# High Efficiency gas furnace - return air temperature

Page 2 of 2
• posted on January 23, 2007, 6:47 pm

snipped-for-privacy@optonline.net writes:

That's an interesting definition of "efficiency". In this context, they must mean that the heat transfer is higher *per unit area* or *per unit volume* of heat exchanger.
That's unrelated to the efficiency of a furnace, which is a measure of how much of the theoretical heat energy in the fuel gets transferred to the house.
You can have two furnaces, one with a thin-wall heat exchanger and the other with a thick-wall heat exchanger that is somewhat larger, such that both furnaces have the same amount of heat transferred with the same air and flue gas inlet and exhaust temperatures. Both *furnaces* will have the same efficiency at heating the house, but the thin-walled heat exchanger is more "efficient" because it's smaller.
Dave
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 23, 2007, 7:48 pm

Not much, I'd ween, if the dimensions of a forced air furnace heat exchanger mostly depend on the air passages. With less metal, it would weigh less and cost less, but those are different concerns.
And 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?
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 23, 2007, 8:12 pm
snipped-for-privacy@ece.villanova.edu wrote:

Heh, don't try to obfuscate the facts by spewing a bunch of calcs as usual, trying to cover up. Just admit that you were wrong when you claimed that "making the metal thinner won't help transfer the heat more effectively." I showed you that:
1 - By the laws of physics, the heat transfered by conduction is inversely proportional to the thickness of the metal. Despite your well known love of spewing equations, you just completely ignored the equation I provided, complete with reference, that says you are wrong.
2 - A manufacturer of air heat exchangers states in their heat exchanger data sheet that they offer a metal thickness of .024 for high efficiency applications and an increase to .050 thickeness for applications where durability is more important.
And what's the crap about poor air film conductance on both sides of a heat exchanger in a modern high efficiency furnace. If it's so damn poor, how come these furnaces are 93%+ efficient? Could it be that manufacturers know how to make heat exchangers that are efficient, including using thinner metal and proper air flow techniques?
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 23, 2007, 9:11 pm

It's 300-year-old physics :-) What's the answer to this simple problem?
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 24, 2007, 9:00 am

Still no clue? Rewrite the steel conductivity as 50 Btu-ft/h-ft^2-F...
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 24, 2007, 1:28 pm
On Jan 24, 4:00 am, snipped-for-privacy@ece.villanova.edu wrote:

Screw you college boy. You claimed making heat exchangers thinner in high efficiency furnaces wasn't a significant factor in improving heat transfer. Actually, it's inversely proportional, per the equation backed by reference I provided you. Yet you go on spewing, like some kind of self proclaimed energy expert, chocked full of formulas and calculations, when you don't even understand the most basic concepts.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 24, 2007, 5:38 pm

Still stuck? Try 600 Btu-inch/h-ft^2-F.
This is an extremely simple heatflow problem :-)
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 25, 2007, 2:00 pm

Another clue: the thick steel conductance is 600/0.050 = 12,000 Btu/h-F-ft^2.
Beginning to understand the basics yet? :-)
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 25, 2007, 8:54 pm

And if we halve the thickness, it becomes 24,000 Btu/h-F-ft^2. Wow!
Got a clue yet? :-)
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 25, 2007, 10:21 pm
On Jan 25, 3:54 pm, snipped-for-privacy@ece.villanova.edu wrote:

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/hbase/thermo/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-Exchanger-Plate-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?
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 26, 2007, 3:23 pm
snipped-for-privacy@optonline.net wrote:

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?
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 26, 2007, 10:25 am

So the steel thermal resistances are 1/12K and 1/24K h-F-ft^2/Btu.
Now what do we do with resistors in series?
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 26, 2007, 3:01 pm
On Jan 26, 5:25 am, snipped-for-privacy@ece.villanova.edu wrote:

steel thermal resistances are 1/12K and 1/24K h-F-ft^2/Btu.

Try shoving them up your ass and get back to us on how many fit.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 27, 2007, 10:23 am

Add them. So the "less efficient" heat exchanger would have a thermal resistance of 1/5+1/12K+1/5 = 0.4000833 h-F-ft^2/Btu vs the "more efficient" 1/5+1/24K+1/5 = 0.4000417 h-F-ft^2/Btu, with 0.01% less thermal resistance :-)
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 23, 2007, 11:28 pm
wrote:

Scrubbing surfaces for better heat transfer IS the industries proverbial _Let's Build a Better Mouse Trap_. Combustion efficiency design and integration is already well established and has many choices to meat a criteria.
What was interesting to watch is the Discovery Channel's Lance Armstrong saga. Specifically, detailing the interaction of air to the surface of his clothing.
It went from researching golf ball dimples to mother natures design of a Tuna! ISTR, dimpled and scaly surfaces were the focal points.
As relating to heat transfer, a couple of years ago there was a program showing the advances of ancient peoples, and how their levels of achievement ranked to modern times. How interesting that a properly hammered Wok was shown to have the best heat transfer of all other kinds of modern designed woks.
My point is there's room for improvement.
-zero
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 24, 2007, 2:55 pm
Well, the verdict is in. I called a Goodman authorized HVAC dealer/contractor (in Denver Metro), and their technician told me that the reason that return air temp must not be less than 55F is because of the possibility of excess condensation.
BTW, I first called the Goodman hotline, but they told me that, for liability reasons, they didn't provide tech support directly to individuals. However, they told me to call one of their authorized dealers with any questions. They gave me three names, and I called one of them.
Cheers.
On Jan 18, 10:12 pm, snipped-for-privacy@yahoo.com wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 24, 2007, 10:57 pm
On 24 Jan 2007 06:55:45 -0800, snipped-for-privacy@yahoo.com wrote:

Gee, where did I hear that answer about a week ago when you posted your question? Oh, thats right. It was from me! Bubba

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on January 24, 2007, 11:05 pm

"Trust, but verify" :)