ME100 Lecture Notes – Introduction to Heat Transfer
S. Lambert, Fall 2005
Objective:
We need to calculate the heat loss for a house, Figure 1.
Q, heat loss
Tin
Tout
Figure 1: Schematic of heat loss model from a house.
Background:
In general, there are several types of heat transfer, including conduction, convection, and
radiation. Conduction is the process we will focus on. It is the transfer of heat energy
through a material, driven by a temperature gradient. Convection involves the transfer of
energy by movement of a fluid, such as air. Radiation involves the transfer of heat energy
through disorganized electromagnetic propagation. All forms are relevant to our
application. However, we can make a very reasonable approximation by considering
conduction and convection only, and by treating both using the same form of equation.
Conduction:
Empirical evidence (testing) suggests that heat transferred by conduction depends on the
temperature difference, material properties, and geometry. The equation we will use is:
 k 
Q  A  T
 x 
where Q is the rate of heat transfer through the material, A is the area through which the
heat is flowing, k is the thermal conductivity of the material, x is the distance through
which the heat must travel, and T is the temperature difference across the material. This
situation is illustrated in Figure 2.
A = hl
h
Q
Q
Tout
Tin
x
l
Figure 2: Heat flow through a wall.
The old-fashioned units for energy are British Thermal Units, or Btu, so that energy flow
rate, Q, can be expressed as Btu/hr. The corresponding unit for temperature difference is
Fo, distance is ft, area is ft2, and thermal conductivity is Btu/(Fohrft).
It is common to combine the thermal conductivity and the thickness of the insulating
material as a thermal conductivity or resistance – thermal resistance is the inverse of
thermal conductivity. In imperial units, the thermal resistance, R-value, is the thickness in
ft divided by the thermal conductivity:
x
R
k
o 2
The units of R are F ft hr/Btu. The heat transfer equation can then be expressed as:
 A
Q    T
R
The resistance values are often tabulated per inch of thickness – simply multiply this
value by the thickness to get the total R-value. Table 1 provides the R-value per inch for
some common insulating materials. These have been taken from Appendix B of
“Modern Hydronic Heating for residential and light commercial buildings”, by John
Siegenthaler, Delmar Publishers, 1995.
Material
Fiberglass insulation
Blown cellulose fibre insulation
Foam in place urethane insulation
Extruded ploystyrene panel insulation
Concrete
Brick
Softwood
Plywood
Drywall
Vinyl siding
R-value
Foft2hr/Btu per inch
3.17 – 3.5
3.1 – 3.7
5.6 – 6.3
5.4
0.10
0.2 – 0.4
0.9 – 1.1
1.24
0.9
0.61 (for all thicknesses)
Table 1: Insulating values (R-value) for common materials (ref: Siegenthaler, 1995).
The total resistance value for a composite structure such as a wall is obtained by adding
the resistance contributions from each layer. The resistances are combined in series since
the heat flow, Q, is the same through each section of the wall. For example, for a typical
wall consisting of 0.5” drywall, 3.5” fiberglass insulation, 0.5” plywood sheathing, and
vinyl siding, the total R-value based on average values would be:
 3.17  3.5 
R  0.5(0.9)  3.5
  0.5(1.24)  0.61  13.35
2


Note that the wall will contain other important elements such as an air infiltration barrier
(towards the outside of the house) and a vapour barrier (towards the inside – warm side),
but these do not contribute significantly to the insulating value. Missing from this
estimate is any consideration of the resistance to heat transfer due to the air film next to
interior of exterior surfaces. They can be approximated using recommended R-values as
given in Table 2, also from Siegenthaler, 1995.
Film Description
Inside Air Films:
Horizontal surfaces with upward heat flow (ceiling)
Horizontal surfaces with downward heat flow (floor)
Vertical surfaces with horizontal heat flow (wall)
45-degree sloped surfaces with upward heat flow
Outside Air Films:
7.5 mph wind on any surface (summer conditions)
15 mph wind on any surface (winter conditions)
R-Value
o
Fft2hr/Btu
0.61
0.92
0.68
0.62
0.25
0.17
Table 2: Equivalent thermal resistance of air films (ref: Siegenthaler, 1995).
For our example, we would add 0.68 for the interior wall and 0.17 for the outside wall
(winter), for a total R-value of 14.20.
In our house model, we need to evaluate the total heat loss from all surfaces. These
surfaces (walls, ceilings, windows, doors) are acting in parallel, so we can determine the
heat loss through each surface and add these values to get the total.
Heat Loss through Walls:
The original walls consist of approximately 20” of stone/concrete on the outside, a 1.5”
air gap, 3/8” of wood lath and 1/2” of plaster. The R-value can thus be approximated as:
Material
Outside Air Film
Stone/Concrete
Air Gap
Wood Lath
Plaster
Inside Air Film
Total
Thickness R/in R
n/a 0.17
0.17
20
0.1
2.00
n/a 0.68
1.36
0.375
1
0.38
0.5
0.9
0.45
n/a 0.68
0.68
5.04
Heat Loss through Ceiling:
Similarly, the R-value for the ceilings can be approximated as:
Material
Thickness
Outside Air Film
n/a
Roof Sheathing
1
Cellulose Insulation
4
Wood Lath
0.375
Plaster
0.5
Inside Air Film
n/a
Total
R/in R
0.17
1
3.4
1
0.9
0.62
0.17
1.00
13.60
0.38
0.45
0.62
16.22
Heat Loss Through Windows:
The house has single pane windows with a separate storm window, for a total of two
panes. The glass itself provides little insulation value, so the effective R-value for each
window can be approximated as four air films in series, one on the outside and three
interior, for a total R-value of 0.17 + 3(0.68) = 2.21.
Heat Loss Through Doors:
The original doors were solid wood, approximately 1.5” thick, so the effective R-value
can be approximated as 1.5(1.0) = 1.5.
Heat Loss Through Basement:
In addition to any insulation on the basement walls, the ground surrounding the basement
will provide some insulation value, which will vary with the depth. Siegenthaler (1995)
suggests the following formulae to estimate the R-value of a basement wall. For the first
2’ below grade:
R = Rfoundation wall + Radded
That is, the first 2’ of wall below grade should not take advantage of the insulating effect
of the ground. For my house, there is no added insulation, so the effective R-value can be
estimated from the resistance due to 20” of concrete foundation (R = 2) and one interior
air film for a total R-value of 2.68. Note also that I estimate that the foundation wall
extends approximately 1’ above ground level; this is reflected in the dimensions provided
in the spreadsheet.
For the next 5’ below grade:
R = 8 + 1.126Radded
Here, no advantage is taken of the foundation wall, although we could perhaps add the
effect of the interior air film. For our case, there is no added insulation, so the effective Rvalue of the wall from 2’ to 7’ below grade is 8.68.
Air Infiltration:
Heat can also be carried out by air moving through the building envelope. The amount of
heat loss will depend on the quantity of air leakage, which is dependent on the quality of
sealing and the size of the house. Siegenthaler (1995) recommends the following values
for the number of air exchanges per hour:
Floor area (ft2):
“Best” Quality
“Average” Quality
“Poor” Quality
900 or less
0.4
1.2
2.2
900 – 1500
0.4
1.0
1.6
1500 – 2100
0.3
0.8
1.2
over 2100
0.3
0.7
1.0
Table 3: Air change rates, N, in exchanges per hour (ref: Siegenthaler, 1995).
In addition to the above factors, add 0.1, 0.2, or 0.6 for each fireplace of best, average, or
poor quality, respectively. The heat loss due to air infiltration, Qi, can then be estimated
using the following equation:
Qi = 0.018NVT
where 0.018 represents the heat capacity of air in Btu/ft3/Fo, N is the number of air
changes per hour, V is the interior volume of the heated space in ft3, and T is the
temperature difference in Fo, i.e., the inside temperature minus the outside temperature.