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If you want to make your house more efficient at repelling the unpleasantness outdoors (whether hot or cold), what should you do first? Insulate the walls? Insulate the ceiling? The roof? Better windows? Draft elimination? What has the biggest effect? While I have regrettably little practical experience tightening up a house (it’s on my bucket list), I at least do understand heat transfer from a physics/engineering perspective, and can walk through some insightful calculations. So let’s build a fantasy house and evaluate thermal tradeoffs at 1234 Theoretical Lane.
There are only three ways for heat to travel: conduction, convection, and radiation. No other options.
The (energy per unit time) flowing across a material by conduction sensibly depends on the material properties (thermal conductivity, κ), the thickness of the material, t, the area, A participating in the conduction (between the cold and hot environments), and the temperature difference, ΔT. Without much thought, you could construct the correct relationship for the power transported by conduction by figuring out how it should scale as we change one variable or the other: Pcond = κAΔT/t, where κ is the thermal conductivity of the material, taking on units of W/m/°C in the metric system. For many building materials, κ is in the range of 0.1–1 W/m/°C. A sheet of plywood, at the lower end of the range (κ ≈ 0.12, measuring 4×8 feet, or 3 m²; t = 0.019 m, or 0.75 inches, thick) would conduct about 19 W per degree Celsius presented across it.
The building industry characterizes materials by their R-value, which in the U.S. has the unfortunate units of ft²·°F·hr/Btu. The SI equivalent is a slightly more tidy m²·°C/W. The R-value builds the thickness, t, into the measure, so the same material in twice the thickness will earn twice the R-value.
Relating to intrinsic properties of the material, κ and t, RUS = 5.7×t/κ in the U.S., or more simply, RSI = t/κ overseas. Our plywood from before would be characterized as R = 0.9 in the U.S., or 0.16 internationally. Note that the R-value is independent of area. To get the power flow across a surface, in Watts, we replace the relation two paragraphs back with Pcond = 5.7×AΔT/RUS, or Pcond = AΔT/RSI.
Convection is at its core just conduction into a moving fluid, which then carries the heat away by simply wafting it along. Adjacent to any surface in a fluid flow is a boundary layer of fluid that clings to the surface, so that the thermal flow is controlled by conduction across the boundary layer. For air, κ ≈ 0.02 W/m/°C, and boundary layer thickness is often in the neighborhood of a few millimeters, putting the effective R-value (US) in the neighborhood of 1.
For an explicit example of how all this works, let’s construct a wall out of a single sheet of plywood (κ = 0.12 W/m/°C; t = 0.019 m; so RUS = 0.9. We’ll have an inside environment with h = 2 W/m²/°C, T = 20°C, and assume the inside wall temperature is close to the same, so that I can use T = 293 K in the radiation approximation term. In this case, I compute R values (US) of 2.85 and 1 for convection and radiation, respectively (for the still air inside, radiation is here the more important channel). In parallel, these add to an effective R-value of 0.74. If the outside of our “wall” is near the ambient temperature of, say, 273 K, and a bit of wind gives us h = 10 W/m²/°C, we have R-values of 0.57 and 1.2 for convection and radiation (note the role reversal in more active air, so that convection dominates). The outside combination is R = 0.39.
Our total transfer through the wall therefore has three R-values in series: 0.74 to get heat into the wall, 0.9 to get heat through the wall, and 0.39 to get it off the outside surface. Summing these, we have RUS ≈ 2.03 in total. For an inside-outside ΔT = 20°C, each square meter of this wall would conduct 5.7×20/2.03 ≈ 56 W.
Now that we have some sense for how to handle conduction, convection, and radiation in the R-value context, we can find and use relevant R-values for common building materials. I get most of my information from this , many values also being available at the .
To compute the effective R-value for a composite surface like a wall with studs inside, one simply combines paths in parallel, weighted by the fractional area of each. For instance, a wall with studs has 15% of the area covered by studs, with a total end-to-end R-value (including convection/radiation, called “air film”) of 7.1. The other 85% is insulated bay, with an R-value of 15.7. The effective R-value is given by 1/R = (0.15/Rstud + 0.85/Rbay), calculating to R = 13.3. If I left out the insulation, I would replace the R=13 fiberglass batting with two “air film” layers carrying values of 0.68 (very similar to our value of 0.74 from above). In this case, we have 1/R = (0.15/7.1 + 0.85/4.1), or R = 4.3. Note that for uninsulated walls, the studs are more insulating than the air space between.
Let’s now assemble a table of values for relevant building blocks. Divide RUS by 5.7 to get RSI.
For the sake of simplicity, we’re going to make a one-story house with a square footprint. We’ll have a pitched roof with attic space, and will look at raised foundations with a crawl space underneath, and also slab foundations. We’ll adorn each side of the house with two moderate-sized windows and a front and back door. For size, we’ll go with something close to the American average of 2700 ft² and take the opportunity to go metric by making our house 15 m on a side, resulting in an area of 225 m² or 2422 ft². The walls will be 2.5 m (8 ft) high. For windows, we’ll make each one 1.5 m² (equivalent to 16 ft², or 4×4 feet). Our doors will take up 2 m² each.
The wall area therefore totals 134 m², floor and ceiling each 225 m², windows 12 m², and doors 4 m².
We will compute the thermal snugness of a house in terms of W/°C, and call this thermal admittance. Each component adds some bit of thermal admittance according to Q = P/ΔT = 5.7×A/RUS. These can then be added for each component of the house.
Using the uninsulated values for everything and single-pane windows, I get Q values, in W/°C, for the walls of 186; ceiling (assumes ample attic ventilation puts it at ambient temperature): 777; raised floor: 513; single-pane windows: 75; doors: 8. The total is 1560 W/°C.
Let’s pause to put this number in perspective. Maintaining room temperature when the outside is at freezing would require 31 kW of power, or 20 space heaters. A furnace rated at 75,000 Btu/hr is equivalent to 22 kW and would not be able to keep up. And we have not even considered drafts yet.
Now we’ll look at the other extreme and put R-13 insulation in the walls, ceiling, under the floor, and use the best windows we can buy. We will again let the attic be fully ventilated and at the outdoor ambient temperature. Now we get walls: 57; ceiling: 99; floor: 103; windows: 17, and doors still at 4. The total is 280 W/°C, and about a fifth of what it was previously. The cost of heating/cooling will likewise improve by at least a factor of five (won’t be needed as often in milder conditions). In our case, 53% of the improvement came from insulating the ceiling, 32% from the floor, 10% from the walls, and 5% from the windows. This suggests an order of priority. Of course even larger gains are possible with greater amounts of insulation—until other factors dominate.
The floor loss is slightly exaggerated here, as the simple numbers assume the crawl space is as cold as the exterior. To the degree that this is not true, the numbers soften a bit, in proportion to the relative temperature rise. It is also the case that the air near the floor is likely to be cooler than the air near the ceiling, unless the interior air is being well mixed. This also reduces heat loss through the floor in the case that it’s colder outside than inside. Still, it is likely that insulating the floor will bring a pretty noticeable improvement.
Perhaps the assumption of a fully ventilated attic caused consternation. Had I assumed a sealed attic (the other extreme), the ceiling and roof would act in series to produce an R-value of 3.5 in the uninsulated case or 26.2 in the insulated case. The thermal admittance values would then be 366 W/°C and 49 W/°C, respectively. Our totals would go from 1150 W/°C to 232 W/°C. The biggest single gain would then stem from insulating the floor. But in reality, the attic tends to be closer to ambient than to interior, so that ceiling insulation is likely to remain the most important step.
Assuming the attic is ventilated, most of the temperature difference between interior and exterior will appear across the ceiling, rendering the roof’s insulating qualities of secondary importance. But this neglects solar load onto the roof. Anyone who has experienced a hot attic knows that attic ventilation is inadequate to prevent the roof from heating the space. Therefore insulating the roof may become an important step in environments where cooling is a large energy sink. For places where heating is more important than cooling, it may actually be better to leaving the roof insulation off so that the winter sun provides some heating benefit by warming the attic a bit.
For slab floors, the evaluation is somewhat more complicated than for raised floors. A six-inch slab of concrete itself has an R-value of around 0.5. But below the slab is dirt. Cobbling together information from a few sources ( and ), I gather that dry soil has a thermal conductivity around 0.8 W/m/°C, and an effective thermal thickness (length scale over which temperature gradient exists) around 0.2 m. This would give it an R-value around 1.4 for a combined slab-ground R-value of 1.9, or 2.6 once factoring in the radiative/conductive coupling. But all this may not matter because the ground temperature is pretty stable throughout the year, and may reach approximate equilibrium with your house temperature—at least away from the slab edge. To address leakage out the sides of the slab (air and ground), the implies a loss rate of 1.2 W/°C per meter of perimeter, or 72 W/°C for our lovely house, which is not too different from what we computed for the insulated raised floor.
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