-EN-ISO-6946-1997

Final approval on behalf of BRE :  Final approval on behalf of BRE :  Final approval on behalf of BRE:  Final approval on behalf of BRE:  Signed

Signed ______________________________ Date __ Date ________________________________ Mrs H J Cuckow, Director, BRE East Kilbride Mrs H J Cuckow, Director, BRE East Kilbride

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 © Building Research Establishment Ltd 1999  © Building Research Establishment Ltd 1999

Examples of U-value calculations

Examples of U-value calculations

using BS EN ISO 6946:1997

using BS EN ISO 6946:1997

Prepared for: Prepared for: DETR/BR DETR/BR By: By: S M Doran and L

S M Doran and L KosminaKosmina BRE East Kilbride BRE East Kilbride De

Deccemembeber r 11999999 ReRepoport rt No No 7878121299 (revised June 2000)

Examples of U-value calculations using BS EN ISO

Examples of U-value calculations using BS EN ISO 6946:19976946:1997 S M Doran and L Kosmina, BRE East Kilbride 

S M Doran and L Kosmina, BRE East Kilbride  December 1999 

December 1999 

This document illustrates the procedures given in BS

This document illustrates the procedures given in BS EN ISO 6946:1997 EN ISO 6946:1997 11 for for  calculating the U-va

calculating the U-value of opaque elements. lue of opaque elements. The procedures are expThe procedures are explained using lained using  examples of U-value calculations for some typical wall, roof

examples of U-value calculations for some typical wall, roof and floor designs which and floor designs which  contain repeating thermal bridges.

contain repeating thermal bridges.

Contents Contents 1

1.. IInnttrroodduuccttiioonn 11

2

2. . OOuuttlliinne e oof f tthhe e pprroocceedduurree 22 3.

3. CCaavivity ty wwaall ll wiwith th liligghthtwweeigight ht mamasosonrnry y leleaf af anand d iinsnsululaateted d ddry ry llinininingg 33 4

4.. TTiimmbbeerr ffrraammeedd wwaallll 88 5

5.. IInnssuullaatteed d ccaavviitty y wwaalll l wwiitth h mmeettaal l wwaalll l ttiieess 1133 6

6.. WWiidde e ccaavviitty y wwaalll l wwiitth h vveerrttiiccaal l ttwwiisst t ttiieess 1188 7

7. . PPiittcchheed d rroooof f wwiitth h iinnssuullaattiioon n bbeettwweeeen n aannd d oovveer r tthhe e jjooiissttss 2211 8

8. . RRoooom m iin n rroooof f ccoonnssttrruuccttiioonn 2244 9

9. . RRoooom m iin n rroooof f ccoonnssttrruuccttiioon n wwiitth h lliimmiitteed d rraafftteer r ddeepptthh 2277 1

100. F. Flloooor r aabboovve e uunnhheeaatteed d ssppaaccee 3311 1

111. . SSuussppeennddeed d bbeeaam m aannd d bblloocck k fflloooorr 3344 1

122. . SSuussppeennddeed d ttiimmbbeer r ggrroouunnd d fflloooorr 3388 A

Appppeennddiixx:: DaDattaa ttaabblleess 4411 R

Reeffeerreenncceess 4455

1. Introduction 1. Introduction

For building elements which contain repeating thermal bridges, such as timber joists For building elements which contain repeating thermal bridges, such as timber joists between insulation in a roof, or

between insulation in a roof, or mortar joints around lightweight blockwork in a wall,mortar joints around lightweight blockwork in a wall, the effect of

the effect of thermal bridges should be taken into account when thermal bridges should be taken into account when calculating thecalculating the U-value.

U-value. At present, Building ReguAt present, Building Regulations specify that U-valulations specify that U-values should be calculatedes should be calculated using the Proportional Area Method, which is described in

using the Proportional Area Method, which is described in the CIBSE Guide, Sectionthe CIBSE Guide, Section A3

A322. . Future regulationsFuture regulations, however, are likely to be based upo, however, are likely to be based upon the method forn the method for

calculating U-values defined in BS EN ISO 6946:1997, which includes the Combined calculating U-values defined in BS EN ISO 6946:1997, which includes the Combined Method for repeating thermal bridges and correction procedures for the

Method for repeating thermal bridges and correction procedures for the effects ofeffects of metal fixings, air gaps and unh

metal fixings, air gaps and unheated spaces. eated spaces. This paper illustrates the use ofThis paper illustrates the use of BS EN

BS EN ISO 6946:1997 for some typical wall, roof and floor designs.ISO 6946:1997 for some typical wall, roof and floor designs.

Thermal conductivity values for common building materials can be obtained from the Thermal conductivity values for common building materials can be obtained from the CIBSE Guide Section A3, 1999 Edition

CIBSE Guide Section A3, 1999 Edition (especially for masonry) or from prEN 12524(especially for masonry) or from prEN 1252433.. For specific insulation products, however, data should be obtained from

For specific insulation products, however, data should be obtained from manufacturers’ de

manufacturers’ declared values. clared values. A table is provided at the end of this documenA table is provided at the end of this documentt giving typical conductivities for some common building materials.

Examples of U-value calculations using BS EN ISO

Examples of U-value calculations using BS EN ISO 6946:19976946:1997 S M Doran and L Kosmina, BRE East Kilbride 

S M Doran and L Kosmina, BRE East Kilbride  December 1999 

December 1999 

This document illustrates the procedures given in BS

This document illustrates the procedures given in BS EN ISO 6946:1997 EN ISO 6946:1997 11 for for  calculating the U-va

calculating the U-value of opaque elements. lue of opaque elements. The procedures are expThe procedures are explained using lained using  examples of U-value calculations for some typical wall, roof

examples of U-value calculations for some typical wall, roof and floor designs which and floor designs which  contain repeating thermal bridges.

contain repeating thermal bridges.

Contents Contents 1

1.. IInnttrroodduuccttiioonn 11

2

2. . OOuuttlliinne e oof f tthhe e pprroocceedduurree 22 3.

3. CCaavivity ty wwaall ll wiwith th liligghthtwweeigight ht mamasosonrnry y leleaf af anand d iinsnsululaateted d ddry ry llinininingg 33 4

4.. TTiimmbbeerr ffrraammeedd wwaallll 88 5

5.. IInnssuullaatteed d ccaavviitty y wwaalll l wwiitth h mmeettaal l wwaalll l ttiieess 1133 6

6.. WWiidde e ccaavviitty y wwaalll l wwiitth h vveerrttiiccaal l ttwwiisst t ttiieess 1188 7

7. . PPiittcchheed d rroooof f wwiitth h iinnssuullaattiioon n bbeettwweeeen n aannd d oovveer r tthhe e jjooiissttss 2211 8

8. . RRoooom m iin n rroooof f ccoonnssttrruuccttiioonn 2244 9

9. . RRoooom m iin n rroooof f ccoonnssttrruuccttiioon n wwiitth h lliimmiitteed d rraafftteer r ddeepptthh 2277 1

100. F. Flloooor r aabboovve e uunnhheeaatteed d ssppaaccee 3311 1

111. . SSuussppeennddeed d bbeeaam m aannd d bblloocck k fflloooorr 3344 1

122. . SSuussppeennddeed d ttiimmbbeer r ggrroouunnd d fflloooorr 3388 A

Appppeennddiixx:: DaDattaa ttaabblleess 4411 R

Reeffeerreenncceess 4455

1. Introduction 1. Introduction

For building elements which contain repeating thermal bridges, such as timber joists For building elements which contain repeating thermal bridges, such as timber joists between insulation in a roof, or

between insulation in a roof, or mortar joints around lightweight blockwork in a wall,mortar joints around lightweight blockwork in a wall, the effect of

the effect of thermal bridges should be taken into account when thermal bridges should be taken into account when calculating thecalculating the U-value.

U-value. At present, Building ReguAt present, Building Regulations specify that U-valulations specify that U-values should be calculatedes should be calculated using the Proportional Area Method, which is described in

using the Proportional Area Method, which is described in the CIBSE Guide, Sectionthe CIBSE Guide, Section A3

A322. . Future regulationsFuture regulations, however, are likely to be based upo, however, are likely to be based upon the method forn the method for

calculating U-values defined in BS EN ISO 6946:1997, which includes the Combined calculating U-values defined in BS EN ISO 6946:1997, which includes the Combined Method for repeating thermal bridges and correction procedures for the

Method for repeating thermal bridges and correction procedures for the effects ofeffects of metal fixings, air gaps and unh

metal fixings, air gaps and unheated spaces. eated spaces. This paper illustrates the use ofThis paper illustrates the use of BS EN

BS EN ISO 6946:1997 for some typical wall, roof and floor designs.ISO 6946:1997 for some typical wall, roof and floor designs.

Thermal conductivity values for common building materials can be obtained from the Thermal conductivity values for common building materials can be obtained from the CIBSE Guide Section A3, 1999 Edition

CIBSE Guide Section A3, 1999 Edition (especially for masonry) or from prEN 12524(especially for masonry) or from prEN 1252433.. For specific insulation products, however, data should be obtained from

For specific insulation products, however, data should be obtained from manufacturers’ de

manufacturers’ declared values. clared values. A table is provided at the end of this documenA table is provided at the end of this documentt giving typical conductivities for some common building materials.

2.

2. Outline of Outline of the prthe procedureocedure The following is an outline of

The following is an outline of the calculation procedure:the calculation procedure: 1.

1. Calculate Calculate the uthe upper pper resistance resistance limit limit (R(Rupperupper) by combining in parallel the total) by combining in parallel the total

resistances of all possible heat-flow paths (i.e. sections)

resistances of all possible heat-flow paths (i.e. sections) through the buildingthrough the building element.

element. 2.

2. Calculate Calculate the the lower lower resistance resistance limit limit (R(Rlowerlower) by combining in parallel the) by combining in parallel the

resistances of the heat flow paths of each layer separately and then summing the resistances of the heat flow paths of each layer separately and then summing the resistances of all layers of

resistances of all layers of the building element.the building element. 3.

3. Calculate Calculate the the total total thermal thermal resistance resistance (R(RTT) from) from

2 2 R R R R R

R_TT

==

upperupper

++

lowerlower

4.

4. Calculate, Calculate, where awhere appropriate, ppropriate, corrections corrections for air for air gaps (gaps (

∆∆

UUgg) and mechanical) and mechanical

fasteners (

fasteners (

∆∆

UUff). Examples of corr). Examples of corrections for air gapections for air gaps are shown in ses are shown in sections 3, 4,ctions 3, 4,

10 and 12 and

10 and 12 and examples of corrections for mechanical fasteners are shown inexamples of corrections for mechanical fasteners are shown in sections 5, 6 and 9.

sections 5, 6 and 9. 5.

5. Calculate Calculate the the U-value U-value fromfrom U = (1 / R

U = (1 / RTT) +) +

∆∆

UUgg ++

∆∆

UUff

The standard permits

The standard permits

∆∆

UUggandand

∆∆

UUffto be omitted if, taken together, they amountto be omitted if, taken together, they amount

to less than 3% of the U-value.

3.

3. Cavity wall with ligCavity wall with lightweight masonrhtweight masonry leaf and insulay leaf and insulated dry-liningted dry-lining In this example

In this examplea)a)there are two bridged layers - there are two bridged layers - insulation bridged by timber andinsulation bridged by timber and example).

example). The construction consists of outer The construction consists of outer leaf brickwork, a clear leaf brickwork, a clear cavity, 100 mmcavity, 100 mm AAC bl

AAC blockwork, 38 ockwork, 38 89 mm 89 mm timber stutimber studs with ds with insulation insulation between between the studs the studs and onand onee sheet of 12.5 mm

sheet of 12.5 mm plasterboard.plasterboard.

The thickness of each

The thickness of each layer, together with the thermal conductivities of tlayer, together with the thermal conductivities of the materials,he materials, are shown

are shown below. below. The external and The external and internal surface resistances used are internal surface resistances used are thosethose two thermal conductivities are given for each layer to reflect the bridged part and the two thermal conductivities are given for each layer to reflect the bridged part and the bridging part in each cas

bridging part in each case. e. For each homogeneoFor each homogeneous layer and for each sectionus layer and for each section (expressed in metres) by the

(expressed in metres) by the thermal conductivity.thermal conductivity.

L Laayyeerr TThhiicckknneessss ((mmmm)) ccoonndduuccttiivviittyy (W/m·K) (W/m·K) resistance resistance (m²K/W) (m²K/W) -- --1

1 oouutteerr lleeaaff brriicb ckk 00..7777 00..113322 a aiirr ccaavviittyy 5500 00..118800 3 3((aa)) 110000 00..1111 3 3((bb)) mmoorrttaar r ((66..66%%)) 00..8888 00..111144 m miinneerraal l wwooool l ((9900..55%%)) 8899 22..334422 4 4((bb)) ((8899)) 00..1133 5

5 ppllaasstteerrbbooaarrdd 00..2255 00..005500

--

--a) a)

Due to requirements for sound insulation this wall construction may only be suitable for Due to requirements for sound insulation this wall construction may only be suitable for

102 mm brick (conductivity 0.77 W/m·K) 102 mm brick (conductivity 0.77 W/m·K)

50 mm air cavity (thermal resistance 0.18 m²K/W) 50 mm air cavity (thermal resistance 0.18 m²K/W) 100 mm AAC blocks (conductivity 0.11 W/m·K) 100 mm AAC blocks (conductivity 0.11 W/m·K) bridged by mortar (conductivity 0.88 W/m·K) bridged by mortar (conductivity 0.88 W/m·K)

mineral wool (conductivity 0.038 W/m·K) mineral wool (conductivity 0.038 W/m·K) between 38

between 38××89 mm timber studs89 mm timber studs

(conductivity 0.13 W/m·K) at 400 mm centres (conductivity 0.13 W/m·K) at 400 mm centres 12.5 mm plasterboard, 12.5 mm plasterboard, (conductivity 0.25 W/m·K) (conductivity 0.25 W/m·K) heat flow heat flow T Toottaal l tthhiicckknneessss 33554 4 mmmm U U--vvaalluuee 00..331 1 WW//mm··KK

Both the upper and the lower limits of thermal resistance are calculated by combining as illustrated below. The method of combining differs in the two cases.

Upper resistance limit

considered to consist of a number of thermal paths (or sections). In this example there are four sections (or paths) through which heat can pass. The upper limit of

upper, is given by 4 4 3 3 2 2 1 1 upper R F R F R F R F 1 R

+

+

+

=

where F1, F2, F3and F4 are the fractional areas of sections 1, 2, 3 and 4 respectively

and R1, R2, R3 and R4 are the corresponding total thermal resistances of the sections.

A conceptual illustration of the method of calculating the upper limit of resistance is shown

below:-Figure 3.2 : Conceptual illustration of how to calculate the upper limit of resistance 

Resistance through section containing AAC blocks and mineral wool

External surface resistance = 0.040

Resistance of bricks = 0.132

Resistance of air cavity = 0.180

Resistance of AAC blocks (93.4%) = 0.909 Resistance of mineral wool (90.5%) = 2.342

Resistance of plasterboard = 0.050

Internal surface resistance = 0.130

Total thermal resistance (R1) = 3.783 m²K/W

Fractional area F1= 0.845 (93.4%

×

90.5%) external surface internal surface 1 4(a) 2 1 5 3(b) F1 F2 3(a) 4(b) 5 F3 F4 3(b) 3(a) 4(a) 4(b) 5 5 1 1 2 2 2

Resistance through section containing mortar and mineral wool

External surface resistance = 0.040

Resistance of bricks = 0.132

Resistance of air cavity = 0.180

Resistance of mortar (6.6%) = 0.114 Resistance of mineral wool (90.5%) = 2.342

Resistance of plasterboard = 0.050

Internal surface resistance = 0.130

Total thermal resistance (R2) = 2.988 m²K/W

Fractional area F2= 0.060 (6.6%

×

90.5%)

Resistance through section containing AAC blocks and timber

External surface resistance = 0.040

Resistance of bricks = 0.132

Resistance of air cavity = 0.180

Resistance of AAC blocks (93.4%) = 0.909 Resistance of timber (9.5%) = 0.685

Resistance of plasterboard = 0.050

Internal surface resistance = 0.130

Total thermal resistance (R3) = 2.126 m²K/W

Fractional area F3 = 0.089 (93.4%

×

9.5%)

Resistance through section containing mortar and timber

External surface resistance = 0.040

Resistance of bricks = 0.132

Resistance of air cavity = 0.180

Resistance of mortar (6.6%) = 0.114 Resistance of timber (9.5%) = 0.685

Resistance of plasterboard = 0.050

Internal surface resistance = 0.130

Total thermal resistance (R4) = 1.331 m²K/W

Fractional area F4 = 0.006 (6.6%

×

9.5%)

Combining these resistances we obtain:

=

+

+

+

=

+

+

+

=

331 . 1 006 . 0 126 . 2 089 . 0 988 . 2 060 . 0 783 . 3 845 . 0 1 R F R F R F R F 1 R 4 4 3 3 2 2 1 1 upper 3.450 m²K/W.

Lower resistance limit

When calculating the lower limit of thermal resistance, the resistance of a bridged layer is determined by combining in parallel the resistances of the unbridged part and the bridged part of the layer. The resistances of all the layers in the element are then added together to give the lower limit of resistance.

shown below:

Figure 3.3 : Conceptual illustration of how to calculate the lower limit of resistance 

The resistance of the bridged layer consisting of AAC blocks and mortar is calculated using: mortar mortar blocks blocks first R F R F 1 R

+

=

and the resistance of the bridged layer consisting of insulation and timber is calculated using: timber timber insul insul second R F R F 1 R

+

=

The lower limit of resistance is then obtained by adding together the resistances of the layers:

External surface resistance = 0.040

Resistance of bricks = 0.132

Resistance of air cavity = 0.180

Resistance of first bridged layer

114 . 0 066 . 0 909 . 0 934 . 0 1 R_first

+

=

= 0.622

Resistance of second bridged layer

685 . 0 095 . 0 2.342 0.905 1 R_second

+

=

= 1.904 Resistance of plasterboard = 0.050

Internal surface resistance = 0.130

Total (Rlower) = 3.058 m²K/W

Total resistance of wall

The total resistance of the wall is the average of the upper and lower limit of resistances: 4(a) 2 1 5 internal surface 3(a) external surface 4(b) 3(b)

=

+

=

+

=

2 058 . 3 450 . 3 2 R R R_T upper lower 3.254 m²K/W

Correction for air gaps between the timber studs

Since the insulation is entirely between studs (ie. there is no continuous layer of insulation) a correction should be applied to the U-value in order to account for air gaps. The overall U-value of the wall should include a term

∆

Ug, where

∆

Ug =

∆

U’’

×

(RI/ RT)²

and where

∆

U’’ = 0.01 (referred to in BS EN ISO 6946 as correction level 1), RIis the

thermal resistance of the layer containing the gaps and RT is the total resistance of

the element.

∆

Ug is therefore

∆

Ug = 0.01

×

(1.904 / 3.254)² = 0.003 W/m²K

U-value of the wall

The effect of air gaps or mechanical fixings should be included in the U-value unless they lead to an adjustment in the U-value of less than 3%.

U = 1 / RT +

∆

Ug (if

∆

Ug is not less than 3% of 1 / RT)

U = 1 / RT (if

∆

Ug is less than 3% of 1 / RT)

In this case

∆

Ug = 0.003 W/m²K and 1 / RT = 0.307 W/m²K. Since

∆

Ug is less than

3% of (1 / RT),

U = 1 / 3.254 = 0.31 W/m²K.

Note

1. Since the cavity wall ties do not penetrate any insulation no correction need be applied to the U-value to take account of them.

2. In the above calculation it is assumed that the noggings (or dwangs) do not penetrate the whole of the insulation. If the noggings (or dwangs) do penetrate the whole of the insulation thickness they should be included as part of the timber percentage used in the calculation.

4. Timber framed wall

In this exampleb)there is a single bridged layer in the wall, involving insulation bridged by timber studs. The construction consists of outer leaf brickwork, a clear ventilated cavity, 19 mm plywood, 38

×

140 mm timber framing with 120 mm of insulation between the timbers and 2 sheets of plasterboard each 12.5 mm thick.

Figure 4.1 : Timber framed wall construction 

The thicknesses of each layer, together with the thermal conductivities of the materials in each layer, are shown below. The external and internal surface

resistances are taken from Table 1 of this document. Layer 4 is thermally bridged and two thermal conductivities are given for this layer, one for the unbridged part and one for the bridging part of the layer. For each homogeneous layer and for each section through a bridged layer, the thermal resistance is calculated by dividing the thickness (in metres) by the thermal conductivity.

Both the upper and the lower limits of thermal resistance are calculated by combining the alternative resistances of the bridged layer in proportion to their respective areas, as illustrated below. The method of combining differs in the two cases.

b)

This construction provides satisfactory sound insulation

Layer Material Thickness (mm) Thermal conductivity (W/m·K) Thermal resistance (m²K/W) external surface - - 0.040

1 outer leaf brick 102 0.77 0.132

2 air cavity 50 - 0.090

3 plywood 19 0.13 0.146

4(a¹) mineral wool between timber studs

120 0.038 3.158

4(a²) air space next to mineral wool 20 - 0.180

4(b) 38 mm

×

140 mm timber studs at 400 mm centres (140) 0.13 1.077 5 plasterboard 25 0.25 0.100 internal surface - - 0.130 Total thickness 336 mm U-value 0.31 W/m·K

102 mm brick outer leaf (conductivity 0.77 W/m·K)

2×12.5 mm plasterboard (conductivity 0.25 W/m·K)

heat flow 50 mm ventilated cavity (thermal resistance 0.09 m²K/W)

19 mm plywood

(conductivity 0.13 W/m·K) mineral wool (conductivity 0.038 W/m·K) between 38×140 mm timber studs (conductivity 0.13 W/m·K) at 400 mm centres

Upper resistance limit

When calculating the upper limit of thermal resistance, the building element is

considered to consist of two thermal paths (or sections). The upper limit of resistance is calculated from: 2 2 1 1 upper R F R F 1 R

+

=

where F1 and F2 are the fractional areas of the two sections and R1 and R2 are the

total resistances of the two sections. The method of calculating the upper resistance limit is illustrated conceptually

below:-Figure 4.2 : Conceptual illustration of how to calculate the upper limit of thermal  resistance 

Resistance through the section containing insulation

External surface resistance = 0.040

Resistance of bricks = 0.132

Resistance of air cavity = 0.090

Resistance of plywood = 0.146

Resistance of mineral wool (90.5%) = 3.158 Resistance of air space next to mineral wool = 0.180

Resistance of plasterboard = 0.100

Internal surface resistance = 0.130

Total (R1) = 3.976 m²K/W Fractional area F1 = 0.905 (90.5%) external surface internal surface 1 2 4(a) 2 1 5 3 F1 F2 3 4(b) 5

Resistance through section containing timber stud

External surface resistance = 0.040

Resistance of bricks = 0.132

Resistance of air cavity = 0.090

Resistance of plywood = 0.146

Resistance of timber studs (9.5%) = 1.077

Resistance of plasterboard = 0.100

Internal surface resistance = 0.130

Total (R2) = 1.715 m²K/W

Fractional area F2 = 0.095 (9.5%)

The upper limit of resistance is then:

=

+

=

+

=

715 . 1 095 . 0 976 . 3 905 . 0 1 R F R F 1 R 2 2 1 1 upper 3.533 m²K/W

Lower resistance

limit:-When calculating the lower limit of thermal resistance, the resistance of a bridged layer is determined by combining in parallel the resistances of the unbridged part and the bridged part of the layer. The resistances of all the layers in the element are then added together to give the lower limit of resistance.

The resistance of the bridged layer is calculated using:

timber timber insul insul R F R F 1 R

+

=

The method of calculating the lower limit of resistance is illustrated conceptually below.

Figure 4.3 : Conceptual illustration of how to calculate the lower limit of thermal  resistance  4(a) 2 1 5 F1 F2 internal surface 3 4(b) external surface

The lower limit of resistance is then obtained by adding up the resistances of all the layers:

External surface resistance = 0.040

Resistance of bricks = 0.132

Resistance of air cavity = 0.090

Resistance of plywood = 0.146

Resistance of bridged layer =

077 . 1 095 . 0 180 . 0 158 . 3 905 . 0 1

+

+

= 2.783 Resistance of plasterboard = 0.100

Internal surface resistance = 0.130

Total (Rlower) = 3.421 m²K/W

Total resistance of wall (not allowing for air gaps in the insulation)

The total resistance of the wall is the average of the upper and lower resistance limits: 477 . 3 2 421 . 3 533 . 3 2 R R R_T

=

upper

+

lower

=

+

=

m²K/W

Correction for air gaps

If there are small air gaps penetrating the insulating layer a correction should be applied to the U-value to account for this. The correctionc) for air gaps is

∆

Ug, where

∆

Ug =

∆

U’’

×

(RI / RT)²

and where RId) is the thermal resistance of the layer containing gaps, RT is the total

resistance of the element and

∆

U’’ is a factor which depends upon the way in which the insulation is fitted. In this example RIis 2.783 m²K/W, RT is 3.477 m²K/W and

∆

U’’ is 0.04. The value of

∆

Ug is then

∆

Ug = 0.04

×

(2.783 / 3.477)² = 0.026 W/m²K

U-value of the wall

The effect of air gaps or mechanical fixings should be included in the U-value unless they lead to an adjustment in the U-value of less than 3%.

U = 1 / RT +

∆

Ug (if

∆

Ug is not less than 3% of 1 / RT)

U = 1 / RT (if

∆

Ug is less than 3% of 1 / RT)

c)

Using Table D.1 of BS EN ISO 6946

d)

In this example RI is the same as the resistance of the bridged layer used in the calculation

In this case

∆

Ug = 0.026 W/m²K and 1 / RT = 0.288 W/m²K. Since

∆

Ug is not less

than 3% of (1 / RT),

U = 1 / RT = 1 / 3.477 + 0.026 = 0.31 W/m²K.

Note

1. In the above calculation it is assumed that the noggings (or dwangs) do not penetrate the whole of the insulation. If, however, the noggings (or dwangs) penetrate the whole of the insulation thickness they should be included within the timber percentage used in the calculation.

2. In this example correction level 2 is appropriate because air may circulate on the warm side of the insulation. If 140 mm of insulation was used instead of 120 mm so as to fill the space between the studs, correction level 1 would be appropriate. 3. The additional timbers at the junctions of plane elements, for example wall/wall,

wall/floor, and wall ceiling junctions, and the additional timbers surrounding openings are taken account of in the treatment of such details and so are not taken into account in the calculation of the U-value of the wall.

4. The Standard (BS EN ISO 6946) states that if the insulation is fitted in such a way that no air circulation is possible on the warm side of the insulation then

∆

U’’ is set to 0.01 W/m²K. If, on the other hand, air circulation is possible on the warm side then it should be set to 0.04 W/m²K. The possible correction levels are summarised as follows:

Description of air gap Correction

level

∆

U’’ W/m²K

∆

Ug

W/m²K Insulation installed in such a way that no air

circulation is possible on the warm side of the insulation. No air gaps penetrating the entire insulation layer.

0 0.00 0.000

Insulation installed in such a way that no air circulation is possible on the warm side of the insulation. Air gaps may penetrate the insulation layer.

1 0.01 0.006

Air circulation possible on the warm side of the insulation. Air gaps may penetrate the insulation.

5. Insulated cavity wall with metal wall ties

In this examplee) an insulated cavity wall has stainless steel double triangle wall ties penetrating the insulation layer. The construction consists of outer leaf brickwork, a cavity filled with mineral wool batts, 100 mm of AAC blockwork and 13 mm of

lightweight plaster. The wall ties are spaced 900 mm horizontally and 450 mm vertically.

Figure 5.1 : Insulated cavity wall (fully-filled) with metal wall ties 

The thicknesses of each layer, together with the thermal conductivities of the materials in each layer, are shown below. The external and internal surface

resistances used are those given in Table 1 of this document. The metal ties are not treated as repeating thermal bridges but instead are accounted for at the end of the calculation. The third layer contains AAC blockwork bridged by mortar with the mortar occupying 6.6% of the cross-sectional area.

Layer Material Thickness (mm) Thermal conductivity (W/m·K) Thermal resistance (m²K/W) external surface - - 0.040

1 outer leaf brick 102 0.77 0.132

2 mineral wool batts 75 0.038 1.974

3(a) AAC blocks (93.4%) 100 0.11 0.909

3(b) mortar (6.6%) (100) 0.88 0.114

4 lightweight plaster 13 0.18 0.072

internal surface - - 0.130

e)

This construction provides satisfactory sound insulation from neighbouring dwellings Total thickness 290 mm

U-value 0.32 W/m·K

102 mm brick (conductivity 0.77 W/m·K)

75 mm cavity filled with mineral wool (conductivity 0.038 m²K/W)

13 mm lightweight plaster, (conductivity 0.18 W/m·K)

100 mm AAC blocks (conductivity 0.11 m²K/W) bridged by mortar (conductivity 0.88 W/m·K)

Upper resistance

limit:-When calculating the upper limit of thermal resistance, the building element is considered to consist of two thermal paths (or sections). The upper limit of resistance is calculated from:

2 2 1 1 upper R F R F 1 R

+

=

where F1 and F2 are the fractional areas of the two sections and R1 and R2 are the

total resistances of the two sections. A conceptual diagram of the upper limit of resistance is shown immediately below

Figure 5.2 : Conceptual illustration of how to calculate the upper limit of resistance 

Resistance through section containing concrete blocks External surface resistance 0.040

Resistance of bricks 0.132

Resistance of mineral wool slabs 1.974 Resistance of AAC blocks (93.4%) 0.909 Resistance of lightweight plaster 0.072 Internal surface resistance 0.130

Total (R1) 3.257

Fractional area F1 = 0.934 (93.4%)

Resistance through section containing mortar External surface resistance 0.040

Resistance of bricks 0.132

Resistance of mineral wool slabs 1.974 Resistance of mortar (6.6%) 0.114 Resistance of lightweight plaster 0.072 Internal surface resistance 0.130

Total (R2) 2.462

Fractional area F2 = 0.066 (6.6%)

Combining the resistances in their appropriate proportions the upper limit of resistance (Rupper) is given by:

=

+

=

+

=

462 . 2 066 . 0 257 . 3 934 . 0 1 R F R F 1 R 2 2 1 1 upper 3.189 m²K/W external surface internal surface 1 2 4 2 1 3(a) F1 F2 3(b) 4

Lower resistance

limit:-To calculate the lower resistance limit the resistance of the bridged layer is

determined by combining in parallel the resistances of the unbridged part and the bridged part of the layer. The resistances of all the layers are then added together to give the lower limit of resistance. A conceptual illustration of the method of

calculating the lower limit of resistance is shown

below:-Figure 5.3 : Conceptual illustration of how to calculate the lower limit of resistance 

External surface resistance = 0.040

Resistanceof bricks = 0.132

Resistance of mineral wool slabs = 1.974

Resistance of AAC blocks & mortar =

114 . 0 066 . 0 909 . 0 934 . 0 1

+

= 0.622

Resistance of lightweight plaster = 0.072

Internal surface resistance = 0.130

Total (Rlower) = 2.970

Total resistance of wall (ignoring wall ties)

The total resistance of the wall is the average of the upper and lower resistance limits:

=

+

=

+

=

2 970 . 2 189 . 3 2 R R R_T upper lower 3.080 m²K/W

Correction for cavity wall ties

The method of calculating U-values as given in BS EN ISO 6946 requires that mechanical fixings, such as cavity wall ties, be taken into account. The following describes how the effect of the wall ties is incorporated into the U-value. The method of correction is the same for both fully filled and partially filled cavity walls. The

corrections are only applied where wall ties actually penetrate the insulation. Since wall ties of low conductivity, such as plastic ties, do not affect the U-value

significantly, the Standard (BS EN ISO 6946) only requires a correction to be made if the conductivity of the tie, or part of it, is more than 1 W/m·K. In practice this means that plastic wall ties can be ignored in the U-value calculation but metal wall ties generally need to be included.

In this example the wall ties are of stainless steel (double triangle) and are 3.7 mm in diameter giving a cross-sectional area of 10.75 mm². They are arranged at 900 mm

3(a) 2 1 4 F1 F2 internal surface 3(b) external surface

horizontal centres and 450 mm vertical centres. Using the procedure in BS EN ISO 6946, the adjustment to the U-value,

∆

Uf, is given by

∆

Uf =

α λ

fnfAf

where

α

is the scaling factor for mechanical fixings, which is 6 for wall tiesf), and

λ

fis

the conductivity of the fixings, which is 17 W/m·K in this example. The number of wall ties per square metre which penetrate the insulation, nf, is calculated as follows

using the above information on the wall tie spacing.

47 . 2 450 900 000 000 1 n_f

=

×

=

Afis the cross-sectional area of the wall tie, expressed in m², which is

0.00001075 m². A value for

∆

Ufof 0.003 W/m²K is obtained using the above formula:

∆

Uf = 6

×

17

×

2.47

×

0.00001075 = 0.003 W/m²K

U-value of the wall

The effect of air gaps or mechanical fixings should be included in the U-value unless they lead to an adjustment in the U-value of less than 3%.

U = 1 / RT +

∆

Uf (if

∆

Uf is not less than 3% of 1 / RT)

U = 1 / RT (if

∆

Uf is less than 3% of 1 / RT)

In this case

∆

Uf= 0.003 W/m²K and 1 / RT = 0.3247 W/m²K. Since

∆

Uf is less than

3% of (1 / RT), U = 1 / RT = 1 / 3.080 = 0.32 W/m²K

Note

1. By adding

∆

Uf(0.003 W/m²K) to the U-value which would be obtained without any

correction for wall ties (0.325 W/m²K) this would imply a U-value of 0.328 W/m²K. The Standard (BS EN ISO 6946), however, permits the effects of mechanical fixings to be ignored if they lead to an increase of less than 3% in the U-value. Since the

∆

Uf correction (0.003 W/m²K) is less than 3% of (1 / RT) the correction

need not be applied. The final quoted U-value, obtained by rounding the (uncorrected) U-value to two significant figures, is 0.32 W/m²K.

2. If vertical twist wall ties are used instead of double triangle ties the correction to the U-value can be considerably larger than that shown above, due to their greater cross-sectional area.

3. If instead of stainless steel ties, galvanised steel ties of conductivity 50 W/m·K are used, this will increase

∆

Uffrom 0.003 W/m²K to 0.008 W/m²K.

4. If the thermal conductivity of the tie, or part of it, is less than 1 W/m·K, no

correction is applied and

∆

Ufis taken to be zero. This would apply, for instance,

in the case of plastic wall ties.

f)

The following is a conceptual diagram showing how the effect of the wall ties, where applicable, is incorporated into the overall U-value calculation:

Figure 5.4 : Conceptual diagram illustrating how the U-value is corrected for the 

presence of wall ties. The U-value calculation is firstly carried out ignoring the effects  of the wall ties and an adjustment is then applied in order to obtain the final U-value.

effect of wall ties U-value in absence of wall ties

6. Wide cavity wall with vertical twist ties

In this exampleg)a wide cavity wall is fully filled with mineral wool insulation with stainless steel vertical twist wall ties in the filled cavity. To obtain the U-value allowing for the wall ties the thermal resistance (RT) should first be calculated

ignoring the effect of the wall ties and then a correction should be made for the presence of the ties. The wall ties are spaced 750 mm horizontally and 450 mm vertically.

Figure 6.1 : Insulated cavity wall (fully-filled) with metal wall ties 

The thicknesses of each layer, together with the thermal conductivities of the

materials, are shown below. The external and internal surface resistances used are those given in Table 1 of this document. The vertical twist wall ties have a cross sectional area of 60.8 mm².

In this example there is no distinction between the upper and lower limit of resistance because all of the layers are considered to be sufficiently homogeneous (for the purposes of thermal calculations). Strictly speaking, the mortar joints between the bricks and concrete blocks could be taken into account, however since the

resistances of the mortar parts do not differ from the brick or block parts by more than 0.1 m²K/W the mortar parts may be ignored.

g)

Due to requirements for sound insulation this construction may only be suitable for detached dwellings

Total thickness 335 mm

U-value 0.30 W/m·K

102 mm brick (conductivity 0.77 W/m·K)

120 mm cavity filled with mineral wool (conductivity 0.038 m²K/W) 100 mm concrete blocks,

(conductivity 1.13 m²K/W)

13 mm dense plaster (conductivity 0.57 W/m·K)

The wall construction may be summarised as follows: Layer Material Thickness (mm) Thermal conductivity (W/m·K) Thermal resistance (m²K/W) externalsurface - - 0.040

1 outer leaf brick 102 0.77 0.132

2 mineral wool batts 120 0.038 3.158

3 concrete blockwork 100 1.13 0.088

4 denseplaster 13 0.57 0.023

internal surface - - 0.130

Total (RT) 3.571

Correction for cavity wall ties

A correction has to be applied to allow for the additional heat loss due to the wall ties. In this example the wall ties are of metal (vertical twist) and have a cross-sectional area, Af, of 60.8 mm². Using the procedure in BS EN ISO 6946, the correction to be

applied,

∆

Uf, is given by

∆

Uf =

α λ

fnfAf

where

α

is the scaling factor for mechanical fixings, which is 6 for wall tiesh), and

λ

fis

the conductivity of the fixings, which is 17 W/m·K for stainless steel. The number of wall ties per square metre which penetrate the insulation, nf, is calculated to be 2.96 / 

m². Note that Af, the cross-sectional area of the wall tie, is expressed in m². The

correction to be applied is therefore

∆

Uf = 6

×

17

×

2.96

×

0.0000608 = 0.018 W/m²K

U-value of the wall

The effect of air gaps or mechanical fixings should be included in the U-value unless they lead to an adjustment in the U-value of less than 3%.

U = 1 / RT +

∆

Uf (if

∆

Uf is not less than 3% of 1 / RT)

U = 1 / RT (if

∆

Uf is less than 3% of 1 / RT)

In this case

∆

Uf= 0.018 W/m²K and 1 / RT = 0.280 W/m²K. Since

∆

Uf is not less than

3% of (1 / RT),

U = 1 / 3.571 + 0.018 W/m²K = 0.30 W/m²K

Note

1. If galvanised steel ties (with a conductivity of 50 W/m·K) are used instead of stainless steel ties, the value of

∆

Uf will be 0.054 W/m²K. This will give a final

U-value of U = 1 / 3.571 + 0.054 = 0.33 W/m²K

h)

2. If the thermal conductivity of the tie, or part of it, is less than 1 W/m·K (eg. plastic ties) the value of

∆

Uf may be taken to be zero and the U-value will be

U = 1 / RT = 1 / 3.571 = 0.28 W/m²K

3. Strictly speaking, the mortar joints between the bricks and concrete blocks could be taken into account in the U-value calculation, however it is permissible to ignore the mortar in both of these layers because the resistances of the mortar  joints differ from the resistances of the bricks or concrete blocks by less than

7. Pitched roof with insulation between and over the joists

Figure 7.1 : Insulation between and over joists at ceiling level 

A pitched roof has 100 mm of mineral wool tightly fitted between 48

×

100 mm timber  joists spaced 600 mm apart (centres to centres) and 100 mm of mineral wool over

the joists. The roof is tiled with felt or boards under the tiles. The external and internal surface resistances used are those given in Table 1 of this document. The ceiling consists of 12.5 mm of plasterboard. The roof construction is summarised below.

Layer Material Thickness

(mm) Thermal conductivity (W/m·K) Thermal resistance (m²K/W) external surface - - 0.040

1 roof space beneath tiled roof with felt or boardsi)

- - 0.200

2 continuous layer of mineral wool 100 0.042 2.381

3(a) mineral wool between 48

×

100 mm timber  joists with 600 mm between centres

100 0.042 2.381

3(b) 48

×

100 mm timber joists between insulation (100) 0.13 0.769

4 plasterboard 12.5 0.25 0.050

internal surface - - 0.100

Upper resistance

limit:-A conceptual illustration of how the upper limit of resistance is calculated is shown immediately below

Figure 7.2 : Conceptual illustration of how to calculate the upper limit of resistance 

i)

Using Table 3 of BS EN ISO 6946 insulation  joist

U-value 0.20 W/m·K (loft space and pitched roof above)

external surface internal surface 1 2 4 2 1 3(a) F1 F2 3(b) 4

Resistance through section containing both layers of insulation External surface resistance = 0.040

Resistance of roof spacei) = 0.200 Resistance of mineral wool over joists = 2.381 Resistance of mineral wool between joists = 2.381 Resistance of plasterboard = 0.050 Inside surface resistance = 0.100

Total (R1) = 5.152 m²K/W

Fractional area F1 = 0.92 (92%)

Resistance through section containing timber joists External surface resistance = 0.040 Resistance of roof space = 0.200 Resistance of mineral wool over joists = 2.381 Resistance of timber joists = 0.769 Resistance of plasterboard = 0.050 Inside surface resistance = 0.100

Total (R2) = 3.540 m²K/W

Fractional area F2 = 0.08 (8%)

The upper resistance limit is given by

971 . 4 540 . 3 08 . 0 152 . 5 92 . 0 1 R F R F 1 R 2 2 1 1 upper

=

+

=

+

=

m²K/W

Lower resistance

limit:-A conceptual illustration of the method of calculating the lower limit of resistance is shown

below:-Figure 7.3 : Conceptual illustration of how to calculate the lower limit of resistance 

External surface resistance = 0.040 Resistance of roof space = 0.200 Resistance of mineral wool over joists = 2.381 Resistance of bridged layer

769 . 0 08 . 0 381 . 2 92 . 0 1 R F R F 1 timber timber insul insul

₊

=

+

=

= 2.039 Resistance of plasterboard = 0.050 Inside surface resistance = 0.100

Total (Rlower) = 4.810 m²K/W

Total resistance of roof

=

+

=

+

=

2 810 . 4 971 . 4 2 R R R_T upper lower 4.891 m²K/W

U-value of the roof U = 1 / RT = 0.20 W/m²K

Note

1. Since there are two layers of insulation, one between joists and the other as a continuous layer covering the first layer, a correction for air gaps need not be applied.

2. Since the nails or fixings do not penetrate any insulation, a correction for mechanical fixings need not be applied.

3(a) 2 1 4 F1 F2 internal surface 3(b) external surface

8. Room in roof construction

An existing loft is converted to a habitable space by inserting tightly fitted insulation between the rafters in the roof. Timber packing pieces of the same width as existing 100 mm deep rafters are attached beneath the rafters in order to provide additional room for insulation. Plasterboard, laminated to insulation, is then attached below the rafters. A 50 mm space is reserved for ventilation above the insulation.

The construction consists of roof tiles, felt, a 50 mm air gap between rafters and 100 mm of insulation between rafters and spacers. Beneath the rafters and spacers there is an insulation laminate consisting of 15.5 mm of insulation bonded to 9.5 mm of plasterboard. In this example the rafters are 100 mm deep but 50 mm timber spacers have been attached below the rafters in order to extend the total rafter depth to effectively 150 mm.

Figure 8.1 : Roof construction shown as two cross-sections (fixing nails not shown) The construction may be summarised as

follows:-Layer Material Thickness

(mm) Thermal conductivity (W/mK) Thermal resistance (m²K/W) external surface* - - -1 tiles* 19 - -2 roofingfelt* 1 -

-3 ventilated airspace between rafters and spacers*

50 - 0.100

4(a) insulation board j) occupying 88% of face area (between rafters and spacers)

100 0.025 4.000

4(b) rafters (beneath ventilated area) occupying 12% of face area

(100) 0.13 0.769

5 insulation board j) 15.5 0.025 0.620

6 plasterboard 9.5 0.25 0.038

internal surface - - 0.100

*All layers to the cold side of the well ventilated airspace are ignored in the U-value calculation and the surface bounding this airspace is taken to have the same resistance as an internal surface. The internal surface resistance is taken from Table 1 of this document.

 j)

For example, phenolic foam or polyurethane, where the conductivity has an allowance for ageing and variation in manufacture

Plan at A-A

U-value 0.27 W/m·K

A

Since the airspace between the rafters is well ventilated, all layers above the airspace are ignored in the thermal calculation and the airspace is treated as a surface resistance of 0.10 m²K/W.

Conceptual diagrams of the methods of calculating upper and lower limits of resistance are shown

below:-Figure 8.2 : Conceptual diagrams of how to calculate the upper and lower limits of  resistance 

Upper resistance

limit:-Resistance through the section between the rafters Effective external surface resistance = 0.100 Resistance of insulation between rafters = 4.000 Resistance of insulation beneath rafters = 0.620 Resistance of plasterboard = 0.038 Internal surface resistance = 0.100

Total thermal resistance (R1) = 4.858 m²K/W

Fractional area F1 = 0.88 (88%)

Resistance through the section through the rafters Effective external surface resistance = 0.100

Resistance of rafters = 0.769

Resistance of insulation beneath rafters = 0.620 Resistance of plasterboard = 0.038 Internal surface resistance = 0.100

Total thermal resistance (R2) = 1.627 m²K/W

Fractional area F2 = 0.12 (12%)

The upper limit of resistance is then obtained from:

923 . 3 627 . 1 12 . 0 858 . 4 88 . 0 1 R F R F 1 R 2 2 1 1 upper

=

+

=

+

=

m²K/W F1 F2 external surface insulation plasterboard internal surface rafters insulation external surface insulation plasterboard internal surface rafters insulation

Lower resistance limit

Effective external surface resistance = 0.100 Resistance of bridged layer

769 . 0 12 . 0 000 . 4 88 . 0 1

+

=

= 2.659

Resistance of insulation beneath rafters = 0.620 Resistance of plasterboard = 0.038 Internal surface resistance = 0.100

Total thermal resistance (Rlower) = 3.517 m²K/W

Total resistance of roof

The total resistance is the average of the upper and lower limits

RT = 3.720 2 517 . 3 923 . 3 2 R R_{upper lower}

=

+

=

+

m²K/W

U-value of the roof U = 1 / RT = 0.27 W/m²K

Note

1. This example assumes that the rafter depth is 100 mm and that 50 mm timber spacers can be attached below the rafters. In instances where the rafters are insufficiently deep (e.g. only 75 mm) there may be practical problems in achieving the required U-value due to a lack of space being available for the insulation. In such cases the insulation beneath the rafters may need to be thicker in order to compensate for the limited rafter depth.

2. In this example the effects of the fixing nails may be ignored since they do not penetrate the main insulating layer.

3. Since there are two layers of insulation, one between rafters and the other as a continuous layer covering the first layer, a correction for air gaps need not be applied.

9. Room in roof construction with limited rafter depth

This roof is similar to that shown in the previous example except that the existing rafters, which are only 75 mm deep in this case, are not extended in depth but

instead a thicker plasterboard-insulation laminate is attached below the rafters. The construction consists of roof tiles, felt, a 50 mm air gap between rafters and 25 mm of insulation between rafters and spacers. As in the previous example the insulation is tightly fitted between the rafters. Beneath the rafters and spacers there is an

insulation laminate consisting of 57.5 mm of insulation bonded to 12.5 mm of plasterboard. The insulation laminate is nailed to the rafters and the nails have a horizontal spacing of 400 mm and a vertical spacing of 150 mm. The external and internal surface resistances used are those given in Table 1 of this document. To calculate the U-value a calculation is first carried out ignoring the nails and then a correction is applied to account for the nails.

Figure 9.1 : Roof construction 

The construction may be summarised as

follows:-Layer Material Thickness

(mm) Thermal conductivity (W/mK) Thermal resistance (m²K/W) external surface* - - -1 tiles* 19 - -2 roofingfelt* 1 -

-3 ventilated airspace between rafters and spacers*

50 - 0.100

4(a) insulation board occupying 88% of face area (between rafters and spacers)

25 0.025 1.000

4(b) rafters (beneath ventilated area) occupying 12% of face area

(25) 0.13 0.192

5 insulation board 57.5 0.025 2.300

6 plasterboard 12.5 0.25 0.050

internal surface - - 0.100

*All layers to the cold side of the well ventilated airspace are ignored in the U-value

calculation and the surface bounding this airspace is taken to have the same resistance as an internal surface. The internal surface resistance is taken from Table 1 of this

document.

Plan at A-A

A

A

Since the airspace between the rafters is well ventilated, all layers to the cold side of the airspace are ignored in the thermal calculation and the airspace is treated as a surface resistance of 0.10 m²K/W.

Conceptual diagrams of the methods of calculating upper and lower limits of resistance are shown

below:-Figure 9.2 : Conceptual diagrams of how to calculate the upper and lower limits of  resistance 

Upper resistance

limit:-Resistance through the section between the rafters Effective external surface resistance = 0.100 Resistance of insulation between rafters = 1.000 Resistance of insulation beneath rafters = 2.300 Resistance of plasterboard = 0.050 Internal surface resistance = 0.100

Total thermal resistance (R1) = 3.550 m²K/W

Fractional area F1 = 0.88 (88%)

Resistance through the section through the rafters Effective external surface resistance = 0.100

Resistance of rafters = 0.192

Resistance of insulation beneath rafters = 2.300 Resistance of plasterboard = 0.050 Internal surface resistance = 0.100

Total thermal resistance (R2) = 2.742 m²K/W

Fractional area F2 = 0.12 (12%)

The upper limit of resistance is then obtained from:

429 . 3 742 . 2 12 . 0 550 . 3 88 . 0 1 R F R F 1 R 2 2 1 1 upper

=

+

=

+

=

external surface plaster board internal surface rafters insulation insulation F2 F1 external surface insulation plasterboard internal surface rafters insulation

Lower resistance

limit:-Effective external surface resistance = 0.100 Resistance of bridged layer

192 . 0 12 . 0 000 . 1 88 . 0 1

+

=

= 0.664

Resistance of insulation beneath rafters = 2.300 Resistance of plasterboard = 0.050 Internal surface resistance = 0.100

Total thermal resistance = 3.214 m²K/W

Total resistance (without correction for the fixing nails) The total resistance is the average of the upper and lower limits

RT = 3.322 2 214 . 3 429 . 3 2 R R_{upper lower}

=

+

=

+

m²K/W

Correction for the presence of fixing nails

The method of calculating U-values as given in BS EN ISO 6946 requires that

mechanical fixings, such as nails or screws for example, be taken into account. The following describes how the effect of the fixing nails is incorporated into the U-value. In this example, the plasterboard-insulation laminate is fixed to the rafters using nails. The nails are arranged at 150 mm vertical centres and since the rafters are 400 mm apart the number of nails per square metre of sloping ceiling will be nf where

7 . 16 150 400 000 000 1 n_f

=

×

=

 / m²

The nails are made of steel with a their thermal conductivity,

λ

f, of 50 W/m·K. Their

cross-sectional area, Af, is 5 mm² or 0.000005 m². The adjustment to the U-value is

∆

Uf, where

∆

Uf =

α λ

f nfAf= 5

×

50

×

16.7

×

0.000005 = 0.021 m²K/W.

where

α

is 5 for all roof fixingsk), Rfis the thermal resistance of the insulation

penetrated by the nails and RTis the total thermal resistance of the roof.

U-value of the roof

The effect of air gaps or mechanical fixings should be included in the U-value unless they lead to an adjustment in the U-value of less than 3%.

U = 1 / RT +

∆

Uf (if

∆

Uf is not less than 3% of 1 / RT)

U = 1 / RT (if

∆

Uf is less than 3% of 1 / RT)

k)

In this case

∆

Uf= 0.021 m²K/W and 1 / RT = 0.301 W/m²K. Since

∆

Uf is not less

than 3% of (1 / RT),

U = 1 / RT +

∆

Uf= 1 / 3.322 + 0.021 W/m²K = 0.32 W/m²K.

Note

Since there are two layers of insulation, one between rafters and the other as a

10. Floor of heated room above an unheated space

In this example a floor has insulation between timber joists. The floor is situated above an unheated space such as a garage or an unheated corridor.

Figure 10.1 : Floor construction over an unheated space 

The construction consists of 19 mm of plywood over timber joists with mineral wool insulation (of conductivity 0.040 W/m·K) between the joists and 12.5 mm of

plasterboard over the unheated space. The total area (Ai) of components between

the internal environment and the unheated space is 35 m² and the total area (Ae) of

components between the unheated space and the external environment is 35 m². Using the procedure in BS EN ISO 6946 for unheated spaces, an additional thermal resistance, Ru, is added as if it were an additional homogenous layer, where

Ru = 0.09 + 0.4 Ai/ Ae

giving Ru = 0.490

The floor consists of 19 mm plywood over 150 mm timber joists with 150 mm glass mineral wool between the joists. Below the joists is 12.5 mm plasterboard forming the ceiling of the garage. The external and internal surface resistances used are those given in Table 1 of this document.

Layer Material Thickness

(mm) Thermal conductivity (W/m·K) Thermal resistance (m²K/W) internal surface - - 0.170 1 plywood 19 0.13 0.146

2(a) glass mineral wool 150 0.040 3.750

2(b) timber joists (occupying 12%) (150) 0.13 1.154

3 plasterboard 12.5 0.25 0.050

4 external - - 0.040

A conceptual illustration of the calculation of the limits of resistance is shown below:

12.5 mm plasterboard above unheated area 19 mm plywood next to heated area

150 mm timber joists with mineral wool between the   joists. Timber fraction is

0.12 (i.e. 12%)

Total thickness 182 mm

Figure 10.2 : Conceptual illustration of how to calculate the upper and lower limits of  resistance 

Upper resistance

limit:-Resistance through the section containing the insulation:

Internal surface resistance = 0.170

Resistance of plywood = 0.146

Resistance of mineral wool insulation = 3.750

Resistance of plasterboard = 0.050

Ru = 0.490

External surface resistance = 0.040

Total thermal resistance (R1) = 4.646 m²K/W

Resistance through the section containing joists:

Internal surface resistance = 0.170

Resistance of plywood = 0.146

Resistance of timber joists = 1.154

Resistance of plasterboard = 0.050

Ru = 0.490

External surface resistance = 0.040

Total thermal resistance (R2) = 2.050 m²K/W

The upper limit of resistance is then obtained from: 033 . 4 050 . 2 12 . 0 646 . 4 88 . 0 1 R F R F 1 R 2 2 2 1 upper

=

+

=

+

=

m²K/W Ru internal surface plywood 2(a) plaster-board external surface 2(b) F1 Ru internal surface plywood 2(a) plaster-board external surface 2(b)

Lower resistance

limit:-Internal surface resistance = 0.170

Resistance of plywood = 0.146

Resistance of bridged layer

154 . 1 12 . 0 750 . 3 88 . 0 1

+

=

= 2.953 Resistance of plasterboard = 0.050 Ru = 0.490

External surface resistance = 0.040

Total (Rlower) = 3.849 m²K/W

Total resistance of floor

The total resistance of the wall is the average of the upper and lower resistance limits

=

+

=

+

=

2 849 . 3 033 . 4 2 R R R_T upper lower 3.941 m²K/W

Correction for air gaps

Since the insulation is entirely between the joists a correction should be applied to the U-value in order to account for air gaps. The overall U-value of the floor should include a term

∆

Ug, where

∆

Ug =

∆

U’’

×

(RI/ RT)²

and where

∆

U’’ = 0.01 (referred to in BS EN ISO 6946 as correction level 1), RIis the

thermal resistance of the layer containing the gaps and RT is the total resistance of

the element.

∆

Ug is therefore

∆

Ug = 0.01

×

(2.953 / 3.941)² = 0.005 W/m²K

U-value of the floor

The effect of air gaps or mechanical fixings should be included in the U-value unless they lead to an adjustment in the U-value of less than 3%.

U = 1 / RT +

∆

Ug (if

∆

Ug is not less than 3% of 1 / RT)

U = 1 / RT (if

∆

Ug is less than 3% of 1 / RT)

In this case

∆

Ug = 0.005 W/m²K and 1 / RT = 0.254 W/m²K. Since

∆

Ug is less than

3% of (1 / RT),

11. Suspended beam and block floor

A beam and block floor consists of blocks of lightweight concrete which are 100 mm thick and 440 mm wide suspended on T-beams which are 70 mm wide. Above the beams and blocks is 65 mm of flooring screed and 100 mm of polystyrene insulation. Beneath the beams and blocks there is an underfloor space over sandy soil. The beams protrude below the blocks by 75 mm. The perimeter of the ground floor is 35.6 metres and its area is 79.1 m² giving a perimeter to area ratio of 0.45.

In order to calculate the U-value, BS EN ISO 6946 is applied to determine the thermal resistance between the dwelling and the underfloor space.

The construction of the floor deck can be summarised as follows:

Layer Material Thickness

(mm) Thermal conductivity (W/m·K) Thermal resistance (m²K/W) internal surface - - 0.170 1 screed 65 0.41 0.159 2 polystyrene 100 0.040 2.500

3(a) light concrete blocks, 440 mm wide 100 0.18 0.556 3(b) concrete beams, 70 mm wide (100) 1.13 0.088 4 lowersurface - - 0.170*

* The internal surface resistance is taken from Table 1 of this document. The surface resistance for the lower side of the floor deck is taken to be 0.17 m²K/W, as this is the value that applies for downwards heat flow in a non-external environment.

Figure 11.1 : Beam and block suspended floor 

U-value 0.24 W/m·K 440

100 70

Figure 11.2 : Conceptual diagram of how to calculate the upper and lower limits of  resistance 

Since the conductivity of the beams is less than 2.0 W/m·K the part of the beam which protrudes below the blocks is ignored, as indicated in BS EN ISO 6946. The U-value between the dwelling and the underfloor space is calculated using BS EN ISO 6946, as follows:

Upper resistance limit (of floor deck)

Resistance through section containing lightweight blocks

Internal surface resistance = 0.170

Resistance of screed = 0.159

Resistance of polystyrene = 2.500

Resistance of light concrete blocks = 0.556 Resistance of lower surface of floor deck = 0.170

Total thermal resistance (R1) = 3.555 m²K/W

Fractional area F1= 0.863 (i.e. 86.3%)

Resistance through section containing concrete beams

Internal surface resistance = 0.170

Resistance of screed = 0.159

Resistance of polystyrene = 2.500

Resistance of beams = 0.088

Resistance of lower surface of floor deck = 0.170

Total thermal resistance (R2) = 3.087 m²K/W

Fractional area F2= 0.137 (i.e. 13.7%)

The upper limit of resistance is then obtained from:

483 . 3 087 . 3 137 . 0 555 . 3 863 . 0 1 R F R F 1 R 2 2 1 1 upper

=

+

=

+

=

internal surface beams lower surface insulation screed blocks

underfloor space & soil

internal surface beams lower surface insulation screed blocks

Lower resistance limit (of floor

deck):-Internal surface resistance = 0.170

Resistance of screed = 0.159

Resistance of polystyrene = 2.500 Resistance of bridged layer

088 . 0 137 . 0 556 . 0 863 . 0 1

+

=

= 0.322

Resistance of lower surface of deck = 0.170

Total (Rlower) = 3.321 m²K/W

Total resistance of floor deck

The total resistance of the floor deck is the average of the upper and lower resistance limits 402 . 3 2 321 . 3 483 . 3 2 R R R

=

upper

+

lower

=

+

=

m²K/W Uf= 1 / R = 1 / 3.402 = 0.294 W/m²K

This gives a U-value (Uf) for the floor deck of 0.294 W/m²K. It should be borne in

mind that Uf includes the surface resistances for the upper and lower sides of the

deck.

Resistance of the remainder of the floor

For determining the resistance of the remaining part of the floor, Table 4 of the Appendix is used. This table gives the U-value of an uninsulated suspended floor, U0, where the U-value of the floor deck has been calculated using standard

assumptions about the thermal resistance of the floor deck and the surface

resistances at the upper and lower sides of the deck. Since Uf, calculated above,

already includes surface resistances the surface resistances need to be subtracted from U0.

The overall U-value of the suspended floor is then calculated using the following:

 

 

 

 



 

 

−

−

−

+

=

lower , si d uninsulate , deck upper , si 0 f R R R U 1 U 1 1 U

where Rsi,upper is the surface resistance of the upper side of the floor deck, equal to

0.17 m²K/W (see Table 1 of this document), Rdeck,uninsulated is the thermal resistance of

a notional uninsulated floor deck, equal to 0.20 m²K/W (see CIBSE Guide A3, part 3.5.5.2), and Rsi,loweris the surface resistance of the lower side of the floor deck,

The remaining calculation is now carried out below, showing how the resistance of the remaining part of the floor is combined with the U-value of the floor deck

calculated above.

The U-value of the floor in the absence of floor insulation is U0 = 0.76 W/m²K (see Table 4 of this document)

U-value of the floor

The U-value of the suspended floor is therefore

(

)

(

)

K ² m  /  W 24 . 0 17 . 0 2 . 0 17 . 0 76 . 0 1 294 . 0 1 1 R R R U 1 U 1 1 U lower , si d uninsulate , deck upper , si 0 f

=

+

+

−

+

=

+

+

−

+

=

Explanatory note: 

The value of 0.2 used in the above equation is based on the CIBSE Guide A3

(3.5.5.2) and represents the thermal resistance of a notional uninsulated floor deck. Rsi,upper and Rsi,lower, which represent the surface resistances of the upper and lower

surface resistances of the (notional) floor deck are obtained from Table 1. Summary of floor details

exposed perimeter (P) 35.6 m floor area (A) 79.1 m² perimeter to area ratio (P/A) 0.45 m-1 wall thickness (w) 0.3 m

soil type sandy

12. Suspended timber ground floor

Figure 12.1 : Suspended timber floor 

A suspended timber ground floor consists of 19 mm of chipboard over timber joists. The timber joists are 150 mm

×

48 mm at 400 mm centres giving a 12% timber fraction. Between the joists there is 150 mm of tightly fitted mineral wool (with a conductivity of 0.040 W/m·K) suspended on netting. Beneath the floor deck there is an underfloor space over clay soil. The perimeter of the ground floor is 40 metres and the area is 100 m². BS EN ISO 6946 is applied to obtain the thermal resistance of the floor deck.

Layer Material Thickness

(mm) Thermal conductivity (W/m·K) Thermal resistance (m²K/W) internal surface - - 0.170 1 chipboard 19 0.13 0.146

2(a) glass mineral wool on netting 150 0.040 3.750 2(b) timber joists (occupying 12%) (150) 0.13 1.154

lowersurface - - 0.170*

*The internal surface resistance is taken from Table 1 of this document. The surface resistance for the lower side of the floor deck is taken to be 0.17 m²K/W, as this is the value that applies for downwards heat flow in a non-external environment.

The methods of calculating the upper and lower limits of resistance are illustrated conceptually

below:-Figure 12.2 : Conceptual illustration of the methods of calculating the upper and  lower limits of thermal resistance 

U-value 0.22 W/m·K

underfloor space and soil

lower surface mineral wool chipboard  joists upper surface underfloor space and soil

lower surface mineral wool chipboard  joists upper surface

Upper resistance limit (for floor

deck):-Resistance through section containing mineral wool on netting Internal surface resistance = 0.170

Resistance of chipboard = 0.146 Resistance of mineral wool = 3.750 Resistance of lower surface of floor deck= 0.170

Total thermal resistance (R1) = 4.236 m²K/W

Fractional area F1 = 0.88 (i.e. 88%)

Resistance through section containing timber joists Internal surface resistance = 0.170 Resistance of chipboard = 0.146

Resistance of timber = 1.154

Resistance of lower surface of floor deck= 0.170

Total thermal resistance (R2) = 1.640 m²K/W

Fractional area F2 = 0.12 (i.e. 12%)

The upper limit of resistance is:

560 . 3 640 . 1 12 . 0 236 . 4 88 . 0 1 R F R F 1 R 2 2 1 1 upper

=

+

=

+

=

m²K/W

Lower resistance limit (for floor deck)

Internal surface resistance = 0.170 Resistance of chipboard = 0.146 Resistance of bridged layer

= 154 . 1 12 . 0 750 . 3 88 . 0 1

+

= 2.953

Resistance of lower surface of floor deck= 0.170

Total (Rlower) = 3.439 m²K/W

Overall resistance of floor deck

The resistance of the floor deck is the average of the upper and lower resistance limits 500 . 3 2 439 . 3 560 . 3 2 R R R_T

=

upper

+

lower

=

+

=

m²K/W