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Old 06-11-2015, 10:25 PM   #31
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Haha.. whoops.. Did anyone notice my error? 14%+76% equals 90%. That was supposed to be 14% and 86%.

By the way, I get those numbers by saying that the ribs are 4" wide (in reality it's actually more like 2.5" because of the way they're formed) and the distance between is 25". That was a guess, but I've measured those spaces enough in my bus that I think it's pretty close.

I just corrected my previous post
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Old 06-12-2015, 01:09 AM   #32
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It honestly doesn't make a difference to me which is right, but I'm sincerely curious to see which way it works out. While in school I managed to dodge the thermodynamics class so many of my peers were required to take. At the time I thought myself lucky, but now I wish I understood it better!

It "seems obvious" (dangerous as that is!) that the ribs are so conductive that the air space between them might not make much difference. But the point is made that the length where those ribs exist is small compared to the length of the entire space, and I guess the question is whether so much heat can flow through that rib connection and the metal interior facing that what's in the air space between is irrelevant. In the illustration I made earlier I implicitly assumed this was the case by shading the whole wall cavity in red. That might have been incorrect.

We could estimate R value of the fiberglass and the conductivity of the steel, and figure out the surface area and thickness of the webs on the rib and so forth... but what about the thermal resistance of the joints between the rib and the two skins? I don't have any idea how to estimate that. I poked around online a little, and it seems the conductance of that kind of joint is highly variable. Is it reasonable to only consider the heat conducting through the webs of the rib vs through the fiberglass, completely ignoring the steel skin on both sides?

IMO

We need someone with a inferred heat gun to take some real life shots.

I don't have a IR gun at this time. When I get one, I will run some numbers on the new 84 passenger that has not had the interior skin removed VS the 64 passenger that has been completely stripped to bare ribs and outer skin.

Both of my full buses are still yellow on the top. The short 8 foot section from the "Haul All" build is a light blue.

I would like to see pics of temp shots from the ribs, 1/4 the way to the center, and at center.

Readings from both before and after demo are welcome.

Even after the build is done, it would be interesting to see readings at various points on the completed ceiling.

So let's see what you all can come up with?

Nat
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Old 06-12-2015, 01:49 AM   #33
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I just bought one and I live in a bus yard! Ive got my bus with 3" of spray foam on my celing, in addition to a slew of stock buses. If we get a sunny day out here, ill do it. The difference will be interesting. If it were me, Id take off the inner skins. It will only cost you a day of time and get you a lot more options. You could even recycle the steel for a few bucks. Thermally bridged steel inside the living space is going to seriously dininish the liveablity of you bus. I can promise you as I've experienced it.
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Old 06-12-2015, 07:58 AM   #34
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Calculations; as in quantifiable number crunching? Care to share?



Incorrect.. The ribs are more thermally conductive than the rest of the wall and roof, bringing heat from the exterior to the interior, but they only make up 14% of the surface. In a typical bus the majority of the heat transfer will be through the other 86% of the roof, though the thermal transfer at the ribs will be most prominently noticed.



This is partially true. In between is also important. Maximum R-value is attained by insulating between the ribs and over them as much as makes sense for what you are trying to accomplish and your budget.



Incorrect. As Nat said, the interior sheet metal is a heat sink which is connected to an exterior heat sink by the ribs. The metal is thin, but it makes a big difference..


Seriously Austin, it sounds like your twisting the information to best suit what YOU want to do. If you were to stand in your bus in varying states of deconstruction, insulation and reconstruction you would have a very different opinion than what you stated above. Of course, you're welcome to do whatever you want, but some of us who have done the deed may have something valuable to say about it.

If you don't care about having the bus for a long time and want things done quick and dirty then go for it, but I'll tell you right now: insulating over the existing walls is an inferior solution. It'll work, but removing the inner walls, dealing with rust, sealing leaks and re-insulating results in a superior product with the best insulating qualities while retaining a decent wall-to-wall distance.
R = L / kA

L is the same (2"). Area of the ribs is about 1/6th that of the insulation between the ribs (my calc used an even smaller area of 1/7th). k value of the metal ribs is over 1000 times greater for the metal than of the best insulation money can buy. And to top that off, it's a thermal circuit in parallel, so the equivalent R value is 1/((1/R.metal)+(1/R.insulation)).

L = 0.0508m; A.ribs=0.125m^2; A.insulation=0.875m^2; k.ribs=43W/mK; k.insulation=0.04W/mK;

R.ribs=0.0508/(43*0.125) = 0.0508/5.375 = 0.009 (K*(m^2))/W

R.insulation=0.0508/(0.04*0.875)=0.0508/0.035=1.451 (K*(m^2))/W

R.eq=1/((1/R.ribs)+(1/R.insulation))=1/((1/0.009)+(1/1.451))=1/(111.1+0.69)=1/111.8=0.009 (K*(m^2))/W)

Insulation between the ribs does virtually nothing.
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Old 06-12-2015, 08:48 AM   #35
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R = L / kA

L is the same (2"). Area of the ribs is about 1/6th that of the insulation between the ribs (my calc used an even smaller area of 1/7th). k value of the metal ribs is over 1000 times greater for the metal than of the best insulation money can buy. And to top that off, it's a thermal circuit in parallel, so the equivalent R value is 1/((1/R.metal)+(1/R.insulation)).

L = 0.0508m; A.ribs=0.125m^2; A.insulation=0.875m^2; k.ribs=43W/mK; k.insulation=0.04W/mK;

R.ribs=0.0508/(43*0.125) = 0.0508/5.375 = 0.009 (K*(m^2))/W

R.insulation=0.0508/(0.04*0.875)=0.0508/0.035=1.451 (K*(m^2))/W

R.eq=1/((1/R.ribs)+(1/R.insulation))=1/((1/0.009)+(1/1.451))=1/(111.1+0.69)=1/111.8=0.009 (K*(m^2))/W)

Insulation between the ribs does virtually nothing.

you missed convection.....
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Old 06-12-2015, 08:49 AM   #36
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you missed convection.....
We're talking about conduction, not convection.....
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Old 06-12-2015, 08:55 AM   #37
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We're talking about the thermal qualities of the wall as a whole and how to deal with the situation.

Convection is another piece of the puzzle. Like any wall with a cavity and very little insulating qualities (or even fiberglass insulated wall cavities), convection comes into play.

Outside wall heats up, air circulates in dead space and transfers heat to inner wall.
Fourier knows...
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Old 06-12-2015, 08:57 AM   #38
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I'm bringing it up especially because you jumped to this conclusion:

Quote:
Originally Posted by austin1989us View Post
Insulation between the ribs does virtually nothing.


while doing only part of the mathematical process.
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Old 06-12-2015, 09:38 AM   #39
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We're talking about the thermal qualities of the wall as a whole and how to deal with the situation.

Convection is another piece of the puzzle. Like any wall with a cavity and very little insulating qualities (or even fiberglass insulated wall cavities), convection comes into play.

Outside wall heats up, air circulates in dead space and transfers heat to inner wall.
Fourier knows...
For simplicity, let’s assume that there's only air between the panels (no moldy fiberglass to get in the way of that convection).

R.conv=1/hA

A=0.875m

h=10.45 - V + 10*sqrt(V)

That’s an easy empirical equation for calculating the convective heat transfer coefficient, but it’s only valid for 2m/s - 20m/s.

At 2m/s (nearly 5mph) the overall R value is reduced by about 15%. But I seriously doubt the air will be moving fast enough to make the convection significant. When that air isn’t moving there’s just conduction.
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Old 06-12-2015, 09:51 AM   #40
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Also, the outside wall is being heated on the vertical edge. The bottom edge of the wall is cooler than the vertical side of the wall. It won't have the same effect as if the wall is being heated from the bottom (which would induce maximum convection).
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