Section 6: Heat Transfer - Part II

ENV-2D02 (2006):Energy Conservation – power point versions of lectures. Will be available on WEB later in Week

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 Values7 Heat Losses from Buildings – Effect of Built Form: Dynamic Effects

8 Introduction to Energy Management - order may be swapped with section 9

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: Concluding Remarks

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

• But heat is also lost by:• Convection• Radiationfrom the wall surfaces

k

dResistance to Heat Flow

Estimated by

where

k is conductivity of material

d is thickness

brick

plaster

External boundary layer

Internal boundary layer

k

d

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 = 0.123 internal surfaceRext = 0.08 sheltered external surface 0.06 normal 0.03 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 U-values 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

Fig. 6.6 Heat flow through wall of 1950's construction

conductivity of brick = 1.0 Wm-1 oC-1 conductivity of plaster = 0.7 Wm-1 oC-1

External surface resistance

brick brick

cavityplaster

Internal surface resistance

6.7 Derivation of 'U'-values for 3 types of wall

• Resistance =

where k = conductivity d = length of heat flow paths (thickness in this case)

• resistance of brick = = 0.11 m2 oC W-1

• resistance of plaster = = 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

===========

k

d

0.1

11.0

0 013

0 7

.

.

6.7 Derivation of 'U'-values for 3 types of wall

Total Resistance = 0.598 m2 oC W-1

• since U =

• 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].

R

1

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 = 0.598 + 0.76 - 0.11 = 1.248 m2 oC W-1 • so U-value = = 0.80 Wm-2 oC-1

i.e. a 50% saving in the heat lost through the walls of

a house.

R

1

blockbrick

cavityplaster

brick

Example 3• As example 2 except that cavity is filled with

insulation

• conductivity of insulation = 0.04 Wm-1 oC-1

• resistance of cavity fill = 0.05 /0.04

= 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 = = 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].

R

1

cavity

blockbrickplaster

Filled cavity

k

d

R

1

Example 4: Single Glazing• conductivity of glass = 1 Wm-2 oC-1

• i.e. resistance = 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

• Assumes internal temperature is 20 oC and external temperature is 0oC

• Temperature gradient is highest in insulating materials

• Greatest in polystyrene layer

• External Surface Temperature = 1.17

• If this falls below 0oC – danger of ice forming and causing bricks to crumble.

0

5

10

15

20

0 100 200 300 400 500

Distance (mm)Te

mpe

ratu

re (d

eg C

)

Without polystyrene, surface temperature would be around 1 0C lower

And PMV would be approx -0.15 lower

Example 6: Pitched RoofComponent

Resistance

( m2 oC W -1)

internal surface resistance 0.11plasterboard 0.06fibre glass (25mm) 0.72fibre glass (50mm) 1.43fibre glass (100mm) 2.86fibre glass (150mm) 4.29air-space to underside of felt 0.18felt (4mm) 0.11air-space between felt and tiles 0.12tiles 0.04external surface resistance) 0.04

Heat Flow:

Internal surface resistance > plasterboard > Loft space >

felt > Felt – tile airspace > tiles > External surface resistance >

Example 6: Pitched Roof

Resistance to heat flow

Vertical m2 oC W-1

Internal surface resistance = 0.11

Plasterboard = 0.06

Loft space = 0.18

Total vertical = 0.35

Inclined

Felt = 0.11

Air space felt – tiles = 0.12

Tiles = 0.04

External surface = 0.04

Total Inclined = 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

A

Doubling insulation does not half heat loss

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

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

Example 7 Double Glazing

• U values 3mm glass =

• 4mm glass =

• little difference irrespective of what thickness of glass is used.

• Double glazing: • U value (3mm glass) =

• Note the U value depends on the thickness of the air-space, which is optimum at about 18-20mm.

m2 0C W -1

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 0.18

CmW52.5055.0003.0123.0

1 02

CmW49.5055.0004.0123.0

1 02

CmW75.2055.0003.018.0003.0123.0

1 02

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

• In example, insulation occupies 8/9th of space (400/450)

• Heat flows are in parallel so use formula

393.09

8.

150.0

04.0

9

1.

100.0

14.0

400 mm

100 mm

50 mm

Insulation (150 mm thick)

joists

total

timber

timber

timbertimber

total

insulation

insulation

insulationinsulation A

A.

k

dr

A

A.

k

dr

insulationtimbereffective rrR

111

i.e. R = 2.45 cf 3.75 if bridging is ignored

• What is effective resistance of joists and insulation?

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