}}{\displaystyle {\int }_{\mathrm{\Omega }}}\rho 𝑑\mathrm{\Omega }$	$=0,$		(2.26)
	${\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{\partial \rho }{\partial t}}𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\partial \mathrm{\Omega }}}\rho u_j{n}_j𝑑A$	$=0,$		(2.27)
	${\displaystyle {\int }_{\mathrm{\Omega }}}\left({\displaystyle \frac{\partial \rho }{\partial t}}+{\displaystyle \frac{\partial (\rho u_j)}{\partial x_j}}\right)𝑑\mathrm{\Omega }$	$=0.$		(2.28)

Equations (2.26-2.28) are integral representations of the law of conservation of mass. The differential representation corresponding to (2.25)

${\displaystyle \frac{\partial \rho }{\partial t}}+{\displaystyle \frac{\partial (\rho u_j)}{\partial x_j}}=0\text{or in vector notation}{\displaystyle \frac{\partial \rho }{\partial t}}+\mathbf{\nabla }\cdot (\rho 𝒖)=0.$

(2.29)

We will now interpret the terms in some of representations of the law of conservation of mass, starting with (2.27). The first term in this equation is the rate of change of mass in a fixed (i.e. Eulerian) volume $\mathrm{\Omega }$. The second term is the rate at which mass exits the volume due to the Eulerian velocity $𝒖$. Conservation of mass imples that the two rates must balance, which is the statement of (2.27).

Next we will discuss the terms in (2.29). The first term is the Eulerian rate-of-change of density the second is the divergence of the convective mass flux $\rho 𝒖$. Another form of (2.29), written using the expression

${\displaystyle \frac{\partial \rho }{\partial t}}+𝒖\cdot \mathbf{\nabla }\rho +\rho \mathbf{\nabla }\cdot 𝒖=0$

(2.30)

as

${\displaystyle \frac{1}{\rho }}{\displaystyle \frac{D\rho }{Dt}}=-\mathbf{\nabla }\cdot 𝒖=-{\displaystyle \frac{1}{V}}{\displaystyle \frac{DV}{Dt}},$

(2.31)

can be interpreted as the relative Lagrangian rate of change of density is balanced by the negative of the relative Lagrangian rate of change of volume. In other words, if volume of an infinitesimal Lagrangian element increases, its density must decrease.

If either the fluid or the flow is “incompressible”, that means that the density of infinitesimal Lagrangian fluid particles does not change with time, i.e., $D\rho /Dt=0$. This implies

$\mathbf{\nabla }\cdot 𝒖\equiv 0\text{ everywhere.}$

(2.32)

The density of momentum is $\rho 𝒖$, and its rate of change is a force as per Newton’s second law of motion. The volumetric source term is a force per unit volume that acts on the fluid and is termed a “body force”, e.g. the force of gravity. We will denote it by $q^B=\rho 𝒈$, where $𝒈$ is acceleration due to the body force. Here the density and the volumetric source of momentum are vectors. The second rank tensor resulting from the surface force $T^B=𝑻$ is the stress tensor, which when acting on an area element $dA$ with normal $n_j$ exerts the force $d{F}_i=T_{ij}{n}_{j}dA$.

The three different integral forms representing conservation of momentum are

	${\displaystyle \frac{D}{Dt}}{\displaystyle {\int }_{\mathrm{\Omega }}}\rho u_i𝑑\mathrm{\Omega }$	$={\displaystyle {\int }_{\mathrm{\Omega }}}\rho g_i𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\partial \mathrm{\Omega }}}{T}_{ij}{n}_j𝑑A={\displaystyle {\int }_{\mathrm{\Omega }}}\rho g_i𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{\partial T_{ij}}{\partial x_j}}𝑑\mathrm{\Omega }.$		(2.33)
	${\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{\partial (\rho u_i)}{\partial t}}𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\partial \mathrm{\Omega }}}\rho u_i{u}_j{n}_j𝑑A$	$={\displaystyle {\int }_{\mathrm{\Omega }}}\rho g_i𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\partial \mathrm{\Omega }}}{T}_{ij}{n}_j𝑑A={\displaystyle {\int }_{\mathrm{\Omega }}}\rho g_i𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{\partial T_{ij}}{\partial x_j}}𝑑\mathrm{\Omega },$		(2.34)
	${\displaystyle {\int }_{\mathrm{\Omega }}}\left({\displaystyle \frac{\partial (\rho u_i)}{\partial t}}+{\displaystyle \frac{\partial (\rho u_i{u}_j)}{\partial x_j}}\right)𝑑\mathrm{\Omega }$	$={\displaystyle {\int }_{\mathrm{\Omega }}}\rho g_i𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\partial \mathrm{\Omega }}}{T}_{ij}{n}_j𝑑A={\displaystyle {\int }_{\mathrm{\Omega }}}\rho g_i𝑑\mathrm{\Omega }+{\displaystyle {\int }_{\mathrm{\Omega }}}{\displaystyle \frac{\partial T_{ij}}{\partial x_j}}𝑑\mathrm{\Omega }.$		(2.35)

The differential form of momentum balance is

${\displaystyle \frac{\partial (\rho u_i)}{\partial t}}+{\displaystyle \frac{\partial (\rho u_i{u}_j)}{\partial x_j}}=\rho g_i+{\displaystyle \frac{\partial T_{ij}}{\partial x_j}}.$

(2.36)

The left-hand side of (2.36) is $D(\rho 𝒖)/Dt$ can be simplified further by using the product rule and mass balance (2.29) as

${\displaystyle \frac{\partial (\rho u_i)}{\partial t}}+{\displaystyle \frac{\partial (\rho u_i{u}_j)}{\partial x_j}}=u_i{\cancel{\left({\displaystyle \frac{\partial \rho }{\partial t}}+{\displaystyle \frac{\partial (\rho u_j)}{\partial x_j}}\right)}}^{0,\text{mass balance}}+\rho \left({\displaystyle \frac{\partial u_i}{\partial t}}+u_j{\displaystyle \frac{\partial u_i}{\partial x_j}}\right).$

(2.37)

This simplifies the differential form of momentum balance (2.36) to the Cauchy momentum balance equation as

$\rho {\displaystyle \frac{D{u}_i}{Dt}}\equiv \rho \left({\displaystyle \frac{\partial u_i}{\partial t}}+u_j{\displaystyle \frac{\partial u_i}{\partial x_j}}\right)=\rho g_i+{\displaystyle \frac{\partial T_{ij}}{\partial x_j}},\text{or}\rho {\displaystyle \frac{D𝒖}{Dt}}=\rho 𝒈+\mathbf{\nabla }\cdot 𝑻.$

(2.38)

The left-hand side of this equation is mass times acceleration per unit volume of an infinitesimal Lagrangian element. The right-hand side contains the net forces on this volume per unit volume. It is obvious that $\rho 𝒈$ is the body force per unit volume. To see that $\mathbf{\nabla }\cdot 𝑻$ is the unbalanced surface force per unit volume, consider forces acting on a infinitesimal cubic fluid element with side $d$ and bottom corner at $𝐱=(x_0,y_0,z_0)$; see figure 2.16. We will start with the $x$-component of this force which will involve us calculating $T_{1j}{n}_j\delta A$ (with $\delta A=d^2$) over all six sides of the cube.

	$\delta F_1=d^2$	$\left(T_{11}(x_0+d,y,z)-T_{11}(x_0,y,z)\right)$		(2.39)
	$+d^2$	$\left(T_{12}(x,y_0+d,z)-T_{12}(x,y_0,z)\right)$		(2.40)
	$+d^2$	$\left(T_{13}(x,y,z_0+d)-T_{13}(x,y,z_0)\right).$		(2.41)

Taylor expanding this expression in small $d$ and keeping the leading order only, we have:

$$\delta F_1=d^3\left[\frac{\partial T_{11}(x,y,z)}{\partial x}+\frac{\partial T_{12}(x,y,z)}{\partial y}+\frac{\partial T_{13}(x,y,z)}{\partial z}\right]=d^3\frac{\partial T_{1j}(x,y,z)}{\partial x_j}$$

where we have used the summation convention (with $j$ the repeated index).

Generalising to the other force components, we have:

$$\delta F_i=d^3\frac{\partial T_{ij}(x,y,z)}{\partial x_j}.$$

symbolically this is the divergence of the stress tensor $\delta 𝐅=d^3\nabla \cdot 𝐓$.

The angular momentum changes at a rate equal to the net external torque applied on a body. For angular momentum, $b=𝒙\times \rho 𝒖$, $q^B=𝒙\times \rho 𝒈$, and $h^B=𝒙\times (𝑻\cdot \widehat{𝒏})$. Note that the sources of angular momentum are related to the sources of linear momentum. We will go straight to the differential form of the angular momentum balance, which reads

${\displaystyle \frac{\partial ({ϵ}_{ijk}{x}_j\rho u_k)}{\partial t}}+{\displaystyle \frac{\partial ({ϵ}_{ijk}{x}_j\rho u_k{u}_m)}{\partial x_m}}={ϵ}_{ijk}{x}_j\rho g_k+{\displaystyle \frac{\partial ({ϵ}_{ijk}{x}_j{T}_{km})}{\partial x_m}}.$

(2.42)

Differentiating each term using the product rule yields

${ϵ}_{ijk}{x}_j\left({\displaystyle \frac{\partial (\rho u_k)}{\partial t}}+{\displaystyle \frac{\partial (\rho u_k{u}_m)}{\partial x_m}}-\rho g_k-{\displaystyle \frac{\partial (T_{km})}{\partial x_m}}\right)+{ϵ}_{ijk}\rho u_j{u}_k={ϵ}_{ijk}{T}_{jk}.$

(2.43)

The first term is the cross product of $𝒙$ with the linear momentum equation (2.36), and the second term is zero due to the anti-symmetry of ${ϵ}_{ijk}$ in the indices $j,k$ and the symmetry of $u_j{u}_k$. Thus the only remaining term is ${ϵ}_{ijk}{T}_{jk}=0$. A direct consequence of this is that $T_{jk}=T_{kj}$, or, in other words, $𝑻$ is a symmetric second-rank tensor.

The differential equations governing a continuum material in an Eulerian description that we have derived so far are mass balance (2.29) and momentum balance (2.38). These equations are partial differential equations for determining the density and velocity by integrating in time, i.e., if suitable initial conditions are available for $\rho$ and $𝒖$, and the volumetric forces $𝒈$ and stress $𝑻$ were known, then one may use these equations to find the rate of change of $\rho$ and $𝒖$, which one then uses to determine the subsequent evolution of $\rho$ and $𝒖$. The situation is similar to the use of Newton’s second law of motion for describing the motion of rigid bodies; knowing the external forces one uses it to integrate for velocity and position. Here we think about integrating for $\rho$ and $𝒖$. Let’s examine quantities we do not know yet; viz. $𝒈$ and $𝑻$.

The simplest one is $𝒈$: it must be prescribed as part of the problem.

The stress $𝑻$ needs to determined based on the fluid and the flow, so that the system of equations (2.29) and (2.38) is closed. The closure of this system is made by the constitutive law for the continuum material. This constitutive law is what distinguishes between a solid and a fluid, and for that matter, between different kinds of “complex” fluids. Next we present some considerations for a constitutive law.