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
1
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.
2
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.
3
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
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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:
Q slab  U  A (Tia  Tg )
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
01 1995
02 1995
03 1995
04 1995
05 1995
06 1995
07 1995
08 1995
09 1995
10 1995
11 1995
12 1995
Ta(F)
24.60
26.92
42.48
51.72
63.36
70.31
75.04
72.77
63.64
51.20
40.94
30.85
Based on these data, the annual average outdoor air temperature is:
5
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 (/365/)0.5] cos {2/365 [tnow - tshift - Depth/2
(365//)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
 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.
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Example:
Calculate the effective ground temperature for Dayton, Ohio on March 1 at a depth of
15 feet using the Kasuda method. Soil diffusivity  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
Input Data
Tmean (F)
Tamp (F)
Thermal diffusivity (alpha) (ft2/day)
depth (ft)
tnow (day)
tshift (day)
Calculations
X1 = cos {2 pi/365 x (tnow - tshift - Depth/2 (365/pi/alpha)
0.5
X2 = Tamp x exp[ -Depth x (pi/365/alpha) ]
Tg,k (F) = Tmean – X2 X1
51.29
25.22
1.617
15
60
30
0.5)
}
0.838
8.44
44.22
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.
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Tia
Rf
Ground Surface
Path 1
Rg1
Rp
Tg
Rg 2
Path 2
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=
Af1
Af2
+
Rf + Rg1 + Rp Rf + 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
Area fraction 2
Effective ground resistance in path 1
Effective ground resistance in path 2
Af1
Af2
Rg1
Rg2
11.40
87.68
4
16
%
%
hr-ft2-F/Btu
hr-ft2-F/Btu
Thus, the equation of conductance for the simplified model is given by:
8
U=
0.1140
0.8768
+
4 + Rf + Rp 16 + 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.
Input Data
Tia (F)
Tg (F)
L (ft)
B (ft)
Rp (hr-ft2-F/Btu)
Rf (hr-ft2-F/Btu)
Calculations
A (ft2) = L B
U (hr-ft2-F/Btu) = [0.1140/(4+Rf+Rp)] + [0.8768/16+Rf]
Q (Btu/hr) = U A (Tia-Tg)
70
42.70
50
30
4
2.08
1500
0.060
2449
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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)
Month
1
1
1
1
2
2
2
2
Week
1
2
3
4
5
6
7
8
Yr
1995
1995
1995
1995
1995
1995
1995
1995
Ta(F)
23.75
20.52
22.74
29.01
31.82
17.98
26.41
30.48
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:
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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.
Ground Surface
Tia
Path 2
Rw
Rg 2
Rf
Tg
Path 1
Rg1
Hence the equation for the conductance can be written as:
U=
Af1
Af2
+
Rf + Rg1 Rw + 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.
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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
Effective ground resistance in path 2
Rg1
Rg2
44.8
4.60
hr-ft2-F/Btu
hr-ft2-F/Btu
Thus, the equation for conductance from a basement for the simplified model is given
by:
U=
Af1
Af2
+
44.8 + Rf 4.60 + 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).
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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.
Input Data
Tia (F)
Tg (F)
L (ft)
B (ft)
H (ft)
Rw (hr-ft2-F/Btu)
Rf (hr-ft2-F/Btu)
Calculations
Afloor (ft2) = L B
Awalls (ft2) = 2(L H) + 2(B H)
Af1 = Afloor/(Awalls+Afloor)
Af2 = Awalls/Awalls+Afloor)
U (Btu/hr-ft2-F) = [Af1/(44.8+Rf)] + [Af2/4.6+Rw]
Q (Btu/hr) = U A (Tia-Tg)
70
42.70
50
30
8
4
2.08
1500
1280
0.54
0.46
0.065
4936
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.
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Heat loss per unit area of the slab (Q/A) versus perimeter insulation Rp
Even with very thick perimeter insulation, heat will still travel from under the slab to the
ground. Thus, significantly reducing ground heat loss requires adding insulation above or
under the floor. The figure below shows that the heat loss per unit area continues to
decline as the floor insulation is increased. Thus, in contrast to perimeter insulation,
high levels of floor insulation are effective.
Heat loss per unit area of the slab (Q/A) versus floor insulation Rf
The figure below shows a detail of perimeter insulation for slabs with effective drainage.
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Source: http://www.energysavers.gov/your_home/insulation_airsealing
Reducing Heat Loss from Basements
Results from this method increase insight into how to effectively insulate basements.
The rate of heat loss from basements is greatest where the thermal resistance is
smallest. Thus, heat loss from basements is greatest at the upper part of the basement
walls. Based on this observation, heat loss can be cost-effectively reduced by first
insulating the walls of the basement. 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 basement were computed. The figure below show that as wall insulation increases,
heat loss from basements is significantly reduced; hence, insulating the walls is top
priority when it comes to basements.
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Heat loss per unit area of the basement (Q/A) versus wall insulation Rw
The figure below shows that the heat loss per unit area remains almost constant as the
floor insulation is increased. This signifies that, in contrast to slabs, insulating the floor
of basements has little effect on basement heat loss.
Heat loss per unit area of the basement (Q/A) versus floor insulation Rf
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New construction offers the opportunity to use insulated concrete, insulated concrete
block or insulated concrete forms in the basement walls. Insulated concrete is made
with a mix of low-density additives that reduce the conductivity of the concrete.
Insulated concrete blocks are made with low-density concrete and/or insulation in the
hollow cores of the block. Insulated concrete forms can be constructed on site or in a
factory and delivered to the site. Insulated concrete forms sandwich insulated foam
board between concrete and can have overall thermal resistances of up to R = 17 hr-ft2F/Btu. In existing basements, it is generally less costly to add insulation to the interior of
the walls. Two common options are shown below.
Source: http://www.energysavers.gov/your_home/insulation_airsealing
Reducing Heat Loss from Floors above Crawl Spaces
Heat loss through floors to crawl spaces is driven the temperature difference between
the indoor air temperature Tia and crawlspace air temperature Tcs. Heat loss through
floors above crawlspaces can be reduced by minimizing air leakage into/from
crawlspaces, insulating crawlspace walls both above and below grade, and covering the
crawlspace floor with plastic sheets that act as moisture barriers. Crawlspace walls can
be insulated on the inside or outside of the wall. The thermal network below shows the
four major paths of heat transfer to/from crawl spaces.
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The conductance (Btu/hr-ft2-F) of a 6-in above-grade concrete crawlspace wall, Ucsw, is:
Ucsw 
1
R csw

1
1
1


R hi  R 6"concrete  R ho  R csw,ins .68  (6  .08)  .20  R csw,ins 1.36  R csw,ins
To calculate the temperature of the crawl space, perform an energy balance on the
crawl space. Heat gain/loss into the crawl space occurs from the house through the
floor, through the above-grad crawlspace walls, from the crawlspace through the
ground and through air infiltration into/from the crawl space. An energy balance on the
crawl space with air temperature Tcs gives:
(UA)floor (Tia – Tcs) - (UA)csw (Tcs – Toa) – UAgrnd (Tcs – Tg) – V pcp (Tcs – Toa) = 0 (SS)
Tcs 
(UA)floor Tia  (UA)csw Toa  (UA)grnd Tg  V pcp Toa
(UA)floor  (UA)csw  (UA)grnd  V pcp
Example
Calculate the rate of heat loss through a 30 ft x 50 ft, R = 5 hr-ft2-F/Btu floor over a crawl
space. The interior air temperature is 70 F, the outside air temperature is 30 F, and the
effective ground temperature is 42.7 F . The un-insulated crawlspace walls are 2 feet
high, and the rate of air leakage into crawlspace is 50 ft3/min.
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Input Data
Tia (F)
Toa (F)
Tg (F)
Lf (ft)
Wf (ft)
Hcsw (ft)
Rfloor (hr-ft2-F/Btu)
Rcsw,ins (hr-ft2-F/Btu)
V (ft3/min)
Rp (hr-ft2-F/Btu)
Rf (hr-ft2-F/Btu)
pcp (Btu/ft3-F)
Calculations
Ufloor (Btu/hr-ft2-F) = 1 / Rfloor
Afloor (ft2) = Lf Wf
UAfloor (Btu/hr-F) = Ufloor Afloor
Ucsw (Btu/hr-ft2-F) = 1/(1.36+Rcsw,ins)
Acsw (ft2) = Perimeter Hcsw
UAcsw (Btu/hr-F) = Ucsw Acsw
Ugrnd (Btu/hr-ft2-F) = [0.1140/(4+Rf+Rp)] + [0.8768/16+Rf]
UAgrnd (Btu/hr-F) = Ugrnd Afloor
Vpcp (Btu/hr-F) = V pcp 60 min/hr
Tcs (F) = (UAfloor Tia + UAcsw Toa + UAgrnd Tg + Vpcp Toa) / (UAfloor + UAcsw + UAgrnd + Vpcp)
Qfloor (Btu/hr) = UAfloor (Tia-Tcs)
70
30
42.70
50
30
2
5
0
50
0
0
0.018
0.2
1500
300
0.735
320
235.29
0.083
124.95
54
49.02
6293
Recommended practices for reducing floor heat loss in homes with crawl spaces are
shown below.
New homes:
 Un-insulated floor
 Unventilated, tightly sealed crawl space
 Insulated crawl space walls
 Taped and sealed polyethylene vapor barrier over dirt floor
Existing homes with ventilated crawl space:
 Seal holes and cracks in floor to prevent air from leaking into house
 Insulate between the floor joists with rolled fiberglass. Install it tight against the
subfloor. Seal all of the seams carefully to keep wind from blowing into the
insulation. Vapor barrier should be on inside of insulation in cold climates and
outside of insulation in extremely hot and humid climates.
 Tape and sealed polyethylene vapor barrier over dirt floor
Existing homes with unventilated crawl space:
 Tightly seal crawl space to eliminate exterior air leakage
 Insulate crawlspace walls and floor near perimeter
 Tape and sealed polyethylene vapor barrier over dirt floor
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Source:http://www.energysavers.gov/your_home/insulation_airsealing/
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