**Streamlines for irrotational flow around a cylinder:** plotted for various values of the swirl *B=UaA*

B=0 B=1

B=2 B=3

**Streamlines around an aerofoil:** if we start with the flow around a cylinder and use a technique called conformal mapping (involving various coordinate transformations), we can calculate the flow around other shapes. We’re not going to go into this in detail here- instead we’ll just look it schematically for an aerofoil.

**Boundary layers:** even when *ν* is small, the viscosity is still important in a thin ‘boundary layer’ near a solid surface.

Thickness of boundary layer ~

**Boundary layer separation:** in practice, low viscosity flows around objects are not quite as smooth as those shown above. The boundary layer tends to break away from the object, giving rise to a turbulent wake, as shown schematically below.

Turbulent wake -> dissipation, drag, inefficiency…

Aerofoil shapes are designed to give separation at the trailing edge only, so that they are more efficient.

**Circulation, Γ:** circulation

*Γ*around a closed curve

_{C}*C*lying within a fluid is defined as the line integral

Stokes’ theorem

*S* is a surface enclosed by *C*; it must lie wholly within the fluid (i.e. same conditions everywhere) for Stokes’ theorem to apply/

For an irrotational flow, *Γ _{C}=*0.

If we have inviscid swirling vortex flow around a cylinder of radius *a*, , then **w**=0, but

This may seem contradictory, but it’s actually fine as the irrotational flow is not across the whole surface bounded by *C* (we have the cylinder in there too), so our Stokes’ theorem arguments no longer hold.

**Lift force**

Normal force on a unit length element of cylinder *adθ*

Lift force is in *y*-direction

Use the potential we used in the aerodynamic extension’ section:

(the last term is clockwise swirl)

On cylinder

from Bernoulli’s equation

Total lift force for a unit length of cylinder (at *r=a*)

but of course the first two terms in the integral will just integrate to zero, making life much easier for us.

So

Circulation

taking *u _{θ}* at

*r=a*

->

**Travelling waves on deep water (2D):** still assuming inviscid, irrotational, incompressible flow

Once again, put

$latex \frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=0

We’ll also need Bernoulli’s equation

At the air-water interface, take *p=p _{0}*=const (1 atm)

Suppose |**u**|, *φ*, *η* are all ‘small’ so that terms like |**u**|^{2} are negligible.

Remember, constants like can be gotten rid of by adding to *φ* (we already discussed this earlier).

Also, we can do a Taylor expansion on

(all higher order terms are negligible).

at *y*=0

Also, at *y=η* but also

Then, doing a Taylor expansion and saying that these variable terms are approximately the same at *y=*0 as they are at *y=η*, we get

Now look for a travelling wave solution

By convention, *k>*0; is called ‘pomega’.

Try

f is bounded as , so *D*=0

and

at *y*=0

Which gives us and respectively.

->

**dispersion relation**

Now we can calculate the path of fluid blobs.

Where *X*, *Y*, 0 are displacements from the mean positions.

ignore the constant of integration because we only care about displacements from the mean and no absolute positions.

Similarly

Circles, radii

Phase speed

Group velocity

Crests and troughs moving faster than packet, so they advance through the packet; if there is a mix of wavelengths, longer waves appear towards the outer edge.

**Very viscous (Stokes) flow:** going back to the Navier-Stokes equation, neglect *g*, assume steady flow, take *ρ, ν* constant

->

We can neglect the non-linear term compared with the viscous term when the Reynolds Number i.e. when viscosity is ‘large’ *ν>>UL*, length scales are ‘small’ *L<< ν/U*, velocities are ‘small’ *U<< ν/L*, or any combination of the above.

Neglecting the non-linear term gives (1)

And since the flow is incompressible

These are the Stokes flow equation.

Now take the div of (1)

where we have used the vector identity

Now we have Laplace’s equation for pressure, but actually this isn’t terribly useful since we don’t usually know the boundary conditions on *p*.

Now let’s take the curl of (1)

as *μ≠*0

Here we have Laplace’s equation for **w**; this is actually three equations- one for each component.

**Reversibility:** the solutions of the Stokes flow equations are reversible- if **u**(**r**) is a possible flow, then so is **u**(-**r**), provided the boundary conditions allow it (*p*-> –*p+*const to reverse the pressure gradient). For example, if a fish tries to swim in treacle, then if a flap of the tail to the right drives it forward, a flap to the left drives it back again. Therefore, it’s just as well that fish aren’t usually found trying to swim in treacle.

**Stokes flow past a sphere:** consider a very viscous fluid moving past a stationary sphere of radius *a*. A long way from the sphere, the flow velocity is equal to *U***k**. If , we expect the Stokes flow equations to hold.

Boundary conditions:

**u**=0 on the sphere

**u**=*U***k** at large distances from the sphere

(const) far from the sphere

To get a formula for the drag, we can use dimensional analysis. Assume dependence of drag, *D*, goes as (*l* does not appear in the Stokes flow equations).

Dimensions (in terms of mass, length and time)

Equate dimensions

Mass:

Length:

Time:

So *α,* *β* and *γ* are all equal to 1 and drag

We can actually go further and find the constant *κ*, but it does take a fair amount of calculation. Here is an outline of the procedure.

Working in spherical polars, assume axisymmetry- , latex u_{\phi}=0$

Boundary conditions

at *r*=0

, as

Plug through the algebra, and we get

*p* is a ‘dipole’ solution of Laplace’s equation, but we can’t see this by inspection because we don’t know the boundary condition on *p* at *r=a* in advance.

For , (downstream)

For , (upstream)

Vorticity is in the +ve *φ* direction.

**Viscous stresses on the sphere**

Components of stress

At *r=a*

The problem is symmetric in all directions except the direction of *U* (taken to be the *z*-direction), so the only component which doesn’t integrate to zero is

Total drag on sphere

**Stokes Law**

So the constant *κ=*6*π*.

**Using Stokes Law:** drop a ball-bearing of density *ρ _{b}* into treacle. At terminal velocity

i.e. Drag=gravity-upthrust

**Flow in a thin film of fluid** e.g. lubricating oil between two solid surfaces.

In *x, y* plane, horizontal flow speed ~ *U*, horizontal length scale ~ *L*

**u=**(*u, v, w*)

If *h<<L*, we have a ‘thin film’

$latex \nabla^2 u= \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}+ \frac{\partial^2 u}{\partial z^2} \simeq \frac{\partial^2 u}{\partial z^2} \sim \frac{U}{h^2}

The fluid is incompressible, so

-> and

(ignoring numerical prefactors)

Compare inertial and viscous terms

We can neglect non-linear terms if Re(\frac{h}{L})^2 \ll 1$; this is less stringent than since we’ve already assumed *h<<L*.

Similarly, and are small, so we can neglect , and

Our Stokes flow equation from earlier becomes latex \frac{\partial^2 \bold{u}}{\partial z^2} $ (1)

So we can neglect vertical pressure gradients and say , i.e. pressure is a function of *x* and *y* only.

*x-*component of (1) gives

Integrate to get

*f *and *g* are arbitrary functions of *x* and *y* (functions of integration).

Similarly, the *y*-component of (1) gives

**The slider bearing**

Flat boundary at *z*=0 moves with speed *U* past a fixed block with a thin film between.

Take this to be a 2D problem: **u**=(*u*,0,*w*); *p=p(x)*

At *z*=0, *u=U* -> *U=g *(remember that’s the function of integration *g*, not acceleration due to gravity)

At *z=h*, *u*=0 ->

So

Volume flux

As the fluid is incompressible, this must be independent of *z*.

Now let’s try to find an expression for the pressure, starting with

Integrate wrt *x*, using boundary conditions *p=p _{0}* at the edges of the block (

*x=*0,

*x=L*).

If *x=L, p=p _{0}* and

We’ll work this out using a simple example: ,

To work out the integrals, make the substitution ,

Similarly

So

Also

But remember we can substitute for *α* using

Now put the contents of the squiggly bracket under a common denominator to get

Which, when cleaned up, gives

As *h* lies between *h _{1}* and

*h*, if

_{2}*h*<

_{2}*h*then

_{1}*p-p*>0, implying a net upward force on the block if the width of the film decreases in the direction of the flow.

_{0}**Syrup on a rotating rod:** a thin layer of viscous fluid, thickness *h(θ)* covers the outer surface of a rotating cylinder whose circumference rotates at a speed *U* under gravity.

User local Cartesian coordinates.

Going back to and this time including gravity:

*x*-component

z-component

For a thin film

So we can neglect and

->

and

Boundary conditions:

*u=U* at *z=*0 (no slip)

at *z=h* (zero stress)

Integrate to get

Now integrate again

*B=U*

Volume flux across any section *θ* =constant, *Q*

Put ,

->

As , *H(θ)* can only exist for all values of *θ* if (as must be

This means that the fluid can only be kept on the cylinder if

where is the mean thickness (not the Planck constant!).

when , leading to the numbers given above.

**Hele Shaw Cell:** we use the same procedure as we did for the slider bearing, except now the flow is between two flat, rigid horizontal boundaries at *z=*0 and *z=h*, so we adjust our boundary conditions accordingly (*u=*0 at *z=*0 and *z=h* as per the good old ‘no slip’ condition). Armed with this information, I’m sure you can plug through the maths yourself.

**Additional- ship waves:** waves in the wake of a ship (or even a duck) lie within a wedge of arcsin 1/3 (19.5°). At each instant, the ship generates a spectrum of waves of different wavelengths (and hence phase and group velocities) which propagate in all directions. As the ship moves on, a constructive/destructive interference pattern is formed.