Dear alll,

From a preliminary review, we received a comment regarding the Energy Modeling saying that the slab on grade floor of the proposed should be modeled using a F-factor, and not using the U-value as it was done. This is the first time we get this comment, even though in all previous certifications we always model the slab on grade floor of the proposed design using the U-value. For the baseline design we do use the F-factor. Any suggestions or ideas on how to reply to this?

Thanks in advance

## Marcus Sheffer

LEED Fellow7group / Energy Opportunities

LEEDuser Expert5680 thumbs up

April 8, 2013 - 5:04 pm

The ASHRAE 90.1-2007 envelope table require the Baseline to be an F-factor in the Baseline. If your software can model F-factor, calculate it for the Proposed and model it as designed. If it can't model F-factor then the review comment generally contains recommendations for what to assume in the U-value calculation for the baseline.

## Alicia Villanueva

CEORevitaliza consultores

75 thumbs up

April 8, 2013 - 5:18 pm

We are using Energy Plus, which allows to model the F-factor. Your suggestion is to use the F-factor for proposed design, using table A6.3 of Ahsrae?

## Marcus Sheffer

LEED Fellow7group / Energy Opportunities

LEEDuser Expert5680 thumbs up

April 8, 2013 - 6:23 pm

Yep that works! The Proposed model's envelope is always calculated according to Appendix A.

## Jean Marais

b.i.g. Bechtold DesignBuilder Expert827 thumbs up

April 9, 2013 - 4:32 am

I'm not sure I understand...are we talking about the baseline or the designcase model...e+ can calculate a real F-factor but only per slab. That means if your "slab" is split by different zones, you'll need to do a seperate simulation with the model adapted so that only one slab surface exists.

As far as I'm concerned the F-factor for your design case project slab for ASHRAE Appendix A, can still be read off the table and is subject to your constructions for comparison. You should be splitting your slab if the insulation thickness varies for example at the perimeter. The modelling software e+ can make a F-factor calculation using an input F-factor, but the heat-transfer simulation should use equivalent u-values. And if that is not good enough, I would use the 3D slab pre-simulation to capture perimeter effects, shadowing on the ground, etc. Check the e+ auxillary programs documentation.

This is an excerpt from e+ inputoutput reference guide:

"Site:GroundTemperature:FCfactorMethod

Site:GroundTemperature:FCfactorMethod is used only by the underground walls or slabs-on-grade or underground floors defined with C-factor (Construction:CfactorUndergroundWall) and F-factor (Construction:FfactorGroundFloor) method for code compliance calculations where detailed construction layers are unknown. Only one such ground temperature object can be included."

An example of the construction input:

Construction:FfactorGroundFloor,

slabconst,

0.12, !F-factor in W/m-K

232.26, !Area in m2

61.0; !Exposed perimeter in m

See also p 113 (139 absolute) of EngineeringReference.pdf.

At the end of the day, the program creates an equivalent concrete and insulation construction with u-values using F-factor (or C-factor) inputs.

I see three paths:

1) proposed design

use the auxillary "Slab" program and use the constructions as designed paying attention to detail such as vertical insulation depth, horizontal insulation, etc.

2) proposed design

find out the equivalent constructions that equate to the F-factor as read off the ASHRAE tables as per the design constructions by:

Modelling an equivalent model with one combined slab-zone extra with F-factor method, adjusting constructions until I come to the F-factor read off the table in Appendix A (based on real constructions), and then use those constructions for the normal model you have.

3)baseline case

It should be noted that F-Factor is the heat transfer through the floor, induced by a unit temperature difference between the outside and inside air temperature, on the per linear length of the exposed perimeter of the floor. The SI unit is W/m·K.

Q = Area * (Tair_out – Tair_in) / Reff = (Tair_out – Tair_in) * (Pexp * F-factor)

Therefore,

Reff = Area / (Pexp * F-factor)

...

Pexp = exposed perimeter

So...you can read the F-factor off the table, and calculate the Reff...then make sure your slab construction has the same R-value (surface to surface without film coeffs)

## Marcus Sheffer

LEED Fellow7group / Energy Opportunities

LEEDuser Expert5680 thumbs up

April 9, 2013 - 10:26 am

Yea, what he said! Good advice Jean especially about checking the results from the modeling software against Appendix A as a quality control procedure.

## Jean Marais

b.i.g. Bechtold DesignBuilder Expert827 thumbs up

April 10, 2013 - 5:22 am

Sorry...I see it is indeed WITH film coeffs.

So,

Rfic = Reff – (Rfilm_in + Rfilm_out) - Rcon

Where,

Rfic is the thermal resistance of the fictitious insulation layer in m2·K/W.

Rcon is the thermal resistance of the concrete layer in m2·K/W.

Rfilm_in and Rfilm_out are the air film resistance of the inside and outside surfaces, respectively.

As ASHRAE 90.1 does not specify film coeff for the slab to earth film...I would note that e+ uses:

The outside air film resistance Rfilm_out = 0.03 m2·K/W. The inside air film resistance Rfilm_in = 0.125 m2·K/W, which is the average of the 0.14 m2·K/W for heat flow up and 0.11 m2·K/W for heat flow down.

...which I don't totally aggree with as there is no air film under the slab, but an soil layer (this may be a documentation error...I'll check up on it).

The Winkelmann’s slab-on-grade model uses a resistance of the12 inch soil layer (Rsoil) assumed as 0.176m2-K/W (1 hr-ft2-F/Btu). Which I am more inclined to use...see http://www-esl.tamu.edu/docs/terp/2012/ESL-TR-12-02-01.pdf

See also the tips and tricks from DOE-2 (see p6 in pdf below)

http://simulationresearch.lbl.gov/dirpubs/un_articles02.pdf

This makes more sense, as here a soil layer is assumed:

A 1-ft(0.3-m) layer of soil (resistance=Rsoil=1.0hr-ft2-F/Btu [0.18m2-K/W])

I can't actually find out how the f-factors were calculated, so any more input will be appreciated.

## Jean Marais

b.i.g. Bechtold DesignBuilder Expert827 thumbs up

April 30, 2013 - 10:31 am

After quite a bit of digging around I got into contact with Dr. Huang who was involved with the F-factor deturmination back in 1988 which consiquently found its way into ASHRAE 90.1. I won't go into detail as to the why, but here is the how (NB not all slabs are "slab-on-grade" which is the only slab type regulated by standard 90.1...all other slabs against ground should be modelled as per the proposed case for the baseline):

My goal is to reverse engineer an equivalent slab-on-grade construction for simulations as per ASHRAE 90.1 Appx G.

F * Pexp = Ueff * A

= 1/Reff * A

This gives

Reff = A / (F * Pexp)...the user inputs

...therefor for a insulation rating value that matches the standard's rated insulation requirement for the required F-factor and

substituting for the construction build up as defined by ASHRAE 90.1 A6

Reff = Rairfilm_in + Rcarpetpad + Rconcreate + Rrequired_rated_insulation + Rsoil + Rother + Rairfilm_out

where

*********************************

Rairfilm_in = 0.77 hr-ft2-F/Btu (0.14 m2-K/W)

is the average of the air film resistance for heat flow up 0.61 hr-ft2-F/Btu (0.11 m2-K/W) and heat flow down 0.92 hr-ft2-F/Btu (0.16 m2-K/W), as per ASHRAE 90.1 A9.1.4 and Winkelmann

**************************************************

Rcarpetpad = 1.23 hr-ft2-F/Btu (0.216726 m2-K/W)

*************************************

Rconcreate = 0.07 m²K /W

as per ASHRAE 90.1 A6 for 0.5ft (150 mm) concrete from ASHRAE 90.1 Table A3.1B, and density 2304 kg/m³

**********************************

Rrequired_rated_insulation can be found by fitting a curve through the values in standard 90.1 Table A6.3 for fully insulated slab-on-grade:

ASHRAE 90.1 F-Factors for fully insulated slabs and Unheated as per Appendix G

SI Units

R-values 0 0.9 1.3 1.8 2.6 3.5 4.4 5.3 6.2 7 7.9 8.8 9.7

F-factors na 0.8 0.71 0.62 0.52 0.45 0.4 0.37 0.34 0.32 0.3 0.29 0.28

we can find the curve fit equation as

Rrequired_rated_insulation = 0.6065 * ( F )-2.165

****************************************

Rsoil is assumed 1 ft (0.3 m) at k=1.3 W/m²-K as per Table A9.4D, which corrisponds to recommendations in UNDERGROUND SURFACES: HOW TO GET A BETTER UNDERGROUND SURFACE HEAT TRANSFER CALCULATION IN DOE-2.1E, Winkelmann

therefore

Rsoil = 0.22 m²K /W

**************************************

Rairfilm_out = 0.0299387 m²K /W as per ASHRAE 90.1 A9.4.1

*************************************

Therefore the only unknown, Rother can be calculated.

Rother = A / (F * Pexp) - (0.14) - (0.216726) - (0.07) - (0.6065 * ( F )-2.165) - (0.22) - (0.0299387)

it is assumed Rother has material properties for density and specific heat, the same as that of soil.

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Now I can build up an equivalent construction for the "slab" espectially honed to exclusively "slab-on-grade" for Appendix G of ASHRAE 90.1. If Rother is negative, Rrequired_rated_insulation must be "derated" to effect the difference.

PS other material properties one can use to build up this construction in e+ could be as follows

Concrete 150 mm Construction as per A6.1:

Properties as per A3.1B for 150 mm, 2304 kg/m³ --> Rc = 0.07, HC = 294 -->

specific heat = 850.694444 J/kg-K

Insulation R-

0.36521482 m²K/W

Properties as per HOF Expanded Polystyrene extruded smooth skin 40 kg/m³, 0.030 W/m²-K

thickness = 0.01095644 mm

Soil to account for thermal mass effects of the soil as per DOE-2 Articles from the Building Energy Simulation User News, January 2003 and ASHRAE 90.1 table A9.4D

k = 1.3 W/m²-K

thickness = 0.3 m

density = 1840 kg/m³ as per DOE-2 referenced above