By Dr. Robert E. Buxbaum, June 3, 2013

It’s common to have companies call during dinner offering to blow extra insulation into the walls and attic of your home. Those who’ve added this insulation find a small decrease in their heating and cooling bills, but generally wonder if they got their money’s worth, or perhaps if they need yet-more insulation to get the full benefit. Here’s a simple approach to comparing your home heat bill to the ideal your home can reasonably reach.

The rate of heat transfer through a wall, Q^w, is proportional to the temperature difference, ∆T, to the area, A, and to the average thermal conductivity of the wall, k; it is inversely proportional to the wall thickness, ∂;

Q^w = ∆T A k /∂.

For home insulation, we re-write this as Q^w = ∆T A/R^w where R^w is the thermal resistance of the wall, measured (in the US) as °F/BTU/hr-ft². R^w = ∂/k.

Lets assume that your home’s outer wall thickness is nominally 6″ thick (0.5 foot). With the best available insulation, perfectly applied, the heat loss will be somewhat higher than if the space was filled with still air, k=.024 BTU/ft^. hr°F, a result based on molecular dynamics. For a 6″ wall, the R value, will always be less than .5/.024 = 20.8 °F/BTU/hr-ft^2.. It will be much less if there are holes or air infiltration, but for practical construction with joists and sills, an R^w value of 15 or 16 is probably about as good as you’ll get with 6″ walls.

To show you how to evaluate your home, I’ll now calculate the R value of my walls based on the size of my ranch-style home (in Michigan) and our heat bills. I’ll first do this in a simplified calculation, ignoring windows, and will then repeat the calculation including the windows. Windows are found to be very important. I strongly suggest window curtains to save heat and air conditioning,

The outer wall of my home is 190 feet long, and extends about 11 feet above ground to the roof. Multiplying these dimensions gives an outer wall area of 2090 ft². I could now add the roof area, 1750 ft² (it’s the same as the area of the house), but since the roof is more heavily insulated than the walls, I’ll estimate that it behaves like 1410 ft² of normal wall. I calculate there are 3500 ft² of effective above-ground area for heat loss. This is the area that companies keep offering to insulate.

Between December 2011 and February 2012, our home was about 72°F inside, and the outside temperature was about 28°F. Thus, the average temperature difference between the inside and outside was about 45°F; I estimate the rate of heat loss from the above-ground part of my house, Q^u = 3500 * 45/R = 157,500/R^w.

Our house has a basement too, something that no one has yet offered to insulate. While the below-ground temperature gradient is smaller, it’s less-well insulated. Our basement walls are cinderblock covered with 2″ of styrofoam plus wall-board. Our basement floor is even less well insulated: it’s just cement poured on pea-gravel. I estimate the below-ground R value is no more than 1/2 of whatever the above ground value is; thus, for calculating Q^B, I’ll assume a resistance of R^w/2.

The below-ground area equals the square footage of our house, 1750 ft² but the walls extend down only about 5 feet below ground. The basement walls are thus 950 ft² in area (5 x 190 = 950). Adding the 1750 ft² floor area, we find a total below-ground area of 2700 ft².

The temperature difference between the basement and the wet dirt is only about 25°F in the winter. Assuming the thermal resistance is R^w/2, I estimate the rate of heat loss from the basement, Q^B = 2700*25*(2/R^w) = 135,000/R^w. It appears that nearly as much heat leaves through the basement as above ground!

Between December and February 2012, our home used an average of 597 cubic feet of gas per day or 25497 BTU/hour (heat value = 1025 BTU/ ft³). Q^U+ Q^B = 292,500/R^w. Ignoring windows, I estimate R^w of my home = 292,500/25497 = 11.47.

We now add the windows. Our house has 230 ft² of windows, most covered by curtains and/or plastic. Because of the curtains and plastic, they would have an R value of 3 except that black-body radiation tends to be very significant. I estimate our windows have an R value of 1.5; the heat loss through the windows is thus Q^W= 230*45/1.5 = 6900 BTU/hr, about 27% of the total. The R value for our walls is now re-estimated to be 292,500/(25497-6900) = 15.7; this is about as good as I can expect given the fixed thickness of our walls and the fact that I can not easily get an insulation conductivity lower than still air. I thus find that there will be little or no benefit to adding more above-ground wall insulation to my house.

To save heat energy, I might want to coat our windows in partially reflective plastic or draw the curtains to follow the sun. Also, since nearly half the heat left from the basement, I may want to lay a thicker carpet, or lay a reflective under-layer (a space blanket) beneath the carpet.

To improve on the above estimate, I could consider our furnace efficiency; it is perhaps only 85-90% efficient, with still-warm air leaving up the chimney. There is also some heat lost through the door being opened, and through hot water being poured down the drain. As a first guess, these heat losses are balanced by the heat added by electric usage, by the body-heat of people in the house, and by solar radiation that entered through the windows (not much for Michigan in winter). I still see no reason to add more above-ground insulation. Now that I’ve analyzed my home, it’s time for you to analyze yours.