R-Value: The higher the R-Value, the better the insulating properties of the subject materials. R-Values are most often used to express the thermal resistance (ability to stop heat flow) of a building wall, ceiling or floor. Because of this, most R-Values are calculated at normal temperatures of approx. 75 F. R-Values are easy to add together so calculating the total R-Value of a wall is simply done by adding the values for the sheetrock, insulation, sheathing and siding.
K-value is a measure of heat conductivity of a particular material. Specifically, it is the measure of the amount of heat, in BTUs per hour, that will be transmitted through one square foot of material that is one inch thick to cause a temperature change of one degree Fahrenheit from one side of the material to the other. The lower the K-value for a material, the better it insulates. If the K-value of the material is known, the R-value per inch can be determined by dividing 1 by the K-value (R-value per inch = 1/K value). The LOWER a K-Value, the better its performance as an insulator.
R or K values have nothing to do with whether a material is flame proof, flame resistant or combustible. Styrofoam, cork, wood and polyester are just some examples of materials which are good insulators but will burn or smoke dangerously when exposed to excess heat.
Technical – For those who desire to calculate their own K or R values, please use the following formulas:
1. R value can be calculated by dividing the thickness by the K value.
For US calculations, we use inches as the unit of measurement.
“In the inch-pound units, thermal resistance is measured in degrees F times square feet of area times hours of time per Btus of heat flow.”
R-value = thickness / K-value
2. K value is the inverse of the R-Value. If one is known, the other can be calculated.
“units of Btu-inch/hour per square foot per degree F”
Thickness/k value = R value
Divide the inches of thickness by R.
k= inches of thickness / R
K-Value Example: A wood stove may call for a floor which has a K factor of 1 or less. A product such as Micore 300 Board from USG has a K-Value of approx .43 per inch. Therefore a 1/2” thickness of this board would have a K-Value of .86, which meets the requirement of our example stove.
R-Value Example: A stove or fireplace may call for an floor with an R-Value of 1.5. The same board above is rated as having an R-Value of 2.33 for a one inch thickness. Therefore, 3/4” of the Micore 300 Board would meet the specifications for this stove.
Summary: R and K values are related, but K is the value commonly used for specifying materials for use with stoves and fireplaces. Be sure that your choice of insulating material for high temperature applications is noncombustible.
With K values, the lower value is a better insulator. With R Values, the highest number is better.
For low profile hearths, it is best to use manufactured materials such as Micore and Cement Board (Durock, Wonderboard, etc.) as these will allow hearth thicknesses of from 1/4” to 2” with most stoves and fireplaces. Most other common building materials will require at least 3” of thickness and usually much more.
See the information and additional links at our companion article below:
Example of Hearth Calculations – this is for a Hearth requirement of approx R=1.15 (figures taken from Ceramic Tile manufacturers trade association)
The assembly that we will evaluate is a layer of Micore 230 and a layer of ½” Util-A-Crete. The first step is to convert the k values of the materials in question into R so that we may add them up and determine if they will provide the necessary insulation value required by the manufacturer.
Micore 230 has a k value12 of .43 so –
1 divided by k = 2.32 times the thickness .375 (3/8”) = 0.87
Util-A-Crete (cement tile backer board) has a k value of 1.6 so – 1 divided by k = .625 times the thickness .5 (1/2”) = 0.3125
Add the values together 0.87 plus 0.3125 = 1.1825 This R- value is an acceptable assembly.
What if we decide to use only one material? In this example, only Util-A-Crete cement board?
We could use the published R-Value of Util-A-Crete13 which is .31 in the ½” material and add them up to the value of the minimum required which is R=1.16
1.16 divided by .31 = 3.74 This assembly would require 3.74 layers of ½” Util-A-Crete to reach the necessary R-value required. Obviously, you would have to round up to the next layer, which would mean that you would have two inches of Util-A-Crete.