Marc Rosenbaum's picture

Heat Loss To The Ground (Part 2)

by Marc Rosenbaum

BE Boston 19 RegistrantsBE Boston 20 RegistrantsBE Boston 21 RegistrantsBE Boston 22 RegistrantsBE Boston 23 RegistrantsBuildingEnergy 16 Full ConferenceMemberWebform Results Access
Thursday, October 23, 2014

Editor's note: This is part two of a blog post on Heat Loss to The Ground by Marc Rosenbaum. If you missed part one, you can get it here.

The THERM model also can yield an assembly U factor. In the case modeled, the assembly U factor is 0.36 BTU/hr-˚F per lineal foot of perimeter of the foundation. If the basement had 120 lineal feet of perimeter, the heat loss coefficient of the foundation would be 43 BTU/hr-˚F. At 68˚F inside and 5˚F outside, the heat loss is 43 BTU/hr-˚F x (68-5)˚F = 2,924 BTU/hr. If 5˚F is the design heating temperature of the location, this represents an estimate of the peak heat loss of the basement.

Annual heat loss of this basement can be estimated by using heating degree days (HDD). For Boston, the annual heat load of the basement is 43 BTU/hr-˚F x 5,650 HDD x 24 hr/day = 5.83 million BTU/yr.

How about a quick reality check? Let’s assume the temperature below the slab is 50˚F and the average temperature against the foundation wall is 35˚F. Our basement is 24 ft x 36 ft. Doing the math, we get a heat loss of 2,161 BTU/hr, 26% below what we calculated using the U factor THERM gave us. We’re in the same ballpark.

In my heat loss calculations I only take the time to do the THERM modeling if the situation is unusual, or more often, to compare the U factors that result from trying different insulation strategies. Baked into my spreadsheet-based heat loss and energy modeling calculators are algorithms that were published a long time ago in the Passive Solar Design Handbooks by the Los Alamos National Lab. For a heated basement, the calculation is: Heat flow (BTU/hr) = 10.67 x Perimeter (ft) x ∆T (˚F) / (R + 8), where R is the thermal resistance of the foundation wall. Note there is no specified R value for the sub-slab – I assume it to be a minimum of half of the basement wall R value.

For the example we worked in THERM, the heat loss is 10.67 x 120 ft x (68-5)˚F / (24+8) = 2,520 BTU/hr. This is 14% lower than the value we calculated in THERM, and 17% above the value we calculated with the simple 1D model assuming the soil temperatures. That 14% is about 400 BTU/hr, or the equivalent of 6 CFM of infiltration. Estimating infiltration to within 6 CFM is a trick, so I think this is a pretty good method of estimating ground heat loss.

The Los Alamos method for a slab on grade is similar: Heat flow (BTU/hr) = 4.17 x Perimeter (ft) x ∆T (˚F) / (R + 5) I use the R value of the perimeter insulation on the frost wall, or if it is a floating slab, the R value of the first few feet of sub-slab insulation. Slab foundations can be trickier in real life too, because it is often difficult to get full insulation value at the actual slab edge. Also, insulation levels can be traded off between perimeter and sub-slab, which is where THERM comes in handy.

Walk-out basements

Here I mix the Los Alamos algorithms, using the basement wall algorithm for the perimeter that is mostly below grade, and the slab algorithm for the walk-out side. If there are significant portions of the basement wall above grade (usually as the grade slopes from the high side to the walk-out side), add the conduction through those walls to the above grade calculation, and split the foundation perimeter they encompass 50:50 between the basement and slab algorithms. If there are two sides of the house 24 ft long that go from fully below grade to a walk-out condition, count 24 ft as basement, and 24 ft as slab, in addition to adding the above grade area into the above grade calculations.

Insulation in the basement ceiling

Not a good idea if there is mechanical equipment in the basement, or if there are mold-susceptible materials stored in the basement. If there are no mechanicals down there throwing off waste heat, to calculate design heat loss, assume the coldest temperature the basement will reach, and do a simple conduction calculation: Heat loss (BTU/hr) = Floor area (ft2) x ∆T (˚F) / R floor For example, if we take a 24 ft x 36 ft house, with an R-19 floor over the basement, and assume the basement can get to 45˚F, the heat loss is: 24 ft x 36 ft x (70-45) ˚F / 19 ft2-hr-˚F/BTU = 1,137 BTU/hr

What if there is no insulation in the basement floor, or on the foundation itself? This is the hardest to calculate, because small changes in the assumed R values make big changes in heat loss. So insulate the foundation and isolate the building thermally from the cold wet ground!

This has been a sample lesson from my BuildingEnergy Masters Series course Zero Net Energy Homes, which I do twice a year with NESEA and Heatspring. It regularly sells out, so I'd encourage you to visit the course page and get signed up today. You can try out the course in what the kids are calling a "test drive" before enrolling, if that suits you best.

Our Mission

NESEA advances sustainability practices in the built environment by cultivating a cross-disciplinary community where practitioners are encouraged to share, collaborate and learn.

Old building geek.

Marc Rosenbaum's picture
BE Boston 19 RegistrantsBE Boston 20 RegistrantsBE Boston 21 RegistrantsBE Boston 22 RegistrantsBE Boston 23 RegistrantsBuildingEnergy 16 Full ConferenceMemberWebform Results Access

Add comment

Log in or register to post comments