I’m modeling the proposed design model and in order to compare it with the baseline model I want to calculate the F-factor of the slab-on-grade floor according to table A6.3 of ASHRAE 90.1-2007. I’m using the SI edition.
It is a heated/cooled slab, the entire floor surface is insulated horizontally, and the thermal resistance R-valued of insulation is equal to 4.1. Therefore I would consider “heated slab” -> “1200 mm horizontal” and since there is no F-factor for R-value equal to 4.1 as F-value I would consider the minimum value of that line, i.e. 1.92. Am I right?
Regards
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Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5909 thumbs up
July 26, 2013 - 2:00 pm
If you have a heated slab (contains piping or other heating/cooling elements) and it is entirely insulated you would use the row labeled "Fully insulated slab" at the bottom (at least that is what is in the IP edition). If you R-value is under R-5 I would interpolate based on the increments surrounding R-5.
Francesco Passerini
engineer90 thumbs up
July 29, 2013 - 4:32 am
I read that in a "fully insulated slab" "insulation extends downward from the top of the slab". The project team has not developed the final version of the design and I don't know exactly how the connection between vertical insulation and horizontal insulation will be. However, I'm going to consider "fully insulated slab" now and I could change the model in the future, when I'll get more information. Actually, I think that the most precise solution to evaluate a detail would be the finite element method, but in order to maintain coherence between baseline and proposed model I'm considering tab. A6.3. Thank you!