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[Phys-L] Re: siphon

Yes JohnD, it was only an "artificial game," on using the "energy only"
approach. I think it would be very difficult to develop a
pedagogically acceptable sequence (writing a book for students and
teachers) in which energy is introduced without relying on the concept
of force, or other concepts (such as work and pressure) that are now
defined in terms of force. Yes, pressure can be viewed as energy per
unit volume. But that is not obvious, unless a traditional sequence is
followed up to the point at which the P*dV is shown to be the same
thing as familiar F*x. And I also see the rho*g*h in your definition of
"head," where g stands either acceleration or force per unit mass.

Yes, we did inherit a traditional path from g to F, then to F*x and
eventually to P*dV. The reversed path from P*dV to F*x (and then to F
and g) can probably be constructed. But that would not be easy.

Ludwik Kowalski
Let the perfect not be the enemy of the good.

On Tuesday, Aug 23, 2005, John Denker wrote:

On 08/22/05 19:49, Ludwik Kowalski wrote:
How to explain siphoning of water using the "energy only" approach?
Yes, each kilogram of water, in the final analysis, loses some
potential energy. But why is its potential energy increasing before
decreasing? Borrowing to pay back later?

This question is easy to answer, and the answer teaches
some simple yet powerful concepts.

Remember that the concept of
force (or force per unit area) can not be used to explain the process.

Why should pressure not be used?
1) Important point: saying that energy is an important idea
doesn't mean it is the *only* idea you are allowed to have
in your head.
2) Even if we decide to play an artifical game in which we
limit ourselves to energy ideas, pressure is still in-bounds.
Pressure is not just force per unit area; it is also
energy per unit volume. In thermo books P is _defined_
in as - partial E / partial V at constant S.

There is a wide class of problems that can be solved either way,
either in terms of energy or in terms of force. Sometimes one
is a little easier than the other; sometimes they're about the
same. If you're smart, you'll master both ways.

The sipon in particular is a veritable poster child, good for
showing off the advantages of the energy approach.

why is its potential energy increasing before decreasing?

Before we get started, let's be careful about what we call
"the" potential energy.
a) GPE of the water alone?
b) GPE of water + column of air above it?
c) total PE of the water, including not just GPE but
other types of PE, including P dV contributions?

Notions (b) and (c) are numerically equal; they are two
different ways of looking at the same thing. The siphon is
hard to understand if you fixate on (a) alone; otherwise
it's easy.

In fluid dynamics, the concept of _head_ is very heavily
used. It is defined as:
head := pressure + rho g h
where rho is the density and h is the height. Head has
units of potential energy per unit volume. You can also
think of it as the _enthalpy per unit volume_ of the water.

It is edificational to calculate the delta(head) between the
two reservoirs. You can then distribute this over the length
of the siphon tube, to get an average value of "head per unit
length" that is really what drives the flow; this average
value gives you an estimate of the actual head-versus-position
function. You can make a graph of head versus position,
which leaves nothing to the imagination:

_________________
/ upper reservoir
/
/
________________/
lower reservoir

While you're at it, you can (on the same abscissa) draw pictures
of pressure versus position, and rho g h versus position.

Remember point (1) above. Head is important, but it is not the
only important thing. The plot of pressure versus position is
also important, because that's what tells you whether there will
be a siphon-break. The break occurs if/where the pressure goes
to zero (*).

You should plot head for one reason, and plot pressure for another.

Returning to the graph of head: The water lowers its potential
energy (notion (b) or (c)) if/when it moves in the direction of
lower head. You can see from the graph that in the siphon, every
single parcel of water is moving toward lower head at all times;
there is no "increasing before decreasing".

=============

Generalizing a bit: fluid dynamics is a poster child, illustrating
the value of emphasizing the conservation laws. In particular,
if you formulate Newton's third law in terms of forces, applying
it to a parcel of fluid is a nightmare. Students get tangled up
trying to figure out what's the force on the left side of the
inter-parcel boundary, and what's the force on the right side of
the same boundary; these are not the same. Since the boundary
has zero thickness, you can't express "the" force as a function
of position ... which is painful, because everything else in the
problem is expressed as a function of position. You're much better
off recognizing N3 as nothing more or less than conservation of
momentum, and then just keeping track of the momentum-flow across
the boundary.

======================
Footnote:
*) You can apply a correction for surface tension if you want.

While we're on the subject: surface tension can be considered
a force per unit length ... or as an energy per unit area. If
you're smart, you'll master both ways of looking at it.
Do you remember the toroidal perpetual-motion machine?
http://lists.nau.edu/cgi-bin/wa?A2=ind0006&L=phys-l&P=R10825