eQuest requires an input of a U-Value when modeling a slab on grab layer. Typically, you enter a F-Factor to account for the loss per perimeter length. I did some research but could not find any solid conclusions on what is suppose to be done to convert the F-Factor to a U-value or configure eQuest to correctly model this. Any suggestions?
Thanks!
William Mak
F-factors can't be entered directly, but are still pretty easy to handle in eQuest.
Open your copy of 90.1, and add a tab to Table A6.3, and read preceding section A6.
90.1 is explicit in what physically constitutes a floor slab with a given F-factor. Typically, I need an unheated slab with F-0.73. From this table and accompanying text I know that means simply to define a 6in. thick slab on grade with zero insulation while in the wizards. eQuest does the rest.
I generally utilize the default library material properties (mass/conductivity) built into eQuest. Those properties came under fire once from a nitpicky reviewer for being slightly different from Appendix A's final tables (see archives), but I learned after-the-fact that the material properties listed in 90.1 are derived from DOE2's libraries, so it was a moot issue.
NICK CATON, E.I.T.
I ran into this problem too because in wizard mode, the menu of insulation configurations for slab-on-grade floors did not have the particular combination I needed. Here's what you can do in detailed edit mode instead:
1. On a space-by-space basis (or for the whole floor if it is all heated or unheated slab) calculate the R-value of the fictitious layer of insulation covering the slab:
Equivalent R-value = ((F-factor * Exposed Perimeter length)/Area of space or floor)^-1
(hint for determining exposed perimeter length: just add up all the widths of the exterior walls for that space. Double-click on the first exterior wall in the component tree and then press the down arrow in the menu to page through the list of just exterior walls)
2. Create a custom material of type Resistance with this R-value.
3. Create (or modify) the underground floor construction using the following layers (outside to inside):
? Material you just made
? Light soil damp 12 in
? 6 in HW (140lb) concrete
Kelsey Van Tassel
Attached please find the research paper that eQUEST Wizards (and EE4
for you Canucks) use as a basis to convert F-factor to U-factor for a
variety of foundation wall insulation configurations.
Hmm, now that I am looking at it again, I am not sure I am doing it right either. I thought the F-factor presented in the tables was already correcting for the R-value of the slab and soil, but now I'm not sure. I may be double-counting.
I ran into this problem too because in wizard mode, the menu of insulation configurations for slab-on-grade floors did not have the particular combination I needed. Here's what you can do in detailed edit mode instead:
1. On a space-by-space basis (or for the whole floor if it is all heated or unheated slab) calculate the R-value of the fictitious layer of insulation covering the slab:
Equivalent R-value = ((F-factor * Exposed Perimeter length)/Area of space or floor)^-1
(hint for determining exposed perimeter length: just add up all the widths of the exterior walls for that space. Double-click on the first exterior wall in the component tree and then press the down arrow in the menu to page through the list of just exterior walls)
2. Create a custom material of type Resistance with this R-value.
3. Create (or modify) the underground floor construction using the following layers (outside to inside):
? Material you just made
? Light soil damp 12 in
? 6 in HW (140lb) concrete
Kelsey Van Tassel
Hope its not too late for me to join this discussion.
If I look at 90.1 table A6.3 F = 0.73 is equal to 150mm concrete poured
directly on earth that is unheated. However, if you look at the article
"Underground Surfaces" by Fred Wilkenmann *
http://esl.tamu.edu/pub/DOE%2D2%5FOCR%5FManuals/*
* (DOE2 articles ground coupling models) page 10 *and look at Table 1 F
factors of an uninsulated and uncarpeted concrete slab on grade, F factor is
1.10.
Am I misinterpreting the article? Is the 0.73 of ASHRAE table A6.3 carpeted?
Because if it is then the F-factor in the table 1 will be 0.77 which is
close to ASHRAE.
Thanks,
Ashok