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Estimating temperature gradients and dew point temperatures for
building envelopes
Scheuneman, E. C.
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ESTIMATING TFNPEBATURE
GRADIENTS
ANDDEW
POINTTEMPERATURES
FOR BUILDING M E L O P E S
Division
of Building
Research, National
R ~ S ~ T C ~
Council of
Canada
ESTIMATING TEMPERATURE GMDIEMTS AND DEW PQINT TEMPERATURES FOB
BUILDING
ENVELOPESE.C. Scheuneman
It is useful and often essantfal t o h o w the temperature a d moisture conditions that exist throughout buildlng assemblies and
components. Such temperature and mofsture information may be utilized when d e s i g d u g a
.new
building, investigating problemsin
an existingbuilding, and planning t o retrofit a building. The temperature gradient and smkr/winter temperature extremes are of
ten
useful data tod e t e d n e the materials t o be used in a building envelope (exterior
s h e l l ) . A knawledge of air-mo$~ture/temperature conditions and
relationships I s essential in locating the vapour barrier d t h respect
to other components in the building assembly.
The following sections give simplified versions of temperature and dew point calmlations that are generally useful for a variety of
purposes. Results
from
these calculations p l i l l be on the conservative s i d e , that is, they may indicate a problem whereas a more accurate analysis would show no problem. For such borderline cases or for c o q l i c a t e d and c r i t i c a l problem, it m y be necessary t o do a more precise analysia or consult a specialist in thermal analysis.TEMPERATURE Gl?ADmm
During c o l d weather the temperature at the interior surface of the building envelope, except for w i n d m s , will be close to t h e room
temperature. Moving outward through t h wall
or
ceiling the temperaturebecomes increasingly colder u n t i l 1 t reaches the outdoor tersgerature j u s t beyond the exterior fFnish. It: is poesXble to calculate and chart these tenperature changes through the various building caponents; this is
known
as the temperature profile or gradient. During warm or hotweather the outdoor temperature may be greater than the indoor temperature, h t the same general rules and procedures apply.
( 5 ) Temperature Change for Components
Before performing such calculations,
we
need t o examine b r i e f l yt h e basic relationshfp between thermal resistance of the building components and temperature drop {change) through each component.
The information in t h i s Note is based on reports given by DBR personnel
If we assttme that heat flows through each component under steady-
s t a t e parallel heat f l o w conditions,
then
the follawing f o r m l a can be uaed t o calculate the temperature change across each component.where
AT =
temperature change across a component R = thereal resistance o f the caqonent%
= total thermal resistance of a l l cornpadentsATT = total temperature change from interior to exterior
This assumption of steady-state condition means that the calculation w f l f be subject to errors, especially for rapidly changing outside a i r
temperatures. Nonetheless, i t provides useful information.
The forms, Ffgures 1
and
2, have been designed to a s s i s t with t h ecalculation
and
visualization of the temperature p r o f i l e through the building assembly. Figure 1 is for vertfeal assemblies (such as walls); Figure 2 is for horizontal assemblies (suchas
ceilings or roofs),(These £ o m may be reproduced t o carry out your awn s t u d i e s , ) The general procedure
i a
t o list the b u i l d i n g components in the table (left side) and draw lines representing their thickness on the graph (right slde) after choosinga
s u i t a b l e scale. The t h e m 1 resistance values(Bus) are found from Appendix A and used with the known interior and exterior teqexatures t o calculate the values (R, R/%, AT,T) t o
complete the cable. The temperatures are then plotted an the graph to give a p i c t o r i p l repreeedtation of the calculations. Note that. the Winter AT and T columns
are
uaed when the Exterior T is less than theInterior T; the S m r
AT
and T columne are used when t h e ExteriorT
is greater than the InteriorT.
TEM P E R A T U
RE
G
R A
I)
I
E N T
T A B L E
COMPONENT
G R A P H NAMEDATE
ADDRESSASSEMBLY
FIGURE 2H O R I Z O N T A L A S S E M B L Y
TEMPERATURE,OC
We w i l l now use the simple e q l e shown in Figure 3 to follow thtough the step-by-step procedure, The results are
shown
In F i g u r e 4.Determine the temperature gradient for the i l l u s t r a t e d wall assembly
(Fig. 3) with an interior temperature of 2 0 . 0 ° ~ and an exterior temperature of -15.U°C.
13 mrn G Y P S mrn M E T A L S I D I N G mm P L Y W O O D
Figure 3,
A
W a l l AssemblyStew 1
List the components in appropriate sequence in t h e table (Figure 4).
Step 2
Using a ruler or other straight-edge, draw the assembly on the
graph a f t e r choosing a s u i t a b l e scale. Designate the coupanents by
number. Step 3
Using t h e appropriate data from Appendix A, f i l l in the lt column in the t-kble.
S t e p 4
Add up a l l the R's t o give
%
( 2 . 5 5 ) and enter in the TOTAL r o w atbottom of t h e R column. Step 5
Calculate t h e ratio K/RT for each component. Total up the ratios
in this column and enter them in the TOTAL row at bottom of the R/% column. Note that t h i s total should always be very c l o s e or equal to
1.00; if not, recheck your c a l c u l a t i o n s for t h i s column.
S t e p 6
Write the i n t e r i o r ( 2 0 , 0 ° ~ ) and exterior ( - 1 5 . 0 " ~ ) temperatures in she appropriate p l a c e s under the Winter
T
column (or under the Summer T column if appropriate).S t e p 7
Calculate
ATT
by subtracting the Exterior T from the Interior T.Enter t h i s figure 135.0) in the TOTAL row at bottom of he AT column. (If the Exterior T is greater than the Interior T, subtract the Interior
T from t h e Exterior T . ) Step 8
Calculate AT for each component by m u l t i p l y i n g each R / K ~ times
ATT
(equation 1) and enter the figures. Note that the total of these A T ' Sshould equal, or be very close ta, ATT. Lf not, recheck the calculations.
S t e p 9
F i l l in the Winter T column by adding the first component No. 1,
AT (0.4) to t h e T in the row above (Exterior T of -15.0) and p l a c i n g t h i s new T ( - 1 4 . 6 ) i n t h e row below component No. 1. Continue t h i s sequence of adding each AT to t h e T above and entering the new T i n t h e r w below. (When Exterior T is greater than Interior
T,
each AT is subtracted from the T in the row above.)As
a check, the f i n a lcoqonent A T added to the preceding T (1.8
+
18.4 = 20.2) s h o u l d yield the Interior T (20.0) or be very close.By glancing down t h e T column in the completed table, one sees the
temperatures on each s i d e (interface) of each component.
On the graph, mark the temperatures ( T ' s ) at each coqonenr edge
( i n t e r f a c e ) , and then draw s t r a i g h t lf nes joining e a c h p o i n t to t h e next. This completes both the arithmetic and graphic representations of t h e temperature or thermal gradient.
( i f ) Temperature at a S p e c i f i c P o i n t and Location of a Certain Temperature
We
may want t o knowthe
temperature at a specific point in the assembly (besides the fnterfaces), or where in the assembly a certaintemperature w i l l occur. Et is possible t o deternine these temperatures or points
by
using either the table or graph.Suppose we wish to know
(1)
the temperature in the fibreglass batt 35 m from its cold s i d e and (2) where in the assembly the temperature w i l l be 5*.(a) Graphical Method
'Phis is the easiest and quickest method. Use a straight-dge
r u n d a g vertically t o l i n e up with 35 mm from the cold side of the batt (73 mm from interior) and note the point of interaeckion with the
teqerature gradient line. This gives a value of about -lo. See vertical dashed line on Figure 4.
Use a straight-edge ruaaing horfzwtally t o line up with 5' and note the point of intersection with the temperature gradient l i n e . This gives a value of about 41 nua into the batr
from
the warm s i d e ( 5 4 ram from interior). (See dashed horizontal l i n e on F i g u r e4.)
(b) Arithmetic Method
This method may be preferred by many as it gives more exact values
than the graphical me tho& h e f ollawf ng
f
0-1 a y i e l d s the temperature ata
s p e c i f i c point:where
T
= temperature at the specific po%ntTw
= teqerature at w a r m interface of conpoaent conLaining the specffic pointL
= distance of specific p o l n t f ram warm side of component LT
= total thickness of componentA = temperature change across component
The following f o r m l a w i l l yield the location of a certain
temperature:
where
L
= location of certain temperature from warm side of component Ifw = temperature at: warm interface of eomp~aent containing thespecific temperature
T
= temperature at the specific pointAT
= temperature change across component$
= total thickness of coqonentFor this e-ple dealfng with component No.
4,
we havewhich yields
The above two results are t h e same as those obtafned graphically (Figure 4 ) .
Note: The temperature gradient calculations for the Table qnd the
k i t h t i c Method are s h m w i t h a precis1 on of 1 or 2 decimal places; thig is done to maintain the relatfonship between the
d i f f e r e n t Rts, R / R ~ ~ s , AT'S and T's, It is m a t h a t i c a l l y convenient with electrouic calculators t o use t h i s degree of precision. The final results of calculaeions are rounded to the nearest whole unft of degrees C e l s i u s (OC) or m f l l t ~ t r e s (mm).
As previously mentioned, the temperature gradient
calculations assume steady-state conditioa~ which means that the rasulte are usually leas accurate than the nearest whole u n i t . However, the calculated values are generally close enough to be
useful, e s p e c i a l l y when looking at extreme or worst-case
situations.
Further explanation and examples can be found
in
references 1, 2a, and 3.(iii) Extreme Temperatures at Outer Surfaces
Qulte often the temperature of the outer surface of a b u t l d i n g assembly can be significantly higher or lower than the outside air
temp eratare. Higher temperatures result from absorption of solar (short-wave) radiation by the surface. Lower temperatures result from the surface g i v i n g o f f infrared (long-wave) radiation t o a clear n i g h t
sky or to o t h e r b o d i e s .
The following formulae have been developed to account for these
two effects on roof assemblies ( 4 , 5 ) . The references use degrees Fahrenheit (OF); this paper uses the formulae in degrees Celsius
(*C)-
They may also be used for walls that are exposed to the sky. Shaded walls would use the outdoor temperature as the e x t e r i o r surfaceterrrperature.
(a) High Heat Capacity Substrate
(such as concrete under the surface material)
HIGH DAY-TIME TEMPERATURE
TS =
TA
+
42a (sunny conditions)TS = TA
+
55a (sunny conditions p l u s reflected sunshinefrom other surf aces) LOW NIGHT-TIME TEMPEUTURE
TS =
TA
- 5 (clear night sky) (b) Low Heat Capacity Substrate(such as insulation under the surf ace mater-lal)
HIGH DAY-TIME TEMPEMTURE
TS =
TA
+
55a (sunny conditions)TS = TA
+
72a (sunny c o n d i t i o n s p l u s reflected sunshinefrom other s u r f a c e s )
LOW MZGHT-TIME TFXPERATURE
(clear night sky)
The deftnitions far the symbols are as follows: TS= temperature of surface ( O C )
TA = temperature of air (OC)
a = solar absorption coefficient (no units)
Appendix B gives a t a b l e of solar absorption coefficients for d i f f e r e n t
materzals. The value can be hfghly variable if the s u r f a c e
deteriorates, accumulates d i r t , or i s covered with water.
These extreme temperatures and temperature g r a d i e n t s s h o u l d b e u t i l i z e d when choosing materials requfred to withstand extreme
Example 2
A f l a t roof assembly built of a 100 m concrete deck (high density) with 25 mm extruded polystyrene on t o p covered w i t h built-up raaflng covered with grey gravel. Determine the extreme temperature gradients for this roof that receives no reflected sunshine for a summer high air temperature of 35% and a d n t e r low air temperature of -30°C. The following temperatures are u t i l i z e d :
-
interior T = 20°C-
winter minimum TA = -30°CReace, mintmum surface temperature T = TA
-
10 givingTS = -30°
-
lo0
= -40°CHence, maximum surface temperature
TS
= T A + 55awhere a = 0.75 {from Appendix
B)
givingSince the max5mm and mfnimum temperatures occur on the surface material rather than i n the
air,
the external air f i l m is omitted from theR,
R/RT andAT
calculations. A t a b l e is completed and the graph is drawnaccording
to the procedure described previously. The results are shown in Figure 5.For comparison, the procedure has been carried out in Figure 6 with no surface temperature effects ( s o l i d line)
and
the graphical results f r o mFigure 5 added (dashed lines). There is
a much
larger difference between the t w o maximum temperature gradients than between the twom i n i m temperature gradients. This illustrates that the heating effect
of solar radiation is nuch larger than the cooling efEecr produced by long-wave radiation, as shorn also by the temperature equations. For
further explanation
and
exaqles see ref erencea 1, 2a, 3, 4 , and 5.8 . DEW POINT TEIfFEBATURES
Relative humidity Ls the percentage ratio of the actual mount of water vapour h e l d in the air compared t o the maxtmrm I t could hold a t a given temperature. Far example, a relative humidity
CR.H.)
of 50% meansthat the air is carrying o n e h a l f the total amount of water vapmr it is capable of holding at that temperature. Air with 100% R.H. has reached i t s saturation point; the water vapour I t holds will begin t o condense if the temperature drops or moxe moisture is added. The water vapour
wfll
condense as l i q u i d if T is greater than O°C or as ice crystals ifT
is
lese than O°C. The temperature a t which condensation takes p l a c e is called the "dew point" temperature.Understanding the dew p o i n t and the moiature capacity of air is c r u c i a l to understanding condensation/moie ture problem f n bu i l d i n g a . When the temperature af air decreases, the m a d m u m amount of water vapour that the a i r can carry is reduced. For example, if air at 2 1 * C and 50% R.H. is cooled t o 1l0C, the RtR, w i l l reach 100% (the dew p o i n t ) . If the tenperature is lowered further, water w i l l condense out
of the a i r to maintan the R.H, at 100%. hferences 2b, 6 and 7 give further explanation and d e t a i l .
A
psychrometric chart (Figure 7) shaws the relationship between the relative humidity (R,E.) and temperature of air.me
curvedlines
beginning on the r i g h t v e r t i c a l scsle i n d i c a t e the, variation o f the r e l a t i v e humidity ( X ) . The straight l i n e s running verticallyfram
the bottom borlzoatalscale
to the saturation curve give the air temperature ( "C)(f) Finding the Dew Point Temperature
Par a Laom value of air temperature and relative humidity (R.H.]
one can use the psychromtric chart to f i n d the dew p o i n t temperature. The procedure is as folluws:
Step 1
Determine the air temperature inside the house by using a
thermonreter (or an assumed value),
Step 2
D e t e d n e the relative humtdfty i n s i d e the house using a sling psychrometer or a hygrometer (ar an assumed value). The R.R. values
recorded by hygrometers may be in error by as rsuch as
20%
so a c a l l b r a t e d instmnment should be used.S t e p 3
On
the psychrometric chart, locate the interior air temperature of the house an the horizontal (bottom) scale and mark the p o i n t where the vertical projection from t h i s temperature intersects the curve which represents the B.H. of the house air.Starting at the intersection p o i n t , run a straight-edge parallel to the bottom OC scale and left to cross the curved lOOX R.H. line. The crossing point gives the dew point temperature.
Example 3
Three examples are s h m in Fl.gure 8:
3(a) The house interror a i r temperature is 2S°C and the R.B. is
60%. The dew point teqerature at: which water v a p w r w i l l
condense as l i q u i d for this situation is 16.5OC.
Water vapour w i l l condense out as l i q u i d at 10.5V
3(c) T = 17%
R.H. = 25%
Water vapour will condense o u t as ice a t the dew point
temperature of -3*C.
Cii) Finding the New B.H. After a Temperature Change
As previously noted, the capacity of a i r t o hold water vapour 3 s
different fur different temperatures. The hfgher the
air
temperature, the greater the amount of water vapour that can be held.The procedure to f i n d the new R.H. after a temperature change is
as follows:
S t e p 1
Locate the point on the psychrametric chart that corresponds to the o r i g i n a l teuperature and relative humidity.
Step 2
U s i n g a straight-edge, p r o j e c t a horfeontal l i n e to the new
temperature. The point of intersection with the curved
R,H.
line gives Lhe new relative humidity.Example 4
Three examples are shown in Figure 9 :
4(a)
?he householda i r
is at; 20°C and 40%R.H.
What is the newR.E1. when the a i r Is cooled to 10°C? (R.H. becomes 75%,) 4(b)
T
= 2 S 0 CR.H. = 80%
What is the R.B. when t h i s
warm,
humid air cools t o 20'C? (R.H. becomes 100% and same water condenses to maintain the R.H. at 100%-)4(e)
T
= -15'C R.H. = 95%:What is the R.H. when t h h c o a l outside air is brought into
the house and warmed to 20"C? {R.B* becomes 8X which is very d q .
1
LOCATION OF DEW
POINT
TEMPERATURES FOR BUILDING ASSEMBLIES Three examples are g l m n t o i l l u s e r a c e the combined use of temperature gradients with the psycbrome.tric chart to locate the dewA double-glazed window has temperature gradients as shown in
F i g u r e 10. When t h e house T = 22°C and
R.H.
= 5 0 X , will water vapour condense on the inside glass when the outside T a -22"C?Ue calculate the values to complete the t a b l e and draw the
temperature profile. Referring to the psgchroretric chart ( F i g . 7 ) , we
f i n d that air at 22%
and
50X R.H. has its dew point at l l . S ° C . Either rhe t a b l e or gradient shows that the inside glass tenperature is 6.6"C.As the glass l a colder than the dew point, water vapour will condense as
l i q u i d on the wf n d w .
Since the i n s i d e glass of a window is usually the coldest surface in a room, it determines the m a d u u m R.H. that can be maintained without condensation. U s i n g t r i p l e glazing raises the inside glass temperature and, thus, the maintained R,H.
T E M P E R A T U R E G R A D I E N T
T A B L E
dNAME
DATE
ADDRESSASSEMBLY
GRAPHFIGURE 10.
E X A M P L E 5
Example 6
We have a partly-insulated wall as s h m fn Figure: 11. Where
will
water vapour condense in the wall assembly if there is no vapourbarrier? The conditions are as follows: Interior T = 18OC
Exterior T = -35OC inside R.H. = 30%
The table and profile are calculated and drawn, The psychrometric
thart reveals that the dew point temperature is 0°C- (1) Graphical Method
We can see at a glance that: O°C lies within the rack wool insulation. By drawing a horizontal line from the O 0 point on the teqerature scale, we find that 0' occure at about 22
mm
from theInterior s i d e of t h e batt (35 mm f r o m interior surface), Arithmetic Hethcd
using $pu.ation (33,
Results from both ti) and (ii) agree.
Note: This discussion applies when the temperature og the
-
iasulation has reached
or is
colder than the dew point temperature.Uuder steady-stalte conditions the water vapour would
continue to d i f f u s e through those tnsulatLons that have little resistance t o moisture flow; the water vapour would condense on the next colder interface such as the exterior sheathing which has a
much
larger resistance. However, the water vapour can condense i n the i n ~ u l a t i o n under norrsteady-state conditions which u s u a l l y occur more than 80X ~f each day. When theinsulation or other materiah in the assembly have a high resistance to moisture flow Cdlffusion), a vapour pressure
gradient s h w l d be drawn (see R e f erenee 71.
When
water
vapour is transported into a building assembly bywarm a i r f l o w rather than vapour diffusion, condensatton can occur within the insulation unless the velocity of the aZr flow ia high enough t o warm up the insulation and carry the vapaur
T E M P E R A T U R E G R A D
1
E N T
T A B L E
I
COMPONENT
I
I
I
SUMMER
I
WINTER
I
G R A P H
NAME
DATE
ADDRESS
ASSEMBLY
A c e i l h g is going to be reinsulated with
new
152 mm f r i c t i o n fftf i k r g l a s s batts over the origxual
63
mm rock wool batts. Canwe
place a 0.15mm
poLyethyleue vapvur barrier over the original insulationbefore adding the new i~aulation?
Lf
the dew point temperature w i l l occur on the colder s i d e of the vapour barrier, then we may use the polyethylene.The c o d i tioas are as follaws: inside T = 20%
attic
T
= -30aCinside
R.H.
= 30XThe table and temperature profile are completed as in Pigtrre 12.
Using the psychrometric chart w e find that condensation should occur at about 2°C. h o k i n g at the table and p r o f i l e we see that the vapour barrier temperature is 4°C. Since this 2s alightly higher than the dew p o i n t temperature, no condensation should occur under these condirions as the vapwr barrier w i l l prevent water vapour from reaching the dew p o i n t temperature which is located within the new insulation.
It should be noted that changing conditions can alter the s i t u a t i o n .
(i) If the R.H. of the house rises t o 3 5 % , the dew poFnt teuperature rises to 4OC. Water vapour can condense at the vapoux barrier (if vapour can pass through the c e i l i n g ) and then d r t p or d i f f u s e back d w n to the ceiling.
Iii) If the a t t i c temperature drops to -35°C instead of -30°C, then the vapour barrier temperature becomes 2.4' which is close to t h e dew point temperature,
When the vapour barrier temperature fs below the dew polnt temperature as a result of very cold weather a x high indoor R.H., there may nat be a condensation problem if t h i s situation occurs for only a few days each heating season s i n c e the building assembly can hold
certain
quantities ofmoisture with no
resultantproblem- This analysis assumes that the vapour barrier is camplete and that there are no air leakage paths through the polyethylene vapatr barrier.
T E M P E R A T U R E
G R A D I E N T
T A B L E
COMPONENT SUMMER
WINTER
R R/RT *
A T I
T
A T I
r
NAME
DATE
ADDRESSASSEMBLY
G R A P HT E M P E R A T U R E ,
OCFIGURE
1 2 ,
E X A M P L E
7
The following references . e v e more detailed and technical
explanat I o n of the concepts discussed in this paper,
( 1 ) LatEa, J.K. and Garden, G.K., Temperature Gradients Through
Building Envelapea, Matima1 Research Council of Canada, D i v i s i o n of Build- Research, CXD 36, 1962 (corrected 1968).
(2) Latta, J8K-b Walls, Windows and Roofs for the Canadfan Climate,
National Research Council of Canada, Divlsion of Building Research, Special Technical Publication No. 1, 1973.
NRCC 13487.
(3) Baker, M.C., Roofs: Design, Application a d Maintenance,
National Research Council of Canada, (Multiscience Publications Limited,
Montreal),
1980,
Chapter 5 .( 4 ) Stephenson, D.G., Extreme Temperaturea at the Outer Surfaces of Buildings, Nat ianal Research Council of Canada, M v i s i o n o f Building Research, CBD
47,
1963.(5) Garden,
G.K.,
Thermal Considerations in Rsnf Design, National Research Council of Canada, Divfsion of Building Research, CBD 70, 1965.( 6 ) Hutcheon, N.B., Humidity id Canadian Buildings, National
Research Council of Canada, Divf sion of Buildt ng Research,
CBD
1 ,1960 (corrected 2968).
(7) Lettta, J.K. and Beach, R.K., Vapour Diffusion and Condensation,
N a t i o n a l Research Council of Canada, Division of Building
THERMAL
RESISTANCE
OF SQPEE COMMON BUILDING MATERIALS* Thermal Resistance B u i l d i n g Material or Component S o l .-
a/-
For Imperial W i n . Fo r Thickness Thickness L i s t e d L i s t e d Insulation Fibreglass Batt Eeockwool Batt Fibreglass Loose F i l l ( B l m - I n ) Fibreglass Loose Fill (Poured)Rockwool Loose Fill (Blown-In)
Rodcwol Loose F i l l (Poured) Cellulose Pibre (Blown) Cellulose Fibre (Poured)
Expanded Mica
(Vermiculite, Zonollte, etc.)
Polystyrene Loose Pill
Expanded PoIystyrene { U g t d ) Extruded Polystyrene (figid] Polyurethane (Rigid)
Polyurethane ( F o a i ~ d i n Place)
Fibreglass Sheathing
Urea Formaldehyde (Foamed in Place, After Curiug)
Wood Fibre
Wood Shavings Cork
Glass F i b r e Roof Hoard
Mineral Aggregate Board Coupressed Straw Board Fibrebaard
Structural Haterials
Saf m o d Lumber (except Cedar) Cedar Logs and Lumber
Concrete 2400 kg/m3 (150 l b / f t 3 ) Concrete 176U k g l m 3 (1 10 l b / f t3)
Concrete 480 kg/m3 (30 l b / f t 3 1
* H E E n Scheuneman, S .
Moff a t t and M. Adelaat, Canadian General Standards Board. CGSB 51-GP-42MP, Ottawa, July 1980.
THERMAL RESISTANCE OF SOME COMMON BUILDING MATERIALS (Cont'd) Thermal Res 1 stance
Building Material or Component S.I. R/mm For I m p e r i a l Rlin. Po r Thickness Listed Thickness L i s t e d Structural Materials ( C o n t ' d )
Concrete Block (3 Oval Core) Sand and Gravel Aggregate
100 mm (4") 200 mm ( 8 " ) 300 mm (12") Cinder Aggregate
loo
mm ( 4 " ) 200 m (8") 300 mm (12") Lightweight Aggregate 100 mm ( 4 " ) 200 TJIIO ( 8 " ) 300 mm ( 1 2 " )Conrmon Brick, Clay, 100 nrm ( 4 " )
Common Brick, Concrete, 100 mm ( 4 " )
Stone ( L i m e or Sand)
S t e e l
Aluminum
Glass (No Air F i l n r s ) 3
-
6 mm(1/8"
-
1 / 4 " )Air
-
Enclosed Air Space (Non-Reflective) Heat F l o w Up, 25
-
100 mm(1'. - 4 " ) ---
Heat Flow Down, 25
-
100 mm(1"
-
4 " )--
Heat Flow Horizontal,25 - 100 mm (1" - 4 " )
--
6.17---
Air Surface F i l m sOutside Air Film (Pioving A i r ) -- 0.03
---
I n s i d e M r Film (Still Air)Horizontal, Heat F l o w Up
-
0.11--
S l o p i n g 4 5 " , Heat Flow Up
-
0.11-
Vertical, Heat Flow Horizontal --- 0.12--
Horizontal, Hear Plow Down --- 0.16-
THERMAL
RESISTANCE OH -SOME COMMUNBUILDING
MATERIALS (Cont'd) Thermal Resistance Building Material or Component: Roofing S . I .-
ImperialB/mm For
in.
ForThickness Thickness
Listed L i s t e d
Asphalt Roll Roofing
--
0 -03Asphalt Shingles -.- 0 -08
Wood Shingles (Cedar Shakes)
--
0 -17 Built-up Membrane (Hot Mopped)--
0.06Crushed Stone ( N o t Dried) 0.00055
--
Sheathing MaterialsSaftwoad Plywood
Mat-Formed Particle Board
Insulating Fibreboard Sheathing Gypsum Sheathing
Sheathing Paper
Asphalt-Coated W a f t Paper
Polyethylene Vapour
Barrier
Cladding Haterials 0,0087-
0.0087--
0.017---
O.UO62---
0.00040--
Negli ,-
N e g l i .--
Ffbreboard Sfding Medium I k n s i t yHardboard,
11 mm (7116")High Density Hardboard,
11 mm (7116")
Softwood Siding (Lapped)
Drop, 18 x 184 m (3/4" x 7
t " )
--
B e v e l , 12 x 184 aan ( * * I x 7 )")-
Bevel, 18 x 235 nm (3/4" x
9
f " )--
Plywood, 9 tam (3/8")
--
Woad Shingles
_-
Brick (Clay or Shale) 100
mm
(4")
-
Brick (Concrete, Sand-Lime)loo
mm
(4")---
Stucco, 25 mm (1") 0.0014 1.25 1-25 2.45 0.89 0.06 N e g l i.
N e g l i *THERMAL RBSISTANCE OF SOW
COMMON
BUfLDING HATEgIALS.(Cont'd) Thermal ResistanceBuilding Material
or Component
S.I. I m p e r i a l
R/w For
in.
FoxThf ckness Thicknes s
Listed L f s t e d
Cladding MaterLals (Conr'dj Hecal Siding
Horizontal Clapboard Profile
Horizontal Clapboard w i t h
Backing
Vertical V-Groove Profile
Vertical Board and Batten Profile
Interior Finishes
Gypsum Boaxd, Gypsum Lath,
Drywall, 13 mm
(4")
0.0062 0.08 0 . 8 9 Gypsum P l a s t e r Sand Aggregate, 13 mm( 3 " )
0.014 0.02 0.20 Lightweight Aggregate, 13 (+") 0.0044 0.06 0.63 Plywood, 7.5nrm
(5116") 0.0093 0.07 1.34 Hardboard, (Standard), hm0")
0.0053 0.03 0.76 Insulating Fibreboard, 25 mm (1") 0.017 0.42 2 . 4 5 FlooringMaple or Oak (hardwood), 19 mm ( 3 1 4 " )
Pine or Fir ( s o f t w o o d ) ,
19 mm ( 3 1 4 " )
Plywood, 16 nrm (518") Mat-formed Particle Board,
16 mm (5/8")
Wood F i b r e Tiles, 13 ma
Linoleum, Tile ( r e s i l i e n t ) ,
3 mm ( 1 1 8 " )
Terrazzo, 25 rmrt (I")
Carpet, Typical Thickness
with F i b r o u s Underlay
T H E W
RESISTANCE
UP SOME COMMON BUILDING MATERIALS (Cont'd) Thermal Resistance BuTldirtg Material or ComponentS.I.
Imperial R / m For R/fn.
For Thickness Thickness Llsced L i s t e d Windows(including i n s i d e and outside
air
f i l m ) Single GlassInsulated Glass (.Double Pane)
5 mm (3116") Air Space 6 mm
E*")
Air Space 13 mm (*") Air Space 19 m ( 3 1 4 " ) Air Space Insulated Glass {Triple Pane)6 mm
(i")
Alr Space13 mm
{i")
Air S p a c e19
aun
( 3 1 4 " ) Air SpaceS t o m Wtndows
Single Pane
+
25-
100 mm(1"
-
4 " ) Air Space 19 mm ( 3 1 4 " ) S e a l e d U n i t+
T a b l e B-1. Roofing Materials
Copper
-
new, p o l i s h e d (brown)-
tarnished (brown)-
weathered patina (green) Aluminum-
weatheredGalvamized Iron
-
new- weathered S t e e l
-
white
enameled-
green enameled-
dark red enameled-
blue enameledLead Sheeting
-
weathered~ s b e s t o s / ~ e w n t :
-
new white-
weatheredTable B-2. Masonry Bricks
-
glazed white-
glazed ivory Lo cream-
common light red-
coramun red Limestone - l i g h t-
darkSandstune
-
lightfawn
-
light grey-
red 0.79 Marble-
white-
dark0 -42
0.8U Granite
-
reddish Smooth Surf ace Asphalt-
weathered 0.93Grey Gravel - weathered 0.75 White Gravel
-
weatheredTable B-3. Faints
(can vary for different shades)
-
black-
white-
green-
yellow - orange-
dark brmn-
dark blue*
Computed by averaging values given by a rangeof
different references.References for Appendix 3
Baker, M.C., Roofs: Design, Application and Maintenance, Multiscience
Publications Limited, Montreal, 1980, page 134.
M f i e , J.A., and W.A. B e c h n , Solar Energy Thermal Process, Wiley- Interscfence, John Wiley & Sons, Toronto, 1974, page 97.
Garden, G.K., Therraal Considerations in Roof Design, Canadian Building Digest 70, National Research Council, O t t a w a , October 1965,
page 70-3.
Kreider, 3 . F . , and F. Kreith, Solar Heating and Cooling: Engineering,
Practical
Revised First Edition, HemispherePub1,isMng Corporation, McGraw-Hill, Toronto, 1977, page 247.
S t r o c k , C., and R.L. Koral, Handbook of A i r Conditioning Heating and Ventilating, Second Edition, Industrial P r e s s , New York, 1965, p. 1