Hybrid Car – More Fun with Less Gas

Help Request - Page 2

register ::  Login Password  :: Lost Password?
Posted by Simon D on February 24, 2006, 6:54 pm

It's the cross sectional area of the plate.

Don't let it worry you though, the calculations are based on linear
temperature gradients, which will not exist in this scenario. As a result,
the answer is not to be trusted.

Besides, I'd defy you to hold your hand over a pot of rapidly boiling water
for any significant time, with or without a copper plate.

You need to look up things like partial differential equations and  Laplace
to get a grip on this kind of problem. See
section 2.

Even in 1 dimension it's a bit tricky, with your problem more so. You also
need to allow for the fact that heat is not just being applied to the part
of copper that is submerged, but the section above water that is being
heated by the steam and condensation.

How long?? I wouldn't like to say.


Posted by Alan Combellack on February 25, 2006, 2:04 pm
  Thanks again but I'm still confused.  Comments are inserted below.
  Alan C

The plate was 1 ft square or 929 cm^2.  Thickness was 1/16 inch, or .15875
cm. Cross section area is thus147.48 cm^2.  Am I still screwed up?

I realise that.  The actual temperature change will be very complex and I am
actually trying to find a reasonably accurate but reasonably simple
expression to determine how fast I can get heat to flow from hot water into
gas in a cylinder.  I am fairly sure that it will turn out to be an
expression similar to the exponential rise (or fall) of the voltage across a
capacitor fed through a resistor such as Vc=Vs(1-e^(-t/Time constant)), but
maybe not!.  The time constant here would be R*C and I need to find the
thermal equivalent of the time constant figure.  I am no mathematician so am
having problems with this.  The gas bit is easy enough to deal with but I
need to evaluate the very significant effects of the mass of metal in the
heat exchanger, which also has to be heated (and cooled) and the speed with
which heat flows through the metal.  This was the reason for my original
simplified question

OK.  Please replace my fingers with some form of thermometer.


I think I can handle the bits about heat going into and out of the parts of
the plate OK but am still unable to figure out how the heat will actually
move with time from the hot end to the cold end, particularly the time
intervals involved

Thats the problem.  I suspect you are better able to guesstimate it than I
am since you took the trouble to reply.
I expect you are getting fed up with this by now but may I impose a little
more on you obvious expertise snd re-define my example a little?
I will use cgs units since you seem to prefer this.
I have two 100 cm squre plates with a thickness of, say 0.2 cm.  These are
connected to each other through a strip of the same gauge copper which is 20
cm long and 2 cm wide.  One plate is in air at 20 deg C.and the air volume
is large so generally, except in the immediate vicinity of the plate, Has a
very slow rise in temperature ( air to surface R value, in the British
system is about 0.67 due to the surface film.  I don't know the value of R
in cgs offhand).  The other plate is also at 20 deg C initially but is now
plaunged into my boiling water.  The plate will heat up at a rate depending
on its mas and specific heat and heat will start to flow along the
interconnecting strip at a rate depending on its thermal conductivity and
cross sectional area.  The incoming heat will start warming the plate in air
at a rate depending on a lot of factors including its mass; specific heats,
thermal conductivity; surface area and temperature... To ask for a complete
equation relating all this stuff is, I think, a bit much but all I really
need might be what you have already given me which is how to relate the
surface areas, specific heat, temperature gradients; and actual temperatures
and themal conductivity to time, particularly through the connecting strip.
If you can give me this I will be most grateful.
  Maybe I will do some actual testing to see what happens.
  Alan C

Posted by Alan Combellack on February 25, 2006, 3:21 pm
 My apologies about the first comment.  I seem to be unable to see the
difference between area and volume.  Your 4.8 cm^2 for the cross sectional
area is, of course, quite correct and I am wrong. Mea culpa.  I still have a
problem with the rest of it though.

  Alan C


Posted by Simon D on February 25, 2006, 7:35 pm

The reason I'm aware of the maths involved is because I did this kind of
thing at University some 15+ years ago. To say I'm rusty on the details
would be putting it kindly <g>. I would recommend you ask your question in a
physics group, where you will get a better answer/answers, and they'll be
checked by others, so you can be more confident of the values you're given.

Good luck.

Posted by andre_54005 on February 24, 2006, 7:25 pm
Alan Combellack wrote:

The copper plate is very likely irrelevant.  Try just holding your hand
6 inches above the surface of a "rapidly" boiling 13 inch diameter pot
of water.
Andre' B.

This Thread
Bookmark this thread:
  • Subject
  • Author
  • Date
---> Re: Help Request Alan Combellack02-23-2006
please rate this thread