Section 6: Heat Transfer - Part II
1
Introduction to organisation of course and Field Course
2
Revision of Simple Economic Analysis
3
Thermal comfort: physical and physiological aspects. What temperatures do we actually need?
4
Energy use by sector:
5
Energy GDP relationships: Energy Balance Tables.
6
HEAT TRANSFER: U Values
7
Heat Losses from Buildings – Effect of Built Form: Dynamic Effects
8
Introduction to Energy Management
9
Energy Management Continued: Energy Targets: Building Regulations
10
Electricity Conservation
11
Thermodynamics
12
Combined Heat and Power
13
The Heat Pump
14
Energy Conservation Measures at UEA
15
Energy Analysis:
- order may be swapped with section 9
Concluding Remarks
ENV-2D02 (2006):Energy Conservation – power point versions of lectures. Will
be available on WEB later in Week
Main Objective of the Lecture
• To apply the basic ideas of Heat Transfer from
previous lecture to estimating the thermal
properties of typical building materials.
• To provide the tools to allow Heat Loss
Estimations from buildings to be made [covered
in next lecture ]
• >> hence to estimate potential savings.
Summary of Last Lecture – Key Point
• Heat is lost from a building by:
• Conduction
Resistance to Heat Flow
Estimated by
where
• But heat is also lost by:
• Convection
• Radiation
from the wall surfaces
d
d
k
k
k is conductivity of material
d is thickness
plaster
External
boundary
layer
brick
Internal
boundary
layer
Surface Resistance
• Analysis of heat flow by convection and radiation is more complex.
• Can be approximated in most situations for buildings by additional
resistance layers.
• Cannot be used if surface temperature are substantially different
from surrounds – e.g. a hot water pipe/radiator.
Consequences of Boundary Layers
• Surface Temperature of window (on room side) is BELOW room
temperature - for single glazing it will be about 70C if outside
temperature is 0oC and internal temperature is 20oC
• External surface temperature will be above surrounding air.
• Internal Surface Temperatures are important as they affect Mean
Radiant Temperature and hence Thermal Comfort.
• External Surface Temperatures can affect weathering properties of
bricks.
6.5 Internal and External Surface Resistances
Typical values for surface resistances (m2 oC W-1):
Vertical Heat Flow
0.11 for upward flow through floors/roof
0.15 for downward flow through floors
Horizontal Heat Flow
Rint =
Rext =
0.123
0.08
0.06
0.03
internal surface
sheltered external surface
normal
severe
Note:
- the orientation of windows is important in heat loss calculations as
the external resistance is a significant proportion of the total
resistance;
6.6 Resistances of Air-Spaces
- thermal conductivity of air-spaces is very small, and heat transfer is
mostly by radiation and convection,
- values are given in tables, but can be divided generally into two
categories:-
• unventilated air spaces (or low ventilation) –
• resistance is about 0.18. Examples:• air-space in modern cavity walls,
• air-space in double glazing,
• air-space between ceiling & underside of felt (post-war
houses).
• ventilated air spaces –
• the resistance is about 0.12. Examples:• older cavity walls and air-space between ceiling and
underside of tiles (pre-war houses).
6.7 Derivation of 'U'-values for 3 types of wall
•
•
•
•
•
•
•
•
•
•
Many standard constructions have Uvalues in Tables
Non-standard constructions do not –
including many new types.
Example 1
6 components:1) external surface layer
2) outer brick layer
3) cavity
4) inner brick layer
5) plaster
6) internal surface layer
Internal
surface
resistance
External
surface
resistance
cavity
brick
Fig. 6.6 Heat flow through wall of 1950's construction
conductivity of brick =
conductivity of plaster =
1.0 Wm-1 oC-1
0.7 Wm-1 oC-1
plaster
brick
6.7 Derivation of 'U'-values for 3 types of wall
d
•
Resistance
=
k
where k = conductivity
d = length of heat flow paths (thickness in this case)
•
•
resistance of brick =
resistance of plaster =
0.11
1.0
0 . 013
0. 7
= 0.11 m2 oC W-1
= 0.02 m2 oC W-1
•
Effective resistances of air spaces are:internal boundary 0.123 m2 oC W-1
external boundary 0.055 m2 oC W-1
air-cavity
0.18 m2 oC W-1
•
•
So total resistance
= 0.055 + 0.11 + 0.18 + 0.11 + 0.02 + 0.123
•
= 0.598 m2 oC W-1
===========
6.7 Derivation of 'U'-values for 3 types of wall
Total Resistance = 0.598 m2 oC W-1
• since U =
1
R
• U = 1.67 W m-2 oC
-1
• Note: that the external resistance is relatively small
< 10% of total resistance
U value for walls varies little with exposure normally [only a
few per cent at most].
Example 2
•
As example 1 except that the inner brick leaf is replaced
by an aerated block wall i.e. construction used from
mid-1960's.
•
•
conductivity for aerated block = 0.14 Wm-1 oC-1
and resistance of such a block = 0.76 m2 oC W-1
•
replaces the inner brick of original wall,
•
new resistance
•
so U-value
= 0.598 + 0.76 - 0.11
= 1.248 m2 oC W-1
=
1
R
= 0.80 Wm-2 oC-1
i.e. a 50% saving in the heat lost through the walls of a
house.
cavity
brick
plaster
brick
block
Example 3
•
As example 2 except that cavity is filled with
insulation
•
conductivity of insulation
•
resistance of cavity fill
= 0.04 Wm-1 oC-1
=
d 0.05 /0.04
1
R
k
= 1.25 m2 oC W-1
•
•
replaces the resistance of 0.18 from the air-cavity
New resistance
=
1.248 - 0.18 + 1.25
=
2.318 m2 oC W-1
•
and U-value =
1
R
= 0.43 Wm-2 oC-1
•
i.e. approximately half of the value in example 2 and
one quarter of the value in example 1.
•
[the U-value for a wall with two brick leaves and
cavity insulation is 0.60 Wm-2 oC-1].
plaster
block
brick
Filled
cavity
cavity
Example 4: Single Glazing
•
conductivity of glass
•
i.e. resistance
= 1 Wm-2 oC-1
= 0.003 m2 oC W-1 (for 3mm glass)
– internal surface resistance = 0.123
– external surface resistance = 0.055
•
Thus total resistance
= 0.123 + 0.003 + 0.055
•
•
= 0.181 m2 oC W-1
and U-value
=
5.5 Wm-2 oC-1
Note
•
resistance of glass makes very little contribution to the overall
resistance
•
if the external resistance changes (from exposure) then the U-value will
also be affected significantly.
•
[Compare this with the situation for the walls (see note to example 1)].
Temperature Profile through a wall
• Example 2 with polystyrene
layer on inside
• Temperature gradient is highest
in insulating materials
Temperature (deg C)
• Assumes internal temperature is
20 oC and external temperature
is 0oC
20
15
10
5
• Greatest in polystyrene layer
• External Surface Temperature =
1.17
0
0
100
200
300
400
Distance (mm)
500
• If this falls below 0oC – danger
of ice forming and causing
Without polystyrene, surface
bricks to crumble.
temperature would be around 1 0C lower
And PMV would be approx -0.15 lower
Example 6:
Pitched Roof
Resistance
( m2 oC W -1)
Component
internal surface resistance
plasterboard
fibre glass (25mm)
fibre glass (50mm)
fibre glass (100mm)
fibre glass (150mm)
air-space to underside of felt
felt (4mm)
air-space between felt and tiles
tiles
external surface resistance)
0.11
0.06
0.72
1.43
2.86
4.29
0.18
0.11
0.12
0.04
0.04
Heat Flow:
Internal surface resistance > plasterboard >
felt >
Felt – tile airspace >
tiles >
Loft space >
External surface resistance >
Example 6:
Pitched Roof
Resistance to heat flow
Vertical
Internal surface resistance
Plasterboard
Loft space
Total vertical
Inclined
Felt
Air space felt – tiles
Tiles
External surface
Total Inclined
m2 oC W-1
= 0.11
= 0.06
= 0.18
= 0.35
= 0.11
= 0.12
= 0.04
= 0.04
= 0.31
Total Resistance (if A = 45o)
=
0.35 + 0.31 cos 45 = 0.57
A
U – value = 1 / R
= 1.75 W m-2 oC-1
Pre-war houses do not have felt
Some houses in extreme weather
areas have boards instead of felt
Example 6:
Pitched Roof
Simple Way to examine effects of insulation
U value without insulation = 1.75 W m-2 oC-1
Resistance = 1 / 1.75 = 0.57 m2 oC-1
Add 50 mm of insulation conductivity 0.04
Additional resistance = 0.05 / 0.035 = 1.43
A
New total resistance = 1.43 + 0. 57 = 2.00
New U-Value = 1 / 2.0 = 0.5 W m-2 oC-1
With 100 mm
New total resistance = 2.86 + 0. 57 = 3.43
New U-Value = 1 / 3.43 = 0.29 W m-2 oC-1
With 150 mm
New U-Value = 0.21 W m-2 oC-1
Doubling insulation does not half
heat loss
Example 7 Double Glazing
m2 0 C W
3mm single pane - resistance
0.003
4mm single pane - resistance
0.004
internal surface resistance
0.123
external surface resistance
0.055
air-space resistance
• U values 3mm glass =
•
4mm glass =
-1
0.18
1
0.123 0.003 0.055
1
0.123 0.004 0.055
 5.52Wm
2 0
 5.49Wm
C
2 0
C
• little difference irrespective of what thickness of glass is used.
• Double glazing:
• U value (3mm glass) =
1
0.123 0.0030.18 0.003 0.055
 2.75Wm
2 0
C
• Note the U value depends on the thickness of the air-space, which is
optimum at about 18-20mm.
6.9 Problems associated with thermal bridging
Thermal bridging leads to:
• cold spots on the internal surfaces
• condensation
• discolouration where the presence of bridges can be
seen.
A thermal imaging camera can be used to identify such
bridges,
but these are often expensive.
Insulating the loft in UK houses
• Place fibre glass between the joists.
• As the thickness of insulation increases problems of
thermal bridging appear.
• timber joists create a thermal bridge
Thermal bridging – an example
• What is effective resistance of joists and insulation?
d insulation A insulation
rinsulation 
.
k insulation
A total
rtimber 
d timber
k timber
• In example, insulation occupies
8/9th of space (400/450)
• Heat flows are in parallel so use
formula
1
Reffective

1

0.14
rtimber
.
1
0.100 9

.
A timber
A total
Insulation (150 mm thick)
400 mm
1

100 mm
rinsulation
0.04
.
8
0.150 9
joists
 0.393
50 mm
i.e. R = 2.45 cf 3.75 if
bridging is ignored
Next Lecture
Heat Losses from a House
• Need to work out U-values and area for:
–
–
–
–
Walls
Windows
Roof
Floor
• Other sources of heat Loss
– Ventilation
Remember:
• you cannot eliminate heat losses – you can only
reduce them.
• Heat lost must be replaced by heating device