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# How Much Heat Really Leaks Out of the Attic Hatch of an Insulated Ceiling?

Dr. Bailes does the math on two ways heat flows through building assemblies.
January 30, 2024
Read time: 7 mins

Introduction to heat flow: how much heat leaks through gaps in the insulation vs. thermal bridges, like framing?

This is Video Number 10 in our Building Science module, and I am Allison Bales of Energy Vanguard here to guide you through this lesson on series and parallel heat flow.

## Series heat flow adds R-values

Let's start with the easy case of series heat flow. This is where you have an assembly made up of different layers. The layers separate the hot side from the cool side, so the heat has to travel through every single layer from the hot side to the cool side. That means that the amount of heat flow is going to be the same no matter which path it takes, and the R values, in this case, will add.

So, the total R-value is going to be the R-value of Layer A in this case plus the R-value of Layer B. That's our total R-value. It doesn't matter how many layers we have; we could have a hundred layers here, and we would just add up all the R values. Let's look at a quick example.

Let's say we have a foam board that has an R-value of 10, and we put that on a concrete foundation wall that has an R-value of 1. It's very simple; the total R-value is 10 plus 1 or 11. We've got an R-11 foundation wall now by putting that R-10 foam board on it.

## Parallel heat flow averages R-values

Rather than layers, we have pathways, and the pathways are choices that the heat has. The heat can go either through one path or another. In this case, we've got two different pathways; even though the two pathways labeled number one are in different places, they're the same type of material, so we call them the same pathway. We can just add them together.

The amount of heat that travels through the assembly depends on which path it takes.

The way we handle this is that we average the U values, and we average the U values using this formula that you see down here. In this case, we have only two types of pathways, 1 and 2, so we're going to multiply the U value for pathway 1 times its area and then add it to the multiplication of U2 and A2. To do all that math on the top, then we divide by the total amount of area, and that will give us the average U value.

Okay, this works for any number of pathways. I've shown it for only two pathways here, but if we had 50 pathways, it would still work. We would just have to add more terms into our equation. It would be U average equals U1 times A1 plus U2 times A2 plus U3 times A3 plus U4 times A4, dot, dot, dot, all the way out to 50, and the area would be the total area.

## Parallel heat flow example

Let's look at an example here, a simple parallel heat flow problem. Let's say we've got a ceiling and it's a thousand square feet. A total of 990 square feet of it has R-38 insulation, but 10 square feet of attic stairs are uninsulated, so we're going to call that R1.

We apply this formula: Uaverage = [(U1)(A1)] + [(U2)(A2)] / Atotal

Remember that we're given R values here, but we have to put U values in the equation, but we know the relation between R and U. If we want U, and we know U is equal to 1/R, so here's what we've got:

• Pathway one is the insulated part, and U1 would be 1 over 38.
• A1 would be 990.

So we do this first, and here's a little tip for you: instead of doing 1 over 38 in your calculator and then multiplying that by 990, the easy way to do this is just 990 divided by 38 gives you the same exact answer. Do that first.

So 990 divided by 38 gives us the first part.

Then we go over here, and we do this part.

• 1 over 1 is 1
• times 10 is 10

So we've got 10 here.

• When we do 90 divided by 38, we get 2.36
• and 2.36 plus 10 is 12.36
• 12.36 divided by 1000 is 0.01236

So that is our average U value. If we want to know what the average R-value is, we just do 1 over the average U-value. So 1 divided by 0.01236, put that in your calculator, you get 81.1.

The takeaway here is that 1% uninsulated area can have a huge effect on your heat flow because that knocks your R value of that R-38 attic down to R-28, basically.

## Series and parallel heat flow

Now, let's look at the series and parallel heat flow problem. This one is more complex. This is a more realistic case because most of the time we do have both series and parallel heat flow. So let's see what we've got. The first thing we need to do is look at our pathways, and we're going to add all the R values in each pathway. So, if this is path 1, we're going to add R A to R B in that path. Then in path 2, we're going to add the R value of this to the R value of that, and that will give us the two R values that we need to average. But of course, in averaging those R values, we're going to turn them into U values first because the equation works with U values. It's the same basic formula, works for any number of layers, any number of pathways. Let's look at an example.

## Ceiling heat flow

This is a ceiling assembly, and we've got 1,000 square feet of area. We've got 2x10 joists, 16 inches on center, with R-30 blown insulation in between. Each joist has an R-value of 11.5, and then the bottom layer, the drywall, has an R-value of 0.5.

The next thing we need to know is the area.

So, the joists are 1.5 inches wide. A 2x10 is 1.5 by 9.25, so the width is 1.5. We've got 16 inches on center, so the cavity is 14.5 inches. The relative areas then would be 1.5 divided by 16. So out of every 16 inches across, we're going to have 1.5 inches of the ceiling joist.

If we do that division, multiply by 100%, that's 9.4% of the area will be ceiling joist, and 14.5 out of 16 is 90.6% of all the area will be the insulation.

If we multiply those percentages out, 9.4% of 1,000 square feet is 94 square feet, and 90.6% of 1,000 square feet is 906 square feet. Those are the areas of the ceiling joists and the insulation, respectively.

If you want to give this a stab, see if you can try to find the average U-value and the average R-value.

Alright, let's work through this now. The first thing is we've got two pathways.

We've got the pathway where the heat travels through the ceiling joist, and then the drywall. Now, this is what we're calling Path 1, and we've got the pathway where the heat travels through the insulation, and then the drywall. That's Path 2.

If it goes through the ceiling joists, the R value it encounters is 11.5 through the wood, and then 0.5 in the drywall. That's the layers, the series heat flow. So the total R-value is the sum of those; it's 12.0. If it goes through the insulation, it encounters an R-30 through the insulation and then 0.5 through the drywall. So the total would be 30.5. Here are the two areas, so these are the numbers that we need for the equation.

Here's our equation.

Now, the average U value is U1 times A1.

So U1 is R1 over R12, and the area is 94.

U2 is 1 over 30.5, and the area 2 is 906.

If we work that out, 94 divided by 12, and remember that little shortcut there is 7.83, and 906 divided by 30.5 is 29.7.

If we add those two together and divide by 1,000, we get 0.038, and that's our average U value. Our average R value would be 1 divided by that or 26.6.

So, in this case, our ceiling with 94 square feet of joists and 906 square feet of insulation has an average R value of 26.6.

You can see the framing in the assembly here does have an effect. It lowers the R-value a little bit from 30 to 26.6. Overall, it's not nearly as bad as having that 1% uninsulated area that we had in the previous problem.

So that's a quick introduction to series and parallel heat flow, a very important concept to understand, and in energy modeling, we use this stuff all the time.

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