Energy Efficient Buildings
Roofs
Introduction
Heat transfer through ceilings and roofs are important components of cooling and
heating loads in most buildings. Understanding how to calculate heat gain and losses,
including the effect of solar radiation on the roof, provides insight into how to design
energy efficient roofs and ceilings. This chapter begins by showing how to model the
effect of solar radiation on roofs using the ‘solair’ temperature. Progressively more
detailed methods of calculating heat transfer, which enable the designer to consider
more design options, are then described. Finally, examples of energy efficient roofs are
presented.
Including Effect of Solar Radiation: Tsolair
Solar radiation on walls and roofs warms the surfaces and affects the rate of conduction
heat transfer through the wall and roof. To account for the effect of solar radiation,
consider the following model of heat flows on the exterior surface of a wall. IT is the
total solar heat flux on the wall, Toa is the outside air temperature, Ts is the
temperature of the exterior surface, Tia is the inside air temperature, Ro the thermal
resistance between the outside air and the exterior surface of the wall caused by the
exterior convection coefficient and Rw is the thermal resistance between the exterior
surface of the wall and the interior air.
IT
Toa
Ro
Ts
Rw
Tia
Noting that the thermal resistance between the outside air and exterior surface of the
wall is caused by the outdoor convection coefficient, ho, an energy balance on the
exterior surface of the wall during steady state conditions gives:
IT  + ho (Toa – Ts) + (1/Rw) (Tia – Ts) = 0
In this equation, , is the fraction of solar radiation absorbed by the wall. The equation
can be rewritten as:
IT  + ho (Toa – Ts) = (1/Rw) (Ts - Tia)
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The right side of this equation represents the heat transfer though the wall. The left side
of the equation represents the energy gain of the surface of the wall from solar
radiation and convection from the outside air. It is useful to reformulate the left side of
the equation in terms of an effective outdoor air temperature, called solair
temperature, Tsa, that includes the effect of the solar heat gain on the wall. Defining
Tsa in this way, allows the left side of the equation to be expressed as:
IT  + ho (Toa – Ts) = (1/Rw) (Ts - Tia)
ho (Tsa – Ts) = (1/Rw) (Ts - Tia)
Setting the left sides of the two equations to be equal gives:
IT  + ho (Toa – Ts) = ho (Tsa – Ts)
Solving for solair temperature, Tsa, gives:
Tsa = Toa + (It  / h)
When Tsa is defined in this way, heat transfer through a wall or roof absorbing solar
radiation can be calculated as:
Qwall = (Awall / Rwall) (Tsa – Tia) = UAwall (Tsa – Tia)
Example
Calculate the heat transfer through a black, flat, R = 10 hr-ft2-F/Btu roof (including
surface convection coefficients) with and without solar radiation if:
Tia = 70°F Toa = 82 F absorbance of roof = 0.90 Ih = 244 Btu/hr-ft2
ft2-F
Neglecting Solar Radiation

 Btu 
Q c Toa  Tia 82  70


 1 .2 
2
Ac
Rc
10
 hr ft 
h = 4 Btu/hr-
Including Solar Radiation
Tsa  Toa 
I
244  .90
 82 
 137 F
h
4

 Btu 
Q c Tsa  Tia 137  70


 6.7 
2
Ac
Rc
10
 hr ft 
In this example, neglecting the effect of solar radiation would underestimate heat
transfer through the roof by more than 500%!
2
The effect of solar radiation on heat gain through building structures is greatest when IT
is high, α is high and ho is low, Thus, it is especially critical to include the effect of solar
radiation on flat, dark roofs during summer, when all three of these conditions are
frequently present.
Heat Loss Gain through Roofs with No Attic
If a building has no attic, then the thermal resistance of the roof and ceiling can be
combined, and heat gain/loss through a ceiling can be calculated using the same
method as heat gain/loss through walls.
Tsa
Tsa
Rc
Tia
Vaulted Ceiling / Roof
Tia
Flat Ceiling / Roof

Q ceil  UA ceil Tia  Tsa 
Heat Gain and Loss through Roofs with Attics
Many buildings have unconditioned attics with insulated ceilings. The heat gain through
a ceiling depends on the attic ventilation rate, the solar radiation incident on sloping
roofs, and on radiant heat transfer from the roof to the ceiling. Three methods for
calculating heat gain/loss through the ceilings are presented below.
Simplified Method
Because of the difficulty of calculating the differential solar gain on multiple sides of a
roof and because the natural attic ventilation rate is highly variable, the simplified
method for approximating heat gain/loss through ceilings in buildings with attics is to:


model the attic as an air space with R = 2 hr-ft2-F/Btu
model the sloped roof as a flat roof with area equal to the ceiling
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Example
Consider a building with:
Tia = 70°F Toa = 82 F absorbance of roof = 0.90 Ihorizontal = 244 Btu/hr-ft2
hext = 4 Btu/hr-ft2-F Rroof = 3 hr-ft2-F/Btu Rceil = 10 hr-ft2-F/Btu Aceil = 1500 ft2
where Rroof and Rceil include convection coefficients.
Calculate the heat transfer through a sloped roof, attic and ceiling using the simplified
method.
Tsa = Toa + Ih  / hext = 82 F + (244 Btu/hr-ft2 0.90) / 4 Btu/hr-ft2-F = 137 F
Qc = [Aceil / (Rroof + Rattic + Rceil)] (Tsa – Tia)
Qc = [1,500 ft2 / (3 + 2 + 10) hr-ft2-F/Btu] (137 – 70) F = 6,700 Btu/hr
Attic Temperature Method
Attic air temperature is affected by heat transfer through the roof, heat transfer
through the ceiling, and the rate of ventilation air entering and leaving the attic. The
figure below shows these major heat flows. In this model, heat loss/gain through the
ceiling is driven by the difference between the indoor air temperature and attic
temperature. Thus, the attic temperature must be calculated prior to calculating
heat/loss gain through the ceiling.
Gable vents
Tsa
Rr
oo
f
R ven
Ta
Rceil
t
Toa
Soffit vents
Tia
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To calculate attic temperature, Ta, write an energy balance on Ta.
Qroof + Qinf + Qceil = 0 (SS)
UA roof Tsa  Ta   UA vent Toa  Ta   UA ceil Tia  Ta   0
Solve the energy balance to get:
Ta 
UA roof Tsa  UA vent Toa  UA ceil Tia
UA roof  UA vent  UA ceil

where
UA vent  V ρc p
After Ta is calculated, the heat gain through the ceiling into the building is:
Qc = UAceil (Ta- Tia)
As in the case of the simplified method, model the sloped roof as a flat roof with area
equal to the ceiling.
Example
Consider the following building.
Tia = 70°F Toa = 82 F absorbance of roof = 0.90 Ihorizontal = 244 Btu/hr-ft2
hext = 4 Btu/hr-ft2-F Rroof = 3 hr-ft2-F/Btu Rceil = 10 hr-ft2-F/Btu Aceil = 1,500 ft2
Vattic = 100 cfm pcp = 0.018 Btu/ft3-F where Rroof and Rceil include convection
coefficients.
Calculate the heat transfer through a sloped roof, attic and ceiling using the attic
temperature method.
Tsa = Toa + Ihorizontal  / hext = 82 F + (244 Btu/hr-ft2 0.90) / 4 Btu/hr-ft2-F = 137 F
UAroof = Aroof / Rroof = 1,500 ft2 / 3 hr-ft2-F/Btu = 500 Btu/hr-F
UAvent = Vattic pcp = 100 cfm 60 min/hr 0.018 Btu/ft3-F = 108 Btu/hr-F
UAceil = Aceiling / Rceil = 1,500 ft2 / 10 hr-ft2-F/Btu = 150 Btu/hr-F
Ta = [UAroof Tsa + UAvent Toa + UAceil Tia] / [UAroof + UAvent + UAceil ]
Ta = [500 Btu/hr-F 137 F + 108 Btu/hr-F 82 F + 150 Btu/hr-F 70 F] /
[500 Btu/hr-F + 108 Btu/hr-F + 150 Btu/hr-F]
Ta = 116 F
Qc = UAceil (Ta – Tia) = 150 Btu/hr-F (116 – 70) F = 6,900 Btu/hr
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Attic Temperature and Radiation Method
In the summer heat gain through attics and ceilings into buildings can be significant.
Three strategies to reduce heat gain through the ceiling during summer are light colored
shingles, attic ventilation fans and low emissivity radiant barriers. The effectiveness of
these strategies can be explored by developing a model of ceiling heat gain which
explicitly considers:



radiation heat transfer from the sun onto the exterior of the roof
radiation heat transfer from the under side of the roof to the attic floor
attic ventilation
The figure below shows the major heat flows through a roof, into an attic, and into a
room.
Qs
Tsa
Toa
1
Rv 
Vc
Rr
p
Roof
T1
R1 
Q12
1
h1
Ta
1
R2 
h2
T2
Attic
Ceiling
Rc
Tia
The radiation heat transfer from surface 1 to 2, Q12, is a function of the StephanBoltzman constant, , emissivities of the surfaces, , areas of the surfaces, A, and the
fraction of radiation leaving surface 1 that is incident on surface 2 (view factor), f12.
Q12 =  [ T14 - T24 ] / [ (1-1) / (1 A1) + 1 / (A1 f12) + (1-2) / (2 A2) ]
A steady state model of these heat flows includes energy balances on the nodes
representing the solar temperature Tsa, the attic air Ta, attic ceiling T1 and attic floor
T2.
E-bal on Tsa: Tsa: Tsa = Toa + Qs alpharoof / hext
E-bal on 1: A1 / Rr (Tsa - T1) + -Q12 + A1 h1 (Ta - T1) = 0
6
E-bal on 2: A2 / Rc (Tia - T2) + Q12 + A2 h2 (Ta - T2) = 0
E-bal on attic air: A1 h1 (T1 - Ta) + A2 h2 (T2 - Ta) + Va pcp (Toa - Ta) = 0
After solving the system of equations, the heat gain into the attic Qc can be calculated
as:
Qc = A2 / Rc (T2 - Tia)
Example
Consider a base case with a dark roof (alpharoof = 0.90), non-reflective surfaces on the
underside of the roof (1 = 0.8) and top of the ceiling (1 = 0.8), and limited natural air
flow through the attic (Va = 50 cfm). Other input data are:
"Input data"
Qs = 1000 "Btu/hr-ft2"
A1 = 100 "ft2"
A2 = 100 "ft2"
Tia = 72 "F"
Toa = 92 "F"
h1 = 1/.92 "Btu/hr-ft2-F"
h2 = 1/.92 "Btu/hr-ft2-F"
hext = 4 "Btu/hr-ft2-F"
Rr = 2 "hr-ft2-F/Btu"
Rc = 10 "hr-ft2-F/Btu"
Stephan Boltzman = 0.1714*(10^(-8))
f12 = 1 "view factor"
pcp = 0.018 "Btu/ft3-F"
Solving the system of equations, gives the heat flux into the room for the base case
Qc = 1,142 Btu/hr.
Adding light-colored shingles (alpharoof = 0.4), a low-emissivity radiant barrier on the
underside of the roof (1 = 0.2) and an attic ventilation fan (Va = 500 cfm) reduces the
heat flux into the roof to
Qc = 284 Btu/hr.
Thus, these measures reduce ceiling heat gain by 75%.
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Field Inspection of Roof/Ceiling Insulation
An effective way to gauge roof and ceiling insulation is to compare snow cover on roofs
with similar sun exposures. The following set of pictures compares several roofs the
morning after a light snow.
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Energy Efficient Roofs
Based on these principles, many strategies can be employed to reduce heat transfer
through roofs.
Use Light Colored Roofs and Shingles to Reduce Cooling Loads
Light colored roofs and shingles reflect more sunlight than dark roofs and shingles, and
reduce heat transfer through roofs into buildings. In cooling dominated climates, this is
clearly an advantage. For climates with both heating and cooling, this creates a tradeoff between reduced cooling energy use in the summer and increased heating energy
use in the winter. The methods described in this chapter can be used to analyze this
situation.
White singles on a net-zero energy use house in Industrial facility roof with both white
and dark gravel.
Colorado. Photo: NREL.
The following graph shows an analysis of a particular case in which a white roof
substantially reduced heat flux in the summer months when the sun is high in the sky
and shines directly on the flat roof, but had a much smaller affect during winter months
when sun is low in the sky.
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Use Radiant Barriers in Attics
Radiation heat transfer from the underside of the roof to the upper surface of the ceil
can be cost-effectively reduced by lining the surfaces with shiny low-emissivity material.
Radiant barriers are frequently placed on underside of the roof rather than the top of
the ceiling so that dust does not gather on the surface and reduce the efficacy of the
radiant barrier.
Source: ORNL
Use High Levels of Ceiling Insulation
U.S. Department of Energy recommended levels of insulation are shown below.
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Zone
Attic
Cathedral Ceiling
1
2
3
4
5
6
7
8
R (hr-ft2-F/Btu)
30 to 49
30 to 60
30 to 60
38 to 60
38 to 60
49 to 60
49 to 60
49 to 60
R (hr-ft2-F/Btu)
22 to 38
22 to 38
22 to 38
30 to 38
30 to 60
30 to 60
30 to 60
30 to 60
Wall
Cavity
Insulated Sheathing
R (hr-ft2-F/Btu)
R (hr-ft2-F/Btu)
13 to 15
None
13 to 15
None
13 to 15
2.5 to 5
13 to 15
5 to 6
13 to 21
5 to 6
13 to 21
5 to 6
13 to 21
5 to 6
13 to 21
5 to 6
Floor
R (hr-ft2-F/Btu)
13
19 to 25
25
25 to 30
25 to 30
25 to 30
25 to 30
25 to 30
Source: http://www.energysavers.gov/tips/insulation.cfm
In colder climates, the recommended ceiling insulation is up to R = 60 hr-ft2-F/Btu.
R=60 insulation requires about 18-inches of fiberglass or blown insulation. To
accommodate this much insulation, builders use special roofing details such as those
shown below that raise the roof to accommodate the insulation.
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Photos: NREL
Green Roofs
Roofs can be covered with soil and vegetation to create a “green roof”.
Thermodynamically, green roofs:



have lower solar absorptivity than black roofs
add thermal mass, which reduces peak cooling and heating loads
allow evaporation from the vegetation, which cools the roof.
In addition, green roofs store and evaporate rain water, which decreases peak flow
through a city’s storm sewer system. Green roofs may also create a pleasant space for
building occupants, especially in urban environments where garden and lawn space is
limited. Because of the density of buildings in highly urban areas, lower roof-top
temperatures contribute to a lower air temperature for the entire city.
Roof of Chicago City Hall with 22,000 ft2 garden. Cross section of green roof. Source:
FEMP. FEMP http://www1.eere.energy.gov/femp/pdfs/fta_green_roofs.pdf
The impact of the additional thermal mass from a green roof on peak cooling and
heating loads can be estimated using a simple single-node finite difference approach.
Consider the following model of a green roof. For simplicity, the only thermal mass
considered is the thermal mass of the soil.
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Tsa
Rs
Ts
Rsr
Soil
Roof
Tia
Tsa is the solair temperature
Ts is the temperature of the soil in the green roof
Tia is the inside air temperature
Rs is the thermal resistance of the soil above Ts, the middle of the soil.
Rsr is the thermal resistance of the soil below Ts plus the thermal resistance of the roof.
A is the area of the roof
An energy balance on Ts gives:
(A/Rs)(Tsa – Ts) + (A/Rsr)(Tia – Ts) = m cp (Ts+ - Ts) / dt
Where:
m is the mass of the soil
cp is the specific heat of the soil
Ts+ is the temperature of the soil at the end of time interval dt
Ts is the temperature of the soil at the beginning of time interval dt
Noting that m = A dx  where dx is soil thickness and  is soil density, the energy balance
equation can be solved for Ts+ to give:
Ts+ = Ts + dt /(dx r cp) [ (Tsa – Ts)/Rs + (Tia – Ts)/Rsr ]
In an hour-by-hour finite difference simulation, the initial value of Ts would guessed and
subsequent values of Ts+ would be calculated using Ts from the previous hour. After Ts
is known, the transient heat transfer into the building, Q, which includes the thermal
energy storage of the green roof, is:
Q = (A/Rsr) (Ts – Tia)
The steady state heat transfer into the building, Qss, which ignores the thermal energy
storage of the green roof, is simply:
Qss = A/(Rs + Rsr) (Tsa – Tia)
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Example
Calculate the hourly steady-state and transient heat fluxes (Btu/hr-ft2) into a building
through a 6”inch green roof on top of a R = 10 traditional roof. Assume the solair
temperature varies sinusoidally over a 24-hour period about a mean temperature of 80
F with an amplitude of 10 F. Soil properties are:
dx = 0.5 ft
 = 109 lb/ft3
cp = 0.191 Btu/lb-F
Rsoil = 1.846 hr-ft2-F/Btu per ft
Thus:
Rs = Rsoil x dx / 2 = 1.846 hr-ft2-F/Btu per ft x 0.25 ft = 0.462 hr-ft2-F/Btu
Rsr = Rroof + Rsr = (10 + 0.462) hr-ft2-F/Btu = 10.462 hr-ft2-F/Btu
Hourly values of Tsa over a 24 hour period are calculated as:
Tsa = Tsa,mean + Tsa,amp x Sin (2 pi hr / 24)
For hr = 1, Ts is set to 70 F as an initial condition. In all subsequent hours, Ts+, which is
soil temperature at the end of the hour, is calculated from:
Ts+ = Ts + dt /(dx  cp) [ (Tsa – Ts)/Rs + (Tia – Ts)/Rsr ]
After Ts is known, the transient and steady state heat fluxes into the building are
calculated as:
Q = (A/Rsr) (Ts – Tia)
Qss = A/(Rs + Rsr) (Tsa – Tia)
The simulation is run for four 24-hour periods to remove the effects of the initial
condition. The graph below shows that after the effects of the initial conditions
become negligible, peak soil temperature is about 4 F less that peak solair temperature
and lags behind peak solair temperature by about 3 hours. Thus, the temperature
driving heat into the building is both reduced and lagged by the thermal mass of the soil.
14
The graph below shows that after the effects of the initial conditions become negligible,
peak heat flux into the building is reduced from 1.83 to 1.54 Btu/hr-ft2, a reduction of
about 16%. In addition, the peak heat flux has been delayed by about 3 hours. This
delay may shift the peak into hours during which the building is unoccupied. This
combination could result in lower electrical demand charges and reduce the required
capacity and cost of the cooling system. In addition, peak cooling may now occur when
the outside air temperature is cooler. This would enable the cooling equipment to
operate at a higher efficiency.
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