Appendix C: Outline Notes on Fluid Mechanics¶

Abstract¶

The governing equations of motion of isothermal flow of an incompressible, viscous fluid are the continuity and Navier-Stokes equations:

Continuity:

Navier-Stokes:

which, together, provide four (horribly) coupled (horribly non-linear) partial differential equations in four unknown quantities - the components of the fluid velocity vector, \(\mathbf{v}\), and the scalar pressure, \(p\).

It is the purpose of these notes to provide you with the essential background to these four equations, which feature so much in our lectures and tutorials on Mesoscale Modelling. Hope this material helps.

The Continuity Equation and Flow Field Kinematics¶

The Continuity Equation¶

The central equation in mathematical fluid mechanics expresses the physical principle of conservation of matter. It is:

The above equation and a range of its variants are derived in this section.

Let \(\mathbf{v} (\mathbf{r}, t)\) denote the velocity field of a fluid.

The ‘increment of mass’, density \(\rho\), carried by flow, in unit time, through an area element \(d\mathbf{A} = dA \hat{n}\) is given by the expression:

Let \(M\) denote the mass of an enclosed volume of fluid, \(V_0\). Then, the mass of fluid flowing into our volume \(V_0\) enclosed by closed surface \(S\), in unit time, i.e. the rate of increase of the mass of volume \(V_0\), is obtained by integrating expression (1) over the surface \(S\):

where the negative sign arises because the unit normal vector, \(\hat{n}\), is conventionally taken to point outwards over a closed surface.

Now, the increase in unit time of the mass of fluid enclosed inside volume \(V_0\) is also obtained by the expression:

where \(d^3 \mathbf{r}\) is the appropriate volume element.

Physical conservation of material requires that, if volume contains no sources or sinks for fluid, expressions (2) and (3) must always be equal:

Taking the partial time derivative inside the integral and re-arranging:

and transforming the first term using the divergence theorem:

which has to be true for whatever volume we choose. The only way in which an integral can always vanish, for all choices of limits, is if the integrand is zero:

Equation (4) is the most fundamental equation of fluid mechanics. It is the continuity equation.

Example 1.1. An incompressible fluid (e.g. a liquid) is a melted solid. A typical volume expansion on melting is about 12%. So the atoms/molecules in a liquid are not much further apart than they were in the solid. Hence a liquid is about as resistant to compression as a solid. That is, pretty resistant! Show that, for an incompressible fluid the continuity equation (4) simplifies to:

The Material Derivative, Fluid Parcels and ‘Advection’¶

The material derivative is sometimes called the total derivative. The material derivative of a quantity \(f(\mathbf{r},t)\) associated with a moving parcel corresponds to the time rate of change of \(f(\mathbf{r},t)\), measured in the particle’s rest frame.

Consider a fluid particle - not an atom or molecule, but an infinitesimal volume element associated with the volume element \(d^3\mathbf{r}\): as the fluid particle advects (moves along in flow), its position and time change increments \(d\mathbf{r} = (dx, dy, dz)\) and \(dt\) respectively. The value of \(f(\mathbf{r},t)\) therefore changes, by an amount \(df\):

Dividing by time increment \(dt\) we obtain:

and noting that e.g. \(v_y = \frac{dy}{dt} = \frac{Dy}{Dt}\) it is immediately clear that:

in which the total time derivative, \(\frac{df}{dt}\), is the material derivative. In a more compact vector notation:

Tutorial Exercises¶

1. Show that the ‘velocity fields’:

1. \(\mathbf{v} = y {\hat{e}}_{x} ~~~~~ 0 \le y \le W\)

2. \(\mathbf{v} = y (W - y) {\hat{e}}_{x} ~~~~~ 0 \le y \le W\)

correspond to a legitimate incompressible flow. (In these examples, \(y=0\), \(y=W\) are the flow boundaries.)

Sketch the flow fields in i. and ii.

Inviscid Fluids¶

The Euler Equation/Momentum Equation¶

The equation of motion for a fluid without friction (viscosity) is the Euler equation. It is a vector partial differential equation: that is, it is three partial differential equations which may be written in a compact form:

We will now derive Euler’s equation(s). As we go, look out for the point at which a new variable is introduced into our description: the (scalar) fluid pressure.

Experience of diving to the bottom of a pool shows that even a fluid at rest contains a pressure gradient. Using Archimedes’ principle, the resultant force, \(\mathbf{F}\), acting on a finite volume of fluid (or, indeed, a submerged football) of total volume \(V_0\), is obtained by integrating the pressure in the fluid acting over the surface \(S\):

We need to get equation (7) into differential form. We do this using an adapted form of the famous Divergence theorem.

Exercise 1 Use the Divergence theorem on vector \(\phi (\mathbf{r}) \hat{a}\), where \(\hat{a}\) is constant, to obtain the following identity:

Using (7) and (8), we may re-express the force as:

which, if we think in terms of a force per unit volume, \(\mathbf{F}^{\prime}\), of fluid, immediately gives

and hence we obtain the differential form of Archimedes’ principle:

Now, Newton’s second law relates the net (unbalanced) force on a body to the acceleration it suffers:

in which the material derivative appears, note. Unit volume of fluid has a mass which is, by definition, equal to the fluid density, \(\rho\). Accordingly, from (9) and (10) we obtain:

which, with a little help from equation (6) for the material derivative, allows us to obtain a very important partial differential equation, Euler’s equation(s):

Euler’s equations (11) are a set of three non-linear, coupled partial differential equations (see exercises below). The equation does not have the linearity property you will have used to great effect in solving other partial differential equations. This is a central problem in mathematical fluid mechanics.

Exercise 2 In deriving (11) what important piece of physical experience have we ignored? (Think about pushing an object along a surface).

Tutorial Exercises¶

1. Unpacking Euler’s equation. Show that Euler’s equation (11) may be written-out ‘component-by-component’ as three coupled partial differential equations:

   \(\frac{\partial v_x}{\partial t} + \left( v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} + v_z \frac{\partial}{\partial z} \right) v_x = -\frac{1}{\rho} \frac{\partial p}{\partial x}\)

   \(\frac{\partial v_y}{\partial t} + \left( v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} + v_z \frac{\partial}{\partial z} \right) v_y = -\frac{1}{\rho} \frac{\partial p}{\partial y}\)

   \(\frac{\partial v_z}{\partial t} + \left( v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} + v_z \frac{\partial}{\partial z} \right) v_z = -\frac{1}{\rho} \frac{\partial p}{\partial z}\)

   How many unknown functions are there in the above set of equations? What extra information would you need to find a solution to the above - that is, to ‘close’ the description?

2. Show that the flows:

   1. \(p = const, \mathbf{v} = y {\hat{e}}_{x} ~~~~~~~~~~~~~~~~~ 0 \le y \le W\), and

   2. \(p = const, \mathbf{v} = y (W - y) {\hat{e}}_{x} ~~~~~ 0 \le y \le W\)

   are possible solutions of the Euler and continuity equations.

3. For an ideal, incompressible fluid, density, subject to a gravitational field, Newton’s second law applied to a fluid particle gives:

   \(\rho\frac{D}{Dt}\mathbf{v} = - \nabla p + \rho\mathbf{g}\)

   in which all symbols have their usual meaning.

   1. Show the above equation may be written:

      \(\frac{\partial}{\partial t} \mathbf{v} + \left( \mathbf{v} \cdot \nabla \right) = - \frac{1}{\rho}\nabla p + \mathbf{g}\)

      making clear all your steps.

   2. Liquid is contained in a uniform solid cylinder with its axis in the \({\hat{e}}_z\) direction and is subject to gravity, which acts in the \(- {\hat{e}}_z\) direction. The cylinder is rotated at a constant angular velocity \(\Omega {\hat{e}}_{z}\) radians per second, until the flow reaches steady state.

   1. Show that the fluid velocity is \(\mathbf{v} = - \Omega y {\hat{e}}_{x} + \Omega x {\hat{e}}_{y}\).

   2. Show the continuity equation, \(\nabla \cdot \mathbf{v} = 0\), is satisfied.

   3. Show the system of equations you found in part a. is:

      \(x \Omega^{2} = \frac{1}{\rho}\frac{\partial p}{\partial x},\) \(y\Omega^{2} = \frac{1}{\rho}\frac{\partial p}{\partial y},\) \(0 = \frac{1}{\rho}\frac{\partial p}{\partial z} + g.\)

   4. Hence that the pressure distribution in the liquid is:

   \(p = \tfrac{1}{2}\rho\Omega^2 \left( x^{2} + y^{2} \right) - \rho gz + C,\)

   where \(C\) is a constant of integration.

   22. Determine the shape of the surface of the rotated liquid.

The Euler Equation for a Fluid Under Gravity¶

The equation of motion for a fluid without friction (viscosity) is the Euler equation. For a fluid under gravity it becomes:

where the vector \(\mathbf{g}\) is the acceleration due to gravity (in the jargon of theoretical physics an example of an external body force) which may be written:

where \(\phi^{'}\) (note the slightly unconventional prime, here) is the gravitational potential. Over short distances, taking the energy origin at \(z=0\), the latter may be approximated:

To derive the above form of the Euler equations is a straightforward generalisation of the derivation in Inviscid Fluids. We won’t actually perform it all - you will be able to see that the equation we derived in this section, which is essentially Newton’s Second Law:

will, in the presence of an additional gravitational force, need to account for the additional contribution to the force which comes from gravity:

Using equation (6) from The Material Derivative, Fluid Parcels and ‘Advection’ we obtain straightforwardly:

You probably won’t take too much convincing that \(\mathbf{g}\) could be any external body force acting on the fluid - electrostatic or magnetic, for example.

Fluid Vorticity¶

The fluid vorticity vector is defined as the curl of the velocity field:

It may be shown that the vorticity vector represents the angular velocity of a fluid particle as it moves in flow. The magnitude of the vorticity, \(| \omega |\), gives the size of the angular velocity (in radians per second, note) and the direction \(\hat{\omega}\) defines the direction of the axis around which the fluid particle is rotating.

Exercise 3 (important result). Use the vector identity:

(where \(v^2 = \mathbf{v} \cdot \mathbf{v}\)) and standard properties of the vector field to show that, for an incompressible fluid, the Euler equation may be written as follows:

Hence show that, if at \(t=0\) an inviscid flow is free from vorticity, then it will remain free from vorticity at all later times.

Steady Flow, Streamlines and Bernoulli’s Equation¶

Bernoulli’s Equation¶

Along a streamline:

Consider an incompressible fluid under gravity, in steady flow (when the velocity field is unchanging). We can now write:

The Euler equation describes the flow. Replacing the acceleration due to gravity with the gravitational potential, using \(\mathbf{g} = -\nabla \phi^{'}\), we obtain:

and using (14) we obtain:

Now comes the clever part. We use the vector identity introduced in example 2.3, namely \(\left(\nabla\times\mathbf{v}\right)\times\mathbf{v} = \left(\mathbf{v}\cdot\nabla\right)\mathbf{v} - \tfrac{1}{2}\nabla\left(v^2\right)\), in the left hand side of the equation above, which therefore becomes:

which is easily re-arranged to give:

and noting that the gradient operation ‘distributes over addition’ (which simply means \(\nabla \left(a + b\right) = \nabla a + \nabla b\)) we obtain:

Substituting for the vorticity, \(\mathbf{\omega} = \nabla \times \mathbf{v}\), we have:

Nearly there now. Take the inner product of the equation with \(\mathbf{v}\). We have:

Now, the term in square brackets in the left hand side of the above equation is a vector which is perpendicular to \(\mathbf{v}\), by definition of the vector or cross product, so the scalar product in the left hand side is evaluated between one vector and another perpendicular vector: that is, the left hand side is zero. So we have:

where we have defined:

In The Material Derivative, Fluid Parcels and ‘Advection’ we considered the material derivative. Recall, for a quantity \(\Phi\) (say) the material derivative measures the time rate of change of \(\Phi\), in the rest frame for the material fluid particle, or element, as it passes through the flow.

The material derivative is:

If we are talking about steady flow, \(\frac{\partial \Phi}{\partial t} = 0\), and the last equation may be written:

So, using equations (17) and (19) we find that for material fluid element, as it passes through a flow:

Put another way, the quantity \(\Phi = -\frac{1}{2}\left(v^2\right) - \frac{1}{\rho} p - \phi^{'}\) evaluated for a fluid particle in a steady-state flow, in the ‘rest frame’ of that particle does not change as the fluid particle moves, i.e.

along a fluid particle path. In theoretical physics’ jargon \(\Phi = -\frac{1}{2}\left(v^2\right) - \frac{1}{\rho} p - \phi^{'}\) is a constant of the motion.

A streamline is the term used for the path traced by a fluid particle in a steady flow. Think of introducing a constant stream of dye (using e.g. a hypodermic needle) into a steady, non-turbulent flow of (strictly) an ideal fluid. The path traced by the dye defines a fluid particle path. It is a streamline. Hence, we can say:

Along a streamline, \(\Phi \equiv -\frac{1}{2}\left(v^2\right) - \frac{1}{\rho} p - \phi^{'} = const\) - which is Bernoulli’s Equation:

which is what we set out to prove.

Potential Flow¶

A potential flow is a flow which has \(\mathbf{\omega} = \nabla\times\mathbf{v} = \mathbf{0}\) everywhere and so its velocity vector may be considered to derive from a scalar velocity potential:

where this velocity potential is not the gravitational potential of The Euler Equation for a Fluid Under Gravity (sorry!). The difference is in the prime.

Potential flows are particularly important in a range of aerodynamic applications.

We shall consider (solve) a potential flow problem in a case study at the end of this section of notes. Let us first consider how the concept of potential flow arises. Some timely reminders first:

• The Euler equation is not the only equation which is important in determining fluid flows - that is, it is not the only equation which the fluid flow velocity field must satisfy. There is the continuity equation to worry about. For an incompressible fluid, we recall that this takes the simplified form, namely \(\nabla\cdot\mathbf{v} = 0\).

• We have seen in equations (14) and (13) of this section of notes, that the Euler equation may be written as follows (repeated here as a reminder):

We can derive other useful perspectives on inviscid fluid flow from equation (14).

First, as a bit of an aside, from (14) it follows that, if, at \(t=0\) an inviscid flow is free from vorticity, i.e. that flow initially has \(\mathbf{\omega} = 0\), then it will remain free from vorticity at all times (since vorticity can never develop if \(\frac{\partial}{\partial t}\mathbf{\omega} = \mathbf{0}\)). We write:

Using equations (14) and (13) and the fact that the vector (cross) product does not commute, ‘it is immediate that’ (i.e. takes quite a bit of thinking about):

‘Clearly’ (another word I see a lot in advanced maths texts and have learned to hate), one solution of the above equation is:

and since the curl of any scalar gradient is identically zero this encourages us to write equation (21):

Now, for an incompressible fluid, we must have \(\nabla\cdot\mathbf{v} = 0\) and hence it is immediate (ha!) that:

Expanding the divergence in the last equation:

which means that the velocity potential in equation (21) must satisfy Laplace’s equation:

Put another way, potential flows may be solved by solving a single, linear partial differential equation for a single scalar, velocity potential, \(\phi(\mathbf{r})\).

Put still another way, in potential flow, Euler’s equation and the all-important continuity equation have been combined into a single, one-line partial differential for a scalar velocity potential.

Mathematicians and theoretical physicists have a lot of experience solving (22) as it arises a lot in electrostatic problems. As a practical approach to solving certain classes of flow problem, therefore, potential flow approximation is very useful indeed.

We will do an example on potential flow at the end of this section of the notes. For the moment, let us proceed to use the concept of velocity potential to establish another important theorem.

Bernoulli’s Theorem for steady and unsteady potential flow¶

Denoting gravitational potential by \(\phi^{'}\), recall now …

Along a streamline, in unsteady flow:

A bit of revision. We saw, in Potential Flow, that, for an ideal fluid, with negligible friction:

and in The Euler Equation for a Fluid Under Gravity we saw the vector identity:

Suppose we have a fluid in which, at \(t=0,\mathbf{\omega}=\mathbf{0}\), so that \(\mathbf{\omega} = \mathbf{0} \forall t\) (say a fluid initially at rest) then, if \(\mathbf{\omega} = \left(\nabla\times\mathbf{v}\right) = \mathbf{0}\), the last identity becomes \(\left(\mathbf{v}\cdot\nabla\right)\mathbf{v} = \tfrac{1}{2}\nabla\left(v^2\right)\) and the Euler equation for a fluid under gravity becomes

and while we are supposing that \(\mathbf{\omega}\equiv\left(\nabla\times\mathbf{v}\right)=\mathbf{0}\), it is still valid to use the velocity potential \(\mathbf{v}=\nabla\phi\) (not to be confused with the gravitational potential, note). Hence the last equation becomes:

Therefore:

The last equation integrates (‘has a first integral’) as follows:

which is Bernoulli’s Theorem for unsteady potential flow.

An example of the use of Bernoulli’s Theorem is an end in itself (although very rewarding and illuminating) which will have to have its own section of lecture notes.

There now follows a tutorial (i.e. an old exam question) as illustrative of the solution of potential flows. There are several interesting and important points to note in this solution - lots of mathematical cunning - the important of which has to be the way in which physical insights are borrowed from electrostatics.

Tutorial Example¶

1. Use the first item of given information to show that any fluid with a density \(\rho\) and velocity \(\mathbf{v}\) is described by the continuity equation:

   \(\frac{\partial}{\partial t} \rho + \nabla \cdot \left( \rho \mathbf{v}\right) = 0.\)

   For an ideal, incompressible fluid with velocity potential, \(\phi\), is such that:

   \(\mathbf{v} = -\nabla \phi.\)

2. Use the second item of given information (the definition of the material derivative) to show that the first equation in part 1 may always be written:

   \(\frac{D}{Dt} \rho + \rho \nabla \mathbf{v} = 0.\)

3. Use the second equation in part 1 to show that \(\phi\) satisfies the Laplace equation:

   \(\nabla^2 \phi = 0\)

A sphere, radius \(R\), with its centre at the origin of a polar coordinate system, moves with a constant velocity \(\mathbf{u}\) through an ideal, incompressible fluid, which is at rest at a large distance from the sphere. The velocity potential, \(\phi\), has the following boundary condition on the spherical surface:

where \(\hat{e}_r\) is the unit radial vector.

4. Assuming without proof that \(\left(\frac{1}{r}\right)\) is a solution of the Laplace equation, show \(\frac{\partial}{\partial x} \left(\frac{1}{r}\right)\) is also a solution of Laplace’s equation.

5. Hence show that, for the problem of the sphere:

   \(\phi = -\frac{R^3}{2 r^2} \mathbf{u} \cdot \hat{e}_r.\)

   Hint. Exploit the linearity property of Laplace’s equation.

6. Indicate how you would find the velocity field for this problem.

Viscous Flow¶

The Momentum Flux tensor¶

Using Cartesian tensor or suffix notation, the momentum flux tensor is defined as follows:

in terms of which the Euler equation of inviscid fluid flow may, in the absence of gravity, be written:

What is the significance of the momentum flux tensor and how do we arrive at the Euler equation in the form of (24)?

First some revision. As we have seen, flow of a fluid under gravity may be shown to be governed by the (scalar) continuity and (vector) Euler equation(s):

where vector \(\mathbf{g}\) is the acceleration due to gravity (in the jargon of theoretical physics and example of an external body force) \(\mathbf{g} = -\nabla\phi^{'}\) and \(\phi\) is the gravitational potential energy. Over short distances, taking the energy origin at \(z=0\), the latter may be approximated \(\phi = gz\).

Exercise 4 Show the continuity and Euler equations may be expressed in Cartesian vector notation respectively as follows:

where the summation convention applies to repeated subscripts as follows:

Exercise 5 Using equations (25) and (26) and the product rule of differentiation, show that the Euler equation(s) (26) may be written:

and hence use the Kronecker Delta symbol to show that:

For a while we will work neglecting the influence of gravity. Supposing \(\mathbf{g}=\mathbf{0}\), the last equation is immediately recognisable as equation (24) simply by invoking the definition of momentum flux in equation (23).

So what was the point? Good question: read on …

Interpretation of the Momentum Flux Tensor¶

Let us do a volume integral of equation (24) over some control volume of fluid \(V\) with surface \(S\):

Re-ordering the time and space variable operations in the left-hand side and unpacking the summation convention in the right hand side:

or

in the right hand side of which we recognise what is basically the divergence of the vector:

So the last but one equation may be written:

and the divergence theorem applied in the right hand side to convert the volume integral to a surface integral now gives:

Equation (27) may now be interpreted. The left hand side is the rate of change (in time) of the \(i\)-component of the momentum of all the fluid contained inside the volume \(V\). The right hand side is the flux of the vector \(\mathbf{Q} \equiv \left(\Pi_{ix}, \Pi_{iy}, \Pi_{iz}\right)\) evaluated over the surface, \(S\) , of volume \(V\).

So the vector \(\mathbf{Q} \equiv \left(\Pi_{ix}, \Pi_{iy}, \Pi_{iz}\right)\) measures the ‘flux’ - the transport - of the \(i\)-component of momentum, per unit time, per unit area. Let us put this in another way.

Choose \(i=2\) and think about a unit area with normal \(\hat{z}\), so \(d\mathbf{A} = \mathbf{1} \hat{z}\). The quantity \(\left(\Pi_{2x}, \Pi_{2y}, \Pi_{2z}\right) = \left(\Pi_{yx}, \Pi_{yy}, \Pi_{yz}\right)\) when ‘dotted’ with \(\hat{z}\) gives:

which measures the amount of the \(y\)-component of momentum transported by flow across an area of one square metre, orientated with its normal in the \(z\)-direction.

Viscous Stress tensor and the Navier-Stokes Equation of viscous flow¶

Friction in any physical system introduces dissipation of energy and ‘loss’ of momentum. All real fluids are subject to ‘viscous dissipation’. Sometimes its effect is small and may be neglected (as we have done until now). But in general this is not so.

Friction shows up in relative motion. Consider fluid elements A and B in contact with each other: suppose A is moving fast relative to B. Any friction in the fluid will result in transfer of momentum between elements A and B – I’m certain your physical experience will tell you that the faster-moving element A will slow down (lose momentum). The slower moving element, B, may pick up some of A’s lost momentum but, in general, momentum is lost from the flow overall. For the physically-minded, lost momentum shows up in ‘viscous heating effects’: tracking this dissipation of momentum to transformation to heat is another story. As a modelling step we take the following statement as a robust start-point in our inquiry into the effects of friction:

A fluid element moving in a fluid which has internal friction (or viscosity) loses momentum.

To model this physical fact we remember that the momentum flux tensor \(\Pi_{ij} \equiv -p \delta_{ij} + \rho v_i v_j\) measures the amount of momentum transported by flow in a fluid and that its components must therefore be reduced by the action of friction in the fluid:

where the viscous stress tensor \(\sigma^{\prime}_{ij}\) is, like \(\Pi_{ij}\), a second rank tensor.

Making this replacement in the Euler equation (24) makes it (the Euler equation) account for the effects of friction or fluid viscosity and results in the Navier-Stokes equations:

The above form of the Navier-Stokes equation is its most general. We shall now adapt (29) to so-called Newtonian fluids and stay there for the remainder of this course.

Fluid Constitutive Equations and Transport Coefficients¶

To make the equation (29) apply to a given fluid we must be able to model the viscous stress tensor in terms of the fluid motion, that is, its velocity field, \(\mathbf{v} (\mathbf{r},t)\). The relationship between the viscous stress tensor and the motion of the fluid is one example of a constitutive relationship. In this course we shall consider only ‘simple’ Newtonian fluids. ‘Non-Newtonian’ fluids are formidably difficult to model and must leave them entirely alone.

We suppose that the viscous stress tensor’s components are functions of the fluid velocity and its spatial gradients:

Let us try to simplify the form of this function.

Viscous effects must vanish when the fluid is in uniform translation so:

and we shall assume that we need only worry about first velocity derivatives (even when the fluid is in turbulent flow - this works and it beats me (and many others) how but turbulence is the last unsolved problem of classical physics, so we’ll certainly give that one a wide berth):

and we now assume that the fluid has no memory of its previous states (this is where visco-elasticity is ruled out and we specialise to Newtonian fluids):

where \(a_{ij}\) are constants (i.e. don’t depend on time). We are now quite a way towards our constitutive equation for a Newtonian fluid. We have one final card to play …

Fluids in uniform rotation are made up of elements which are not moving relative to each other and so friction effects must vanish for the same reasons as they do in uniform rectilinear flow.

Exercise 6 The velocity field of a fluid in uniform rotation may be expressed as \(\mathbf{v} = \mathbf{\Omega}\times\mathbf{r}\) (here \(\hat{\Omega}\) is the axis around which the rotation is occurring and \(\mathbf{\Omega}\) is the angular velocity). Show that the following combinations of velocity gradients vanish:

1. \(\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right)\)

2. \(\frac{\partial v_k}{\partial x_k}\)

where the summation convention applies in the second expression. Note that the second expression is essentially \(\left(\nabla\cdot\mathbf{v}\right)\).

Exercise 7 Try to think of other expressions linear in velocity spatial derivatives (like i. and ii. of the previous exercise) which vanish when \(\mathbf{v} = \mathbf{\Omega}\times\mathbf{r}\). (Trick question: there are none.)

In fact you can’t find any other expressions linear in velocity spatial derivatives which vanish when \(\mathbf{v} = \mathbf{\Omega}\times\mathbf{r}\). We can apply this fact to our deliberations around equation (30). The constants, \(a_{ij}\), in (30) must be such that:

in which \(a\) and \(b\) are simple (scalar) constants. The last equation is essentially the constitutive equation of a Newtonian fluid. Note that the Kronecker delta had to be inserted in the second term on the right hand side to make a second rank tensor out of the scalar \(\frac{\partial v_k}{\partial x_k}\). We have a little more work to do before putting a number on the last equation: we re-arrange its terms (the terms in the last equation):

and re-name the constants to obtain the traditional form of the constitutive equation of a Newtonian fluid:

where \(\eta\) is the fluid’s shear viscosity and \(\zeta\) its bulk viscosity. These constants are molecular properties of the fluid and are termed transport coefficients. It’s the business of a lot of molecular dynamicists to determine these from direct simulations. You probably know (a lot) more about that than me.

Exercise 8 Show that

Exercise 9 Show that, for a liquid (incompressible fluid)

which result shows that the bulk viscosity of an incompressible fluid is irrelevant.

The Navier-Stokes equation revisited¶

The Navier-Stokes equation for a compressible, Newtonian fluid is:

Let us derive the Navier-Stokes equation - the equation of motion of a viscous, Newtonian fluid - from equations (23), (29) and (31), in its traditional form:

Substitute (23) and (31) into (29):

and using the summation convention to tidy up a little we have equation (32):

which is the Navier-Stokes equation as it applies to any Newtonian fluid – compressible or otherwise. Scary, isn’t it?

The Navier-Stokes equation for an incompressible fluid¶

This is the form of the Navier Stokes equation which we shall be using in all our calculations of incompressible flow fields:

where \(g\) is the acceleration of gravity (if necessary). In fact, we will neglect gravity in what follows - for the moment.

From (32) it is relatively easy to specialise our description to incompressible fluids, that is to obtain (33). First we recall that the continuity equation for an incompressible fluid takes the form \(\nabla\cdot\mathbf{v} = 0\), the Cartesian tensor form of which is:

whence equation (32) immediately loses two terms in the right hand side:

Expanding the ‘viscous terms’ in the right hand side, and re-ordering the differentiations in the third term on the right hand side, now emphasised with a bracket:

and again using the continuity equation in the bracket term we have:

The remaining viscous term is the tensor form of \(\nabla^2 v_i\). Use this fact and a product rule expansion of the derivatives in the left hand side now:

and grouping terms in the left hand side:

and, finally, using the continuity equation in its compressible form (to kill the second bracket term in the left hand side) and dividing by the density, we obtain the result.

Exercise 10 Show that the incompressible Navier-Stokes equation may be expressed in vector equation form as follows:

Kinematic Viscosity¶

This is important mainly for an incompressible fluid; it is a molecular property and is defined as follows:

Anatomy and nomenclature of the Incompressible Navier-Stokes Equations¶

There are very few of these. Before we proceed to consider them, let us consider the explicit system of four partial differential equations in (34), which we have to solve for the four unknown flow velocity and pressure fields …

The terms \(\left(v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} + v_z \frac{\partial}{\partial z}\right) v_x\) in the above equation (35) are referred to as the non-linear, advective, convective or inertia terms. The terms \(\nu \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) v_x\) are the diffusive or viscous terms.

Exact solutions of the Navier-Stokes Equation for incompressible flow¶

The good (bad?) news is that there are not many (exact) solutions. We will need all the help we can get. So we need first to think about simplifying strategies. Basically that means exploiting the symmetry of the problem and considering special circumstances.

Steady Flow¶

If flow is steady, the partial time derivative of the velocity field in equations (35) is zero.

Unidirectional Flow¶

Unidirectional flow arises when the boundary conditions on the flow allow us to infer that the flow in always and everywhere (sounds like a boy band song title: sorry) in the same direction. Let us so align our coordinates that this special direction is the \(z\)-direction. Then we can say:

Exercise 11 Show that for steady, unidirectional flow, equations (35) reduce to

and hence show that, for steady, unidirectional flow:

in which equations the non-linear or inertia term has vanished, note.

Exercise 12 Integrate the equations (36) to obtain, still for unidirectional flow:

Incompressible flow between infinite parallel plates (ignorable coordinates)¶

This flow has a translational invariance and, as a result, an ignorable co-ordinate, as we shall see.

The symmetry of the problem implies that the flow must be unidirectional, parallel to the boundary planes everywhere. Without loss of generality, let the \(z\)-axis point along the direction of flow and let the infinite planes be situated at \(y = 0,h\). Hence, the boundary conditions on this flow are that:

With this choice of coordinates, the vector velocity field of the flow has translational invariance in \(x\) and \(z\) - whatever values the \(x\) and \(z\) coordinates take the system looks the same. For this to be true:

• the flow field cannot depend on the \(x\) and \(z\) coordinates, and

• the pressure field must be a uniform gradient.

If you’re wondering about the last point, suppose any pressure gradient varied from point to point. From the gradient of the pressure you would be able to tell where in the flow domain you were situated which contradicts the assumption of translational invariance.

Then, from the boundary conditions and our choice of coordinates we can again say:

with

Now comes the neat bit. From the first equation, the fact that any pressure gradient must be constant and the fact that the \(x\) coordinate cannot enter the problem (the \(x\) coordinate is ignorable) we immediately see that:

where \(G\) and \(P_0\) are constants. It is immediate that:

which has a first integral:

(Note I’m using ordinary differentiation now, as there is only the \(y\) coordinate to think about and I’ve replaced the kinematic viscosity with the shear viscosity). Integrating again:

where \(C\) and \(D\) are integration constants to be determined now.

Exercise 13 Show the velocity boundary conditions (38) mean \(f(0) = f(h) = 0\).

Exercise 14 Use the boundary conditions on \(f(y)\) to show that:

Problems¶

Decay of velocity in duct flow (used in tutorial exercises)

Fluid is confined between infinite, parallel planes \(y=0\) and \(y=h\), and is induced to flow in the \(x\)-direction, by the steady pressure field:

with \(G\) and \(P_0\) as constants. For this unidirectional flow the Navier-Stokes equations take the form:

and \(v_x = v_y = 0\) everywhere.

1. Verify that, in the steady-state, the flow is:

   \(v_x (y) = \frac{4 V_0}{h^2} y \left(y - h\right),\)

   Provided \(V_0 = \frac{G h^2}{8 \nu}\), where \(V_0\) is the flow velocity at \(y = h/2\).

2. At \(t = 0\) the pressure gradient is removed and hence, for \(t>0\), \(v_x (y,t)\) satisfies the time-dependent one-dimensional Navier-Stokes equation:

   \(\frac{\partial}{\partial t} v(y,t) = \nu \frac{\partial^2}{\partial y^2} v(y,t)\)

   and the condition found in part a. represents the initial condition on the velocity field. Seek a separable solution for the above Navier-Stokes equation and use the information given below to show, for \(t>0\):

   \(v_{x}{\left( y,t \right) = \frac{32V_{0}}{\pi^{3}}\sum_{m}{\frac{1}{{(2m + 1)}^{3}}\sin\left( \frac{(2m + 1)\pi y}{h} \right)}e}^{- \frac{{(2m + 1)}^{2}\pi^{2} \eta t}{h^{2}\rho}}\).

Information

\(\int_{0}^{h}{\sin\left( \frac{m \pi x}{h} \right)\sin\left( \frac{n \pi x}{h} \right)}dx = \frac{h}{2}\delta_{\text{nm}} ~~~~~\) \(m\), \(n\) are integers.

\(\cos\left( m \pi \right) = \left( - 1 \right)^{m} ~~~~~\) \(m\) is an integer.

Decay of velocity in shear flow

Fluid of kinematic viscosity \(\nu\), initially at rest, is confined between infinite parallel planes \(y = 0\) and \(y = h\) both at rest. The top plate is moved at velocity \(u_0 \hat{x}\) from time \(t \ge 0\). The fluid velocity is unidirectional in the \(x\)-direction, \(\mathbf{v} = v (y,t) \hat{x}\), where \(v(y,t)\) satisfies the one-dimensional linearised Navier-Stokes, or diffusion, equation:

The inhomogeneous boundary and initial conditions on velocity which describe this problem are:

and the solution to the above Navier-Stokes equation is to be sought in the form:

where \(V(y)\) is the steady state condition.

1. Show that the steady-state solution is the uniform shear:

   \(V(y) = \frac{u_0}{h} y.\)

2. Find the equation satisfied by \(u(y,t)\) and state the boundary and initial conditions on \(u(y,t)\). Use a separable trial solution for \(u(y,t)\) and the information given to show that:

   \(v(y,t) = u_0 \bar{y} + \frac{2 u_0}{\pi} \sum_m \frac{(-1)^m}{m} \sin \left(m \pi \bar{y}\right) e^{-m^2 \pi^2 \bar{t}},\)

   where:

   \(\bar{y} = \frac{y}{h} ~~~~~~~ \bar{t} = \frac{\nu t}{h^2}\)

If you have plenty of spare time, e.g. on a dark winter evening and/or while self-isolating, you could measure the lattice fluid viscosity using the above result and compare it with the results you obtained previously.