The following shows compliments of a Design Engineer. He was recently asked to comply with the ASHRAE 90.1 standard for Super Therm coating to be considered as usable in a new mid-rise construction. Further research guided him to find the rules to be met in order to claim compliance with the ASHRAE 90.1 standard. His communication with this organization established that independent approved laboratories tests are all that is needed to be able to claim compliance with the ASHRAE 90.1, (**provided that these tests reveal the required insulation properties for the application**). The **ASTM C236-89** *Standard test method for Steady-State Thermal Performance of Building Assemblies by Means of a Guarded Hot Box*, and the **ASTM E1269** *Standard Test Method for Determining Specific Heat Capacity by Differential Scanning Calorimetry* gives Super Therm precisely the requirements to meet ASHRAE 90.1 compliance. With 10 mils DFT of Super Therm having a “U” (Thermal Flow) value of **0.052**, Super Therm is less than the maximum “U” value of **0.061** allowed, (remember lower is better). Because the “U” value is used to measure “area-weighted average”, insulated walls or roofs will have a less favorable “U” value than the one suggested by the type of insulation used, here is why:

Metal or wood studs and rafters have less resistance “R” value, as well a higher “U” (Thermal Flow) making the assembly having a lower “R” value than the one proposed by the insulation fillers, glass, foam, etc.

Which would you prefer?

When Super Therm is applied in lieu of the wall filler, or as enhancement for the existing insulation, the “U” value is homogeneous through the whole surface, thus the “U” value given for Super Therm will be the same for the whole assembly. Super Therm has a more consistent “R” **value equivalence** for the assembly. The key, is the lower “U” **value for the “area-weighted average”**. The chart below will show you how to calculate any values you may need to make your case, remember SuperTherm thickness and “U” value never changes.

**INSULATION: VALUES, TERMS, AND CALCULATIONS **

'R' Value'R' = d/K |
Thermal Resistance |
m² °C / W or m² K / W |

i.e. the "R" value can be calculated by dividing the thickness in meters by the K value |
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'K' Value 'K' = d/R |
Thermal Conductivity |
W/mK |

i.e. the "R" value can be calculated by dividing the thickness in meters by the K value |
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Required thickness'd' = (t°C x K)/Q |
To calculate the required thickness of insulation | |

i.e. Required thickness can be calculated by multiplying the temperature differential in °C (t °C) by the insulation material's "K" value, and dividing the result by Q, (a number between 8 and 10) that relates desired efficiency where 8 is efficient and 10 is less efficient. |
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Required thickness'd' = "R" x K |
To calculate the required thickness of insulation | |

i.e. a material's required thickness can be calculated by multiplying the known "R" value by the material's "K" value. |
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Note: d = thickness in meters,i.e. 25mm = 0.025m. "R" factors may be added together to determine the total "R" or thermal resistance |
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Note: a lower K value (conductivity) = better or conversely a higher "R" value (thermal resistance) = better.*Glass wool batts and blankets' stated "R" values have +/- 10% tolerance. Super Therm has +/- 1% |

Below are images of how heat in the winter can transfer (thermal bridging) through the wooden studs to the outside. The heat from the inside comes into contact with the surface of the wallboard. The fiberglass between the studs loads with heat and dissipates through the wall system to the outside. Keeping in mind that warm air molecules always move towards the cold. The studs will be much colder as they are not insulated and offer little thermal resistance. Wall Shadowing is caused by Condensation on the wooden studs.