Uneven Insulation R-Values and Heat Flow
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Like a thumb partially covering a garden hose, if 90% of an attic is R-50 and 10% is R-10, is the flow of heat thru the R-10 greater than if everywhere was R-10?
1. Is the average R-value R-46 (90% x 50 + 10% x 10)? Or is the flow of heat thru the R-10 greater than if everywhere was R-10?
2. Can the average R-value be calculated?
3. A few years ago I glimpsed an article with a diagram showing this effect. I have tried to find the article again, without success. I think it was on greenbuildingadvisor.com, and I think it was by Martin Holladay, but I may be wrong. Anyone know the article?
Thanks
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Insulating some areas more than R10 doesn't make the R10 area lose any more heat than it would if the entire surface was insulated to R10. The R10 area loses exactly how much heat the R10 area would always have lost by being only R10. What happens is that you can't use a simple average over the entire surface since it's a little more complicated than that. The actual "whole surface" R value CAN be calculated though.
I think the article you are thinking of is the one written by Allison Bailes at Energy Vanguard. Martin may well have linked to it from one of his own articles. Allison Bailes is well known to regulars here on GBA. The article you're probably thinking of is this one: https://www.energyvanguard.com/blog/how-much-does-your-insulation-r-value-decrease-with-poor-installation/
or possibly this one that shows how the calculation works:
https://www.energyvanguard.com/blog/Flat-or-Lumpy-How-Would-You-Like-Your-Insulation
Bill
Convert your R values to U values, do the math and then convert back to R value.
I don’t know the article you are referencing but here is a recent related article from Allison Bailes that gets at the same question: Does Poorly Installed Insulation Decrease R-Value?
R-Values are useful when talking about insulation because they can be added -- if you layer R-10 over R-6 you have R-16.
U-Values are useful because they can be averaged to find the U-Value of a wall or other assembly. R=1/U and U=1/R.
So your R-50 has a U-value of 0.02 and the R-10 has a U-value of 0.1. If you have 90% R-50 and 10% R-10 the U-value is (0.02)*0.90 + (0.1)*0.1= .018 + 0.01=0.028. U of 0.028 is an R-value of 35.7.
That's not too bad. But the thing to know about insulation is that it's all about cost-effectiveness: add insulation until it stops making financial sense. In this roof you're paying what it would cost to have R-45 over the whole roof, but you're getting R-35. That may or may not work for you.
Total energy loss of a surface also has to do with the percentage of total area insulated at each particular R value. If you only have, for example, a single, small spot of R10 (maybe a squished foot print in the cellulose...), with the rest of the 1,000 square feet at R50, you're not really losing much energy through that R10 footprint. If one side of the attic is deeper than the other because the installers made a ski slope in your attic, then you have problems -- R93 at one end gradually tapering to R12-1/4 at the other doesn't average out as high as you'd think...
Bill
Thank you @DCContrarian R-35.7 is shockingly low when 90% is R-50. Good to know.
Are you able to say: Is the flow of heat thru the R-10 part greater than if everywhere was R-10?
>"Are you able to say: Is the flow of heat thru the R-10 part greater than if everywhere was R-10?"
No, you can't say that. The loss through the R10 part is the same as it would always be at R10. You are still better off with insulation that varies between R50 and R10 in terms of total heat lost compared to what would be lost if everything was only R10. Basically it doesn't ever hurt to add more insulation, even if you only add it in random places, but a randomly upgraded insulating job well never be optimum or as good as an insulation job that tried to get an even layer installed to the correct value -- even if that correct value is somewhat lower than the areas of highest R value in the poorly installed, uneven job.
Bill
Two conflicting statements seem true:
1. Insulation that has u=0.1 is u=0.1 insulation, whatever insulation surrounds it.
2. If you put your thumb over the end of a garden hose, the flow rate thru the still-open part will increase, and the overall flow-rate will stay the same. In the insulation context, I’m sure the overall flow-rate will be reduced, but I "sense" that the analogy is partially correct, that there will be some increase in flow thru the more-open area.
Let's revise my initial question slightly, to better match the thumb-on-hose analogy: If an attic is flat R-10, and then R-40 is added to 90% of the attic, will the heat flow thru the R-10 area increase?
If the house is heated to 70°F before and after adding the extra insulation, and if there is no difference in wind or humidity or exterior temperature, then is this equivalent to a thumb-on-hose context where the water pressure is constant however much of the hose opening is covered?
If the same [[pressure]] of heat wants to expand/equilibriate/rise, then does it not make sense that adding R-40 to part of the attic will act like a funnel / partial-dam channeling more heat thru the R-10 area than when the whole attic was R-10. As Allison Bailes writes, "Heat will take the path of least resistance".
The average R-value would certainly be improved by adding the R-40, but one of my two logics still says that the flow thru the R-10 portion would increase.
Help?
Thanks
> "Heat will take the path of least resistance"
This isn't true - heat takes all available paths, inversely proportional to resistance.
> where the water pressure is constant
This isn't true either.
There is no such thing as "heat pressure". Your water analogy isn't really correct, either -- when you block part of the hose, the pressure in the hose DOES increase, limited by the supply pressure and the pressure drop in the length of hose. What happens is the VELOCITY of the water increases in the smaller opening, while overall flow rate will drop somewhat compared to what it was with the end of the hose fully open.
A more appropriate analogy is a sunny window with window film. Imagine you have sun blasting into your home through a window, and you want to cut down on the glare. You cover part of the window with window film to reduce the light getting through by 50%, but you run out and leave a few square inches of the window uncovered. Does all the extra light blocked by the film squeeze through those few square inches that are uncovered? Nope, it doesn't work that way -- the light hitting the areas covered by the film is reduced by 50% as expected, and the light going through the uncovered areas remains unchanged from how it was before you applied the window film. The light getting through the uncovered area doesn't increase or decrease regardless of of how opaque the window film on the rest of the window is.
What DOES happen is that the amout of light coming through the uncovered area is now 50% more than that coming through the areas covered with window film, but that "50% more" is relative to the SHADED areas, not the original amount of light that was coming through. You never increase the energy from the amount you started with in this example.
Bill
EC,
Certain transport properties of heat, mass, and momentum are analogous in the sense that the mathematical framework describing them is similar. Your water hose analogy is a good thought experiment, but the confusion arises in assigning the proper momentum/heat analogous properties. Tyler Keniston pointed this out below. Pressure is analogous to temperature, and water flow is analogous to heat flow. What are the driving forces in these phenomena? In your water hose example, we can assume a simple model of a constant flow source. The analog for heat transport is a constant heat source, not a constant temperature source. Imagine a poorly insulated room in which you have placed an "ideal" portable radiator running constantly. Conservation of energy requires that the heat generated from the radiator is transferred through the walls, eventually at a constant heat flow. Now cover most of the walls with better insulation. The same total heat flow must occur, but you're right, now the greater portion flows through the poorly insulated portion. But if you posit a constant temperature source, then as others have said, heat flows through the insulation at a rate proportional to the temperature difference. The constant of proportionality is the inverse of the R value. This situation arises in a heated house in which the heater strives to maintain a constant temperature.
Now, you raise the confounding issue of an attic. During the day the attic materials absorb heat, and at night they radiate that heat. In other words, at night the attic acts like a constant heat source! If the attic is poorly insulated, an uncomfortably (in the summer) noticeable portion of that heat will be transferred into the conditioned space. Add additional insulation, and a lot of that heat is transferred back through the roof. Heat flow is indeed increased through the poorly insulated portion.
Your models and your intuition are on the right track, they just need a bit of refinement.
This may be the article?
https://www.greenbuildingadvisor.com/article/how-to-perform-a-heat-loss-calculation-part-1
See part-2 as well.
The answer to 'is the flow greater through the R-10' depends on whether you are heating the space to a constant temperature, or heating with a constant power output (BTU/h). Obviously most of us strive to do more the former.
For a good, albeit imperfect, analogy consider 'constant current' vs 'constant voltage' sources in electronics.
If you heat to a constant temperature, then the energy flow through the R-10 will be the same. Temperature differential is all that matters. But if you heat with a constant power source, the increased insulation in other areas will raise temperature and therefore increase heat flow across the R-10.
GBA needs a like button. Not really needs, but I want one. Thanks
@Zephyr7,
GBA won't let me reply to your latest comment - perhaps the thread is already too many levels deep.
I love the sunlight analogy. Thanks.
Thanks also for correcting my water hose analogy.
And yet! Do you know why the sunlight analogy applies better than the water hose analogy (where the velocity of water increases thru the remaining opening)?
Light is electromagnetic radiation, water is a fluid. Differing physics between the two. Heat is infrared radiation (when radiated), or atomic/molecular motion (when it's "hot stuff" and not just radiation).
Since we're talking about radiated energy here, you have to think in terms of electromagnetic radiation, which doesn't really have a concept of "pressure", since it doesn't really have mass (physical material). If we were dealing with physical material in a fluid, you can have pressure, and pressure will force bits of that fluid to squeeze through any available hole -- "pressure" is the force of all the particles that make up the fluid being squeezed together, and the force is exerted outwards in all directions. Different physics for different things.
Bill
Thanks!
Imagine you are standing in front of a 30’ high, 100’ wide dam that has a big lake behind it. There are three holes in the face of the dam. Each hole is 1/2” in diameter and 25’ below the waterline. A stream of water is shooting out of each of all three holes. You plug one of the holes. The flow of water through the other two holes would not increase.
Now, if the height of the lake increases measurably in the two-holes-open scenario relative to the three-holes-open scenario, then the flow through the two holes will be greater. This would be akin to a situation in which the temperature rose in your house because you had a heat source that was adding a constant amount of heat and you have added partial insulation.
But in most home-heating situations we set the indoor heating system to maintain a target temperature, which is akin to fixing the height of the lake with a toilet tank valve.
Thanks. To ensure that wasn't just a case of insignificance, I had to test that by considering if the holes were somehow 10' in diameter and I plugged two, with no water entering the lake. I see that indeed blocking the other 2 holes does not mean the same amount of flow needs to happen thru a smaller area, because the pressure at any moment shooting water thru the one hole would be the same in that moment as if the water was shooting thru 3 holes. In both cases it's the pressure of the volume of the lake against the 10'-diameter non-wall.
I'm not sure if my expression could have been clearer, but I think I get it. Thanks.
The strange thing is that I'm sure the diagram I saw, made by (and in support of the article by) someone whose name I already recognized as a well-respected name (Martin Holladay? Allison Bailes? ...?) clearly showed more heat rising thru the poorly insulated spot than if everywhere had the lesser level of insulation - not more total heat loss, but more heat loss in that particular spot - something like a funnel. It was surprising to me at the time - I don't think my somewhat-brief-viewing and now memory are being distorted by confirmation bias, because I was surprised by it. But it certainly seems everyone here agrees the heat loss won't increase, and the reasoning all seems right.
Thanks
It's possible the arrows were showing percentages? A higher percentage of heat loss will occur through the lesser insulated section as insulation increases in other sections.
Note that it is not the VOLUME of water that would dictate the pressure in the above dam analogy, only the height (head). In building heat-loss terms, the 'head' is the temperature differential.
Of course a big house and a big wide dam may each move a lot of heat and water respectively, but the 'pressure' across any given area will be determined by the temperature differential and head respectively.
Thanks. Thanks. And thanks. :)