wrote:

On Jan 25, 3:54 pm, wrote:

wrote:

... if the metal is a good conductor, eg steel with 50 Btu/h-ft-F, with
poor airfilm conductances on both sides, eg 5 Btu/h-F-ft^2, thinner steel
won't help much. How much, in this case, starting with 0.050" steel?

Heh, don't try to obfuscate the facts by spewing a bunch of calcs as
usual, trying to cover up...

It's 300-year-old physics :-) What's the answer to this simple problem?

Still no clue? Rewrite the steel conductivity as 50 Btu-ft/h-ft^2-F...

Still stuck? Try 600 Btu-inch/h-ft^2-F.

This is an extremely simple heatflow problem :-)

Another clue: the thick steel conductance is 600/0.050 = 12,000 Btu/h-F-ft^2.And if we halve the thickness, it becomes 24,000 Btu/h-F-ft^2. Wow!

Got a clue yet? :-)

Nick- Hide quoted text -- Show quoted text -



You've established yourself in the group long ago as someone who likes
to try to impress folks by spouting numbers and equations, but having
no common sense when it comes to practical home repair subjects. In
this thread, you claimed I was wrong when I stated that the thickness
of a furnace heat exchanger does directly affect the heat transfer and
efficiency. You posted:

"But metals are such good conductors that making the metal thinner
won't
help much, given high resistance air layers on both sides, and thicker
metal will spread out hot spots and increase efficiency. "

Clearly you are the clueless one, as I provided both physics as well as
practical references that you are wrong:

http://hyperphysics.phy-astr.gsu.edu...mo/heatra.html
Conduction is heat transfer by means of molecular agitation within a
material without any motion of the material as a whole. If one end of a

metal rod is at a higher temperature, then energy will be transferred
down the rod toward the colder end because the higher speed particles
will collide with the slower ones with a net transfer of energy to the
slower ones. For heat transfer between two plane surfaces, such as heat

loss through the wall of a house, the rate of conduction heat transfer
is:


Calculation


Q/t = kA(Thot-Tcold)/d


Q = heat transferred in time = t
k = thermal conductivity of the barrier
A = area
T = temperature
d = thickness of barrier


Clearly from the above, the conducted heat transfer is inversely
proportional to
the thickness of the heat exchanger.


And second, from an industrial company that actually makes air to air
heat exchangers:


http://www.anguil.com/downloads/Heat...ate-Anguil.pdf
In the spec sheet for their air heat exchanger product it says:


"Plate thickness ranges from .024" for high efficiency to a heavy-duty
and durable .050" thick plate"


So, just fess up and admit you were wrong, instead of trying to
obfuscate with one liners and leave people with misinformation. It
must be embarrassing to have been caught in such a blatant lack of
knowledge in your self professed field of expertise. I mean, if you
don't realize that thickness of a material directly affects heat
transfer, which you should have learned in basic physics, what good are
any of your other theoretical pontifications?

while I don't have the math to truly follow along, it would seem, while
ther are valid points fer and agin, the manufacturers would not bother
with potential warranty issues if there were not an advantage, but that
the advantage is relatively small, what with the enormous amount of
square feet in the heat exchanger and the large tmeperature differential
across it.

To avoid the aforementioned warranty issues, they probably have to make
the heat exchanger out of more corrosion resistant stuff, ie add nickel
or chrome, which I would assume negates the advantage to a point, since
IIRC stainless steel is less efficient a conductor than plain steel.

Anyway, cantcha jus git along?