Re: Measuring Power dissipation emperically

On Sun, 13 Apr 2008 14:42:38 -0600, "Jon Slaughter"
<Jon_Slaughter@xxxxxxxxxxx> wrote:

Um, if, say, you put the device in water, the power dissipation will change
the temperature of the water. This is how they measure the power dissipation
of resistors(or used too) and also other things. Don't you ever remember
doing those experiments in physics?

I sure do. I also remember putting it in an insulated (adiabatic)
environment, so that none of the heat generated leaks out to the room.
However, doing the test in an Coleman ice chest caused a problem.
Since the system was insulated, the heat would just accumulate. The
more power we pumped in, the hotter it became. It never reached
equilibrium. If we kept applying power to the resistor, it would
eventually get hot enough to melt the ice chest.

However, running the same test in an uninsulated container yields a
different result. The thermal gradient (degrees-C/watt) between the
resistor and the accompanying heat sink, and that between the heat
sink and the air are easily calculated or measured. Applying power
over a long period of time results in equilibrium, as the air removes
as much of the heat as possible. Move the air past the heat sink, and
even more heat is removed.

However, you do have a point. Your original two sentence question
managed to misuse the definitions of power, dissipation, energy, and
temperature. I have no clue as to what you are trying to actually
measure, empirically or otherwise. That's why I suggested that
perhaps you might want to measure the temperature (empirically with
your finger).

The real problem is that you haven't supplied sufficient information
with which to answer your question. What are you measuring (units of
measure)? What devices are involved? What resources do you have?
What are you really trying to accomplish? We'll be happy to help, if
possible (unless this is a student exam question).

However, you wanted an empirical measurement:
<http://en.wikipedia.org/wiki/Empirical>
which implies that one cannot use test equipment or measuring devices,
and must use your senses (touch, observation, smell, hearing, etc).
Been there, done that. Put your finger on the heat source. If you
can hold it there, the cooling system is working. If you have to
remove your finger after about 15 seconds, it's at the upper border of
acceptable. If you burn the hell out of your finger, you need a
bigger heat sink.


Same page:
"In a second sense "empirical" in science may be synonymous with
"experimental.""

I've seen it misused both ways. If you simply dropped the term
"empirically", it would mean exactly what you're suggesting.
Measurements are done "experimentally" as in standard procedures and
methods. However, adding the term "empirically" implies that the
standard methods of measurement are somehow inadequate, and that it is
"dependent on evidence or consequences that are observable by the
senses". Note the word senses. If you want to emulate a thermometer
with your "senses", I supplied the method to do so.

I'm looking something more precise than 2 or 3 orders of magnitude.

Actually, I know doctors and nurses that can measure temperature with
the back of their hand to within a few degrees. Granted, the range is
limited, but the accuracy is amazing. Incidentally, 2 orders of
magnitude is roughly the difference between freezing and boiling (in
degrees Celsius). With 3 orders of magnitude, your system will
probably ignite.

For example, I know that I could use water. Put the device in the water and
run it for a while. The water will heat up. Since one knows the specific
heat capacity of water,

http://en.wikipedia.org/wiki/Specific_heat_capacity

Its not difficult to measure the initial and final temperature of the water
to get the power dissipation... actually its quite easy(much easier than
measuring the temperature of the device itself IMO... although more work to
setup). The formula is a simpel calculation.

Sure. Please re-read my comments at the beginning of this rant. If
you insulate the system, the heat will accumulate. If you expose your
water bucket to the environment, the environment becomes part of your
thermal calculations. For example, stupid things like someone opening
the door, and creating a draft, will affect your results.

Hmmm.... I smell high skool or college fizzix homework. Waco, Texas?

BTW, empirical doesn't mean no calculation even experiments require
calculations(you might have to add 1 + 1 to get 2.

OF course, you're correct. However, you have apparently also missed
my point. You haven't defined (or controlled) what you're measuring.
By supplying a rather useless and possibly humorous answer, I had
hoped to kick you into some altered reality where people that ask
question also do their homework first, and supply sufficient
information to answer your question. Incidentally, you're asking in
the wrong newsgroup. This is a fizzix problem, not an electronic
design problem.

"...such methods are opposed to theoretical ab initio methods which are
purely deductive and based on first principles."

Yep. Just one problem. You haven't supplied any additional
information sufficient to answer your question or do your homework.
Keep trying but realize that you're not going to learn anything have
people in a newsgroup do your homework.


--
Jeff Liebermann jeffl@xxxxxxxxxx
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
.

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