90% of a home's energy is lost through radiant energy.First of all, this depends greatly on how much air leakage your home has, but even if it's true there's a major problem with the statement: it completely ignores the story of how the energy got to the outside of your home where it's now being radiated out because all warm objects radiate energy.
R-value only measures conductive heat transfer.False. R-value measure thermal resistance due to all three types of heat transfer. This statement is so blatantly wrong that I'm going to repeat it later on just in case you're scanning because you already decided this article is boring
Radiant Barriers are actually a manifestation of a hither o unknown deity and should therefore be worshiped and tithed unto.Alright, you got me, no one actually said that...at least not that I'm aware of. In response I've written a long and tedious article to try to present a bit of something you can't seem to find in many places in regard to radiant barriers, namely facts and scientific evidence. How Heat Works There are three main types of heat transfer: conduction, convection and radiation. Conduction is the movement of heat from one molecule to another that are in direct contact with each other, convection is the movement of heat from liquids or gasses moving from warmer areas to cooler areas, and radiation is the movement of heat by electromagnetic waves emitted from all warm bodies. Calculating Radiant Heat Transfer A black body is a hypothetical object that absorbs all the radiation that falls on it's surface. They don't actually occur in nature but are useful for creating the equations necessary to calculate radiant heat transfer. The equation for calculating the radiation per unit of time from a black body is as follows: q = ? T4 A where q = heat transfer per unit time (W) ? = 5.6703 10-8 (W/m2K4) - The Stefan-Boltzmann Constant T = absolute temperature Kelvin (K) A = area of the emitting body (m2) What we are more interested in for our purposes are 'gray bodies', or objects that absorb only part of the radiation they're exposed to: q = ? ? T4 A where ? = the constant emissivity coefficient of the object or material (a black body would have a value of 1) Here are some emissivity coefficients for some common materials we'll be interested in looking at: Aluminum paint 0.27 - 0.67 Aluminum Foil 0.04 Wood 0.91 Now, using a variation of the above equation we can calculate the net radiation loss rate from a material next to an airspace (or vacuum). For our purposes I'm going to convert everything to imperial units so that we can get BtU hours per square foot: q = ? ? (Th4 - Tc4) Ac where Th = hot body absolute temperature (oR) Tc = cold surroundings absolute temperature (oR) Ac = area of the object (ft2) We'll use a Steffan-Boltzmann constant expressed in imperial units: = 0.1714 10-8 ( Btu/(h ft2 oR4) ) In our first case let's look at the performance of one of the most onerous forms of radiant barriers, reflective paint or a layer of reflective foil on the underside of a roof deck. We'll take a look at the only time when these are remotely beneficial, in the summertime. We have to make some temperature assumptions in order to make the calculations: Roofing assembly(hot body) is at 150 oF (609.67 oR) and the attic air temperature(cold surroundings) is at 130 oF (589.67 oR) q = ? ? (Th4 - Tc4) Ac q=0.27(aluminum paint) x 0.1714 10-8 ( Btu/(h ft2 oR4) ) x (609.67 oR4 - 589.67 oR4) x 1ft2 This gives us a value of 7.99 Btu per hour per square foot of roof deck for aluminum paint. q=0.04(aluminum foil) x 0.1714 10-8 ( Btu/(h ft2 oR4) ) x (609.67 oR4 - 589.67 oR4) x 1ft2 This gives us a value of 1.18 Btu per hour per square foot for aluminum foil. Now here is the calculation for a wood roof deck with no radiant barrier: q=0.91(wood) x 0.1714 10-8 ( Btu/(h ft2 oR4) ) x (609.67 oR4 - 589.67 oR4) x 1ft2 This give us a value of 26.92 Btu per hour per square foot for wood. Big difference! But wait, for a wood roof deck with no radiant barrier we need to adjust our roof deck temperature downward by 5 to 10 degrees Fahrenheit (or degrees Rankine) since the roof assembly won't be as hot. q=0.91(wood) x 0.1714 10-8 ( Btu/(h ft2 oR4) ) x (599.67 oR4 - 589.67 oR4) x 1ft2 We arrive at a value of 13.12 Btu per hour per square foot of roof deck. For the sake of argument let's assume that the attic air is 5 degrees F/R hotter because we aren't using a radiant barrier: q=0.91(wood) x 0.1714 10-8 ( Btu/(h ft2 oR4) ) x (599.67 oR4 - 594.67 oR4) x 1ft2 Now we see a value of 6.64 Btu per hour per square foot of roof deck. Well isn't that interesting, the fact that the radiant barrier is raising the roof assembly temperature has a negative effect on the radiant barrier's emissivity performance. Decreased attic air temperatures also mean greater radiant heat transfer from the roof assembly. As some may have noticed, I'm completely ignoring the reflectivity of the radiant barrier. This is so that I can use disinformation to pseudo-prove my case like certain barrier makers fore mentioned. Just kidding, caught you napping. The reason for this is that in this case we're looking at radiant barriers applied directly to the bottom of the roof deck which is the case with sprayed on paints and foil faced roof decking; the reflective surface is on the wrong side of the airspace to have any significant effect on heat coming down from the roof into the attic space. R-values and You, friends for life Now let's take a look at the effect of insulation on the attic floor. Blown cellulose has an R value of 3.7 per inch. The heat transfer rate in Btu per hour for a given area can be calculated using the following equation:
where h is the inverse of the R value in Btu per degrees Fahrenheit per hour, A is area in square feet and then we multiply by the temperature difference on either side of the insulation.
If the attic is 130 deg F and the house is 70 deg F the transfer rate per square foot of attic insulation is:
1 inch thick insulation = 16.22 Btu/hr @1500 SF = 24,324 Btu/hr
2 inch thick insulation = 8.11 Btu/hr @1500 SF = 12,162 Btu/hr
4 inch thick insulation = 4.05 Btu/hr @1500 SF = 6,081 Btu/hr
6 inch thick insulation = 2.70 Btu/hr @1500 SF = 4,054 Btu/hr
8 inch thick insulation = 2.03 Btu/hr @1500 SF = 3,040 Btu/hr
10 inch thick insulation = 1.62 Btu/hr @1500 SF = 2,432 Btu/hr
Now let's reproduce that table at an attic temperature of 120 deg F and the same 70 deg F indoor temperature:
1 inch thick insulation = 13.51 Btu/hr @1500 SF = 20,270 Btu/hr
2 inch thick insulation = 6.76 Btu/hr @1500 SF = 10,135 Btu/hr
4 inch thick insulation = 3.38 Btu/hr @1500 SF = 5,067 Btu/hr
6 inch thick insulation = 2.25 Btu/hr @1500 SF = 3,378 Btu/hr
8 inch thick insulation = 1.69 Btu/hr @1500 SF = 2,534 Btu/hr
10 inch thick insulation = 1.35 Btu/hr @1500 SF = 2,027 Btu/hr
Let's look at the thermal transfer through a fixed amount of insulation, 8 inch cellulose R-29.6 at different attic temperatures with a constant interior temperature:
90 deg F attic temp = 0.68 Btu/hr @1500 SF = 1,013.51 Btu/hr
100 deg F attic temp = 1.01 Btu/hr @1500 SF = 1,520.27 Btu/hr
110 deg F attic temp = 1.35 Btu/hr @1500 SF = 2,027.03 Btu/hr
120 deg F attic temp = 1.69 Btu/hr @1500 SF = 2,533.78 Btu/hr
130 deg F attic temp = 2.03 Btu/hr @1500 SF = 3,040.54 Btu/hr
140 deg F attic temp = 2.36 Btu/hr @1500 SF = 3,547.30 Btu/hr
We can see from the above that for each 10 deg F the attic tempurature increases we are adding just over 500 Btu per hour to the cooling load. This, however, doesn't account for a difference in thermal performance of our insulation at higher temperatures for which I have no data (R value tests are performed at 70 deg F). It is safe to presume the thermal transfer rate will increase, or the thermal performance will decrease, as the attic temperature increases.
Even though the added insulation gains per each 2 inch were diminishing, going from 8 to 10 inches of cellulose insulation @ 130 deg F still produced a greater energy savings than a 10 deg F lowering in attic temperature by nearly 100 Btu/hr.
I Warned You I Would Repeat This
Some manufacturers of radiant barriers and other sources falsely claim that R-value measures only conductive heat flow while ignoring the other two heat-flow mechanisms: convection and radiation. In fact, R-values, by the method of which the test is performed, include all three heat-transfer mechanisms.
The only significant factor that is not accounted for in R-values is air-tightness. The thermal resistance test does measure the insulation's performance with naturally occurring convective air loops in the material, but the test is performed in still air conditions which give no account of wind driven effects on thermal performance.
We know from practical experience that attic temperatures rarely go above130 deg F on hot sunny days in a properly vented attic. Most homes have some type of venting in the soffits at the bottom of the roof and some type of venting, typically a ridge vent that runs the entire length of the peak, at the top of the roof. This creates air movement, a convective loop, of (relatively) cooler air entering at the bottom and circulating up the bottom side of the roof deck and out the vent at the top pulling heat out of the attic. Uninsulated attics that are not vented properly present lots of problems: higher attic temperatures leading to greater cooling loads in the summer and ice dams backing up on the roof in the winter. In case I have seen an improperly vented attic get so hot it buckled the plywood sheathing up off the rafters leaving a gaping hole and torn shingles.
In Conclusion... Finally... Can Someone Shut This Guy Up?
We now know that there are three important factors to be considered here to reduce the cooling load on the building: the temperature of the roof assembly and thereby how much heat is moving to the attic, the temperature of the attic and thereby how much heat is available to move into the conditioned house, the R-value of the ceiling insulation.
In northern climates the most important thing you can do to reduce your energy usage is to add traditional insulation in the attic and/or reduce air leaks. This impacts both your heating and cooling loads. After insulation and air leaks have been addressed a metal roof or some other type of reflective roofing material will save money during the summer months by reducing the temperature of the attic. They also have excellent durability, lasting 3 to 4 times longer than most shingled roofs. It is also very important to make sure your attic is properly vented.
In southern climates the most important thing you can do to reduce your energy usage is usually to roof your house with a reflective roof coating. For more information on reflective roofing materials follow this link: cool roof. In lieu of doing a re-roofing project a cheaper method would be a radiant barrier attached to the bottoms of the rafters with an airspace between the barrier and the roof deck that is well ventilated. This will most likely produce the most improvement as long as there is at least R-19 insulation at the attic floor. It will most likely reduce the service life of the roof.
It is also important to note that duct work in an unconditioned attic can be a significant energy hog in any climate. If at all possible the duct work should be moved into the conditioned space, or at the least air sealed and insulated as much as possible.
There are two places where I would highly recommend a radiant barrier: against the interior wall directly behind a radiator and in the floor beneath radiant floor heat. In the case of the radiant floor heat I would still use cellulose or fiberglass insulation below a radiant barrier and then an air space between the barrier and the floor above.
One place I would never use one is behind heat sensitive materials, such as behind vinyl siding.