Slab-on-grade heat loss estimation

[LaSalle1940]

This should be an easy one. Assuming I have a fairly sizable slab on grade (let's say 30 ft x 40 ft), I would assume the right way to approach this is to divide the slab into a perimeter zone (a band 10 ft wide around the perimeter) and a center zone.

The loss from the center zone is fairly simple; i.e., I believe ASHRAE cites an overall U value of 0.1 BTU/hr-ft²-°F for the typical 20°F conditioned space-to-earth temperature difference.

Let's assume there is insulation around the perimeter of the slab, and let's say (for discussion) that this insulation is R-6. I believe that the right way to estimate the loss from the perimeter zone is to calculate an overall U (i.e., the reciprocal of the sum of the R values for the insulation and the outdoor air film) and multiply that overall U by the appropriate temperature difference and the exposed area (in the case under discussion, assuming a 4" slab measuring 30 ft x 40 ft, we're talking 400 sq ft).

Now: what did I miss in the U calculation for the perimeter zone? It seems to me that I would need to include the inside air film R. It also seems to me that I would need to account for the resistance to heat flow in the slab. But am I making this harder than it has to be?

 

[CESSNA1]

LaSalle1940:
Here is my 2 cents. The ground temperature under the slab can be assumed to be 50-55 degress F if this is a basement. If it is slab on grade, refer to ASHRAE. Typically most of the heat lost in a normal size slab is through the edge exposed to the air. Foundation (slab) sides are usually insulated for about 2 feet down. I have seen this on basement walls, but in my house I went to the basement floor. Underslab insulation is expensive and is usually glass blocks set in sand and then the concrete is poured over that. This is only reasonable in large slab areas or very cold regions. Film coefficients are given in ASHRAE for the sides. They contribute some to the thermal barrier, but are usually small overall.

Regards
Dave

 

[LaSalle1940]

I have the '89 ASHRAE fundamentals book handy; that shows a chart (table 5 in chapter 25) of factors F for perimeter losses (BTU/hr-ft-°F) for various types of slab-on-grade construction. Typical values are roughly in the range of ~0.5 BTU/hr-ft-°F for construction including R-5.4 insulation from the edge to the footer. What I'm having trouble rationalizing is how a given insulated F value matches up with the corresponding uninsulated F value (typically on the order of ~0.8-0.9, same units); that is, given the uninsulated F value, I'm having trouble deriving the corresponding F insulated value using the usual reciprocal of the sum of the resistances method. What am I not seeing?