Q4MEs: Heat Conduction Q - y no thickness?

Factual Questions
1

You sit through a basic heat transfer course, and it all seems logical until you go to apply it. Then, nothing makes sense. For one, how come the formula for conduction through a plane wall never accounts for the thickness of the material?
(Using q= kAdT -> where A is surface area)

Also, suppose I have insulation between an inner wall and an outer wall. The text simply shows how one might determine “U”, the overall heat transfer coefficient as U = 1/(1/h1 + dx/kDT + 1/h2) …but how the heck does one determine h? As for the middle term, is it assumed that only the thickness of the walls is irrelevant? Doesn’t each layer of construction, and its respectve thickness, create a resistance to heat flowing through it? Yet, every source will say the heat transfer across surface A = heat transfer across surface B, but isn’t that saying it doesn’t matter if you have a metal wall or a styrofoam wall???

Can anyone help me make sense of these things that college professors prefer to cram down your throat than let you digest bite by bite?

Thanks,
Jinx

2

Ok, I found an answer to the first part of my question… The equation should be
q = kAdT/dx where dx=thickness, but what about applying dx when determining the overall heat transfer coefficient, U? It seems dx of the walls is ignored (in example above)? - Jinx

3

I guess I don’t understand this part of your question. You claim “dx” is ignored, but it’s right there in the equation for U. Is the differential notation throwing you off?

As far as “h” goes, I’m assuming from your U equation that it’s a problem with convective boundary conditions. Convective heat transfer is calculated just like conductive, except the formula is q=hADT. The problem is that h must be determined according to the situation, i.e. the fluid in motion, its speed, etc. You can’t simply tabulate it for all possible situations, because the table would be essentially infinite. Does this help at all?

4

If you take your wall and divide it into a whole bunch of itty bitty thin walls each ittle wall is still going to see the same amount of heat passing through it (assuming steady state conduction). However, the temperature gradient across the wall will depend on the thickness of the wall overall - each little mini wall each takes a chunk of the temperature to pass the required amount of heat.