Links to papers discussed in class:

Simpson 1996 JPO

Rippeth JPO 2003

Kundu Chapter 12

Comments on Buckingham PI theorem and Sludge Example:

Question: (1) Why would Vs enter as relevant paremeter?

(2) Why was there no explicit term for mixing?

Also—Atomic Bomb Example:

Atomic bomb blast

A very famous example: when the atomic bomb was tested at Alamagordo, NM in 1945, a number of high speed photographs were taken. After the war, these pictures appeared in a 1947 issue of Life Magazine. At this time, the yield of the bomb was still classified.

However, based on these pictures, British (and Soviet) scientists were able to come up with a very good estimate for the bomb yield!

Let’s give this a try….

Atomic bomb blast

Assume a total amount of energy E dumped into an infinitesimal volume very rapidly. Assume a resulting spherical shockwave of radius r(t) expanding into the surrounding undisturbed air of density .

We have four parameters:

ML2/T2 E bomb yield

L r shock radius

T t time

M/L3 r air density

and three units (L,M,T) , which means there should just be one dimensionless parameter(!) that remains constant during this process!

Atomic bomb blast

Plugging in, we can quickly find that:

Et²/rr⁵

should be constant during the expansion of the shockwave.

Atomic bomb blast

Assuming the constant of proportionality is just 1, and using the scale bar so thoughtfully provided by the Army, one can calculate that the yield of the Trinity test was around 25 kT. Because of the beauty of scaling, one can actually do tests with, say, dynamite to determine the actual value of that numerical coefficient.

Comment on Reynolds Number—Transition to turbulence from Laminar flow to turbulent flow

Show Okubo’s Swimming Reynolds Number Diagram.

Turbulence:

Turbulence is the extraction of energy from the mean flow to a turbulent flow. This energy is extracted at a relatively large scale that is either set by geometry (i.e. boundary layer height) or by stratification (which will limit turbulent eddy size). The energy then cascades down from this large-scale with virtually no dissipation (thus viscosity is not important at the larger turbulent scales) to smaller and smaller scales

In c:\class\movie the following movies are relevant

1_06re1000 –series

2_05005.mov

2_05015.mov

2_inch_fl

W_inch_smlov

Block_re3000 & re40000

Lam-turb

Reynolds 2

Turbulent flows do not allow a strict analytical study. We must therefore rely on physical intuition, dimensional analysis and observations.

According to an apocryphal story, Werner Heisenberg said on his deathbed “When I meet God, I am going to ask him two questions: Why relativity? And why turbulence? I really believe he will have an answer for the first.”

The general problem of fluid flow is, in fact, one of the seven Millennium Prize Problems for which the Clay Institute of Mathematics is offering $1 million

Characterisitics of turbulence:

(1) Irregular, chaotic and unpredictable flows

(2) Highly Non Linear. The nonlinearities serve two purposes. First it causes the relevant nondiimensional parameter to exceed a critical value. U,L is essentially a measure of non-linearity. Secondly the non-linearity results in vortex stretching which drives eddying motion which maintains the eddy-like nature of turbulent flows

(3) Diffusivity: Turbulent flows are characterized by a rapid rate of diffusion of momentum, heat, salt, ect.

(4) Vorticity. Turbulent flows are characterized by high levels of fluctuating vorticity production structures that vaguely resemble eddies that coalesce, divide stretch and spin. The eddy size varies continuously from the largest scale (can be many meters and either set by geometry or stratification) down the smallest size of a mm that is set by viscosity.

(5) The vortex stretching mechanism transfers energy and vorticity to increasingly smaller scales until gradients become so large that they are dissipated by viscosity. Turbulent flows therefore require a constant supply of energy to make up for the viscous losses.

These features of turbulence suggest that many flows that seem “random” such as gravity waves are NOT turbulent because they are not dissipative, vertical and nonlinear.

Turbulence flows are flows. Turbulence is a not a feature of fluids—but a feature of flows.

Draw Picture on Board of growing boundary layer—use data from by dye experiments. The largest eddy size will be the height of the boundary layer—if it is unstratified. But if it’s stratified it will be smaller.

And the eddies get smaller and smaller until the energy gets sucked out by viscosity.

Typically turbulent flows are isolated from the mean flow by removing a time series.

We define the mean as:

And thus the turbulent velocity, u’, is:

Exploiting the commutative property we find

Identical averaging is done to obtain values for v,u pressure,temperature, ect.

Note that the mean of the turbulent quantities is zero. But the mean of the product of turbulent quantities is generally not zero.

Note that the turbulent time series will have some “memory” in that the autocorrelation will fall off to zero only after a certain lag. This is called the integral time scale.

When taking these averaged with data—you want to take it over a time scale that you expect the background flow to remain relatively constant. Thus if you had data from a tidally driven estuary—you would want to average over less than an hour—typically 10-20 minutes are used

Now that we have the mean –we can calculate the variance and the root mean square (the standard deviation) of the turbulent quantity u’

Often the derivation of the equations that govern turbulence utilize a tensor notation to make book-keeping easier when developing these equations. This can be seen by how concise the 3 momentum equations can be written:

Write on board full equations for du/dt, dv/dt and dw/dt

Then write

Sure saves a lot of ink!

Where T is temperature and this term (T-To) incorporates the effects of stratification on vertical accelerations.

We can also write a concise continuity equation for both the instantaneous, mean and turbulent flows, all of which look like:

Now we can write the mean momentum equation as .

Where the capital letters are the mean and the primed numbers are the turbulent quantities.

Taking the mean of the first term we find:

The turbulent quantities also drop out of the mean of the terms on the right hand side.

However—the nonlinear term—we find:

The first term on the right hand side is non-zero

The second and third terms on the right hand size are zero.

We can write the forth them noting the chain rule:

And using the continuity equation we find that the last term on the rhs is zero. Thus

Thus the mean momentum equation reads

END OF LECTURE

This last term is the Reynolds stress term. It enters the mean momentum equation via the non-linear term and the turbulent motion. This extra stress on the mean flow are much larger than the viscous contributions—except very close to a sold surface where the turbulent fluctuations (and Reynolds number) are small.

The tensor ru_iu_j is called the Reynolds stress tensor and as nine components.

This is a symmetric tensor. The diagonal components are normal stresses and the off-diagonal components are shear-stresses.

If the turbulent fluctuations are completely isotropic (draw figure of u’w’) then the off-diagonal components are zero and u^2=v^2=w^2. Here there is no directional preference.

However, it the turbulence is anisotropic the turbulence field has polarity (draw figure u’w’). Positive w tends to show a reduction in u, and negative w results in increase in u.

Draw graph of vertical shear showing how this occurs.

Go back to the last equation and expand it for the u velocity. Here we get the three stresses

d(u’u’)/dx + d (u’v’)/dy + d(u’w’)dz

These represent vertical and horizontal stress divergences and they are parameterized by an eddy viscosity times a mean shear.

t=u’w’=Av du/dz

You can think about this the eddy viscosity represnts the turbulent fluctuations and it is taking energy out of the mean flow. Thus both the turbulence and the mean flow are contained in this parameterization.

What are the units if Av?

L²/T

Think about this as an eddy size times a eddy velocity. And the eddy size is the largest eddy size in the turbulent flow—since that is the one that is taking the turbulence out of the mean flow and cascading it down to smaller and smaller scales.

An identical analysis could be done to the temperature equation from which we would find

F=<w’T’> =Kv dT/dz

Draw examples of isotropic vs anisotropic turbulent fields.

Ask: Which one would have non-zero (u’w’);

Draw example of why this might occur using a sheared flow.

This tends to transport velocity downward and smear out the shear.

Thus the paradox. You need shear to produce turbulence (We’ll see this more explicitly when we discuss the turbulent kinetic energy equation) yet turbulence reduces vertical shear. This is perhaps why it’s intermittent. There is a force that is driving shear in the flow. The shear increases until turbulence develops. The turbulence reduces the shear – which shuts down the production of turbulence. Thus the forcing allows the shear to re-intensify and start the process over again.

The Turbulent Kinetic Energy Equation:

In addition to using tensor notation—derivation of the TKE equation is made more compact using “comma notation”

(1)

Using this notation we can write the momentum equation for the total and mean flow as:

(2)

(3)

Subtracting these two equations we get the equation for the motion of the turbulent velocity ui.

(4)

The Turbulent Kinetic Energy Equation is obtained by multiplying the above equation by the turbulent velocity ui

The first two terms on left hand side give

(5 & 6)

5 represents the local time rate of change of turbulences—while equation 6 is the advection of the turbulence by the mean flow.

Terms 3-5 in equation 4 become:

(4)

(5)

OK the trick to this guy is noting that

And the second term in the above equation is zero because the u_j,j term is the continuity equation

Thus

The fifth term in equation 4 is zero because the stress term (u_iu_i ) is already averaged and thus a constant. Multiplying this constant by the turbulent velocity u_i results and taking the average results in zero.

On the right hand side of 4 we find

??

And

The messiest term is the viscous term (the vicious viscous term!)

We write is as:

The second term on the right hand side is simply a mathematical trick and adds zero to the equation.This it is the doubly contracted product of a symmetric tensor and an anti-symmetric tensor which is zero.

Details of the derivation of this term can be found in Kundu (page 436). It’s messy and does not provide much insight. But the key is that by expanding the above equation and writing it in terms of the fluctuating strain rate

This term represents the deformation of a parcel of fluid associated with gradients in the flow field.

The viscous term can be written as:

Collecting terms we write (equation 34 in Kundo)

The first three terms are associated with the redistribution of turbulence by the turbulence itself. It does not generate or dissipate turbulence.

The last three terms are the

P=Shear production

B=Buoyancy production

e=Dissipation

Simplest balance if no stratification

P= e

If there is stratification then

P= e+B

P is the production of turbulence by stress acting on the mean shear. It takes energy out of the mean flow and inserts into the turbulent flow at the largest turbulent scales

The Buoyancy Flux can be of either sign. IF the density stratification is un-stable it will be a source of turbulence. However if the density stratification is stable, as is often the case it will be a sink of turbulence—and this energy goes into mixing and raises the potential energy of the system.

Finially the last term is the dissipation. This occurs at the smallest scale.

END OF LECTURE

Start with the TKE equation—writing u’ as q.

Scaling suggests the dissipation term is q3/l

Flux Richardson Number

Mixing Efficiency

Turbulent Prandle Number & Richardson Number.

What controls the scale of the biggest turbulent eddy?

(1) length from boundary:

(2) stratification

Eddy sizes:

Kolmogorov Scale—eddy size controlled by dissipation and viscosity

Ozmidov Scale—eddy size controlled by stratification and dissipation rate

Later I’ll make some estimates of these scales.

Turbulence scales and Spectrum

Turbulence is dissipative: Thus it is reducing the kinetic energy of the flow. Therefore dissipation will have the units of the time rate of change of kinetic energy.

What units are those?

L^2/T^3.

The basic idea of turbulence is that the largest eddies are extracting energy at the larges scale and cascading it down to smaller and smaller scales until it the scale gets so small that effects of viscosity become important.

What is the size of the largest eddy?

Could be controlled by boundaries—but in most of ocean controlled by stratification.

Can we make a length scale associated with stratification. The two important parameters are stratification (N) and dissipation. This scale is called the Ozmidov Scale

What is the smallest scale?

Here the important parameters are dissipation and molecular viscosity.

This scale is called the Kolmogorov microscale.

Here all the turbulence is going into heat.

Consider the case of a 100-watt household mixer in 1 kg of water. All the power to generate the turbulence the rate of dissipation is 100 watt/kg= 100 m2/s2. Using viscosity of 10-6 m2/s the Komogorov micorscale is .01 mm.

In between the largest scale and smaller scales of the turbulent motion—there is a region called the inertial subrange—because energy is transferred by forces of inertia (vortex stretching). Both the production and dissipation of turbulence are small in this range. Thus the only thing that will control the turbulent spectrum (S) is the dissipation and of course the wave number (which is a dimension in the spectrum)..

While the large scale turbulence is anisotropic – because it is “aware” of the mean gradients in the flow the smaller scale turbulent motion is nearly isotropic. This we call

S~ f(e,k)

The spectral energy is energy per wave number which has units m^3/s^2.

Thus of S= e^ak^b

And dimensionally we find that a=2/3 and b=-5/3.

Thus the turbulent spectum rolls off with k ^-5/3

This gives us a way to actually measure dissipation—by measuring the turbulent spectrum in the inertial sub range. This is what microstructure profiles do ( the Hartmut Peters papers) or one could measure the production of turbulence (Trowbridge, Stacey papers)

Boundary layer turbulence

Consider case of unstratified fluid.

Example is the boundary layer problem where there is some bottom friction u* and the eddy size is going to scale with u* times the distances from the boundary or

k u*z

where k is von Karman’s constant (one of the founders of fluid mechanics).

So the stress is simply the eddy viscosity times the vertical shear—or can be also written as the friction velocity squared.

Which yields

which can be integrated to yield the famous “Law of the Wall”

where z_o the bottom roughess is a constant of integration obtained by assuming that the velocity goes to zero at height z_o above the bottom.

If you square the above equation and rearrange the terms you find

So like the wind the bottom friction is proportional to the current speed squared/.

Frictional forces on the bottom will tend to slow down the fluid motion.

note that we use u times the absolute value of u so that the bottom frictional force always opposes the flow. While numerous things can impact the value or k it is also tends to fall in the range of .002- .003.

OK lets try to estimate the Komogorov microscale in an unstratified boundary layer.

Given a current speed calculate the shear production:

If this is equal to dissipation (it must be because there is no stratification).

We can estimate the microscale.

Example of a mixing process

DO an example of changing kinetic energy from a linearly stratified water column to a well mixed water column.

Potential energy is

PE=mgh

For a continuously stratified fluid it is:

_{SO if density is constant this becomes:}

PE=r_ogH²/2

Check units and note that this really needs to be multiplied by an area to get PE!

Which is the mass of the water column (rho*H) times the center of mass H/2

However if the water is stratified—say

r(z)=r_o+a(z/H-1/2)

Show that the mean density is r_o

How may a water column get stratified?

Surface Heating.

Straining (shear)

So mixing the stratified water column produces a water column of density r_o

Which one has a higher potential energy?

What is the change in potential energy?

DPE=agH²/6

Check units—and note that a has units of density.

Say that a=2 and H=20 meters

Results in a change of potential energy

=1333 kg m^2/s^2

For each square meter of ocean.

If this happened over a 24 hour period than the work (d/dt PE) would be

=.0154 kg m^2/s^3

Two questions--- how much turbulence production do you need to mix this water column

(Production = rho*(u’w’) du/dz)

Check same units

So home much TKE do you need to produce to mix this water column?

Depends on how Efficient the mixing is:

Do example of linearly sheared 1m/s flow to estimate shear production.

Entrainment: How mixing works. Dk/dz * ds/dz

Etc..