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Heat transfer mechanisms
The equation for heat transmission (or flow) through
a single element by conduction is as follows:
Heat transmission by conduction
where:
q = rate of heat or energy flow (J/s or W)
Q = total heat transmitted (J)
= time during which flow occurs (s)
k = conductivity (W/C m)
A = cross-sectional area of flow path (m2)
t = temperature difference (C)
l = length of flow path (m)
q=
Q
=
kA
t
l
Alternate formula when conductivity and material thickness
are combined, and presented as conductances:
Heat transmission by conduction
C = k/l = conductance (W/C m2)
R = 1/C thermal resistance (C m2/W)
q=
Q
=
kA
t
l
=
k
l
At
=
CAt=
1
RAt
where:
q = rate of heat or energy flow (J/s or W)
Q = total heat transmitted (J)
= time during which flow occurs (s)
k = conductivity (W/C m)
A = cross-sectional area of flow path (m2)
t = temperature difference (C)
l = length of flow path (m)
q=
Q
=
kA
t
l
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Heat transfer
coefficients
Thermal conductivities
material k (W/mC) material k (W/mC)
still air 0.026 plywood 0.115
polyisocyanurate 0.027 softwood lumber 0.12
polyurethane spray foam 0.022 - 0.037 gypsum wallboard 0.16
extruded polystyrene 0.029 - 0.039 common brick 0.72
expanded polystyrene 0.037 window glass 1.00
mineral wool 0.038 concrete 0.77 - 1.32
fibreglass batts 0.042 stone 1.3 3.0
cellulose fibre 0.039 - 0.046 steel 45
straw bale 0.059 aluminum 220
vermiculite 0.066 copper 390
Thermal conductivity values (k) for some common construction materials:
Combining materials into assemblies
Construction assemblies are typically comprised of a number of
components made of different materials, requiring thedetermination of combined thermal conductances or resistances.
In general, components are either combined in a manner whichresults in series heat flow, or inparallel heat flow.
series heat flow parallel heat flow
These heat flows must be equal, or q = q1 = q2Solving each equation for t:
Heat flow in series
When materials are in series, note that heat flow through each component must
be the same, though the temperature drop across each component is different. If the
intermediate temperature is t, heat flow through each component is as follows:
Note that in series heat flow, the total
thermal resistance is simply the sum
of the individual resistance values.solving for heat flow: q=
1
R1+R
2
At
q1=
1
R1A(t
t1)q2=
1
R2A(t
t2 )and
t=R
1
A
At
1
R1
+q
t=R
2
A
At
2
R2
+q
and
t1+
R1q
A
= t2
R2q
A
equating:
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The basic equation for series heat flow is:
where:
q = rate of heat flow (J/s or W)
R = 1/C thermal resistance (C m2/W)
A = cross-sectional area of flow path (m2)
t = temperature difference (C)
q=1
R1+R
2+R
3
At=
1
Rtotal
At
Calculate the overall thermal conductance (and resistance) of a wall assembly
comprised of 200 mm solid concrete, 38 mm thick polystyrene insulation and
12.5 mm thick GWB.
Example: Series heat transmission
Since the heat flow through each element of the assembly is equal:
Rtotal
= R1
+ R2
+ R3
= 1/Ctotal
thermal properties conductances (C) resistances (RSI)
200 mm thick concrete 6.6 W/m2C 0.15 m2C/W38 mm thick polyst yrene in sulat ion 0 .763 W/ m2C 1.31 m2C/W
12.5 thick gypsum wallboard 12.5 W/m2C 0.08 m2C/W
totals: 1.54 m2C/W
R or RSI of the assembly = 1.54 m2C/W (8.74 R-value in Imperial units)
Ctotal
= 1/(1.54 m2C/W) = 0.649 W/m2C
example continued
0.649 W/m2C
Thermal resistance ofair films & cavities Air films
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Air films
configurationstill
air
moving air
(12 km/hr)
moving air
(24 km/hr)
horizontal surface
heat flow up9.3 23 34
45 surface
heat flow up9.1 23 34
vertical surface
heat flow horizontal8.3 23 34
45 surface
heat flow down7.5 23 34
horizontal surface
heat flow down6.1 23 34
Air cavities or spaces
configuration19 mm
airspace
92 mm
airspace
horizontal position
heat flow up6.8 6.3
45 position
heat flow up6.3 6.1
vertical position
horizontal heat flow5.8 5.8
45 position
heat flow down5.7 5.3
horizontal position
heat flow down5.5 4.6
Air cavities or spaces Multiple glazed assemblies
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Example: Thermal Gradients through Assembly
Assuming the following conditions:
in winter, an interior temperature of 20C
and an exterior temperature of -10C (with
12 km/hr wind),
and in summer, the same interior
temperature, but an exterior temperature of30C with the same wind, calculate and plot
the thermal gradient through the insulated
cavity of the following wood frame wall
assembly:
12.7 mm gypsum wallboard, painted
6 mil polyethylene air/vapour barrier
140mm wood studs @ 400mm o/c
with fibreglass batt insulation
15.9 mm plywood sheathing
60 minute building paper
19 mm stucco with 2 mm acrylic finish
Summer & winterthermal gradients
30C
20C
10C
0C
-10C
Impact of insulation on thermal gradients
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Impact of insulation on thermal gradients Impact of insulation on thermal gradients
Impact of insulation on thermal gradients Impact of insulation on thermal gradients
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When materials are in parallel, note that thetemperature difference across eachcomponent is the same, and the total heatflow is the sum of the individual heat flowsthrough each component, therefore thethermal conductances times their respectiveareas are additive:
Heat Flow in Parallel
q= q1+q
2
q= (C1A
1+C
2A
2)t=
A1
R1
+A
2
R2
t
q=CAt=1
RAt
Thermal Bridging of Wood Studs
