Many years ago some guy decided to leave a legacy behind. His goal was to confuse the heck out of us. But with a little patience I will try to clear the muddy water on our way to understanding R-values. R-values are a measure of thermal resistance, or the ability to insulate. You will most commonly find R-values on insulating materials at your local building supply store. R-values range from 1 (for a single-pane window) to 7.2/inch (for polyisocyanurate foil-faced panels) or as high as 10 for high-tech silica aerogel and 30 for vacuum insulated panels. But what does R-value mean? Well, here comes the confusing part: R=ft²-°F-h/Btu (I am using the non-SI definition here because it is most common in the US). What?

Basically, R-value is the amount of energy lost, in BTU per hour, through a square foot of surface, based on the temperature difference between the inside and outside. To really wrap your head around it let’s look at an example… Let’s say we have a hallway that leads to the south pole, where it’s quite cold. We could just run the heater at full blast to try and keep the cold back, but energy is expensive. The other option is to block off this hallway with an insulated wall. Let’s say the cross section of the hallway is 20 ft², it is 0F at the South Pole today, and we want to keep our end of the hallway at 70F. I think I would like to install a three-inch thick wall of the polyisocyanurate foil-faced panels, with an R value of 21.6.

With a little bit of algebra we can get the R=ft²-°F-h/Btu equation to say Btu/h=ft²-°F/R. Plugging in 20 ft², 70°F (70°F – 0°F), and 21.6R, we get 64.8 Btu/h. This answer assumes that the remaining walls of the hallway are hyper-insulated but it tells us how much heat will still be lost through the new wall.

I will do one more example, using an enclosed space with both insulating panels and a glass window to find the overall heat-loss. This result will be useful for next week’s AskPablo.

Let’s say we have a box that is 1x1m, 25cm high and has a 1x1m double-pane window on the front. What will be our heat-loss at various temperature levels (assuming a constant 50°F exterior temperature)? Let’s assume that we are using the 3 inch thick polyisocyanurate foil-faced panels again, which would have a combined surface area of 20,000 cm², or 21.53 ft². The double-pane glass has an R-value of 2 and an area of 10.765 ft². To calculate heat-loss for a composite wall (made up of different materials) we can use this equation: Btu/h=(AreaWall/RWall + AreaWindow/RWindow) x °F. The part of the equation in parenthesis comes out to 6.38. To get BTU/h for any temperature difference, we simply need to multiply it by this factor. If the temperature inside the box is 50F, there is no heat loss; at 60°F the heat loss is 63.8 Btu/h; at 70°F the heat loss is 127.6 Btu/h; and so on.

For those metric fans out there, 63.8Btu/h is equal to 18.7 Watts. So for every 10°F rise in interior temperature we would need to turn on one 20W bulb to maintain that temperature. Neat!

Pablo Päster, MBA

Sustainability Engineer

www.AskPablo.org

Pablo(dot)Paster(at)gmail(dot)com