Products: Multifoil Insulation

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- CommentAuthorbiffvernon
- CommentTimeMay 18th 2010
It's about time someone discussed the effect of fixing magnets to their multifoil.
- CommentAuthorPaul in Montreal
- CommentTimeMay 18th 2010
Posted By: biffvernonIt's about time someone discussed the effect of fixing magnets to their multifoil.
Well, it has to help, especially if the radiation can be polarized by a crystal first. Then all the wavefronts can be aligned and nullified with the correct field orientation. In fact, with judicious use of magnets and crystals, together with some copper matrix weaving the perforations that rbisys deems are necessary, will make the multifoil suck radiation in from the near-zero point and pump it instantaneously into the house in winter, and out again in summer. Kind of like a very large thermos - keeps the place warm in winter and cool in summer. I can't see how it couldn't work - especially if nano-particle magnets are used.

Paul in Montreal.
- CommentAuthorfostertom
- CommentTimeMay 18th 2010 edited
I knew you'd take the pledge in the end, Paul! Once Biff saw the light.
- CommentAuthorrbisys
- CommentTimeMay 18th 2010
Hold on there nuyt.

Your trying to incorporate some quantum principles that I don't believe are valid.

Radiation is the process not the energy. It is the energy that radiates from one surface to another. I'm all for pulling energy out of the vacuum and that energy can be converted to electric to heat or cooling. But, your not going to pull the infra red spectrum directly out of a vacuum and use it as heat. Magnets or no magnets. I wish it was that simple.

Posted By: CWattersRadaition requires a transparent medium which can be the "air" in a bubble in the insulation
This is a correct statement.

Posted By: CWattersRadaition requires a transparent medium to propagate through so presumably you have to establish a gap/cavity for it to radiate across
This is a correct statement.

Quote "I agree - I shouldn't do it. So I'll restate the example and hope you can support:
"For example, one end of a conductive couple may rise in temp but the resultant 'hot front' takes calculable time (a decrement phenomenon) to propagate to the the cooler end of the couple. Until it does, the cool end doesn't 'know' it's about to start receiving more heat flux, to establish a new (increased flow) equilibrium and/or to fill the 'hole'.
Whereas by radiant transfer, heat flux alters instantaneously, as soon as delta-t changes...."

This is a correct statement and presented better.

I believe the problem in trying to under stand heat exchange lies in the muddiness the insulation companies have made.

Here's some things to keep in mind.
If you have a n airspace the primary mode of transfer is radiation.
If you stuff something in the wall you induce conduction. Up to now the ONLY THING conducting was the studs or joists. I covered this in the main article.
For some reason convection is a minor problem in walls insulated with foil. This is most likely from the reduced size airspace. There isn't enough energy to drive convection currents.

In a foil insulate wall system, in winter, the temp difference between floor level and ceiling level is about 1 deg.
- CommentAuthorMike George
- CommentTimeMay 18th 2010
Hi rbisys. You have mentioned an article a few times now. Do you have a copy you can link to please? I would be most interested in reading it.
- CommentAuthorCWatters
- CommentTimeMay 18th 2010
Easy. I install a 1 or 2 layer material between the joist or studs.
I install a single layer across the stud joist surfaces.
I install wood or metal furring strips across the studs or joists .
I install dry wall.

That just tells us how to build one but gives no clues how it works or how well it works. How do we calculate the overall heat loss/R-Value? How do we decide which make of multi foil is best/most cost effective?
- CommentAuthorrbisys
- CommentTimeMay 19th 2010
* Mike George Hi rbisys. You have mentioned an article a few times now. Do you have a copy you can link to please? I would be most interested in reading it.

The article is on the preceding page. The btu chart did not come out too well, you have to correct it as I describe in the post following the article.

CWatters That just tells us how to build one but gives no clues how it works or how well it works. How do we calculate the overall heat loss/R-Value? How do we decide which make of multi foil is best/most cost effective?

I tell you how to calculate heat loss in the article. Reread it and follow the directions from above. You do not have "R" values for RB, that's why I came up with this method. It levels the playing field and gives you a more accurate indication of you requirements.

Let me reiterate what I said in the article about calculations. Commonly used tests/calculations are based on laboratory type conditions that DO NOT reflect installed conditions, especially the effects of condensation. Basically they are a joke. The system I recommend is based on the surface temp of the interior wall/ceiling for summertime. The 110 deg surface temperature represents the typical interior surface temp you find in summer where you have direct sunlight on the exterior surface, especially the ceiling. This is the measurement you want as it is the most severe and you can calculate your a/c from that. You'll note that the btu rate for 0 degs is about the same for the 110 deg. This you can use for heating.

In a home insulated as I describe, the biggest heat gain is from infiltration and windows. Shade your windows.

One thing you have to watch for a/c is over sizing as your run time will about 50% less during the day. If the a/c cycles too fast it will not remove as much humidity.

Speaking of humidity, since RB do not cause condensation the home will be more comfortable.
- CommentAuthorfostertom
- CommentTimeMay 19th 2010
Great to have your fresh input to this long-running topic, George (rbisys). Looks like you've unlocked it (a bit).
- CommentAuthorSteamyTea
- CommentTimeMay 19th 2010
Tom

I was being a bit cheeky earlier.

I think one of the problems here is confusion over units and terminology (there is another thread about this), this makes it hard to know exactly what is being compared.

Do we all agree that the following is correct:

Thermal Conductance = W.K^-1.m^-2

Thermal Resistance = K.m^2.W^-1

Thermal Transmittance (U-Value) = W.K^-1.m^-2

This leads to the EU concept of K-Value = W.K^-1.m^-3

Thermal Radiation
- CommentAuthorrbisys
- CommentTimeMay 19th 2010
SteamyTea I think one of the problems here is confusion over units and terminology (there is another thread about this), this makes it hard to know exactly what is being compared.

EXACTLY. That's why I came up with the new method. Regardless of what insulation material your using, under what conditions, you have the actual temp to calculate YOUR conditions, with simplicity. I guarantee you the insulation mfgrs/deals and utility companies do not like this method because it immediately exposes the the false claims of the mfr and reveals what your energy usage will be. Now you can make an intelligent choice.

Now here is something that's interesting. This chart is based on RB vs FG of which the base material is glass. Glass is about 95% efficient in absorbing and emitting RE. Cellulose is slightly better. Foam according to foam mfgrs assc is about 12-15% better than FG. The avg for foam is about 80%. Since alum is 97% eff you can see why there a large difference in energy savings.

In the USA RB mfrs are required that their material is 97% eff. This is certified by the alum mfr.
- CommentAuthorSteamyTea
- CommentTimeMay 19th 2010
George

I don't quite follow how your method works. Could you put an example up of how your calculate the internal temperature by just knowing the reflective properties of your material. I see it that as having taken a numerical calculation approach (or a look up chart) but cannot use it to work out a temperature without some extra information i.e. the actual make up of the walls, the solar inputs, the temperature differences etc.

Please enlighten me while I go to work.
- CommentAuthorCWatters
- CommentTimeMay 19th 2010
rbisys - Do you mean the article you posted on the Apr 19th 2010? If so I'm also having problems identifying your design method from that. Is there a better article?
- CommentAuthorJohan
- CommentTimeMay 20th 2010
Posted By: Paul in Montreal
Posted By: biffvernonIt's about time someone discussed the effect of fixing magnets to their multifoil.
Well, it has to help, especially if the radiation can be polarized by a crystal first. Then all the wavefronts can be aligned and nullified with the correct field orientation. In fact, with judicious use of magnets and crystals, together with some copper matrix weaving the perforations that rbisys deems are necessary, will make the multifoil suck radiation in from the near-zero point and pump it instantaneously into the house in winter, and out again in summer. Kind of like a very large thermos - keeps the place warm in winter and cool in summer. I can't see how it couldn't work - especially if nano-particle magnets are used.

Paul in Montreal.
Thanks for the post Paul (and Biff), made me chuckle!
- CommentAuthorfostertom
- CommentTimeMay 20th 2010 edited
Posted By: SteamyTeaDo we all agree that the following is correct:

Thermal Conductance = W.K^-1.m^-2

Thermal Resistance = K.m^2.W^-1

Thermal Transmittance (U-Value) = W.K^-1.m^-2

This leads to the EU concept of K-Value = W.K^-1.m^-3
Conductivity = k-value (now renamed lamda-value) = Conductance x thickness = W/m2K x m = W/mk or W.K^-1.m^-1 (not W.K^-1.m^-3).

Transmittance (U-value) has same units as Conductance because it's a composite of a series of Conductances, but unfortunately that series of Conductances can't just be added together.
- CommentAuthorSteamyTea
- CommentTimeMay 20th 2010 edited
Whoops, mistyped, sorry, there must be an alternation character set that we can use here to save using the ^-x.

Maybe I should have been a bit clearer on the last one too.

(m^2.m^3).(W.K^-1.m^-2)=W.K^-1.m^-3

This is K-value as opposed to k (for conductance with the symbol Lamba). All very odd these upper and lower cases especially when the Greeks get involved with their alphabet.

Edit:

Found out how to do it

http://www.planetburrito.com/tb/alt_codes.html

mÂ¯3

Well Almost
- CommentAuthorPaul in Montreal
- CommentTimeMay 20th 2010
(m².m³).(W.K^-1.m^-2)=W.K^-1.m^-3

Judicious use of the HTML <sup> tag

Paul in Montreal.
- CommentAuthordjh
- CommentTimeMay 20th 2010 edited
Posted By: SteamyTeaDo we all agree that the following is correct:

err, no?

I suggest using http://en.wikipedia.org/wiki/Thermal_conductivity as a basis

Thermal Conductance = W.K^-1.m^-2

W/K

Thermal Resistance = K.m^2.W^-1

K/W

Thermal Transmittance (U-Value) = W.K^-1.m^-2

yea!

This leads to the EU concept of K-Value = W.K^-1.m^-3

W/K.m and lambda as Tom says and also the logic goes the other way. This is the fundamental material property from which the others are derived.

Posted By: fostertom
Conductivity = k-value (now renamed lamda-value) = Conductance x thickness = W/m2K x m = W/mk or W.K^-1.m^-1 (not W.K^-1.m^-3).

Transmittance (U-value) has same units as Conductance because it's a composite of a series of Conductances, but unfortunately that series of Conductances can't just be added together.

Transmittance has the same units as conductance because they are both measurements of flux. The difference is that conductance is purely concerned with the conduction of heat, while transmittance includes all transmission by conduction, convection and radiation.

It's all explained on the wikipedia page.
- CommentAuthorfostertom
- CommentTimeMay 20th 2010 edited
Posted By: djhtransmittance includes all transmission by conduction, convection and radiation
each of which respond in various and different ways to given delta-t.
Pure conduction, as in a block of copper, may well be strictly proportional to delta-t.
Convection certainly isn't, in complex ways.
Radiation is roughly proportional to delta-t to the power of 1.2, within the typical range of building physics, because it's the difference between the fourth powers of the two absolute temps.

Yet conventional building science, upon whose phony results insulation manufacturers make their profits, lumps all three of these different, parallel types of heat transmission together and for no good reason treats them all according to the law of pure conduction.
Even the conventional view agrees that, though conduction predominates, convection and radiation within any typical building material play a significant part.
And as stated above, it's becoming slowly accepted that in fact radiation is in real life (as opposed to the steady-state test-rig) the dominant mode.
The only reason the the law of pure conduction (and even the language - 'conduction' through an insulant - ridiculous!) is adopted, is that it's the simplest, simple enough for even blinkered manufacturer-paid routine-technicians masquerading as building scientists to hold in their tiny minds.
- CommentAuthorbot de paille
- CommentTimeMay 20th 2010
"it's becoming slowly accepted that in fact radiation is in real life (as opposed to the steady-state test-rig) the dominant mode."

This is the crux of the matter. This debate could be settled once and for all by linking to one or many research papers that demonstrate to what extent radiation plays a part as a percentage of overall heatloss.

in the vacume of outer space radiation is the primary method of heat transfer, hence the usefulness of mf in this enviroment. Whats the story on Earth?
- CommentAuthorfostertom
- CommentTimeMay 20th 2010
bot, did you a) read b) understand what I wrote above? You don't have to agree but it'd be nice to know that you understand what I said.
- CommentAuthorPaul in Montreal
- CommentTimeMay 20th 2010
Posted By: bot de pailleThis is the crux of the matter. This debate could be settled once and for all by linking to one or many research papers that demonstrate to what extent radiation plays a part as a percentage of overall heatloss.
It doesn't really matter - test houses have been built which use conventional and multi-foil insulation and their actual energy usage monitored over a period of months or years. If "conventional insulation science" was so far off the mark then the measured figures would be far away from theoretical, yet they're not.

In my own case, the hot2000 model of my house has been close to measured performance over 5 years now - consumption has gone down as I've added some insulation here and there and, as I've modified the model, the predictions have pretty much matched the reality (factoring in actual weather conditions (which I do track) versus "typical").

Question for Tom: if radiation is key, what happens when the sun strikes a brick wall - is it conduction that dominates or radiant transfer? If the former, then your "chaotic micro-scale variations" hypothesis is kind of irrelevant isn't it?

Paul in Montreal.
- CommentAuthorrbisys
- CommentTimeMay 20th 2010
For those of you that are interested in this chart, I have aligned it properly.
These numbers came from calculations in Marks Engineering Handbook, 13 ed.
I did not do the cal personally. I borrowed them from an insulation article. Emphasis on borrow as I will return them first chance I get.
I do not want to chance messing up the chart with too much verbiage, so I will give example on following post.

Summer Winter
â€œAâ€ â€œBâ€ â€œCâ€ â€œDâ€ â€œAâ€ â€œBâ€ â€œCâ€ â€œDâ€
150 88 5 83 75 0 0 0
140 75 4 71 70 5 .3 5
130 61 3 58 60 14 1 13
120 49 3 48 50 22 1 21
110 37 2 35 40 31 2 29
100 26 1 25 30 38 2 36
90 15 1 14 20 45 3 42
80 5 .3 5 10 52 3 49
75 0 0 0 0 58 3 55
- CommentAuthorrbisys
- CommentTimeMay 20th 2010
I got wacked again. Had it all straightened out and when I posted here's what we got.

Anyone wanting an original copy e-m me At rbisys@juno.com. In subject "RB copy"

For those of you who wanted a better example. I can't think of one,BUT I can offer a visualization method that might help.

Think about touching a south facing wall or ceiling in the morning. Is that wall or ceiling cool, probably? In fact it might be 75 degs, the same temp that is used as a FLOOR constant in the chart. If so you will not have energy flow. There has to be a Delta T for energy transfer.

Now does it make any difference what insulation is used and how much condensation is present? Yes it does. but you have don't know how much or how much additional energy is being trans. All we know is the ceiling temp of this particular ceiling. And that temp is all we need to know. This is known as "in situ testing".

Now touch that same area, particularly the ceiling in mid to late after noon. If the roof is not shaded then the ceiling is going to be about 110 degs. So you have, according to the chart 37 btu/hr/sf radiating to your 75 deg floor. Now there are other factors such as the amount of energy radiating to the basement floor, etc., but this just to let you COMPARE. For the die hard purists, they have more time than I do and all I want to know IS, what size HAVC units I need, AND what comfort level can I expect, because the less energy you have going in or out the more comfortable you will be and the less you will pay. Comfort also depends on how much energy your body is losing or gaining. If you are in a home with low performing insulation your body will radiate more energy to the colder walls/ceilings so you have to raise the thermostat, more energy. Same for summer.

I believe I mentioned this before, but just in case. The 110 deg ceil temp is about 25 degs hotter than a radiant heating system. That' s like having your furnace on all day while you run the a/c. And you wonder why your costs are so high. Same is true for winter.
- CommentAuthordjh
- CommentTimeMay 20th 2010
Tom: "Yet conventional building science, upon whose phony results insulation manufacturers make their profits, lumps all three of these different, parallel types of heat transmission together and for no good reason treats them all according to the law of pure conduction."

You're trying to scare up a conspiracy theory out of normal scientific practice. It's approximation for convenience! It's got nothing at all to do with building and even less to do with insulation companies.

Let's go to another level of reference source. I think I've mentioned it before but there's a book "A Heat Transfer Textbook", written by two professors at MIT, that's available free on the web

http://web.mit.edu/lienhard/www/ahtt.html

See what it says in section 2.4, where it defines the "overall heat transfer coefficient" U. It goes on to explain a bit about why it is defined and some of the limitations in using it.

You'll also find sections in the book that deal with the dynamic effects of varying temperatures - transient conduction. And radiative transfer, and mass diffusion etc. There's no magic in these subjects; the theories are all well understood and tested.

For that matter the whole theory, even as presented in that book, has a big approximation in it that isn't usually even mentioned. The approximation is that lambda is a single number, a scalar. It is commonly acknowledged that the value of lambda changes somewhat with temperature, but it's still a scalar value. What that implies is that heat is conducted equally in all directions, but that isn't always true. There are materials that conduct heat better in some directions than in others. So really, lambda is a tensor.* It hasn't been of much importance, but some recent results with graphene may turn out to have significant implications in the semiconductor industry. But for most of the world, continuing to believe that lambda's a scalar is much simpler and easier and much more ! useful !

* and here's a couple more references in case you think I'm making it up.
http://books.google.co.uk/books?id=_wUYcEo96IsC&pg=PA172&lpg=PA172&dq=heat+transfer+tensor&source=bl&ots=pgPuUFZNf0&sig=28TO7UYDAZPxR5bVMudtRNjTvBQ&hl=en&ei=f6b1S_okiPM5h6650Qg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CB4Q6AEwAA#v=onepage&q=heat%20transfer%20tensor&f=false
http://iopscience.iop.org/1367-2630/11/9/095012/fulltext
- CommentAuthorCWatters
- CommentTimeMay 21st 2010
rbisys - What part of the world do you live in? Here in the UK we don't really need AC.

I quite understand how radiation from a wall to a person can change how hot/cold that person feels independantly of the room air temperature - but I don't follow the argument that most of the heat loss through the structure is by radiation.

Take a tin roof shack for example. The sun heats up the roof until its really hot. Obviously the tin roof radaites the heat into the room so a person inside is cooked. However tin is opaque to IR energy so ALL the heat passing through the thin layer of tin plate must travel through it by conduction.

If you line the inside with a foil layer you may reflect some (or even 97%) of the radaited of the heat back out, but what about conducted heat? The air between tin and foil will still get hot and eventually so will the foil. It seems to me that a foil covered foam insulation would be better because it blocks both conducted and radiated heat.
- CommentAuthorfostertom
- CommentTimeMay 21st 2010
Posted By: Paul in Montrealwhat happens when the sun strikes a brick wall - is it conduction that dominates or radiant transfer?
Classic case - radiant transfer will dominate highly, within the interstices of the brick, especially when subjected to such a steep 'hot-front' or temp-change wave. In such a clear case, the 'chaotic' might actually be more all-one-way, less chaotic. Even in the absence of sudden temp-shock i.e. as near-steady-state as you can get in the real world, on a micro-scale chaos will reign, even if the nett result approximates to calculability. That is true in other fluid systems, as I'm sure you know - so why maintain that heat transmission is uniquely not so?
- CommentAuthorfostertom
- CommentTimeMay 21st 2010 edited
Posted By: djhIt's approximation for convenience!
Aha! not gospel truth then. Tell that to the 'scientists' incl on this forum - they need to admit to
Posted By: djhsome of the limitations in using it
and a whole lot more, I'm suggesting. Plus the one I hadn't thought of, that you suggest in your next para - v interesting.
Posted By: djhYou're trying to scare up a conspiracy theory out of normal scientific practice
OK, that was cheap - not a shred of truth in it, you'd say?
- CommentAuthorfostertom
- CommentTimeMay 21st 2010
Posted By: CWatterstin is opaque to IR energy so ALL the heat passing through the thin layer of tin plate must travel through it by conduction
Absolutely true - solid metal, unlike just about all other building materials, has no micro-pores (or does it, at the inter-molecular level?) so indeed, all transmission through it is classic conduction. The exception that proves the rule, amongst building materials.
- CommentAuthorCWatters
- CommentTimeMay 21st 2010 edited
So whats the best insulation to use on the inside of a hot tin roof? One that blocks radiant heat or both radiant and conducted?

The point is I don't see how the dominant transport mechanisim in the existing uninsulated structure effects the choice of insulation type being added. I wouldn't dream of saying that transport through the hot tin roof is by conduction so there is no point in using a radiant barrier.
- CommentAuthorSteamyTea
- CommentTimeMay 21st 2010
Tom

Physicists approximate from their data quite a bit, much of it because of the accuracy of the equipment used, mathematicians on the other will take the theoretical process and express it as an equation. This equation may have limits and bounds on it, just depends what is asked of them.

Colin

Go have a look at http://www.greenbuildingforum.co.uk/newforum/comments.php?DiscussionID=5643&page=1

Not quite what your after but has some real life data.

Dave

Thanks for the link, shall read it later.

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