What is a sensible up-charge for better performing windows?
Howdy folks,
This is perhaps a topic that is discussed to death, but never the less, the opinions of GBA members (and experts) would still be be appreciated for my particular situation.
I’m building a Pretty Good House (or slightly better) in NW Ontario. We are currently receiving quotes from 2 Canadian window and door manufacturers. Access Windows (near Winnipeg, MB) manufactures windows using the 86mm Geneo profiles. Innotech (Langley, BC) uses the 88mm Kommerling profile. Both are PH rated although the Innotech windows receive a better PH score.
For our particular build we’ve been quoted between $50,000-$70,000 for windows/doors for various performance levels from these two manufacturers, between U-0.18 to U-0.13.
I’ve seen some folks (much smarter that I) calculate ROI’s based on heat lost difference between a couple U-value points. I do not know our heating loads yet unfortunately. I also realize that ROI’s for high performance windows is pretty much none existent.
What would you consider an “acceptable” markup to go from a U0.18 to U0.15 for a $60,000 order?
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Replies
We say in the classic (old) Porsche world, "buy the best 911 you can afford".
I think the same logic applies here.
There are always other variables so it's hard to say. For whatever percentage of your wall is glass, it's the equivalent to having R-5.6 insulation or R-7.7 insulation over that portion of the wall.
I would build a BEopt model and try to estimate the energy savings between the two options. It doesn't take long to learn, it's completely free, and it's good enough accuracy for this type of problem.
Unfortunately it looks like BEopt is only for windows PC and not MAC.
Total window/door area is 531 sq ft. My total above grade exposed wall area is approx 2620 sq ft meaning windows/doors make up approx 20% of my total wall area.
Perhaps someone can verify I’m on the right track.
I did a basic R per sq ft of wall area calculation comparing R5 vs R7 windows and came up with a value of only a 1% difference in total wall R value. At best it appears that an R7 window would only save a few dollars per month in electrical usage. This was sorta as I expected.
I believe I’ve settled on the Geneo profile windows from Access Windows. They’re manufactured about 4x closer to me than Innotech and they’ve included freight in their quote so pricing will actually come in much closer to the lower end, but their performance is just shy of R7. Seems to be the best balance of performance/cost.
I'm not sure how you calculated that. You have to calculate this like resistors in parallel, not in series. You never told us what your wall assembly R-value is (not center of bay value, but overall value after accounting for framing factor).
Something like this:
1 / R_total = (1 / R_window)*0.20 + (1/R_wall)*0.80
1 / R_total = (1 / 5) *0.20 + (1 / R_wall)*0.80
Christine Williamson has a great video on this: https://www.youtube.com/watch?v=zJRsmiaqh6A
But even if you calculate this, that just tells you changed in overall R-value, not how much energy savings you will have. That part requires weather station data like heating/cooling degree days, etc. There may be a back-of-the-napkin energy cost calculator you can find online.
The customary way to calculate seasonal energy usage is to use heating degree-days and extrapolate from there.
I calculated it as a simple percentage change in total R value per sq ft of exposed wall. Whole wall R value will average out around r40 including framing losses.
My calculation assumes that 1% less r value equals 1% more energy usage.
After watching that video my whole wall r values, with windows (with r40 walls and r5-7 windows):
R5 windows= R16.6 effective
R6 windows= R18.7 effective
R7 windows = R20.5 effective
Deleted
Those numbers seem reasonable, but I'm not sure you can assume an X% change in R-value is equal to X% change in energy usage. I think it would have to be scaled for whatever the "average" delta T across your wall assembly would be. But we're at the edge of my knowledge here. Here is the first link I found when searching for a quick online calculator:
https://cellulose.org/insulation-savings-calculator/
you'll need to know your HDD (heating degree-days) but you can google for that too
With 2620 square feet, a R16.6 wall will lose 3772 BTU per degree-day. Let's say you have 6,000 degree-days per year, the annual energy usage for that wall would be 22 million BTU. A gallon of fuel oil nets roughly 100,000 BTU, so let's say 220 gallons of fuel oil. I paid $4/gallon last time I filled the tank, so $880/year. (Note that this isn't for the whole house, just the walls.)
R6 would be $784 per year. R7 would be $715. So the savings would be $96 and $165 per year.
That gives an order-of-magnitude of the savings. Putting in degree-days for your area and your cost of energy will give a better estimate.
Thanks DC!
Note that 2096 square feet of R40 wall loses 1257 BTU per degree-day, regardless of what the windows in it are, and 524 square feet of R5 windows loses 2515 BTU per degree day, regardless of what the wall around it is.
So you don't have to calculate the r-value of the wall assembly, you can just do the heat loss through the windows themselves.
Doesn't that ignore the fact that heat travels through the path of least resistance? Like if you had two R40 walls, one with no penetrations, and one with a large aluminum or steel pipe going through the middle of it, would you expect those two wall assemblies to have the same heat loss?
If you could just separate the sections of the wall (window vs wall) and add them together, then that would be the "resistors in series" method but in reality I think you have to do the "resistors in parallel" approach.
Matthew, you are correct; you can't just look at a single element, you add the weighted U-factors to get the total R-value, or take the inverse of that to get the overall U-factor, as in the link you posted from Christine Williamson above. That's also why it's important to include the roof and floor in calculations.
Matthew and Michael,
I'm interested in this because it's something that's come up a few times in the past and I'm still very vague on.
Isn't DC making a more limited claim? He is not saying that upgrading the windows is the most efficient use of resources you might use elsewhere. He is saying you can take any part of the enclosure, and evaluate whether the cost of upgrading it makes sense in isolation. There may be a ROI regardless of the overall picture.
So for example, if you had a completely un-insulated basement of R-1 walls, you could decide it made sense to insulate three of them to R-15 by looking at the energy savings vs cost of upgrading each wall in isolation, and ignore the fact that the un-insulated remaining wall meant the aggregate R-value of all the walls was just about R-2?
Or have I got this wrong?
Malcolm,
I see what you're saying but I don't know for certain if that approach is feasible. In your 4-wall example:
1/R_total = (1/1) * 0.25 + (1/15) * 0.75
R_total = 3.33
So if the walls were the only surface experiencing heat flow, you would have improved them from R-1 to R-3.33, but what if the Roof is also R-1 or the floor? I think you probably need to account for all 6 sides and also the fact that some sides (like basement floor/walls) have a different heat flow profile than other (above-grade) surfaces. This is why I told JMR to use BEopt or some other modeling software because all of this makes my head hurt.
Malcolm, think of it this way--if the windows were just holes in the wall covered with a sheet of polyethylene, say R-1, and they made of 20% of the wall, would the walls plus windows still be performing at R30+?
That may not be intuitive enough. (And this is something I struggle with as well, and have provided bad advice on previously here.) If the walls are R-100 and the windows are 50% of the wall and R-1, are the walls plus windows going to perform as if they were R-50?
Or R-200 walls and R-1 windows over 80% of the wall surface. Using the parallel method, that should be an average of R-40. But that's not how the heat loss is going to work.
Michael,
Forgive me - the problem is no doubt mine, but I still think we are talking about two different things.
Forgetting about whole wall values, and dealing with the windows in isolation (as DC suggests). If you have a window that costs you a certain amount in energy flow through it, and you have the chance to replace it with a window that allows less energy to escape (the cost of which was low enough that the savings just for that opening were greater than the cost of the window), why isn't that a legitimate way to look at ROI?
Matthew,
Let me make another similar argument to the one I made replying to Michael, and maybe one of you will be able to set me straight.
The basement wall doesn't know what the R-values of the other ones or ceiling is. It experiences heat flow through it based on its own R-value. If you can reduce the heat flow though it for less than than cost of the energy it was losing, doesn't that in terms of straight ROI that make a case for doing so?
I think it's because the amount of heat flow is dependent on the other surfaces. So if the thing you are considering replacing is already the worse-performing part of the building, then yes you will greatly reduce heat flow. But if the thing you are considering replacing is already one of the better performing parts of the building, you will reduce heat flow a little but you will increase the amount of heat flow occurring at other weaker parts of the assembly so the overall effect is much more muted than you would expect if you were just analyzing the change in heat flow through that one piece of the building. At least this is how I understand it to work. Like electricity, heat wants to take the path of least resistance. When you replace the least-resistant part with something more resistant, the heat then finds the next most least-resistant part, increasing the amount of heat flow through those areas which could be a totally different part of the assembly to what you replaced. I think that's why you have to consider all surfaces.
Matthew,
That's interesting, and all new to me. I had understood the R values of materials or assemblies to be a constant, and heat flow through them to be entirely based on the Delta T between the two sides, independent of what is happening elsewhere. So while in relative terms the remaining un-insulated wall would be the source of more of the total heat loss, in absolute terms the amount of heat it lost would remain the same.
Edit: Malcolm and DC are totally right. I'm not sure where my head was but for some reason this concept has always confused me. I don't think it's even the first or second time that DC has corrected me. That's what's great about this forum--nobody has all the answers, but somebody usually has the right answer.
Michael,
Probably we are confused by it because while true, it isn't usually a good basis for the decisions we make when allocating resources. So to return to the hypothetical basement for one more kick at the can. There may be a case to be made based on ROI for insulating just three of the four walls (say the cost of the foam was such that you get it back in say five years of energy savings), these marginal gains aren't worth spending time on when the remaining wall of the basement is hemorrhaging heat.
Intuition might be failing here because when we think about excessive heat loss through say, poor windows, we naturally imagine the room getting colder and the smaller delta T resulting in less heat loss through the other assemblies. We have to remind ourselves that there is a heating system continually replacing the lost heat and keeping the inside temperature constant.
One thing to keep in mind is that if you just search online for the heating degree days for a location, you will most likely get a value that uses a base temperature of 65° F, commonly used for sizing heating equipment. For calculating heat loss, it will result in an underestimate of the actual heat loss by about 15% in colder climates.
If you wanted to be more precise, you would use your actual wintertime thermostat setting. Some weather sites (ex. https://www.degreedays.net/) allow you to calculate heating degree days based on the temperature of your choosing, such as 70° F if that is what you keep your home at.
The choice of 65F as the baseline for heating degree-days is an attempt to account for the contribution of other heat sources, like occupant behavior.
When designing a cooling system you estimate the number of occupants and how much heat will put off. On the heating side they just assume five degrees and call it good.
In a well-insulated, tight house five degrees is not enough and the standard degree-days calculation will over-estimate annual heating load.
Yes, understood. However, when you are choosing between two different assemblies for the same house, the internal gains don't change. The occupants won't run their appliances more just because the windows are less efficient. Since the internal gains don't change, the additional heat loss on a cold day has to be 100% made up by the heating system.
The definition of base 65° heating degree days is:
∑(65° - avg daily temp) for all days with avg daily temp < 65°
What we really need is:
∑(70° - avg daily temp) for all days with avg daily temp < 65°
I don't know of any site that provides that directly. You could download the data and calculate it yourself. In lieu of that, I'm suggesting that base 70° HDD is a reasonable approximation. At least for a cold climate (with many days below 65° and few days between 65° and 70°), it will give a result much closer to reality than using base 65° HDD.
What changes when the construction of a house changes is the break-even temperature -- the outside temperature where internal gains balance heat loss when the inside is at room temperature. If the outside temperature is breakeven or above you don't need heat. In most climates, houses will have enough heat capacity that if the average daily temperature is at or above break-even you can be comfortable without heat.
I would argue that the most predictive calculation of heating degree-days would thus be:
∑(breakeven - avg daily temp) for all days with avg daily temp < breakeven
Basically HDD65 assumes all houses have a breakeven temperature of 65F. With modern construction it's often quite a bit lower.
You are right that changing the window R-value will change the heating balance point. If you knew the exact heating balance points, then you could use your formula to calculate the two heating loads and then the difference.
The challenge is how to precisely determine the heating balance points. You could build a detailed energy model of the house, but then you are no longer doing a simple calculation.
Using the same balance point for both instead of the actual balance points will result in an error due to either including extra heating days for the better window or omitting some heating days for the worse window. But, that error will be small because the change in balance point will be small, maybe a degree or two at the most. Suppose it changes from 65° to 63°. In a cold climate, there will be relatively few days falling within that narrow range. If you use 65° for both windows, the better window will get an added 5 to 7 HDD per extra day. If you use 63° for both, the worse window will lose 5 to 7 HDD per omitted day. Either way, if there were 30 days in that 63° to 65° range, that would only be about a 180 HDD difference in a 9,000 HDD climate.
DC, it took some pondering but I think there are two major problems with using the heating balance point to analyze a single building component.
First, as you said, using the same balance point for both windows ignores the fact that the better window will give the house a lower heating balance point.
The second problem is that heating balance points are intended to calculate heating system load, not heat loss. That might seem like a fine distinction, but it means you cannot look at the windows in isolation.
The reason is that when the heating balance point gets lowered, it gets lowered for the entire enclosure. The better windows have less heat loss, leaving more of the internal gains available to offset heat losses through the walls, the roof, the floor, etc. So, the heating system load is reduced for those components as well.
Here's a contrived example to illustrate:
The good windows have Uw = 0.2 and balance point Tb = 65°
The better windows have Uw = 0.1 and balance point Tb = 60°
Window area Aw = 400 sf
The rest of the enclosure has Ur = 0.05 and area Ar = 1600 sf
Heating load for a single 50° day considering the windows only:
With good windows:
Q = Uw·Aw·(Tb-50)·24
= 0.2·400·(65-50)·24
= 28,800 BTU
With better windows:
Q = 0.1·400·(60-50)·24
= 9,600 BTU
Reduction in heating load = 28,800 - 9,600 = 19,200 BTU.
Heating for a single 50° day considering the entire house:
With good windows:
Q = (Uw·Aw + UrAr)·(Tb-50)·24
= (0.2·400 + 0.05·1600)·(65-50)·24
= 57,600 BTU
With better windows:
Q = (0.1·400 + 0.05·1600)·(60-50)·24
= 28,800 BTU
Reduction in heating load = 57,600 - 28,800 = 28,800 BTU.
So, the windows account for only 2/3 of the total reduction in heating load.
Mucking about with a year's worth of weather data for Thunder Bay showed that using HDD65 for this window analysis underestimates the heat loss difference by about 14%.
The formula for heat flow through an assembly is (area in square feet)*(temperature difference between the two sides, in F)/ (R-Value)= heat flow, in BTU/hr.
That literally comes from the definition of R-value.
If you accept that the formula is a reasonable approximation of reality, you see that nowhere does it say anything about adjacent assemblies. If you have a room with four walls, the heat flow through any one wall is not affected by the heat flow through any other wall. It would be peculiar, at least to me, if it did. What would be the mechanism for one wall to affect another?
Where the conversation is kind of going off the rails is that people are mixing up heat flows and R-values. If you have adjacent assemblies, there are two ways you can calculate the heat flow through the combined assembly. You can either calculate the R-value and area of each assembly on its own, and calculate the heat flow from the R-value and area, and add them up. Or you can calculate the R-value and area of the combined assembly, and calculate the heat flow from the combined values. Just to be clear, the formula for combined R-value of parallel assemblies is not simple addition, rather it's:
Rc= 1/(A1/(Ac*R1)+A2/(Ac*R2)+ ... + An/(Ac*Rn))
Where Rc is the combined R-value, Ac is the combined area, and A1 to An are the areas of each assembly and R1 to Rn are the R-values.
But that's R-value. The formula for heat flows through parallel assemblies is simple addition.
The point I was trying to make in post #15 is that the two methods give the same result. If they didn't, there's a problem with the assumption. And that it's usually computationally much simpler to calculate the heat flow through each assembly and add them up.
Doh, now it's clear. You're right, of course.
Thank you for clearing this up DC, I follow now
This.
How we typically model heat flow via R-value/U-value is completely linearizable. This is to say, we can look at components in isolation and determine a delta vs. a different variation of that component.
(Whether the model is correct is a completely different question, but let's not get into that.)
I had already emailed Christine Williamson and Allison Bailes before DC's last response, which they both mostly confirmed as the most practical way to do calculations and weigh options. But they both stressed that those are one-dimensional calculations and only work at all for flat surfaces. They said that THERM does a little better, as a 2D model. But in reality, heat flow is a three dimensional phenomenon that one-dimensional analysis cannot accurately represent. For the sake of this discussion, that's not important.
I'm sorry, but that sounds like a bunch of hand-waving. Maybe you're not fairly transmitting their concerns.
The process for Manual J is to break a house into a set of surfaces, calculate the losses for each surface and add them all up. There's a saying I first heard from an economist: "all models are wrong, some models are useful." Manual J is useful because it has good predictive power.
Possibly. I don't think they would mind me sharing their responses.
This is what Christine wrote:
"It depends on what you’re getting at. In one sense, a straight u-value to u-value comparison is the most useful and that’s pretty much all you can really do from a practical perspective.
But in another sense, it is absolutely true that heat pathways are complex and the thermal performance of materials is absolutely affected by adjacent materials and their orientation. For example you’ll get more heat flow through the portion of a wood stud immediately adjacent an exposed concrete slab (like at a floor line) than you’d get through the same stud in the middle of the wall, even though the r-value/u-value is the same.
The simple equation we use to determine the effective r-value of a wall is indeed a simplification, and it doesn’t take these other pathways into account, nor does it really work for anything that isn’t mostly flat.
The person who’s helped me understand this best (apart from Alison through his writing) is Chris Schumacher. He teaches these heat pathways to his engineering students which makes me think he might have some slides or presentation materials he could share. "
Which Allison followed with this: "I’ll add just one thing. The method I wrote about in that article is one-dimensional. Actually heat flow is three-dimensional. The software THERM goes beyond 1-D but still only does 2-D heat flow. In many cases 1-D modeling is sufficient."
[Reply to #37]
I disagree with that analysis. The break-even point doesn't change the interior temperature. The total heating load is going to be the load based on the difference between interior and exterior temperature, less the internal gains. The internal gains don't change when you change the windows*, when you're comparing two different windows the internal gains cancel out and the difference between them is the difference in heat flows at room temperature.
You're essentially counting the improvement in the windows twice, once when you adjust the break-even and again when you calculate the loss.
Now, what does happen is that with the better windows you have fewer degree-days of heating. So when you take my calculation of BTU/degree day, you can multiply by a slightly smaller number. But in order to do that you need to know the break-even for the whole building, so you can't just look at windows in isolation. Since these are crude, back-of-the-envelope calculations anyway I don't think changing annual degree-days is going to affect the thrust of the outcome.
*(I guess if you change the solar characteristics of the glass they might, but that's not what we're discussing here.)
For the sake of illustration, let me throw out some hypothetical numbers. Imagine a house that has a heating load of 35,000 BTU/hr at an indoor temperature of 70F and outdoor 0F. Further imagine it has 2500 Btu/hr of internal gains. The house loses 500 BTU per degree-hour, so its break-even temperature is 65F. Also imagine this house has 500 square feet of windows that are R5. At 0F, those windows lose (500sf)*(70F)/R5= 7000 BTU/hr.
Now imagine those windows are replaced with R7 windows. The new windows lose (500sf)*(70F)/R7= 5000 BTU/hr at 0F. The total heating load of the house drops from 35,000 BTU/hr to 33,000 BTU/hr. And it drops from 500 BTU/degree-hour to 471 BTU/degree-hour. The break-even temperature drops from 65F to 64.7F.
You can convert from BTU/degree-hour to BTU/degree-day by multiplying by 24, so the first house is 12,000 BTU/degree-day and the second is 11,300. You can estimate the annual heating load by multiplying BTU/degree-day by average degree-days. The degree-days you use here should be based on the break-even temperature of the house. However, in this example I would argue that the difference between HDD65 and HDD64.7 is so small as to have a negligible impact on the accuracy of the estimate.
If I had used the same heating balance point for both windows in my example, the result would be even farther off the mark.
Looking into your example, if the house is losing 35,000 BTU/hr and the R5 windows lose 7000 BTU/hr then the rest of the envelope loses 28,000 BTU/hr at ∆T=70° → 400 BTU/degree-hour → 9,600 BTU/degree-day. In a 9,000 HDD climate, a 0.3° difference in the balance point results in roughly 90 HDDs difference. 9,600 BTU/degree-day × 90 HDD = 860,000 BTU.
860,000 BTU is small in the context of discussing the house as a whole, < 1% of the annual heating load. But, the same magnitude error becomes more significant in the context of the windows only, and even more significant when calculating the difference between two window options.
The difference between the windows is 2,000 BTU/hr at ∆T=70° → 28.5 BTU/degree-hour → 685 BTU/degree-day. 685 BTU/degree-day × 9,000 HDD = 6,200,000 BTU.
860,000 BTU ∕ 6,200,000 BTU = 14%. Whether a 14% error matters or not depends on the context, but I wouldn't call that a negligible impact.
When using base 65° HDDs with the data for Thunder Bay, the error is 13% (underestimate), assuming the real balance points are around 60°. When using base 70° HDDs, the error is about 3% (overestimate). So yes, using 65° is good enough for a rough estimate, but you can get a much less rough estimate by using 70° (or whatever the actual inside temperature is).
To address your concern that I am counting the improvement in the windows twice by using both the new U-value and the new balance point, maybe this will help.
The energy savings due to a change in the windows is equal to the difference between the total heating load of the house with the old windows and the total heating load of the house with the new windows.
With everything else held constant, the total heating load Q of each house is a function of the window U-value and the house's heating balance point Tb:
∆Q = Q(U1,Tb1) − Q(U2,Tb2)
The heating load Q for each house is comprised of the heating load Qw due to the windows plus the heating load Qr due to the rest of the enclosure:
Q(U1,Tb1) = Qw(U1,Tb1) + Qr(Tb1)
Q(U2,Tb2) = Qw(U2,Tb2) + Qr(Tb2)
Therefore,
∆Q = [Qw(U1,Tb1) + Qr(Tb1)] − [Qw(U2,Tb2) + Qr(Tb2)]
Rearranging to group the windows together and the rest of the enclosure together:
∆Q = [Qw(U1,Tb1) − Qw(U2,Tb2)] + [Qr(Tb1) − Qr(Tb2)]
You are suggesting calculating it as just: ∆Q = Qw(U1,Tb1) − Qw(U2,Tb1)
Using the same balance point for both windows introduces a small error, while ignoring the lowered heating load of the rest of the enclosure introduces a much more significant error.
"You are suggesting calculating it as just: ∆Q = Qw(U1,Tb1) − Qw(U2,Tb1)"
No, I'm suggesting calculating it using the indoor temperature. I don't see why the balance point is relevant.
Think of it this way. You have the gross heat loss, which is the loss of the building without thinking about internal gains. Then you have the net heat loss, which is the gross loss less internal gains.
I'm saying that the gross loss is calculated from room temperature, you're saying it's calculated from break-even temperature. Here's a thought experiment: what happens when the internal gains change? Let's say you have a house that's being heated. You plug in an air fryer, which consumes 400W of electricity and contributes 400W of heat into the building.
Under my model, the gross heat load doesn't change, the internal gains increase by 400W and the net heat load declines by 400 W. Under your model, increasing the internal gains moves the break-even point, which lowers the gross heating load. So the net heating load decreases by the 400W from the air fryer, plus whatever the gross load declined by. So you're getting a more than 400w decrease in the heating load by using that appliance? Do air fryers have a COP greater than 1.0?
This is why I say you're counting the internal gains twice when you adjust the gross load to the internal gains.
In my previous comment above, I specified "with everything else held constant". The internal gains are assumed to be unchanged...no air fryers getting plugged in.
I've been arguing all along that using base 70° HDDs gives more accurate results than base 65° HDDs or whatever the balance point happens to be. If you're now saying to use the interior temperature then we agree.
In my previous comment above, I specified "with everything else held constant". The internal gains are assumed to be unchanged...no air fryers getting plugged in.
You're completely ignoring the substance of my comment.
To clarify, we're talking two separate but related things here. One is calculating the heating load for a building, which is done to size the heating system so that you're not cold on the coldest days. That's done using Manual J, which calculates gross heating load based on room temperature.
The question that the original poster asked was about the impact of changing windows, which gets to the second question, modeling annual usage from heating load and observed heating degree-days. In post #35 I said that this calculation would be most accurate if the degree-days used were based on the actual breakeven temperature of the house in question.
In post #41 I offered a realistic example, which showed that a change like the OP is suggesting would have a minimal impact on the breakeven temperature -- a 6% change in heating load moves the breakeven by just 0.3F in my example -- which means updating the degree-day calculation to use the new breakeven temperature would have a minimal impact on the annual energy use estimate.
There are 3 methods being discussed for how to calculate the effect of changing windows:
1. Use the actual inside temperature to calculate the heating degree days and use that to calculate the heat loss through the windows. This is the method I recommend to get reasonable accuracy with little effort.
2. Use the heating balance points to compare the heating loads of the two options.
For each window choice, use it's U-value and corresponding balance point to calculate the house's heating load. The difference between the two heating loads is the energy saved by changing the windows.
The problems with method 2 make it impractical:
- You have to know the two heating balance points.
- You have to calculate the heat load of the whole house. That means you have to know the R-values and areas of ALL the envelope assemblies, not just the windows.
Ignoring the difference in heating balance points results in a significant error. The 0.3 °F change in your example may seem minuscule but ignoring it would result in a 14% underestimate of the difference between the two windows.
3. Calculate the HDDs using 65° as the base, or using the actual balance point temperature as the base. Then use that HDD value to calculate heat loss through the windows only. This seems to be what you are suggesting. If not, you'll have to clarify.
This method results in a sizeable error. When using HDD65 with the OP's window selections in a 9,000 HDD climate (northwest Ontario), the error is approximately 13%. Using HDD60 (closer to the actual balance point for a pretty good house) is even worse, doubling the error to 28%.
The reason for the error is easy to see. A single 40° day with an inside temperature of 70° equals 30 HDDs. If you use 65° as the base temperature then you only capture 25 HDDs, underestimating the heat loss through the windows by 17%. If you underestimate the heat loss for both windows by 17%, then you also underestimate the difference between them by 17%.
None of those three methods are what I'm recommending.
It would be helpful if for each method you said who was recommending it and in which post.
DC, then you'll either have to clarify exactly what is your criticism of my recommendation to calculate the HDDs using a base of 70° rather than 65° (comment 31), and what method you are recommending instead.
Previously you have seemed to be arguing for the 3rd method:
"The choice of 65F as the baseline for heating degree-days is an attempt to account for the contribution of other heat sources, like occupant behavior." (comment 32)
"I would argue that the most predictive calculation of heating degree-days would thus be:
∑(breakeven - avg daily temp) for all days with avg daily temp < breakeven" (comment 35)
Interesting discussion. One other consideration here is condensation risk, which is a function of several things: outside temp, u-value, interior humidity and temp, and the height of the glass (tall expanses develop larger convective loops, resulting in lower glass temp.) PHIUS’ WUFI software can be used to model this.
I'm in early stage design of a major renovation project and considering the same tradeoffs as the OP. What I find interesting is the need to evaluate the wall and windows together as a system. It's unavoidable that windows are a much high conductivity path than walls, giving the windows substantially more impact on the composite wall system R-value. Run the calculations on a few wall constructions and window options and you may be surprised at how much your high performing walls are degraded by even the best performing windows that are available. There's no point in going overboard with the wall design unless you pair it with very low conductivity windows.
Making the financial tradeoff is trickier, but given how long good quality windows will last, one shouldn't hinge the decision on a relatively quick payback period. Consider a project being financed with a 20 year mortgage. Do the better performing windows reduce energy cost by more than the incremental mortgage payment increase? If so, that would be an easy decision for me, particularly given escalating energy costs. Harder to factor in is the comfort value in increasing the MRT and the value of reducing carbon footprint, yet these are also considerations in favor of better window performance.
The point I made in post #15 and fleshed out in #26 was that you don't have to calculate the windows and wall together as a system, you get the same result by calculating the heat flow through the window alone.
Yes - I completely agree with your technical analysis. This is a linear system. We can analyze elements independently without concern that they impact other elements, at least to a first order. Analyzing each element independently is also a great way to develop an intuitive understanding of how they work together in a complete house.
The point I intended to make is that a very high performing wall with a moderate performing window is likely a design optimization mismatch. Consider a wall with 25% glazing using U-0.20 windows. The composite R-value is R-14.5 with an R-40 wall and R-11.4 with an R-20 wall. That's only a 21% reduction in composite R-value for a 50% reduction in wall R-value. If we instead use U-0.15 windows, an R-24 wall yields a similar, R-14.6 composite R-value.
R-40 walls seem a common goal here for my New England climate, but I contend that much of their value is circumvented without a low glazing ratio or very high performing windows. Again, this is from a design and cost optimization perspective, not an absolute energy reduction perspective.
DC, I truly appreciate your input in this discussion. Unfortunately much of what has been discussed is simply over my head.
I see your breakdowns in post #13 and #15. Can you run the numbers with the following data for 2 separate options with glazing u-factors of 0.17 and U0.145.
-HDD 10,000
-531 sq ft glazing
-ASHP with COP of 2.0
-Electrical is 9 cents/kWh (backup heat when below -20 will be from a Blaze King woodstove, or resistance heaters if we are away)
The markup for the better performing windows (U0.145 vs u0.17) is around $9,000.
OK, 531 square feet of U0.17 loses 2,166 BTU per degree day. At 10,000 HDD that's 21,660,000 BTU for the season.
The same 531 square feet at U0.145 loses 1,848 BTU per degree day, or 18,480,000 BTU per season.
The difference is 3,180,00 BTU per season. That's equal to 942 kWh. At COP 2.0 that's equal to 471 kWh gross, or $42.39 per year at 9 cents per kWh.
If the better windows cost $9,000 more, the simple payback period is 212 years.
Wonderful, thank-you for that breakdown.
There are other benefits to lower-U windows. They're more comfortable to sit next to and less likely to fog up. So it's not a straight question of payback.