When a whirlpool forms above a tub drain as the tub empties, does the
whirlpool speed up the rate at which the tub drains? (or slow it
down?) (or not affect the rate at all?) I am looking for a web page
that explains the physics behind the answer.

I know that a whirlpool (or vortex, e.g., tornado, waterspout,
whirlpool) is a "dissipative structure", I just haven't seen any
discussion of how much (or whether or not) it increases the rate of
dissipation. An answer which generalized to this overall perspective
would be especially useful.

Is your fluid a liquid flowing through a hole which is initiallied
filled w/ air or the same fluid?

An interesting experiment for you!  If you have two of those two litre
pop (or soda) bottles.  Empty them both.  Then fill one with a fluid,
such as Water.  Keep the other one empty.  Next take the empty one and
turn it upside down and line up the two openings of the bottles.  Next
use tap or some other method to SEAL the two lips together.  So now
you should have two bottles, one empty and one full with the lips
sealed so that no fluid may escape.

Now turn it over so that the fluid flows from one bottle to the other
(kinda like an hour glass).  You will see that bubbles are rising
through the fluid.  This is the first scenario.  Record the time it
takes for the fluid to completely leave the bottle from which it is
flowing.

Now for the next run - start swirling the fluid by spinning it with
your hands very quickly to make a sort of whirlpool.  Then quickly
flip over the bottles into the hour glass fashion once more.  You will
then see that the fluid is flowing down through the sealed lips in a
whirlpool fashion.  get the time to complete.

You will notice that the fluid moves from the top bottle to the bottom
in a much shorter time.

Just a friendly comment. =)

Thanks for the helpful comment. It helps confirms my belief that a
whirlpool does increase draining efficiency. I'll give it a try when I
can find two 2 liter bottles. (I don't drink softdrinks much anymore.)

I still need to know WHY a whirlpool increases draining efficiency. A
friend and I are discussing dissipative structures (DS's) and I
pointed to a vortex as a great example of a self-organizing DS (SODS?)
because it accelerates the dissipation of energy (temperature)
gradients. Then he pointed out that vortices are often associated with
turbulence (e.g., on the tips of wings), NOT more efficient flow. He
claimed that laminar flow is the most efficient flow, not votical
flow. I tentatively agreed, but claimed that vortical flow was still
more efficient that random flow (e.g., water draining from a tub w/o
laminar or vortical flow). He wasn't sure he believed this. Hence the
question.

Laminar Flow is nice most of the time.  But there are many instances
to even CREATE transition or turbulent flow which are highly
beneficial.  To prove my point please see the following journal
articles (which are simply two examples) - (I am not the author of any
of these papers) -

Enhancement of passive mixing tabs by the addition of secondary tabs 
AIAA Paper 96-0545 (AIAA Accession number 18505)
D. G. Bohl and J. F. Foss (Michigan State Univ., East Lansing)
AIAA, Aerospace Sciences Meeting and Exhibit, 34th, Reno, NV, Jan. 15-18, 1996

The Design and Testing of a Winglet Airfoil for Low Speed Aircraft
AIAA 2001-2478
Mark D. Maughmer, Timothy S. Swan, Steven M. Willits (Penn. State
Univ., Univ. Park)

I'm not sure if TECHNICALLY a vortice is a turbulent structure.  Some
people in industry might refer to it as one and some strict academics
might not.  I'm not sure.  But let me reference Kovasznay when he
decomposed the fluctuations into... the VORTICITY, acoustic, and
entropy modes.  So I believe it can be classified as one.

For the winglets on the plane, which amplifies the trailing vortices
from the wingtip - Yes the drag coefficient is reduced at the correct
conditions.  That is why at airports the past years you have seen them
being added to older planes.  Also if you look at some birds they have
them!  A great example of man copying nature... of course I believe
that this discovery was refined and implimented with CFD methods.

Let me ask around today and I'll post a comment later on the subject. 
I don't think I'm going to find an answer in papers but by an
explanation from a professional.  I'm guessing it can best be
described by some kind of math juggling with the navier-stokes
equations and not a turbulent explanation.

Best
Steve.

For the more complex case of a single phase flow (not two phase flow
as my example in my first comment/experiment) The rate will stay the
same.  According to the Reynold's Transport Theorem and the
conservation of mass:

material derivative*(integral(density w/ volume))=partial derivative
w/ time of the integral(density,volume) + integral of
(density*velocity dot Unit normal) over the area.

for simplicity lets reduce this to:
0 = integral of (density*velocity dot Unit normal) over the area
(inlet and outlet).

Since the density is constant (water) we can take that out of the equation...

0 = integral of (velocity dot Unit normal) over the area (inlet and outlet).
since the velocity is dotted with the unit normal of the exit (drain)
you can see that any component in the direction normal to the outlet
will remain and any component of hte velocity flush with the outlet
(eg the swirl/whirlpool) will be thrown out.  Therefore the flow rate
out of the drain will remain constant.
This was for the viscous case.

For the invicid case we can just use the principle of bournoulli to
show that it will flow at the same rate.

You can look up RTT in any basic fluid mechanics or dynamics text book.

Steve.

Any bartender who has filled wine carafes from bottles could tell you
observing the speed improvement from creating a vortex doesn't require
taping two bottles together. When emptying a bottle, there is no air
pressure pushing down from the top of the bottle so liquid leaving the
bottle must be replaced by air from outside the bottle. In the case of
no vortex, air must bubble up through the liquid, creating a great
deal of turbulence at the neck which reduces flow. When a strong
vortex is created in a properly shaped bottle, the the vortex extends
through the neck and air can pass directly through the vortex into the
top of the bottle. The pouring speed increase when this vortex is
created is dramatic, and can actually seem like the liquid is
pressurized when the hole is formed. However, the shape of the bottle
is significant: if the vortex doesn't extend through the neck, then
the speed increase will solely be due to a more streamlined flow and
air must still bubble up through the liquid. It would be interesting
to take two identical bottles and cut the bottom off one bottle, then
empty them with and without vortex twirling to see what the speed
differences were - although the open bottom one would probably drain
so quickly it would be impossible to create a vortex.

In the case of the bathtub, of course, air pressure is already forcing
the water down the drain so this bottle effect wouldn't be of any
significance. Assuming a properly functioning drain, any speed
difference should be attributable to more efficient flow.