**The Navier Stokes equation:** apply Newton’s Second Law to a small ‘blob’ of fluid moving with the flow (imagine we can watch it as it goes).

Normal pressure forces in x-direction

Net pressure force in x-direction

Vector pressure force

Viscous forces: take bulk fluid component *u* in the *x-*direction- assume it is independent of *x* and *y*.

Tangential viscous stress

*μ* is the dynamic viscosity (also known as *η* in kinetic theory, just to stop you getting complacent).

Net contribution to viscous force on blob

Including *x* and *y* variation

when $latex(nabla.\bold{u})=0$ (incompressible)

Gravity force: *mg* downwards

Putting it all together-

acceleration

where is the **kinematic viscosity**.

Now to get the acceleration in terms of *u*:

The **Lagrangian view** is to follow the blob as it moves with the flow. The blob’s position is , velocity , acceleration .

Of course, it’s hardly easy in practice to sit and watch a blob, so it’s also useful to have another way of thinking- the **Eulerian picture**.

Consider a time-varying velocity field referred to fixed points in space

Langrangian velocity $\frac{d\bold{r}}{dt}=\bold{u}$ Eulerian velocity

Differentiate wrt *t*

where is the **material **or **advective** derivative (see below).

Putting it all together, we finally get the **Navier-Stokes equation** for viscous fluid flow under gravity-

**Material derivative: **the time rate of change following the motion of a fluid blob, rather than the rate of change at a fixed point (). Since is a non-linear term, this makes the maths harder, and in three dimensions, the non-linearity allows for chaos (the mathematical kind, rather than the realisation that you can’t work out the equations).

**Mass conservation continuity equation:**

Net mass inflow per unit time = rate of increase of mass in container

Net mass inflow in x-direction per unit time

=

~

= Taylor expansion

Including *y *and *z* inflow -> total inflow =

Rate of increase of mass in container =

For a fixed container, *ΔV*=constant.

using a vector identity

Following a blob of **incompressible** density, . Incompressibility is a good approximation for liquids and gases where , the speed of sound. Under incompressible conditions, sound waves can be neglected.

**Rectilinear flow between parallel plates**

in x-direction

Now use the Navier-Stokes equation, but neglect gravity.

x-component:

-> as the other terms are zero.

If we take the x-derivative of the above, all terms involving *u* must vanish, because

So and is independent of *x*.

*y* and *z* components:

Now make the further assumption that the flow is steady and y-independent.

(steady) and

For steady flow, *p* is also independent of *t*, so is a constant.

Define , the pressure gradient.

Putting this all into the original x-component equation, we end up with

Boundary conditions: for viscous flow, the **no slip** condition means that the fluid comes to rest at the walls, i.e. *u*=0 when *z=±h*.

We get a parabolic flow profile- an example of **Poiseulle flow**.

Volume flux per unit *y* distance for the above flow is

**Flow in a circular pipe**

Assume flow is independent of *θ* and *t* -> *u=u(r)*

taking the radial component of del-squared.

Boundary conditions: *u=*0 at *r=a* (pipe wall- no slip condition again) and no singularity at r=0.

putting *r=*0 -> const=0

$u=\frac{-Gr^2}{4\mu}+const$

As *u=*0 at *r=a*, const=

Volume flux=

Average speed=

**Reynolds number, Re,** is a dimensionless number used to quantify the importance of viscosity. We do this by considering the ratio of inertial forces to viscous forces, and then simplifying with scale analysis.

where *U*=typical velocity scale, *L*=typical length scale of variation

If Re>>1, viscosity is usually unimportant

If Re<<1, viscosity is usually important

For Re>Re_{crit}~1000, flow becomes turbulent (rectilinear flow is unstable). Below this value, the flow remains rectilinear.

**Flow between rotating cylinders**

(**i _{r}**,

**i**and

_{θ}**i**are unit vectors in the respective directions

_{z}*r*,

*θ*and

*z*.)

Inner cylinder is fixed.

Outer cylinder rotates with angular velocity *Ω*.

We want steady, incompressible flow with circular symmetry

Neglect end effects and say there is no variation in *z*-direction.

Take the Navier-Stokes equation in cylindrical polar coordinates, neglecting gravity.

->

The two terms on the LHS are centrifugal and pressure gradient terms respectively, whilst the RHS terms are viscous terms.

**i _{r}** and

**terms vanish separately.**

_{iθ}For *ν≠*0

Try

->

->

Boundary conditions: *u _{θ}=*0 at

*r=a*and

*u*at

_{θ}=Ωb*r=b*

At *r=a*:

At *r=b*:

We can now work out the torque needed to keep the inner cylinder at rest.

Tangential viscous stress on inner cylinder:

per unit area of cylinder

Torque per unit *z*-distance on inner cylinder:

This can be used as a way of measuring *μ* (viscometer)

As *Ω* increases, laminar flow breaks down and we see waves and turbulence.

Finally, to find the pressure distribution, use

**Vorticity, w,** is the curl of the velocity field

Vorticity is the measure of the local ‘spin’ or ‘rotation’ of the fluid, but it is not in general a measure of large scale rotation.

Note: for 2D flow in the *xy* plane, **w** is of course in the *z*-direction.

a) ‘**Solid body’ rotation**

Imagine dropping a ‘vorticity meter’ into the flow

flow gets faster as *r* increases, meter spins.

b) ‘**Line vortex’**

for *r≠*0 -> flow is **irrotational**.

flow gets slower as *r* increases, in just such a way that the meter doesn’t spin.

c) **Linear shear flow**

No large scale rotation, but so there is local ‘spin’.

**The vorticity equation:** start with the Navier-Stokes equation, taking an incompressible fluid with *ρ* and *ν* constant.

Vector identity

Now take the curl of this equation, remembering that always and

Vector identity

The last two terms must vanish because we already established that and of course

With all this in mind, we get

the vorticity equation.

**Conservation of vorticity in 2D inviscid flow:** for inviscid flow, *ν*=0 (Re=∞).

Considering 2D flow, **u**=(*u, v, *0), **w**=(0, 0, *w*)

So

So, following the motion of a fluid blob, the vorticity of each blob is conserved. In particular, if the fluid is irrotational everywhere at *t*=0, then it remains irrotational for *t*>0$

**2D inviscid irrotational flow**

Irrotational, so

Because , we can say that **u** is the gradient of some **velocity potential** *φ*

; ;

Incompressible, so Laplace’s equation.

**Bernoulli’s equation in 3D** (take *y*-axis as ‘up’)

Go back to

Put *ν*=0, **w**=0 (inviscid, irrotational flow), so that we can say

$\latex \frac{\partial}{\partial t}(\nabla \phi)+\nabla(\frac{1}{2}|\bold{u}|^2+\frac{p}{\rho}+gy)=0$

i.e. the contents of the bracket on the LHS must be independent of *x, y, z*, so

But we can absorb F(t) into *φ* (remember, you can arbitrarily add constants to a potential without changing the gradient- all we care about is that

So put

So now Bernoulli’s equation.

We can also set the RHS equal to a constant other than 0.

**2D inviscid, irrotational flow around a cylinder**

Put a cylinder into a uniform flow (*U*, 0, 0); how does the flow go around the cylinder?

In cylindrical polar coordinates, Laplace’s equation becomes

Boundary conditions *u _{r}*=0 on

*r=a*(normal velocity=0) and

*u*≠0 at

_{θ}*r=a*(because there is no viscosity, the fluid can slip on the surface).

As , and

All solutions must have the same *θ* dependence, so the solutions of Laplace’s equation that we need are

for *r≥a*

So

Using Bernoulli’s equation, with and *g*=0

On cylinder

The constant is usually set so that as ,

*The pressure on the cylinder has ‘left-right’ and ‘up-down’ symmetry (i.e. we can replace θ* with *π-θ or –θ* and nothing changes), so because the flow is inviscid, there is no pressure induced drag on the cylinder.

**Aerodynamic extension:** now add an asymmetric ‘swirl’ term to the potential we found earlier.

*A>*0

Now

Where the last term is like a line vortex, leading to additional clockwise swirl but zero vorticity for *r*≠0

Now, on the cylinder

There is still ‘left-right’ symmetry, but not ‘up-down’ symmetry, so we have a lift force (pressure is greater below the *x*-axis than above).

**The streamfunction:** because of incompressibility, , i.e. in 2D.

We can write and

For an irrotational flow, we get Laplace’s equation for *ψ* as well as *φ*.

For our standard 2D flow example, , so the velocity is parallel to the lines of constant *ψ* (and perpendicular to the lines of constant *φ*).