Energy Efficient Buildings
Heat Gain/Loss through Walls
Introduction
To reduce heating and cooling energy use, energy efficient buildings seek to minimize
conductive heat gain and loss through a building’s walls, windows, ceiling and floor. To
understand how to design and construct more energy efficient walls, it is necessary to
understand conduction heat loss through walls. Thus, this chapter begins by presenting the
equations for steady-state conduction heat transfer. The conduction equations are used to
show to importance of minimizing wall area and thermal breaks and maximizing thermal
resistance in order to reduce heat gain and loss. Finally, the use of vapor and moisture barriers
to control moisture migration through walls is discussed.
Principles of Energy-Efficient Walls
Energy Balance Approach
The equation for steady-state conduction through walls is:
Q kA
ΔT
Th - Tc
kA
Δx
Δx
Where
Q = rate of heat transfer
k = conductivity (Btu ft / hr ft2 °F)
A = surface area (ft2)
T = temperature difference between the inside and outside air (F)
x = wall thickness (ft)
Following this equation, the energy efficiency opportunities are:



Reduce wall area
Reduce the conductivity of the wall
Increase the thickness of the wall
This chapter uses engineering principals for quantifying heat loss through walls as guides to
improving the energy efficiency of walls.
1
Calculating Heat Transfer through Walls
A Conceptual Look and Conduction and Conductivity
Conduction is the transfer of heat from through a stationary media. On a molecular scale,
temperature is a measure of the vibrational energy of atoms. Atoms in high temperature
molecules vibrate more than the atoms in low temperature molecules. Conduction heat
transfer is the transfer of this vibrational energy from high temperature molecules to low
temperature molecules. Thus, heat is always transferred from high to low temperature.
In solids, molecules are densely packed and held in lattice structures by molecular bonds. Thus,
vibrational energy is easily transferred from high to low temperature molecules. As a
consequence, solids have high thermal conductivity. In liquids and gasses, the intermolecular
bonds get weaker and the density of molecules get smaller. This makes it harder to transmit
vibrational energy from high to low temperature molecules. Thus, liquids and gasses generally
have progressively lower conductivities.
It follows that conductivity is generally a function of density. This is the reason people are
relatively comfortable in 50 F air, but will become hypothermic in 50 F water. When this
thinking is applied to building materials, dense materials like glass and concrete have high
thermal conductivity and low thermal resistance. They make poor insulators. Low density
materials like foam board or fiberglass batts have low conductivity and make good thermal
insulators.
Fourier’s Law of Conduction
Fourier’s law of conduction states that the rate of heat flow, Q, through a solid media is
proportional to the area, A, and temperature difference T, and inversely proportional to the
thickness of the material, x. The constant of proportionality is called the conductivity, k, which
is a property of the media.
.
Q
Q
T1
.
A ΔT
Δx
Q k
A (T2  T1 )
x2  x1
where k = constant of proportionality = conductivity
T2
Q  k A
X1
X2
ΔT
dT
or Q  k A
Δx
dx
X
In the equation for heat flow, the negative sign can be omitted if ΔT is defined so that heat flows
from high temperature, Th, to low temperature, Tc. Thus, conduction heat transfer can be
calculated as:
2
ΔT
Th - Tc
kA
Δx
Δx
Q  U A ΔT  U A Th - Tc
A ΔT A (Th - Tc)
Q

R
R
Q kA
k = conductivity (Btu ft / hr ft2 °F)
U = k / Δx = conductance (Btu / hr ft2 °F)
R = 1 / U = Δx / k = thermal resistance (Btu / hr ft2 °F)-1
Conduction through Composite Walls
Most building structures such as walls serve multiple purposes including providing structure,
presenting an appealing appearance, retarding vapor and moisture transmission, and providing
thermal resistance. To serve these multiple purposes, walls are typically composites of multiple
materials. These materials are typically arranged such that heat transfer follows both serial and
parallel paths.
Conduction through Serial Paths
When heat is transferred through materials in a serial arrangement, the total thermal resistance
is the sum of the thermal resistances. These thermal resistances should include the thermal
resistances of the convection coefficients on each side of the structure. The figure below shows
the temperature profile through a wall from a warm inside air temperature, Ti, to a cooler
outside air temperature To. The temperature decreases as a result of the thermal resistances of
the convection coefficients and wall materials.
A
Ti
B
To
Ti
Twi
Tmid
Two
To
In steady-state conduction, the rate of heat transfer between each set of temperature nodes
must be equal.
QA
T  Tmid
T  Two
T  To
Ti  Twi
 A wi
 A mid
 A wo
Rhi
RA
RB
Rho
Thus, the rate of heat transfer can be written as:
QA
Ti  To
where RT = Rhi + RA + RB + Rho
RT
3
Example
Consider a 60 ft2 wall made of two materials positioned such that heat transfer through the wall
occurs in series through both materials. The thermal resistances of the materials are 2.18 hr ft2
°F /Btu and 0.45 hr ft2 °F /Btu. The thermal resistance of the convection coefficients on interior
and exterior surfaces are 0.68 hr ft2 °F /Btu and 0.17 hr ft2 °F /Btu. The inside air and outside air
temperatures are 72oF and 50oF. Calculate the heat flow rate through the wall.
R T  Rhi  R A  RB  Rho  0.68  2.18  0.45  0.17  3.48 hrft 2 o F/Btu
QA
Ti  To
72 o F  50 o F
 6 0 ft 2 
 379 Btu/hr
RT
3.48 hrft 2 o F/Btu
Conduction through Parallel Paths
When heat flows through parallel paths, the total thermal resistance can be formulated in terms
of the area fractions and thermal resistances of the paths. Consider the following example of
parallel heat flow through a wall with a window.
Rwin
Ti
To
Window
Rdoor
Wall
The total rate of heat transfer, Q, is:
Q
AT
(Ti  To )
RT
The total resistance RT can be found in terms of the area fractions, FA, and thermal resistances,
R, of the parallel path by noting that:
Q = Qwin + Qwall
= UAwinΔT + UAwallΔT
= (UAwin + UAwall) ΔT
= [[(Awin / Rwin) + (Awall / Rwall)] ΔT = (AT / RT) ΔT
RT 
AT
A win A wall

R win R wall

1
A win /A T A wall /A T

R win
R wall
where FA win  A win /A T
and

1
FA win
R win

FA wall
R wall
FA wall  A wall /A T
4
Parallel paths should be joined at nodes with common temperatures. In most cases this means
that parallel paths should be joined at the nodes representing the air temperatures on either
side of the structure, since the air is typically well mixed. Combining the nodes at common
materials in the structure is less accurate because of two-dimensional heat transfer through the
common materials.
Example
Consider a wall with gross surface area of 80 ft2 that includes 50 ft2 of wall area, 20 ft2 of door
area and a 10 ft2 of window area. The thermal resistance of the wall, door and windows are 15
hr ft2 oF/Btu, 5 hr ft2 oF/Btu and 2 hr ft2 oF/Btu, including convection coefficients. The inside air
and outside air temperatures are 72 oF and 50 oF. Calculate the total thermal resistance of the
wall including the door and window, and the heat flow rate through the wall.
The area fractions of the wall, window and door are:
FAwall = 50 ft2 / 80 ft2 = 0.625
FAwin = 10 ft2 / 80 ft2 = 0.125
FAdoor = 20 ft2 / 80 ft2 = 0.25
The thermal resistance of the wall is:
RT 
FA win
R win
1
1

 6.47 hr ft 2 oF/Btu
FA wall FA door
0.125 0.625 0.25




2
15
5
R wall
R door
The total heat transfer is:
QA
Ti  To
72 o F  50 o F
 80 ft 2 
 272 Btu/hr
RT
6.47 hr ft 2 o F/Btu
Effective Convection Coefficients That Include Radiation
Hot surfaces lose heat to the surroundings via convection and radiation. The equation for heat
loss, Q, to the surroundings at Ta, from a hot surface at Ts, with area A is:
Q = h A (Ts – Ta) +  A  (Ts4 – Ta4)
where h is the convection coefficient, is the Stefan-Boltzman constant (0.1714 · 10-8 Btu/ft2-hrR4, or 5.67 · 10-8 W/m2-K4),  is the emissivity of the surface. Emissivity is the ratio of the
radiation emitted from a surface at a given temperature to the radiation emitted by a black
5
surface which emits the maximum radiation possible at the same temperature. Thus, emissivity
ranges from 0 to 1.0, and radiation heat loss increases as the emissivity increases. The thermal
emissivity of most building materials is between 0.90 and 0.95. To reduce radiation heat loss
surfaces can be painted with aluminum paint that has as an emissivity of about 0.55 or covered
with aluminum foil type material that has as emissivity of about 0.10.
In building applications, the temperature differences between walls and the surrounding
environment and air are typically small. In these cases, the heat loss from convection is
approximately equal to the heat loss from radiation. To simplify calculations, an effective
coefficient, h’, that includes heat loss from both radiation and convection is often used where:
Q = h A (Ts – Ta) +  A  (Ts4 – Ta4) = h’ A (Ts – Ta)
Example
Consider a wall with surface temperature of 75 F and emissivity of 0.90. The temperature of the
air and surroundings is 50 F. Calculate the effective convection coefficient h’.
The natural convection coefficient for air in contact with a vertical wall can be estimated as
(ASHRAE Fundamentals, 1989):
h =0.19 [T]0.33 (Btu/hr-ft2-F) h =0.19 [25]0.33 (Btu/hr-ft2-F) = 0.550 (Btu/hr-ft2-F)
Convection heat loss from the wall is:
Q/A = h (Ts – Ta) = 0.550 (Btu/hr-ft2-F) (75 F – 50 F) = 13.74 (Btu/hr-ft2)
Radiation heat loss from the wall is:
Q/A =   (Ts4 – Ta4) = 0.1714 · 10-8 (Btu/ft2-hr-R4) 0.9 [(75+460)4 – (50+460)4]
Q/A = 22.02 (Btu/hr-ft2)
The total heat loss from the wall is:
Q/A = 13.74 (Btu/hr-ft2) + 22.02 (Btu/hr-ft2) = 35.76 (Btu/hr-ft2)
The effective convection coefficient including radiation is:
h’ = Q/A / (Ts – Ta) = = 35.76 (Btu/hr-ft2) / (75 F – 50 F) = 1.43 (Btu/hr-ft2)
The value of the convection coefficient, h, depends on whether the air movement over the
surface is forced or natural due to the buoyancy of air. For exterior walls and structures, wind
causes forced convection and the convection coefficient is a function of wind speed. Typical
values for effective exterior convection coefficients including radiation are:
6
EXTERIOR
Summer
Winter
h’
4.0 (Btu/hr-ft2-F)
6.0 (Btu/hr-ft2-F)
R’ = (1/h’)
0.25 (hr-ft2-F/Btu)
0.17 (hr-ft2-F/Btu)
For interior surfaces, the natural convection coefficient is a function of the geometry and the
temperature difference between the surface and air. For example, consider a warm floor. The
air near the wall expands and rises as it is warmed by the floor. Air rising off the floor
encounters no friction from the floor and hence rises quickly. The relatively high speed causes
the convection coefficient to be high. Next, consider a warm vertical wall. The air near the wall
expands and rises as it is warmed by the wall. However, air rising along the wall encounters
some friction resistance from the wall and hence rises at a moderate rate with a moderate
convection coefficient. Finally, consider air under a warm ceiling. The air near the ceiling
expands and rises as it is warmed. However, the ceiling impedes movement and the air is
largely stagnant. Thus, the convection coefficient is negligible.
By analogy, cold ceilings also cause adjacent air to cool and drop with relatively high velocities
and high convection coefficients. Similarly, cold floors cause adjacent air to pool with low
convection coefficients. Typical values for effective interior convection coefficients including
radiation are:
INTERIOR
Warm floor or cold ceiling
Vertical wall
Warm ceiling or cold floor
h’
1.64 (Btu/hr-ft2-F)
1.47 (Btu/hr-ft2-F)
1.09 (Btu/hr-ft2-F)
R’ = (1/h’)
0.61 (hr-ft2-F/Btu)
0.68 (hr-ft2-F/Btu)
0.92 (hr-ft2-F/Btu)
Some envelope structures include air spaces. The total thermal resistance of an air space is a
function of the convection coefficients and emissivities on the inside walls of the space. As
before, convection coefficients depend on the geometry. Typical values for effective thermal
resistances, including radiation, across air spaces are:
AIR SPACES
Heat flow up
Horizontal
Heat flow down
R’ = (1/h’)
0.87 (hr-ft2-F/Btu)
1.01 (hr-ft2-F/Btu)
1.02 (hr-ft2-F/Btu)
Thermal Resistance Values
A handy compilation of thermal resistances for building energy applications is shown below.
The thermal resistances for surface air films listed below include radiation heat loss; thus
radiation heat loss does not need to be explicitly calculated when using these coefficients. The
modifications of the surface air film values to include radiation assume that the surroundings
are at the same temperature of the air. However, in cases such as heat transfer from a roof to a
clear night sky, radiation heat transfer may be much larger than assumed by the modified
surface air film coefficients. Thus, in these cases, it may be useful to explicitly calculate
convection and radiation heat loss individually rather than by using a modified surface film
coefficient. ASHRAE Fundamentals includes an even larger set of thermal properties for most
building materials.
7
Exterior Materials
R(
Wood/inch
Wood bevel siding
Stucco/in.
1/2-in. insulated board sheathing
1/4-in. plywood
1/4-in. hardwood
3/4-in. softwood board
Building paper
8-in. concrete blocks
Common brick/in.
Face brick/in.
Sand and gravel concrete/in.
Wood siding shingles
Asbestos-cement shingles
Asphalt roof shingles
Wood roof shingles
Built-up roofing
)
0.83
0.81
0.20
1.32
0.31
0.18
0.94
0.06
2.18
0.20
0.11
0.08
0.87
0.03
0.44
0.94
0.33
Interior Materials
1/2-in. gypsum board
1/2-in. plaster
3/4-in. hardwood finish flooring
Floor tile
Carpet and fibrous pad
Carpet and foam rubber pad
Insulation (per inch)
Rigid Polyisocyanurate
Fiberglass (batts)
Styrofoam sheets
Fiberglass (loose)
Mineral wool (loose)
Cellulose (loose)
Foam (blown)
R(
)
0.45
0.32
0.68
0.05
2.08
1.23
7.2
3.2
4.2
4.2
3
3.3
4.4
Source: Mitchell, J., 1983, “Energy Engineering”, John Wiley and Sons, Inc.
Summary: Calculating Steady State Heat Transfer Through Walls
These values and heat transfer equations can be used to calculate heat loss and gain through
walls.
8
Example
Calculate Rtotal = RT for typical stud wall. Assume winter conditions.
½ inch gypsum board
studs and insulation
½ inch insulated board sheathing
Side view:
wood siding
Ti
To
3
Top view:
5"
8
stud
insu
Rhi = .68 (Btu / hr ft2 °F)-1
Rgypsum = .45
16" O.C.
1
5
1
A stud
 8  .10
AT
16
A ins
 1  .10  .90
AT
5"
8
5
8
Rstud = ( 3 ) 0.833 = 3.02
stud
5
8
insu
Rinsulation = ( 3 ) 3.2 = 11.6
stud
Nominall (2" x 4") stud = 1 5/8 inch x 3 5/8 inch
Rsheathing = 1.32
Rsiding = .81
Rho = .17
Rstud path = Rhi + Rgypsum + Rstud + Rsheathing + Rsiding + Rho = 6.45 (Btu / hr ft2 °F)-1
Rinsulation path = Rhi + Rgypsum + Rinsulation + Rsheathing + Rsiding + Rho = 15.03 (Btu / hr ft2 °F)-1
RT 
1
A ins /A T A stud /A T

R ins
R stud

then Q 

1
.90
.10

15.03 6.45
 13.27 (Btu/hr ft 2 F) 1
A tot
(Ti  To )
RT
Note that the parallel paths are joined at the air temperature nodes on each side of the wall,
and not at the common gypsum or sheathing nodes. This is more accurate, since the
temperature of the gypsum and sheathing over the insulation and studs is not identical.
Reducing Heat Loss by Reducing Exterior Wall Area
From the conduction equation, one of the ways to reduce heat loss through walls is to reduce
exterior wall area. Circular buildings have the largest ratio of enclosed area to wall area. Thus,
circular buildings like Mongolian yurts require less wall material and experiencing less heat loss
through walls than other buildings with the same floor area.
9
Universal Public Domain
In rectangular construction, squares have the largest ratio of enclosed area to perimeter. Thus,
the long narrow houses in the University of Dayton student neighborhood have much greater
heat loss per enclosed area than more square buildings.
Irregularities in buildings add architectural interest, but increase the ratio of wall area to
enclosed floor area. To create interesting buildings without compromising energy efficiency,
architects have extended porches and garages and used multiple room lines on rectangular
building envelopes.
10
Act 2 Project, Pacific Gas and Electric
Reducing Wall Heat Loss by Increasing Insulation
In existing buildings, uninsulated wall cavities can be filled with foam or cellulose. The R-value
of a 4-inch wall cavity increases from R = 1.01 (hr-ft2-F/Btu) to R = 4 x 3.3 = 13.2 (hr-ft2-F/Btu)
after filling with cellulose. The cost of cellulose, which is shredded newspaper treated with a
fire retardant is quite moderate. To reduce the likelihood of water condensation on the outside
of the insulation after passing through the insulation from the building interior, the interior side
of exterior walls can be painted with water-resistant paint.
Photos: NREL http://www.nrel.gov/data/pix
11
Foam insulation has a higher R-value (3.6-6.5 hr-ft2-F/Btu per inch) than fiberglass batt
insulation (3.2 hr-ft2-F/Btu per inch). All closed-cell polyurethane foam insulation made today is
produced with a non-CFC (chlorofluorocarbon) gas as the foaming agent. Polyurethane foam is
water-vapor permeable, remains flexible, fire resistant and provides good air sealing. Soybased, polyurethane liquid spray-foam is also available.
http://www.energystar.gov/ia/new_homes/behind_the_walls/insulation2.jpg
Reducing Wall Heat Loss by Minimizing Thermal Bridging
A thermal bridge is a component of low thermal resistance that allows heat to bypass
components with high thermal resistances. An example of a thermal bridge is a stud in an
insulated frame wall. The studs allow heat to bypass the insulation in the wall cavity. Thermal
breaks reduce the overall thermal resistance of an envelope component, and should be
eliminated or minimized whenever possible.
2" x 4" stud
Insulation
To demonstrate the importance of minimizing thermal bridging, consider the example below.
Both paths of heat loss have equal heat-loss areas, however path 2 is a thermal bridge with
1/10th the thermal resistance of path 1. Intuition may suggest that the overall thermal
resistance would be the average of the two paths, R = 5.5, since the two heat loss areas are
equal. However, calculating the thermal resistance shows that the overall thermal resistance is
R = 1.82. In other words, the overall thermal resistance is dominated by thermal bridges with
low thermal resistances.
12
A1 = 10 R1 = 10
Ti
To
A2 = 10 R2 = 1
Rtot = 1 / [ AF1 / R1 + AF2 / R2 ] = 1 / [ 0.5 / 10 + 0.5 / 1 ] = 1.82
An even more graphic way of gaining intuition about the importance of thermal breaks is to
consider that heat, like water, follows the path of least resistance. Thus, the thickness of the
walls of a dam don’t really matter, if there is a break in the dam wall. The water will flow
through the break and not the dam walls.
In frame walls, thermal bridging due to studs can be reduced by spacing studs farther apart. For
example, 2x4 studs are typically placed 16 inches apart, whereas 2x6 studs are typically placed
24 inches apart. Reducing the “framing fraction” from 1.625”/16” = 0.102 to 1.625”/24” = 0.068
reduces thermal bridging. The calculations below shows that walls with 2x6 studs with 24-inch
spacing reduce heat loss by 24% compared to walls with 2x4 studs with 16” spacing. Part of the
reduction is due to reduced “framing fraction and part to increased wall cavity insulation. Also
note that walls with 2x6 studs with 24-inch spacing use only 3% more wood for studs than walls
with 2x4 studs with 16” spacing. Thus, the overall material cost of 2x6 studs about the same as
2x4 studs. Energy efficient builders are also finding ways to reduce window and door headers
and other framing to reduce thermal bridging.
2x4 stud w/ fiberglass
Input
Rho (hr-ft2-F/Btu)
Rsid (hr-ft2-F/Btu)
Rpoly (hr-ft2-F/Btu)
Stud thickness (in)
Rstud/in (hr-ft2-F/Btu/in)
Ins thickness (in)
Rins/in (hr-ft2-F/Btu/in)
Rgyp (hr-ft2-F/Btu)
Rhi (hr-ft2-F/Btu)
Stud spacing (in)
Stud width (in)
Calcs
Rstud (hr-ft2-F/Btu)
Rins (hr-ft2-F/Btu)
Rt,studpath (hr-ft2-F/Btu)
Rt,inspath (hr-ft2-F/Btu)
frame frac
ins frac
Rt (hr-ft2-F/Btu)
Ut (Btu/hr-ft2-F)
Volume Studs (in3/ft)
0.17
0.81
7.2
3.625
0.833
3.625
3.2
0.45
0.68
16
1.625
3.020
11.600
12.330
20.910
0.102
0.898
19.530
0.051
4.42
2x6 stud w/ fiberglass
Input
Rho (hr-ft2-F/Btu)
Rsid (hr-ft2-F/Btu)
Rpoly (hr-ft2-F/Btu)
Stud thickness (in)
Rstud/in (hr-ft2-F/Btu/in)
Ins thickness (in)
Rins/in (hr-ft2-F/Btu/in)
Rgyp (hr-ft2-F/Btu)
Rhi (hr-ft2-F/Btu)
Stud spacing (in)
Stud width (in)
Calcs
Rstud (hr-ft2-F/Btu)
Rins (hr-ft2-F/Btu)
Rt,studpath (hr-ft2-F/Btu)
Rt,inspath (hr-ft2-F/Btu)
frame frac
ins frac
Rt (hr-ft2-F/Btu)
Ut (Btu/hr-ft2-F)
Volume Studs (in3/ft)
0.17
0.81
7.2
5.625
0.833
5.625
3.2
0.45
0.68
24
1.625
4.686
18.000
13.996
27.310
0.068
0.932
25.657
0.039
4.57
% Decrease
% Decrease
24%
-3%
Steel studs enable more heat loss that wood studs, since metal is about 300 times more
conductive than wood. Despite the poor thermal performance of steel studs, they are
13
frequently used in multifamily residences due to fire codes. The following table shows ASHRAE
correction factors for metal framing. To use the table, multiply the correction factor and the
cavity insulation to get an effective cavity insulation that includes the steel stud. The effective
cavity insulation is then added to the other thermal resistances of the wall. To mitigate the
effect of the thermal bridge caused by the steel stud, insulated sheathing that covers the entire
wall should be used.
Stud
Size
2" x 4"
2" x 4"
2" x 6"
2" x 6"
2" x 8"
2" x 8"
Stud
Cavity Correction Effective
Spacing Insulation Factor
R-Value
16"
R-11
0.50
R-5.5
R-13
0.46
R-6.0
R-15
0.43
R-6.4
24"
R-11
0.60
R-6.6
R-13
0.55
R-7.2
R-15
0.52
R-7.8
16"
R-19
0.37
R-7.1
R-21
0.35
R-7.4
24"
R-19
0.45
R-6.6
R-21
0.43
R-9.0
16"
R-25
0.31
R-7.8
24"
R-25
0.38
R-9.6
New construction with metal studs and insulated sheathing.
Example
Compare heat loss between a wall with 2x6 wood studs placed 24-inches apart with R-19
insulation and a wall with 2x6 metal studs placed 24-inches apart with R-19 insulation.
2x6 wood stud w/ fiberglass
Input
Rho (hr-ft2-F/Btu)
Rsid (hr-ft2-F/Btu)
Rpoly (hr-ft2-F/Btu)
Stud thickness (in)
Rstud/in (hr-ft2-F/Btu/in)
Ins thickness (in)
Rins/in (hr-ft2-F/Btu/in)
Rgyp (hr-ft2-F/Btu)
Rhi (hr-ft2-F/Btu)
Stud spacing (in)
Stud width (in)
Calcs
Rstud (hr-ft2-F/Btu)
Rins (hr-ft2-F/Btu)
Rt,stud (hr-ft2-F/Btu)
Rt,ins (hr-ft2-F/Btu)
AF,stud
AF,ins
Rt (hr-ft2-F/Btu) = 1/[AF,stud / Rt,stud + AF,ins / Rt,ins]
Ut (Btu/hr-ft2-F) = 1 / Rt
0.17
0.81
7.2
5.625
0.833
5.625
3.378
0.45
0.68
24
1.625
4.686
19.001
13.996
28.311
0.068
0.932
26.478
0.038
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According to the table above, the correction factor for a wall with 2x6 metal studs placed 24inches apart with R-19 insulation is 0.37. The correction factor should be multiplied by the
thermal resistance of the cavity insulation to find an effective resistance of the metal studs and
cavity insulation. The total thermal resistance of the wall is then calculated as the sum of the all
the thermal resistances, including the effective resistance of the metal studs and cavity
insulation.
2x6 metal stud w/ fiberglass
Input
Rho (hr-ft2-F/Btu)
Rsid (hr-ft2-F/Btu)
Rpoly (hr-ft2-F/Btu)
Ins thickness (in)
Rins/in (hr-ft2-F/Btu/in)
Rgyp (hr-ft2-F/Btu)
Rhi (hr-ft2-F/Btu)
Correction Factor, CF
Calcs
Rins (hr-ft2-F/Btu)
Rins,eff (hr-ft2-F/Btu) = Rins x CF
Rt (hr-ft2-F/Btu) = Rho+Rsid+Rpoly+Rins,eff+Rgyp+Rhi
Ut (Btu/hr-ft2-F) = 1 / Rt
0.17
0.81
7.2
5.625
3.378
0.45
0.68
0.37
19.001
7.030463
16.030
0.062
Thus, the fraction increase in heat loss through a wall with steel studs compared to a with wood
studs and identical insulation is:
% Increase = (Ut,metal - Ut,wood) / Ut,wood
65%
For comparison, the correction factor applied to the wood stud wall that would give the same
thermal resistance is 0.89.
Structural Insulated Panels (SIPs) completely eliminate framing and the associated thermal
breaks. SIPs are constructed by gluing ‘oriented strand board’ (OSB) panels to foam cores. SIPs
are constructed in factories and cut to shape before shipping, which speeds onsite assembly.
The solid cores also reduce air infiltration.
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(Source: NREL public domain photos)
Typical thermal resistances of SIP walls depend on the type of foam insulation. Expanded
polystyrene (EPS) has a lower thermal resistance than extruded polystyrene (XPS).
EPS
XPS
Polyurethane
4 1/2"
14.4
19.5
21.7
6 1/2"
21.6
29.5
32.9
Thickness
8 1/4"
10 1/4"
27.9
35.1
38.3
48.3
N/A
N/A
12 1/4"
45.9
58.3
N/A
Reducing Wall Heat Loss by Increasing Wall Thickness
Truss and beam houses reduce framing fraction and allow for thick insulation. The Canadian
house below employed truss and beam construction with 11.8-inch thick wall cavities filled with
wet-blown cellulose, and won an ASHRAE 1996 Technology award.
Strawbale houses use stacked straw bales for both structural and thermal integrity. The design
virtually eliminates thermal breaks and results in thick walls with high thermal resistance and
minimal air leakage. .
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Photo 1: http://en.wikipedia.org/wiki/File:Straw-bale-construction-john-cross.jpg (Creative
Commons Licensing). Photo 2: http://www.mo.nrcs.usda.gov/news/MOphotogallery/rcd.html
Reducing Heat Loss by Field Inspection and Measurements
Infrared cameras can identify under-insulated walls and thermal breaks. The picture below
shows an uninsulated wall panel on the left side of the building. It also shows the effectivensess
of window shades at reducing heat loss; the upstairs front windows have R=4 window shades
while the downstairs windows are uncovered.
The picture below shows where wall cavities have insulated (yellow) and the additional heat loss
through the studs (red). The center wall panel is completely uninsulated. In addition, heat is
escaping through the attic vents as a result of warm air leaking from the house into the attic.
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The picture below shows an insulated wood-frame addition (green) on an un-insulated brick
(red) house. The un-insulated brick is much warmer and is losing heat much more rapidly.
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Field Measurement of Thermal Resistance
The thermal resistance of existing walls, Rw, can be estimated from measurements of outside
air temperature, Toa, inside air temperature, Tia, and the temperature of the inside surface, Tis.
Consider the diagram shown below.
Wall
Tia
Toa
Tis
Rw
Rhi
At steady-state conditions heat flow through the wall equals heat flow from the surface of the
wall to the interior air. Solving this energy balance, and substituting an estimate of the thermal
resistance of the interior convection coefficient, Rhi, gives a good estimate of the thermal
resistance of a wall excluding the interior convection coefficient.
Q = A / Rw (Toa – Tis) = A / Rhi (Tis – Tia)
Rw = Rhi (Tis – Toa) / (Tia – Tis)
The total thermal resistance of the wall is then:
Rw,tot = Rw + Rhi
Example
Calculate the thermal resistance of a wall if the outside air temperature is 30 F, the inside air
temperature is 70 F, and the temperature of the inside surface if 65 F.
From the table of property data, the thermal resistance of the interior convection coefficient is
0.68 hr-ft2-F/Btu. The thermal resistance of the wall, excluding the interior convection
coefficient is:
Rw = Rhi (Tis – Toa) / (Tia – Tis) = 0.68 (hr-ft2-F/Btu) (65-30) (F) / (70 – 65) (F)
Rw = 4.76 (hr-ft2-F/Btu)
The total thermal resistance of the wall, Rw,tot is:
Rw,tot = Rw + Rhi = 4.76 (hr-ft2-F/Btu) + 0.68 (hr-ft2-F/Btu) = 5.44 (hr-ft2-F/Btu)
Wall Construction: Vapor and Moisture Barriers
Vapor barriers prevent moisture from migrating from the warm humid inside air to dry cold
outside air and condensing on exterior side of insulation. Vapor barriers are usually a thin
plastic sheet with negligible R-value. In moderate and cold climates, it is important that the
vapor barrier be placed inside the insulation, so that water vapor does not diffuse through the
insulation and then condense on the exterior side of the insulation.
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In addition to vapor barriers on the inside of the walls, most exterior walls include a moisture
barrier just inside of the exterior sheathing. The moisture vapor prevents water from being
blown through the sheathing into the insulation cavity during stormy conditions. However, it is
important to allow any moisture that does enter the wall to escape. Thus, moisture barriers are
typically fabricated to transmit water vapor, while being impervious to liquid water. Moisture
barriers also impede infiltration of air through the wall when all joints are taped and sealed.
Interior wall board
Vapor Barrier
Insulation
Exterior Sheathing
Exterior Siding
Vapor barrier on inside of wall insulation.
Moisture barrier on outside of wall insulation.
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