# “Cost” of uninsulated concrete

I’ve heard a variety of conflicting answers to this question. Trying to understand the _real_ impact, and whether it’s worth the cost to remediate or not.

View Image View Image In the pictures, you see the building (Ohio), which is sprayed concrete with exterior polyurethane foam. Focus on the canopy over the doors and the canopy over the windows. These also are 3″ of foam over 3″ of concrete. But for the canopies, the underside of the canopy is not foamed, and because that concrete connects to the concrete of the shell with no thermal break, there’s heat loss in winter. The net area of all these spots where the concrete canopy connects to the otherwise insulated concrete shell is about 7 sf (28′ x 3″/ft) for each canopy 100 sf. Owner says he can feel those areas as colder than other areas. Now, those areas have windows, too, so it could be difficult to separate the sources of any heat loss one perceives. To foam, sand, shape, finish these areas during construction would cost probably $500-$1000 per canopy (10-15 canopies), mostly in labor, as it’s a real pain in the butt process. How would you analyze the decision to foam or not foam those areas? View Image

## Replies

In the winter time, those canopies will be acting like giant heat sinks pulling the heat out of the surrounding cement structure.

If that makes the cement wall cold enough to form condensate, then something really should be done.

Otherwise - try to figure out a much cheaper way to insulate that area. At $500 a canopy the payback would never be worth it.

No condensation.

Analysis:The Area heat is lost over is the thickness of the concrete (what I called "Loss Area") over the insulated wall times the length.

The equivalent thickness of the material is the length of the dotted line thru the middle of the loss zone; the thickness of the wall plus pi x 1`/4 Loss Area. Ok, I shoulda called "Loss Area" Loss Height or sumpin.

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My solution:

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SamT

Anyone who doesn't take truth seriously in small matters cannot be trusted in large ones either. [Einstein] Tks, BossHogg.

Yeah, that seems about right. One would have to do some arithmetic with thermal loss numbers to figure out how much $$ is being lost through the "leak", and how much insulation would be needed to "neutralize" it.Basically, you need to lengthen the loss path (by insulating out in either direction) to the point where the R value of the loss path is about equal to the R value of the insulation minus the R value of the standard thickness of concrete.

So convenient a thing it is to be a reasonable Creature, since it enables one to find or make a Reason for everything one has a mind to do. --Benjamin Franklin

>The equivalent thickness of the material is the length of the dotted line thru the middle of the loss zone; the thickness of the wall plus pi x 1`/4 Loss Area.Very interesting idea, Sam. Thanks. Never thought of that. Is there a formula or rule of thumb for calculating the loss as the size of the chamfer increases from 0"->x"?

It would be equivalent to applying insulation the width of which was 1/2 the sqrt of the width of the chamfer face and 1/2 smaller again that thick to both the wall and the ceiling.

Umm, if the chamfer was sqrt 2n across the face, it equals 2 pieces 1n wide and 1/2n thick, one on the wall and one on the ceiling.

Effectively, you are just lengthening the heat loss path to past the insulation, so I would recalc the heat loss with the longer path and assume the foam was a perfect insulater. It's not, but I don't have a Cray to do that kind of itterative calculations, and, whatever heat does pass through get through the foam will still reduce the coolth felt past it.

And this is a problem of perception, it is a very small area that doesn't even cause condensation.SamT

Anyone who doesn't take truth seriously in small matters cannot be trusted in large ones either. [Einstein] Tks, BossHogg.

You're a pioneer in one of the most energy efficient and durable structures being made available to the public.

You're going to have to bite the bullet and just do it.

be no stone left unturned

every court needs a jester

>just do it.Just do what?

Trying to understand the _real_ impact, and whether it's worth the cost to remediate or not.

How would you analyze the decision to foam or not foam those areas?

Thought you were just debating the idea of messing with the areas in question as opposed to petitioning the board to barnstorm ideas.

every court needs a jester

Nah, this was a real question. Client's intending to adress this somehow and is looking for ideas. His only thought was adding foam to the whole exposed exterior concrete, which is a lot of work and expense. It'd be effective in minimizing the concrete as a transfer path for heat, but my intuition is that it's not cost effective.I'm trying to put something more objective behind that intuition. For example, if the concrete canopy extends 4' beyond the building shell, is wrapping that 3" of concrete that's 4' away from the shell at all effective in limiting heat loss? What about areas 2' away? 6" away?Sam's is an interesting approach be/c he talks about the heat paths. And his solution is to lengthen the transfer path. He gives a recommendation on length as a function of width. I'm most curious, for example, on the math behind this. Why that length? Why not double? What's the point of diminishing returns? etc, etc.

When the heat lost on that path becomes comparable to the heat lost on a similar cross-sectional area of plain wall then that's diminishing return.

So convenient a thing it is to be a reasonable Creature, since it enables one to find or make a Reason for everything one has a mind to do. --Benjamin Franklin

Cloud Hidden.

I'm not so sure you have fully takin in all of the ramifications..

If we assume that "repairing it" has a fixed cost of X we cannot just assume all future energy costs will also be fixed..

It's a moving target you see that requires a level of forcasting skill.

I would think the formula would be something like.. the projected energy costs unmodified times whatever appreciation numbers seems practical to apply.

Assume this building will remain standing for 100 years.. (use whatever number seems reasonable) then take the increase in energy costs over that same time frame. How much would it have cost to heat that building say in1907 compared to today..

If it's 10 times than whatever ten times todays energy costs are, is the number you should compare the cost of retrofitting to..

The life of the occupant isn't an issue here.. If the building will last say 60 years more than the currant owner can possibly live that improvement will continue to add value for those 60 years.. If he sells it or even gives it away it's more valuable if energy costs are cheaper than if they are more expensive..

He gives a recommendation on length as a function of width. I'm most curious, for example, on the math behind this. Why that length? Why not double?When did I do that?

You mean the isoscles triangle? Effective insulative size of the chamfer?

Splitting an isoscles from mid base to apex results in two triangles, the bases of which were the sides of the original.

If the face of chamfer is the unit measurement, X, then a side is equal to X/sqrt 2. . .ahh, that was my mistake. . ./sqrt2. Doesn't effect my previous post 'cuz I didn't use it 'cuz I knew there was a mistake, just couldn't see it clearly.

Anyway, the height of an isoscles is equal to 1/2 the base and its' area is equal to a rectangle where the long dimension is the same as the triangles base and the short dimension is 1/2 the triangle height.

Chamfer Face = XChamfer Side = X/sqrt 2, also =Split Chamer Base = X/sqrt 2Split Chamfer Height = 1/2 of X/sqrt 2Equivalent Rectangle Width = Split Chamfer BaseEquivalent Rectangle Heigth = 1/2 Split Chamfer Height

For the concrete ceiling and the insulated wall, each:

AproximateEquivalent Insulation Width is Chamfer Face / sqrt 2AproximateEquivalent Insulation Thickness = 1/4 (Chamfer Face / sqrt 2)A differential equationist can get more accurate.

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The assumptions in the above approximation are:

That the heat loss through the dotted line is an average of the heat loss across the cross section of the Heat Loss Path.

That the side of the chamfered foam is the same width as the height of the Heat Loss Path over the insulated wall.

The length of the dotted line is the thickness of the insulated wall plus the length of the two 1/4 circle arcs on each end. The line splits the thickness of the concrete above the wall, so the arcs combined length is 1/2 pi x D (half the circumference) where D = 1/2 thickness of concrete.SamT

Anyone who doesn't take truth seriously in small matters cannot be trusted in large ones either. [Einstein] Tks, BossHogg.

I'm trying to put something more objective behind that intuition. For example, if the concrete canopy extends 4' beyond the building shell, is wrapping that 3" of concrete that's 4' away from the shell at all effective in limiting heat loss?Absolutely not.

What about areas 2' away?Don't think so.

6" away?Maybe. A little bit.

How thick is that overhang? I think any spot more than twice the thickness away from the wall is not part of the Heat Loss Path.SamT

>How thick is that overhang?It's 3"-4".>I think any spot more than twice the thickness away from the wall is not part of the Heat Loss Path.This is the kind of stuff I've been trying to get at. Is there science behind this toward which I can direct the client. He's an EE, if I recall, and has a tendency to ask for details. If I recommend minimizing the heat loss path as you describe (a concept of which I am very fond, btw), he's certain to follow up with "why?" or "how do we know that twice the thickness is not part of the path?" I could say, "Because Sam T said so." That carries weight with me be/c I've read enough of your stuff over the years. With him, not so much. <g>So, where would I begin to document the science behind this?

If he's an EE, take a cross-section and tell him it's an electrical resistance problem. Get the R/inch values for concrete and foam and tell him to treat them like bulk resistance values. Assume one connection to the outer wall and another to the inner. Compute the resistance of the various wall components.

So convenient a thing it is to be a reasonable Creature, since it enables one to find or make a Reason for everything one has a mind to do. --Benjamin Franklin

Tell him SamT was an EE in the navy, before he learned a little about thermodynamics.

Heat is like water and electricity, it always takes the easiest path.

Tell him the heat flows exactly like current between two elements on a resistive plate. The plate is large (about 4',) and so is one element. The elements are copper, and the small one is about 3"-4" away from the big one. The current is moderate.

If you tell him that the dotted line is 70.7% down from the top of the concrete to the wall, he'll think I/you know what I'm/you're talking about.

As to references, I no longer have them. I lost the thermodynamic bookshelf in the '80s and gave away the electronic one in the '90s. I wish I could help.

Hey! I got a computor shelf now, about 16', and a construction one too, about 12', would they help?

0.707 is the square root of 1/2 and is used a lot in both the applicable maths. In this case 1/2 way down =70.7%

SamT

Edited 3/7/2007 10:56 pm by

SamTAccording to

http://www.coloradoenergy.org/procorner/stuff/r-values.htm

the R value for concrete is .08 per inch - which makes just about anything but copper or aluminum better insulation.

If you assume that there are good windows with a U factor of .25, then the concrete would need to be insulated for a total of 50 inches to give you the same equivalent value of insulation.

Check out this page; Monolithic Dome "R" Fairy Tale

I found the below reference material from this google; Heat "flow path" through nonlinear solid

This Article from Electronics Cooling Magazine would be good to show your EE client.

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Figure 2. FEA thermal solutions of a) Slab and b) 45Ă‚Â° pyramid. Models cut midplane to show temperature contours under heat source.

This one, Journal of the Brazilian Society of Mechanical Sciences and Engineering has more pictures and lots more math;

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Note that the university only made a 2d model so they would not overload

theircomputors.SamT

Edited 3/8/2007 10:52 am by

SamTthat NIIIICCCCCEEEEEEEE. hey cold, I got socks on because my feet are cold on the concrete floor. its 75 degrees. just do it.

everybody calculate insulation according to payback factor.Here my way of calculation insulation factor- Are you cold? yes insulate. but cost, dont care, are you cold.

What is that, 250 square feet of glass compared to a 3 inch by 20 foot section of uninsulated concrete?

How about taking sheet foam and covering the glass under the canopy and see if he notices a difference?

Excellent idea! I'm definately not a heat loss expert, but I'm inclined to think that the majority of the loss is through the glass. Should be worth testing anyway.

It would be a good test to isolate the window from the concrete. I suspected the windows as a source of what he's experiencing, but know that the concrete will also have some quantifiable heat loss. I'd like to better understand the theory or science behind all aspects of this.

Do you think the current r-value of the concrete is based primarily on the thickness of the wall (basically the same as if the concrete ended at the outside edge of the wall)? As you add insulation on the outside, you would basically be increasing the r-value of the concrete because there is a longer path to lose heat. Diminishing returns would be when the r-value of the concrete is greater than that of the windows (or the next lowest r-value)? No return if the r-value of the ceiling exceeds the wall r-value?

I'm sure if I could remember college physics equations there would be a way to express the relationship (something with Joules), but I am at a loss at the moment.

Insulating all those last sections of concrete would certainly decrease the heat loss through that concrete, but at a big cost. Insulating inside (Sam's suggestion) is quicker and easier, but I don't have a good grasp on its effectiveness versus the other method.I _was_ hoping for something expressed in Joules...don't disappoint me! <G>

Q=m cdeltaTwhere

Qis the heat energy put into or taken out of the substance,mis the mass of the substance,cis the specific heat capacity, and deltaTis the temperature differential.I googled specific heat, which might be applicable. "C" for concrete is .88, meaning that it takes .88 Joules to raise 1 kilogram of concrete one degree Celsius. For comparison, water takes 4.184 Joules and copper takes .35 Joules to change the temperature by one degree.

Seems like if you insulate on the inside you would be increasing the mass that the outside would have to cool in order to effect inside comfort. Conversely, if you insulate on the outside, you would be increasing the mass of the heated concrete because you are extending the building envelope.

I would imagine that if it is 0 degrees F outside, and 72 inside that a majority of the outside concrete would be 0 and the inside would be 72. Above the exterior wall there is going to be a temperature gradient between those two, and the real question is what do you need to do to keep a gradient line of an uncomfortable temperature from reaching interior space.

Edited 3/8/2007 9:12 am ET by

AaronBut that's the equation for heating/cooling a mass, whereas we're interested in the heat FLOW once a steady state has been reached (the mass is at its final temperature).So convenient a thing it is to be a reasonable Creature, since it enables one to find or make a Reason for everything one has a mind to do. --Benjamin Franklin

I am looking at it as two masses, one unheated mass and a heated mass--it takes energy to keep it at equilibrium.

I think that's too complicated. Once it's at equilibrium you can treat it as a heat flow problem, analogous to the electrical resistance problem. You do have to use some thumb-suck approximation for the path length, but the longer that gets the more it's a simple straight line.If you try to consider masses you'd have to use a discrete element approach, breaking the problem down into small masses and solving the heat flow into and out of each.So convenient a thing it is to be a reasonable Creature, since it enables one to find or make a Reason for everything one has a mind to do. --Benjamin Franklin

How would you analyze the decision to foam or not foam those areas?

We would look at the building as a whole and determine the least cost place to insulate and do that first, unless the lack of insulation is causing cold or hot spots that are a problem.

For instance, the easiest way to "add" insulation to an under insulated wall is usually to over insulate the attic first.

Beer was created so carpenters wouldn't rule the world.