Energy Efficient Buildings
Floors and Basements
Heat Loss through the Ground
Path of Heat Loss
Heat loss from floors and basements occurs between the warm indoor air and the cold outdoor air through the ground. The figures below show the path of heat loss from a slab-on-grade floor and basement. The rate of heat loss is greatest where the thermal resistance is smallest. Thus, heat loss through slab-on-grade floors is greatest along the edge of the slab and heat loss from basements is greatest at the upper part of the basement walls, where the length of travel from the warm inside air to the cold outside air is smallest. Thus, vertical insulation around the perimeter of the slab and along basement walls should always be applied. However, perimeter insulation does not stop heat flow from the center of the slab, through the earth to the surface of the ground. Thus, horizontal floor insulation must also be applied to effectively insulate a slab or basement.
These principles are demonstrated in the figures below, which show results from detailed hour-by-hour finite-difference simulation of heat loss from a slab. In the figures below, a six-inch thick concrete slab extends along the top of each graph from the left edge to about the middle of the graph. The left edge of the graph represents the centerline of the building slab. The slab has a 2-foot deep insulated footing around the perimeter, which marks dramatic change in ground temperature. The figures show heat loss on December 31 when the inside air temperature is 70 F and the outdoor air temperature is 26.1 F. It is clear that the maximum heat loss occurs around the edge of the slab, but that heat also travels from the underneath the slab along an arc to the ground outside the slab.
Ground temperature profile under slab on December 31.
Plots of temperature isotherms and heat flux under slab on December 31.
Similarly, the figures below show results from detailed hour-by-hour finite-difference simulation of heat loss from a basement. In the figures below, the basement is in the upper left corner, and extends 8-feet under the ground. The figures show heat loss on December 31 when the inside air temperature is 70 F and the outdoor air temperature is 26.1 F. It is clear that the maximum heat loss occurs from the basement walls, but that heat also travels from the basement floor along an arc to the ground outside the slab.
Ground temperature profile around basement on December 31
Plots of temperature isotherms and heat flux around basement on December 31.
Ground Energy Storage
Ground temperature is driven by outside air temperature; hence the average annual ground temperature is equal to the average annual outdoor air temperature. However, because the ground stores energy, ground temperature does not rise as quickly as air temperature during the summer or fall as quickly as air temperature during winter. Hence, the annual variation in ground temperature is less than the annual variation in air temperature, and ground temperature lags behind air temperature. The graph below shows the annual variation in ground temperature. The amplitude of the variation decreases with increasing depth and the time lag of the variation increases with increasing depth.
As the heat flows between the warm inside air and cold outside air it passes through the ground. The ground has large thermal capacitance (the product of mass and specific heat); thus, a significant amount of energy is stored in the ground as it responds to changing outdoor air and indoor air temperatures. Because of the energy storage characteristics of the ground, heat loss from slabs and basements lags the indoor outdoor air temperature difference and is less affected by short-term changes in the indoor outdoor air temperature difference. The graphs below, which show hourly indoor-outdoor air temperature difference, and simulated heat loss from a slab and basement, clearly demonstrates the lagging and smoothing effects.
Slab: hourly (Tia-Toa) and heat loss Basement: hourly (Tia-Toa) and heat loss
Simplified Slab Heat Loss Equation
These observations suggest that heat transfer from buildings through the ground can be modeled by an equation of the form:
where U = effective conductance, A = area of the slab, Tia = indoor air temperature, Tg = effective ground temperature.
Effective Ground Temperature for Slab Heat Loss Calculations
The effective ground temperature should lag ambient temperature and have a smaller annual amplitude. Detailed simulations of heat loss from slabs suggest that effective ground temperature for heat loss from slabs, Tg, can be calculated as the weighted average of the average annual outdoor air temperature Toa,yr and the average outdoor air temperature during the previous three months, Toa,3mo:
Tg = (1.7 Toa,yr + 1.0 Toa,3mo) / 2.7
Example:
Calculate the effective ground temperature for slab heat loss for Dayton, Ohio on March 1 using the averaging method.
Monthly WeaTran output from the Dayton, Ohio TMY3 file is shown below.
Toa,yr = 51.29 (F)
Mo Yr Ta(F)
01 1995 24.60
02 1995 26.92
03 1995 42.48
04 1995 51.72
05 1995 63.36
06 1995 70.31
07 1995 75.04
08 1995 72.77
09 1995 63.64
10 1995 51.20
11 1995 40.94
12 1995 30.85
Based on these data, the annual average outdoor air temperature is:
Toa,yr = 51.29 F
The average outdoor air temperature during the previous three months, Toa,3mo is:
Toa,3mo = (Toa,Dec + Toa,Jan + Toa,Feb) / 3 = (30.85 + 24.60 + 26.92) / 3 = 27.46 F
The effective ground temperature on March 1 using the averaging method, Tg,avg, is:
Tg,avg = (1.70 Toa,yr + 1.0 Toa,3mo) / 2.7 = (1.7 51.29 + 1.0 27.5) / 2.7 = 42.46 F
Kasuda (1965) found that the temperature of the ground is a function of the time of the year and the vertical depth from the surface. He formulated a relationship that can be used to find the temperature distribution of the soil at any particular time of the year and at any depth. The equation is given by,
Tg = Tmean – Tamp exp[ -Depth x (p/365/a)0.5] cos {2p/365 [tnow - tshift - Depth/2 (365/p/a)0.5]}
Where:
Tg is the ground temperature, oF
Tmean is the annual outdoor air temperature, oF
Depth is the depth from the surface, ft
a is the thermal diffusivity of the ground,
tnow is current day of the year
tshift is the day of the year of the minimum surface temperature (use 30 days)
Tamp is (Tmo,max – Tmo,min) / 2 were Tmo,max is the maximum monthly temperature and Tmo,min is the minimum monthly temperature computed from TMY3 or EPW data.
It was found that the equation for Tg can also be formulated using Kasuda’s equation at a depth of 15 feet. Correlations between the average temperature method and Kasuda method for formulating Tg for different sites are shown below.
Example:
Calculate the effective ground temperature for Dayton, Ohio on March 1 at a depth of 15 feet using the Kasuda method. Soil diffusivity a is 1.6 ft2/day.
According to TMY3 data, the average annual temperature in Dayton, OH is 51.2 F.
According to TMY3 data, Tamp = (Tmo,max –Tmo,min) / 2 = 25.22
Effective Conductance for Slab Heat Loss Calculations
Based on the preceding argument that slab heat loss can be modeled as the difference between indoor air temperature, Tia, and effective ground temperature, Tg, the resistance to heat transfer occurs primarily along two paths as shown in the figure below.
Part of the heat lost from the house to the ground passes through the floor insulation Rf into the ground directly beneath the floor Rg1 and takes the shortest path through the perimeter insulation Rp to the ground. Part of the heat lost from the house to the ground passes through the floor insulation Rf and through the deeper ground Rg2. The conductance, U is the reciprocal of resistance, R. Hence the equation for the conductance can be written as:
U=Af1Rf + Rg1 + Rp+Af2Rf + Rg2
In the above equation, there are four unknown variables: area fraction for the first path Af1, effective ground resistance encountered in the first path Rg1, area fraction for the second path Af2, and the effective ground resistance encountered in the second path Rg2.
To determine values for these coefficients, a data set of 48 finite difference simulation results were derived for different floor insulation and perimeter insulation levels. Regression was then performed on this dataset to determine the values of coefficients that resulted in the best fit to the data. The following values resulted in a fit that explained 99.99% of the variation (R2 = 0.9999).
Area fraction 1 / Af1 / 11.40 / %Area fraction 2 / Af2 / 87.68 / %
Effective ground resistance in path 1 / Rg1 / 4 / hr-ft2-F/Btu
Effective ground resistance in path 2 / Rg2 / 16 / hr-ft2-F/Btu
Thus, the equation of conductance for the simplified model is given by:
U=0.11404+Rf + Rp+0.876816+Rf
Simplified Equation for Slab Heat Loss
The graph below shows excellent agreement between heat loss from a slab in Dayton, Ohio predicted by a finite-difference simulation, Q finite difference, and the heat loss predicted by the simplified method.
Example
Calculate the rate of heat loss from a slab-on-grade floor for a house with a 30 ft x 50 ft slab, if the interior air temperature is 70 F and the effective ground temperature is 42.70 F. The slab has R= 4 hr-ft2-F/Btu perimeter insulation and R = 2.08 floor insulation.
Simplified Basement Heat Loss Equation
A simplified equation for heat loss from basements can be found using the same method.
Effective Ground Temperature for Slab Heat Loss Calculations
The effective ground temperature for heat loss from basements, Tg, can be calculated as the weighted average of the average annual outdoor air temperature Toa,yr and the average outdoor air temperature during the previous 1.5 months, Toa,1.5mo:
Tg = (0.6 Toa,yr + 1.0 Toa,1.5mo) / 1.6
Example:
Calculate the effective ground temperature for basement heat loss for Dayton, Ohio on Feb 27 (end of week 8) using the averaging method.
Monthly WeaTran output from the Dayton, Ohio TMY3 file is shown below.
Toa,yr = 51.29 (F)
Based on these data, the annual average outdoor air temperature is:
Toa,yr = 51.29 F
The average outdoor air temperature during the previous 1.5 months, Toa,1.5mo is:
Toa,1.5mo = (Toa,wk3 + Toa,wk4+ Toa,wk5+ Toa,wk6+ Toa,wk7+ Toa,wk8) / 6
= (22.74+29.01+31.82+17.98+26.41+30.48) / 6 = 26.41 F
The effective ground temperature on Feb 27 using the averaging method, Tg,avg, is:
Tg,avg = (0.6 Toa,yr + 1.0 Toa,1.5mo) / 1.6 = (0.6 * 51.29 + 1.0 * 26.41) / 1.6 = 35.74 F
Kasuda (1965) found that the temperature of the ground is a function of the time of the year and the vertical depth from the surface. The Kasuda equation is shown in the previous section. Tg for basement heat loss can also be formulated using Kasuda’s equation at a depth of 8 feet.
Effective Conductance for Basement Heat Loss Calculations
Based on the preceding argument that basement heat loss can be modeled as the difference between indoor air temperature, Tia, and effective ground temperature, Tg, such that the resistance to heat transfer occurs primarily along two paths.
Hence the equation for the conductance can be written as:
U=Af1Rf + Rg1 +Af2Rw + Rg2
In the above equation, there are two unknown variables: effective ground resistance encountered in the first path Rg1 and the effective ground resistance encountered in the second path Rg2. Area fraction of path 1, Af1, is the fraction of total basement surface area comprised by the floor. Area fraction of path 2, Af2, is the fraction of total basement surface area comprised by the basement walls.
As in the case of slab heat loss, to determine the effective ground resistance values, a data set of 48 finite difference simulation results was created. Af1 and Af2 were set to their respective values. Regression was then performed to determine the values of coefficients that resulted in the best fit to the data. The following values resulted in a fit that explained 99.84% of the variation (R2 = 0.9984).
Effective ground resistance in path 1 / Rg1 / 44.8 / hr-ft2-F/BtuEffective ground resistance in path 2 / Rg2 / 4.60 / hr-ft2-F/Btu
Thus, the equation for conductance from a basement for the simplified model is given by:
U=Af144.8+Rf +Af24.60+Rw
Af1 = Afloor / (Afloor + Awalls) Af2 = Awalls / (Afloor + Awalls)
Simplified Equation for Basement Heat Loss
The graph below shows excellent agreement between simulated basement heat loss, (Q/A from FD) and the heat loss predicted by the simplified method (Q/A from simplified model).
Example
Calculate the rate of heat loss from a basement if the basement floor is 30 ft x 50 ft and the below grade basement walls are 8 ft high. The interior air temperature is 70 F and the effective ground temperature is 42.70 F. The basement has R= 4 hr-ft2-F/Btu wall insulation and R = 2.08 floor insulation.
Reducing Heat Loss from Slab-on-Grade Floors
Results from this method increase insight into how to effectively insulate slabs. The rate of heat loss through slab-on-grade floors is greatest where the thermal resistance is smallest, and decreases as the thermal resistance increases. Thus, heat loss through slab-on-grade floors is greatest along the edge of the slab. Based on this observation, heat loss from slab-on-grade floors can be cost-effectively reduced by first insulating the perimeter of the slab. Finite difference simulations were performed for different values of Rf and Rp, and the average annual values of heat loss per unit area of the slab were computed. The Figure below shows that perimeter insulation is more important when the floor is un-insulated and less important as floor insulation increases. In addition, adding perimeter insulation in excess of about R = 5 hr-ft2-F/Btu has little affect.