Thermal conductivity

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Final approval on behalf of BRE:

Signed ________________ Date ________________

Mrs H J Cuckow, Director, BRE East Kilbride

BRE East KilbrideKelvin RoadEast Kilbride

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© Building Research Establishment Ltd 1999

Examples of U-value calculationsusing BS EN ISO 6946:1997

Prepared for:DETR/BR

By:S M Doran and L Kosmina

BRE East Kilbride

December 1999 Report No 78129

(revised June 2000)

Examples of U-value calculations using BS EN ISO 6946:1997S M Doran and L Kosmina, BRE East KilbrideDecember 1999

This document illustrates the procedures given in BS EN ISO 6946:19971 forcalculating the U-value of opaque elements. The procedures are explained usingexamples of U-value calculations for some typical wall, roof and floor designs whichcontain repeating thermal bridges.

Contents

1. Introduction 1

2. Outline of the procedure 2

3. Cavity wall with lightweight masonry leaf and insulated dry lining 3

4. Timber framed wall 8

5. Insulated cavity wall with metal wall ties 13

6. Wide cavity wall with vertical twist ties 18

7. Pitched roof with insulation between and over the joists 21

8. Room in roof construction 24

9. Room in roof construction with limited rafter depth 27

10. Floor above unheated space 31

11. Suspended beam and block floor 34

12. Suspended timber ground floor 38

Appendix : Data tables 41

References 45

1. Introduction

For building elements which contain repeating thermal bridges, such as timber joistsbetween insulation in a roof, or mortar joints around lightweight blockwork in a wall,the effect of thermal bridges should be taken into account when calculating theU-value. At present, Building Regulations specify that U-values should be calculatedusing the Proportional Area Method, which is described in the CIBSE Guide, SectionA32. Future regulations, however, are likely to be based upon the method forcalculating U-values defined in BS EN ISO 6946:1997, which includes the CombinedMethod for repeating thermal bridges and correction procedures for the effects ofmetal fixings, air gaps and unheated spaces. This paper illustrates the use ofBS EN ISO 6946:1997 for some typical wall, roof and floor designs.

Thermal conductivity values for common building materials can be obtained from theCIBSE Guide Section A3, 1999 Edition (especially for masonry) or from prEN 125243.For specific insulation products, however, data should be obtained frommanufacturers’ declared values. A table is provided at the end of this documentgiving typical conductivities for some common building materials.

2. Outline of the procedure

The following is an outline of the calculation procedure:

1. Calculate the upper resistance limit (Rupper) by combining in parallel the totalresistances of all possible heat-flow paths (i.e. sections) through the buildingelement.

2. Calculate the lower resistance limit (Rlower) by combining in parallel theresistances of the heat flow paths of each layer separately and then summing theresistances of all layers of the building element.

3. Calculate the total thermal resistance (RT) from

2

RRR lowerupper

T

+=

4. Calculate, where appropriate, corrections for air gaps (∆Ug) and mechanicalfasteners (∆Uf). Examples of corrections for air gaps are shown in sections 3, 4,10 and 12 and examples of corrections for mechanical fasteners are shown insections 5, 6 and 9.

5. Calculate the U-value from

U = (1 / RT) + ∆Ug + ∆Uf

The standard permits ∆Ug and ∆Uf to be omitted if, taken together, they amountto less than 3% of the U-value. This has been done in the examples that follow.

3. Cavity wall with lightweight masonry leaf and insulated dry-lining

In this examplea) there are two bridged layers - insulation bridged by timber and

example). The construction consists of outer leaf brickwork, a clear cavity, 100 mmAAC blockwork, 38 89 mm timber studs with insulation between the studs and onesheet of 12.5 mm plasterboard.

The thickness of each layer, together with the thermal conductivities of the materials,are shown below. The external and internal surface resistances used are those

two thermal conductivities are given for each layer to reflect the bridged part and thebridging part in each case. For each homogeneous layer and for each section

(expressed in metres) by the thermal conductivity.

Layer Thickness(mm) conductivity

(W/m·K)resistance

(m²K/W)- -

1 outer leaf brick 0.77 0.132air cavity 50 0.180

3(a) 100 0.113(b) mortar (6.6%) 0.88 0.114

mineral wool (90.5%) 89 2.3424(b) (89) 0.135 plasterboard 0.25 0.050

- -

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

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

50 mm air cavity (thermal resistance 0.18 m²K/W)

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

mineral wool (conductivity 0.038 W/m·K)between 38 × 89 mm timber studs(conductivity 0.13 W/m·K) at 400 mm centres

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

heat flow

Total thickness 354 mm

U-value 0.31 W/m·K

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 examplethere 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

1upper

RF

R

F

RF

RF

1R

+++=

where F1, F2, F3 and F4 are the fractional areas of sections 1, 2, 3 and 4 respectivelyand 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 isshown 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 = 0.040Resistance of bricks = 0.132Resistance of air cavity = 0.180Resistance of AAC blocks (93.4%) = 0.909Resistance of mineral wool (90.5%) = 2.342Resistance of plasterboard = 0.050Internal surface resistance = 0.130Total thermal resistance (R1) = 3.783 m²K/W

Fractional area F1 = 0.845 (93.4% × 90.5%)

externalsurface

internalsurface

1 4(a)

21

53(b)

F1

F2

3(a) 4(b) 5F3

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 = 0.040Resistance of bricks = 0.132Resistance of air cavity = 0.180Resistance of mortar (6.6%) = 0.114Resistance of mineral wool (90.5%) = 2.342Resistance of plasterboard = 0.050Internal surface resistance = 0.130Total thermal resistance (R2) = 2.988 m²K/W


Resistance through section containing AAC blocks and timber

External surface resistance = 0.040Resistance of bricks = 0.132Resistance of air cavity = 0.180Resistance of AAC blocks (93.4%) = 0.909Resistance of timber (9.5%) = 0.685Resistance of plasterboard = 0.050Internal surface resistance = 0.130Total thermal resistance (R3) = 2.126 m²K/W


Resistance through section containing mortar and timber

External surface resistance = 0.040Resistance of bricks = 0.132Resistance of air cavity = 0.180Resistance of mortar (6.6%) = 0.114Resistance of timber (9.5%) = 0.685Resistance of plasterboard = 0.050Internal surface resistance = 0.130Total thermal resistance (R4) = 1.331 m²K/W


Combining these resistances we obtain:

=+++

=+++

=

331.1006.0

126.2089.0

988.2060.0

783.3845.0

1

RF

R

F

RF

RF

1R

4

4

3

3

2

2

1

1upper 3.450 m²K/W.

Lower resistance limit

When calculating the lower limit of thermal resistance, the resistance of a bridgedlayer is determined by combining in parallel the resistances of the unbridged part andthe bridged part of the layer. The resistances of all the layers in the element are thenadded 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 calculatedusing:

mortar

mortar

blocks

blocksfirst

R

F

R

F1

R+

=

and the resistance of the bridged layer consisting of insulation and timber iscalculated using:

timber

timber

insul

insulsecond

R

F

R

F1

R+

=

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

External surface resistance = 0.040Resistance of bricks = 0.132Resistance of air cavity = 0.180Resistance of first bridged layer

114.0066.0

909.0934.0

1R first

+= = 0.622

Resistance of second bridged layer

685.0095.0

2.3420.905

1Rsecond

+= = 1.904

Resistance of plasterboard = 0.050Internal surface resistance = 0.130Total (Rlower) = 3.058 m²K/W

Total resistance of wall

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

4(a)

21 5 internalsurface

3(a)

externalsurface

4(b)3(b)

=+

=+

=2

058.3450.32

RRR lowerupper

T 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 ofinsulation) a correction should be applied to the U-value in order to account for airgaps. 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), RI is thethermal resistance of the layer containing the gaps and RT is the total resistance ofthe 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 unlessthey 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 than3% 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 beapplied to the U-value to take account of them.

2. In the above calculation it is assumed that the noggings (or dwangs) do notpenetrate the whole of the insulation. If the noggings (or dwangs) do penetratethe whole of the insulation thickness they should be included as part of the timberpercentage used in the calculation.

4. Timber framed wall

In this exampleb) there is a single bridged layer in the wall, involving insulationbridged by timber studs. The construction consists of outer leaf brickwork, a clearventilated cavity, 19 mm plywood, 38 × 140 mm timber framing with 120 mm ofinsulation 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 thematerials in each layer, are shown below. The external and internal surfaceresistances are taken from Table 1 of this document. Layer 4 is thermally bridgedand two thermal conductivities are given for this layer, one for the unbridged part andone for the bridging part of the layer. For each homogeneous layer and for eachsection through a bridged layer, the thermal resistance is calculated by dividing thethickness (in metres) by the thermal conductivity.

Both the upper and the lower limits of thermal resistance are calculated by combiningthe 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 MaterialThickness

(mm)

Thermalconductivity

(W/m·K)

Thermalresistance

(m²K/W)external surface - - 0.040

1 outer leaf brick 102 0.77 0.1322 air cavity 50 - 0.0903 plywood 19 0.13 0.1464(a¹) mineral wool between timber

studs120 0.038 3.158

4(a²) air space next to mineral wool 20 - 0.1804(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


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 (thermalresistance 0.09 m²K/W)

19 mm plywood(conductivity 0.13 W/m·K)

mineral wool (conductivity0.038 W/m·K) between38 × 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 isconsidered to consist of two thermal paths (or sections). The upper limit of resistanceis calculated from:

2

2

1

1upper

RF

RF

1R

+=

where F1 and F2 are the fractional areas of the two sections and R1 and R2 are thetotal resistances of the two sections. The method of calculating the upper resistancelimit is illustrated conceptually below:-

Figure 4.2 : Conceptual illustration of how to calculate the upper limit of thermalresistance

Resistance through the section containing insulation

External surface resistance = 0.040Resistance of bricks = 0.132Resistance of air cavity = 0.090Resistance of plywood = 0.146Resistance of mineral wool (90.5%) = 3.158Resistance of air space next to mineral wool = 0.180Resistance of plasterboard = 0.100Internal surface resistance = 0.130Total (R1) = 3.976 m²K/W

Fractional area F1 = 0.905 (90.5%)

external

surface

internal

surface

1 4(a)2

21

53

F1

F2

3 4(b) 5

Resistance through section containing timber stud

External surface resistance = 0.040Resistance of bricks = 0.132Resistance of air cavity = 0.090Resistance of plywood = 0.146Resistance of timber studs (9.5%) = 1.077Resistance of plasterboard = 0.100Internal surface resistance = 0.130Total (R2) = 1.715 m²K/W


The upper limit of resistance is then:

=+

=+

=

715.1095.0

976.3905.0

1

R

F

R

F1

R

2

2

1

1upper 3.533 m²K/W

Lower resistance limit:-

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

The resistance of the bridged layer is calculated using:

timber

timber

insul

insul

R

F

R

F1

R+

=

The method of calculating the lower limit of resistance is illustrated conceptuallybelow.

Figure 4.3 : Conceptual illustration of how to calculate the lower limit of thermalresistance

4(a)

21 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 thelayers:

External surface resistance = 0.040Resistance of bricks = 0.132Resistance of air cavity = 0.090Resistance of plywood = 0.146

Resistance of bridged layer =

077.1095.0

180.0158.3905.0

1

++

= 2.783

Resistance of plasterboard = 0.100Internal surface resistance = 0.130Total (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 resistancelimits:

477.32

421.3533.32

RRR lowerupper

T =+

=+

= m²K/W

Correction for air gaps

If there are small air gaps penetrating the insulating layer a correction should beapplied to the U-value to account for this. The correctionc) for air gaps is ∆Ug, where

∆Ug = ∆U’’ × (RI / RT)²

and where RI d) 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 whichthe insulation is fitted. In this example RI is 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




c) Using Table D.1 of BS EN ISO 6946d) In this example RI is the same as the resistance of the bridged layer used in the calculationof the lower resistance limit

In this case ∆Ug = 0.026 W/m²K and 1 / RT = 0.288 W/m²K. Since ∆Ug is not lessthan 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 notpenetrate the whole of the insulation. If, however, the noggings (or dwangs)penetrate the whole of the insulation thickness they should be included within thetimber percentage used in the calculation.

2. In this example correction level 2 is appropriate because air may circulate on thewarm side of the insulation. If 140 mm of insulation was used instead of 120 mmso 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 surroundingopenings are taken account of in the treatment of such details and so are nottaken 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 waythat no air circulation is possible on the warm side of the insulation then ∆U’’ isset to 0.01 W/m²K. If, on the other hand, air circulation is possible on the warmside then it should be set to 0.04 W/m²K. The possible correction levels aresummarised as follows:

Description of air gap Correctionlevel

∆U’’

W/m²K

∆Ug

W/m²KInsulation installed in such a way that no aircirculation is possible on the warm side ofthe insulation. No air gaps penetrating theentire insulation layer.

0 0.00 0.000

Insulation installed in such a way that no aircirculation is possible on the warm side ofthe insulation. Air gaps may penetrate theinsulation layer.

1 0.01 0.006

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

2 0.04 0.026

5. Insulated cavity wall with metal wall ties

In this examplee) an insulated cavity wall has stainless steel double triangle wall tiespenetrating the insulation layer. The construction consists of outer leaf brickwork, acavity filled with mineral wool batts, 100 mm of AAC blockwork and 13 mm oflightweight plaster. The wall ties are spaced 900 mm horizontally and 450 mmvertically.

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

The thicknesses of each layer, together with the thermal conductivities of thematerials in each layer, are shown below. The external and internal surfaceresistances used are those given in Table 1 of this document. The metal ties are nottreated as repeating thermal bridges but instead are accounted for at the end of thecalculation. The third layer contains AAC blockwork bridged by mortar with themortar occupying 6.6% of the cross-sectional area.


(mm)

Thermalconductivity

(W/m·K)

Thermalresistance


1 outer leaf brick 102 0.77 0.1322 mineral wool batts 75 0.038 1.9743(a) AAC blocks (93.4%) 100 0.11 0.9093(b) mortar (6.6%) (100) 0.88 0.1144 lightweight plaster 13 0.18 0.072

internal surface - - 0.130

e) This construction provides satisfactory sound insulation from neighbouring dwellings


U-value 0.32 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)

heat flow

Upper resistance limit:-

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

2

2

1

1upper

RF

RF

1R

+=

where F1 and F2 are the fractional areas of the two sections and R1 and R2 are thetotal resistances of the two sections. A conceptual diagram of the upper limit ofresistance is shown immediately below


Resistance through section containing concrete blocks

External surface resistance 0.040Resistance of bricks 0.132Resistance of mineral wool slabs 1.974Resistance of AAC blocks (93.4%) 0.909Resistance of lightweight plaster 0.072Internal surface resistance 0.130Total (R1) 3.257


Resistance through section containing mortar

External surface resistance 0.040Resistance of bricks 0.132Resistance of mineral wool slabs 1.974Resistance of mortar (6.6%) 0.114Resistance of lightweight plaster 0.072Internal surface resistance 0.130Total (R2) 2.462


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

=+

=+

=

462.2066.0

257.3934.0

1

R

F

R

F1

R

2

2

1

1upper 3.189 m²K/W

external

surface

internalsurface

1 42

21

3(a)

F1

F2

3(b) 4


To calculate the lower resistance limit the resistance of the bridged layer isdetermined by combining in parallel the resistances of the unbridged part and thebridged part of the layer. The resistances of all the layers are then added together togive the lower limit of resistance. A conceptual illustration of the method ofcalculating the lower limit of resistance is shown below:-


External surface resistance = 0.040Resistance of bricks = 0.132Resistance of mineral wool slabs = 1.974

Resistance of AAC blocks & mortar =

114.0066.0

909.0934.0

1

+= 0.622

Resistance of lightweight plaster = 0.072Internal surface resistance = 0.130Total (Rlower) = 2.970

Total resistance of wall (ignoring wall ties)

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

=+

=+

=2

970.2189.32

RRR lowerupper

T 3.080 m²K/W

Correction for cavity wall ties

The method of calculating U-values as given in BS EN ISO 6946 requires thatmechanical fixings, such as cavity wall ties, be taken into account. The followingdescribes how the effect of the wall ties is incorporated into the U-value. The methodof correction is the same for both fully filled and partially filled cavity walls. Thecorrections are only applied where wall ties actually penetrate the insulation. Sincewall ties of low conductivity, such as plastic ties, do not affect the U-valuesignificantly, the Standard (BS EN ISO 6946) only requires a correction to be made ifthe conductivity of the tie, or part of it, is more than 1 W/m·K. In practice this meansthat plastic wall ties can be ignored in the U-value calculation but metal wall tiesgenerally need to be included.

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

3(a)

21 4

F1

F2

internal

surface3(b)

external

surface

horizontal centres and 450 mm vertical centres. Using the procedure in BS ENISO 6946, the adjustment to the U-value, ∆Uf, is given by

∆Uf = α λf nf Af

where α is the scaling factor for mechanical fixings, which is 6 for wall tiesf), and λf isthe conductivity of the fixings, which is 17 W/m·K in this example. The number ofwall ties per square metre which penetrate the insulation, nf, is calculated as followsusing the above information on the wall tie spacing.

47.24509000000001

n f =×

=

Af is the cross-sectional area of the wall tie, expressed in m², which is0.00001075 m². A value for ∆Uf of 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


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 than3% 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 anycorrection 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 mechanicalfixings 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 correctionneed 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 tothe U-value can be considerably larger than that shown above, due to theirgreater cross-sectional area.

3. If instead of stainless steel ties, galvanised steel ties of conductivity 50 W/m·K areused, this will increase ∆Uf from 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, nocorrection is applied and ∆Uf is taken to be zero. This would apply, for instance,in the case of plastic wall ties.

f) see Table 3 of this document

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

Figure 5.4 : Conceptual diagram illustrating how the U-value is corrected for thepresence of wall ties. The U-value calculation is firstly carried out ignoring the effectsof 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 ofwall ties

6. Wide cavity wall with vertical twist ties

In this exampleg) a wide cavity wall is fully filled with mineral wool insulation withstainless steel vertical twist wall ties in the filled cavity. To obtain the U-valueallowing for the wall ties the thermal resistance (RT) should first be calculatedignoring the effect of the wall ties and then a correction should be made for thepresence of the ties. The wall ties are spaced 750 mm horizontally and 450 mmvertically.

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

The thicknesses of each layer, together with the thermal conductivities of thematerials, are shown below. The external and internal surface resistances used arethose given in Table 1 of this document. The vertical twist wall ties have a crosssectional area of 60.8 mm².

In this example there is no distinction between the upper and lower limit of resistancebecause all of the layers are considered to be sufficiently homogeneous (for thepurposes of thermal calculations). Strictly speaking, the mortar joints between thebricks and concrete blocks could be taken into account, however since theresistances of the mortar parts do not differ from the brick or block parts by morethan 0.1 m²K/W the mortar parts may be ignored.

g) Due to requirements for sound insulation this construction may only be suitable fordetached dwellings


U-value 0.30 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)

heat flow

The wall construction may be summarised as follows:


(mm)

Thermalconductivity

(W/m·K)

Thermalresistance

(m²K/W) external surface - - 0.0401 outer leaf brick 102 0.77 0.1322 mineral wool batts 120 0.038 3.1583 concrete blockwork 100 1.13 0.0884 dense plaster 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-sectionalarea, Af, of 60.8 mm². Using the procedure in BS EN ISO 6946, the correction to beapplied, ∆Uf, is given by

∆Uf = α λf nf Af

where α is the scaling factor for mechanical fixings, which is 6 for wall tiesh), and λf isthe conductivity of the fixings, which is 17 W/m·K for stainless steel. The number ofwall 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. Thecorrection to be applied is therefore

∆Uf = 6 × 17 × 2.96 × 0.0000608 = 0.018 W/m²K

U-value of the wall




In this case ∆Uf = 0.018 W/m²K and 1 / RT = 0.280 W/m²K. Since ∆Uf is not less than3% 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 ofstainless 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) see Table 3 of this document

2. If the thermal conductivity of the tie, or part of it, is less than 1 W/m·K (eg. plasticties) the value of ∆Uf may be taken to be zero and the U-value will beU = 1 / RT = 1 / 3.571 = 0.28 W/m²K

3. Strictly speaking, the mortar joints between the bricks and concrete blocks couldbe taken into account in the U-value calculation, however it is permissible toignore the mortar in both of these layers because the resistances of the mortarjoints differ from the resistances of the bricks or concrete blocks by less than0.1 m²K/W

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 timberjoists spaced 600 mm apart (centres to centres) and 100 mm of mineral wool overthe joists. The roof is tiled with felt or boards under the tiles. The external andinternal surface resistances used are those given in Table 1 of this document. Theceiling consists of 12.5 mm of plasterboard. The roof construction is summarisedbelow.

Layer Material Thickness(mm)

Thermalconductivity

(W/m·K)

Thermalresistance


1 roof space beneath tiled roof with felt orboardsi)

- - 0.200

2 continuous layer of mineral wool 100 0.042 2.3813(a) mineral wool between 48 × 100 mm timber

joists with 600 mm between centres100 0.042 2.381

3(b) 48 × 100 mm timber joists between insulation (100) 0.13 0.7694 plasterboard 12.5 0.25 0.050

internal surface - - 0.100


A conceptual illustration of how the upper limit of resistance is calculated is shownimmediately below


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

surfaceinternalsurface

1 42

21

3(a)

F1

F2

3(b) 4

Resistance through section containing both layers of insulation

External surface resistance = 0.040Resistance of roof spacei) = 0.200Resistance of mineral wool over joists = 2.381Resistance of mineral wool between joists = 2.381Resistance of plasterboard = 0.050Inside surface resistance = 0.100Total (R1) = 5.152 m²K/W

Fractional area F1 = 0.92 (92%)

Resistance through section containing timber joists

External surface resistance = 0.040Resistance of roof space = 0.200Resistance of mineral wool over joists = 2.381Resistance of timber joists = 0.769Resistance of plasterboard = 0.050Inside surface resistance = 0.100Total (R2) = 3.540 m²K/W

Fractional area F2 = 0.08 (8%)

The upper resistance limit is given by

971.4

540.308.0

152.592.0

1

R

F

R

F1

R

2

2

1

1upper =

+=

+= m²K/W


A conceptual illustration of the method of calculating the lower limit of resistance isshown below:-


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

769.008.0

381.292.0

1

R

F

R

F1

timber

timber

insul

insul +=

+= = 2.039

Resistance of plasterboard = 0.050Inside surface resistance = 0.100Total (Rlower) = 4.810 m²K/W

Total resistance of roof

=+

=+

=2

810.4971.42

RRR lowerupper

T 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 acontinuous layer covering the first layer, a correction for air gaps need not beapplied.

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

3(a)

21 4

F1

F2

internal

surface3(b)

external

surface

8. Room in roof construction

An existing loft is converted to a habitable space by inserting tightly fitted insulationbetween the rafters in the roof. Timber packing pieces of the same width as existing100 mm deep rafters are attached beneath the rafters in order to provide additionalroom for insulation. Plasterboard, laminated to insulation, is then attached below therafters. 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 and100 mm of insulation between rafters and spacers. Beneath the rafters and spacersthere is an insulation laminate consisting of 15.5 mm of insulation bonded to 9.5 mmof plasterboard. In this example the rafters are 100 mm deep but 50 mm timberspacers have been attached below the rafters in order to extend the total rafter depthto effectively 150 mm.

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

The construction may be summarised as follows:-


Thermalconductivity(W/mK)

Thermalresistance(m²K/W)

external surface* - - -1 tiles* 19 - -2 roofing felt* 1 - -3 ventilated airspace between rafters

and spacers*50 - 0.100

4(a) insulation boardj) occupying 88% offace area (between rafters andspacers)

100 0.025 4.000

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

(100) 0.13 0.769

5 insulation boardj) 15.5 0.025 0.6206 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-valuecalculation and the surface bounding this airspace is taken to have the same resistanceas an internal surface. The internal surface resistance is taken from Table 1 of thisdocument.

j) For example, phenolic foam or polyurethane, where the conductivity has an allowance forageing and variation in manufacture

Plan at A-A

U-value 0.27 W/m·KA

A

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

Conceptual diagrams of the methods of calculating upper and lower limits ofresistance are shown below:-

Figure 8.2 : Conceptual diagrams of how to calculate the upper and lower limits ofresistance


Resistance through the section between the rafters

Effective external surface resistance = 0.100Resistance of insulation between rafters = 4.000Resistance of insulation beneath rafters = 0.620Resistance of plasterboard = 0.038Internal surface resistance = 0.100Total thermal resistance (R1) = 4.858 m²K/WFractional area F1 = 0.88 (88%)

Resistance through the section through the rafters

Effective external surface resistance = 0.100Resistance of rafters = 0.769Resistance of insulation beneath rafters = 0.620Resistance of plasterboard = 0.038Internal surface resistance = 0.100Total thermal resistance (R2) = 1.627 m²K/WFractional area F2 = 0.12 (12%)

The upper limit of resistance is then obtained from:

923.3

627.112.0

858.488.0

1

R

F

R

F1

R

2

2

1

1upper =

+=

+= m²K/W

F1 F2

externalsurface

insulation

plasterboard

internalsurface

rafters

insulation

externalsurface

insulation

plasterboard

internalsurface

rafters

insulation

Lower resistance limit

Effective external surface resistance = 0.100Resistance of bridged layer

769.012.0

000.488.0

1

+= = 2.659

Resistance of insulation beneath rafters = 0.620Resistance of plasterboard = 0.038Internal surface resistance = 0.100Total thermal resistance (Rlower) = 3.517 m²K/W

Total resistance of roof

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

RT = 720.32

517.3923.32

RR lowerupper =+

=+

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 timberspacers can be attached below the rafters. In instances where the rafters areinsufficiently deep (e.g. only 75 mm) there may be practical problems in achievingthe required U-value due to a lack of space being available for the insulation. Insuch cases the insulation beneath the rafters may need to be thicker in order tocompensate for the limited rafter depth.

2. In this example the effects of the fixing nails may be ignored since they do notpenetrate the main insulating layer.

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

9. Room in roof construction with limited rafter depth

This roof is similar to that shown in the previous example except that the existingrafters, which are only 75 mm deep in this case, are not extended in depth butinstead a thicker plasterboard-insulation laminate is attached below the rafters. Theconstruction consists of roof tiles, felt, a 50 mm air gap between rafters and 25 mm ofinsulation between rafters and spacers. As in the previous example the insulation istightly fitted between the rafters. Beneath the rafters and spacers there is aninsulation laminate consisting of 57.5 mm of insulation bonded to 12.5 mm ofplasterboard. The insulation laminate is nailed to the rafters and the nails have ahorizontal spacing of 400 mm and a vertical spacing of 150 mm. The external andinternal surface resistances used are those given in Table 1 of this document. Tocalculate the U-value a calculation is first carried out ignoring the nails and then acorrection is applied to account for the nails.

Figure 9.1 : Roof construction

The construction may be summarised as follows:-


Thermalconductivity(W/mK)


external surface* - - -1 tiles* 19 - -2 roofing felt* 1 - -3 ventilated airspace between rafters

and spacers*50 - 0.100

4(a) insulation board occupying 88% offace area (between rafters andspacers)

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 2.3006 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-valuecalculation and the surface bounding this airspace is taken to have the same resistance asan internal surface. The internal surface resistance is taken from Table 1 of thisdocument.

Plan at A-A

A

A

U-value 0.33 W/m·K

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

Conceptual diagrams of the methods of calculating upper and lower limits ofresistance are shown below:-

Figure 9.2 : Conceptual diagrams of how to calculate the upper and lower limits ofresistance


Resistance through the section between the rafters

Effective external surface resistance = 0.100Resistance of insulation between rafters = 1.000Resistance of insulation beneath rafters = 2.300Resistance of plasterboard = 0.050Internal surface resistance = 0.100Total thermal resistance (R1) = 3.550 m²K/WFractional area F1 = 0.88 (88%)

Resistance through the section through the rafters

Effective external surface resistance = 0.100Resistance of rafters = 0.192Resistance of insulation beneath rafters = 2.300Resistance of plasterboard = 0.050Internal surface resistance = 0.100Total thermal resistance (R2) = 2.742 m²K/WFractional area F2 = 0.12 (12%)


429.3

742.212.0

550.388.0

1

R

F

R

F1

R

2

2

1

1upper =

+=

+=

externalsurface

plasterboardinternalsurface

rafters

insulation

insulation

F2F1

externalsurface

insulation

plasterboard

internalsurface

rafters

insulation


Effective external surface resistance = 0.100Resistance of bridged layer

192.0

12.0

000.1

88.0

1

+= = 0.664

Resistance of insulation beneath rafters = 2.300Resistance of plasterboard = 0.050Internal surface resistance = 0.100Total 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 = 322.32

214.3429.32

RR lowerupper =+

=+

m²K/W

Correction for the presence of fixing nails

The method of calculating U-values as given in BS EN ISO 6946 requires thatmechanical fixings, such as nails or screws for example, be taken into account. Thefollowing 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 mmapart the number of nails per square metre of sloping ceiling will be nf where

7.161504000000001

n f =×

= / m²

The nails are made of steel with a their thermal conductivity, λf, of 50 W/m·K. Theircross-sectional area, Af, is 5 mm² or 0.000005 m². The adjustment to the U-value is∆Uf, where

∆Uf = α λf nf Af = 5 × 50 × 16.7 × 0.000005 = 0.021 m²K/W.

where α is 5 for all roof fixingsk), Rf is the thermal resistance of the insulationpenetrated by the nails and RT is the total thermal resistance of the roof.

U-value of the roof




k) see Table 3 of this document

In this case ∆Uf = 0.021 m²K/W and 1 / RT = 0.301 W/m²K. Since ∆Uf is not lessthan 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 acontinuous layer covering the first layer, a correction for air gaps need not be applied.

10. Floor of heated room above an unheated space

In this example a floor has insulation between timber joists. The floor is situatedabove 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 woolinsulation (of conductivity 0.040 W/m·K) between the joists and 12.5 mm ofplasterboard over the unheated space. The total area (Ai) of components betweenthe internal environment and the unheated space is 35 m² and the total area (Ae) ofcomponents between the unheated space and the external environment is 35 m².

Using the procedure in BS EN ISO 6946 for unheated spaces, an additional thermalresistance, 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 glassmineral wool between the joists. Below the joists is 12.5 mm plasterboard formingthe ceiling of the garage. The external and internal surface resistances used arethose given in Table 1 of this document.


Thermalconductivity

(W/m·K)

Thermalresistance

(m²K/W)internal surface - - 0.170

1 plywood 19 0.13 0.1462(a) glass mineral wool 150 0.040 3.7502(b) timber joists (occupying 12%) (150) 0.13 1.1543 plasterboard 12.5 0.25 0.0504 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 withmineral wool between thejoists. Timber fraction is0.12 (i.e. 12%)


U-value 0.25 W/m·K

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


Resistance through the section containing the insulation:

Internal surface resistance = 0.170Resistance of plywood = 0.146Resistance of mineral wool insulation = 3.750Resistance of plasterboard = 0.050Ru = 0.490External surface resistance = 0.040Total thermal resistance (R1) = 4.646 m²K/W

Resistance through the section containing joists:

Internal surface resistance = 0.170Resistance of plywood = 0.146Resistance of timber joists = 1.154Resistance of plasterboard = 0.050Ru = 0.490External surface resistance = 0.040Total thermal resistance (R2) = 2.050 m²K/W


033.4

050.212.0

646.488.0

1

R

F

R

F1

R

2

2

2

1upper =

+=

+= m²K/W

Ru

internal

surface

plywood

2(a)

plaster-

board

external

surface

2(b)

F1

Ru

internalsurface

plywood

2(a)

plaster-board

external

surface

2(b)


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

154.112.0

750.388.0

1

+= = 2.953

Resistance of plasterboard = 0.050Ru = 0.490External surface resistance = 0.040Total (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

849.3033.42

RRR lowerupper

T 3.941 m²K/W


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

∆Ug = ∆U’’ × (RI / RT)²


∆Ug = 0.01 × (2.953 / 3.941)² = 0.005 W/m²K

U-value of the floor




In this case ∆Ug = 0.005 W/m²K and 1 / RT = 0.254 W/m²K. Since ∆Ug is less than3% of (1 / RT),

U = 1 / RT = 1 / 3.941 = 0.25 W/m²K.

11. Suspended beam and block floor

A beam and block floor consists of blocks of lightweight concrete which are 100 mmthick and 440 mm wide suspended on T-beams which are 70 mm wide. Above thebeams 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. Thebeams protrude below the blocks by 75 mm. The perimeter of the ground floor is35.6 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 thethermal resistance between the dwelling and the underfloor space.

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


Thermalconductivity(W/m·K)


internal surface - - 0.1701 screed 65 0.41 0.1592 polystyrene 100 0.040 2.5003(a) light concrete blocks, 440

mm wide100 0.18 0.556

3(b) concrete beams,70 mm wide

(100) 1.13 0.088

4 lower surface - - 0.170*

* The internal surface resistance is taken from Table 1 of this document. The surfaceresistance for the lower side of the floor deck is taken to be 0.17 m²K/W, as this is thevalue 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·K440

100

70

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

Since the conductivity of the beams is less than 2.0 W/m·K the part of the beamwhich 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 usingBS EN ISO 6946, as follows:

Upper resistance limit (of floor deck)

Resistance through section containing lightweight blocks

Internal surface resistance = 0.170Resistance of screed = 0.159Resistance of polystyrene = 2.500Resistance of light concrete blocks = 0.556Resistance of lower surface of floor deck = 0.170Total thermal resistance (R1) = 3.555 m²K/WFractional area F1 = 0.863 (i.e. 86.3%)

Resistance through section containing concrete beams

Internal surface resistance = 0.170Resistance of screed = 0.159Resistance of polystyrene = 2.500Resistance of beams = 0.088Resistance of lower surface of floor deck = 0.170Total thermal resistance (R2) = 3.087 m²K/WFractional area F2 = 0.137 (i.e. 13.7%)


483.3

087.3137.0

555.3863.0

1

R

F

R

F1

R

2

2

1

1upper =

+=

+=

internalsurface

beams

lowersurface

insulation

screed

blocks

underfloor space & soil

internalsurface

beams

lowersurface

insulation

screed

blocks

underfloor space & soil

Lower resistance limit (of floor deck):-

Internal surface resistance = 0.170Resistance of screed = 0.159Resistance of polystyrene = 2.500Resistance of bridged layer

088.0137.0

556.0863.0

1

+= = 0.322

Resistance of lower surface of deck = 0.170Total (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 resistancelimits

402.32

321.3483.32

RRR lowerupper =

+=

+= 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 inmind that Uf includes the surface resistances for the upper and lower sides of thedeck.

Resistance of the remainder of the floor

For determining the resistance of the remaining part of the floor, Table 4 of theAppendix 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 standardassumptions about the thermal resistance of the floor deck and the surfaceresistances at the upper and lower sides of the deck. Since Uf, calculated above,already includes surface resistances the surface resistances need to be subtractedfrom U0.

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

−−−+

=

lower,siduninsulate,deckupper,si0f

RRRU1

U1

1U

where Rsi,upper is the surface resistance of the upper side of the floor deck, equal to0.17 m²K/W (see Table 1 of this document), Rdeck,uninsulated is the thermal resistance ofa notional uninsulated floor deck, equal to 0.20 m²K/W (see CIBSE Guide A3, part3.5.5.2), and Rsi,lower is the surface resistance of the lower side of the floor deck,equal to 0.17 m²K/W (see Table 1 of this document).

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


The U-value of the suspended floor is therefore

( )

( )

K²m/W24.0

17.02.017.076.01

294.01

1

RRRU1

U1

1U

lower,siduninsulate,deckupper,si0f

=

++−+=

++−+=

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 lowersurface resistances of the (notional) floor deck are obtained from Table 1.

Summary of floor detailsexposed perimeter (P) 35.6 mfloor area (A) 79.1 m²perimeter to area ratio (P/A) 0.45 m-1

wall thickness (w) 0.3 msoil type sandyventilation parameter (ε) 0.015 m²/m

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% timberfraction. Between the joists there is 150 mm of tightly fitted mineral wool (with aconductivity of 0.040 W/m·K) suspended on netting. Beneath the floor deck there isan underfloor space over clay soil. The perimeter of the ground floor is 40 metresand the area is 100 m². BS EN ISO 6946 is applied to obtain the thermal resistanceof the floor deck.


Thermalconductivity(W/m·K)


internal surface - - 0.1701 chipboard 19 0.13 0.1462(a) glass mineral wool on netting 150 0.040 3.7502(b) timber joists (occupying 12%) (150) 0.13 1.154

lower surface - - 0.170*

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

The methods of calculating the upper and lower limits of resistance are illustratedconceptually below:-

Figure 12.2 : Conceptual illustration of the methods of calculating the upper andlower limits of thermal resistance

U-value 0.22 W/m·K

underfloorspace and soil

lowersurface

mineralwool

chipboard

joists

uppersurface

underfloorspace and soil

lowersurface

mineralwool

chipboard

joists

uppersurface

Upper resistance limit (for floor deck):-

Resistance through section containing mineral wool on netting

Internal surface resistance = 0.170Resistance of chipboard = 0.146Resistance of mineral wool = 3.750Resistance of lower surface of floor deck= 0.170Total thermal resistance (R1) = 4.236 m²K/W

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

Resistance through section containing timber joists

Internal surface resistance = 0.170Resistance of chipboard = 0.146Resistance of timber = 1.154Resistance of lower surface of floor deck= 0.170Total thermal resistance (R2) = 1.640 m²K/W

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

The upper limit of resistance is:

560.3

640.112.0

236.488.0

1

R

F

R

F1

R

2

2

1

1upper =

+=

+= m²K/W

Lower resistance limit (for floor deck)

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

=

154.112.0

750.388.0

1

+= 2.953

Resistance of lower surface of floor deck= 0.170Total (Rlower) = 3.439 m²K/W

Overall resistance of floor deck

The resistance of the floor deck is the average of the upper and lower resistancelimits

500.32

439.3560.32

RRR lowerupper

T =+

=+

= m²K/W


Since the insulation layer is entirely between joists a correction should be applied tothe floor deck U-value in order to account for air gaps. The overall U-value of thefloor deck should be adjusted by adding a term ∆Ug, where

∆Ug = ∆U’’ × (RI/RT)²


∆Ug = 0.01 × (2.953 / 3.500)² = 0.007 W/m²K

U-value of the floor deck (Uf)


Uf = 1 / RT + ∆Ug (if ∆Ug is not less than 3% of 1 / RT)

Uf = 1 / RT (if ∆Ug is less than 3% of 1 / RT)

In this case ∆Ug = 0.007 W/m²K and 1 / RT = 0.286 W/m²K. Since ∆Ug is lessthan 1 / RT the U-value of the floor deck is

Uf = 1 / 3.500 = 0.286 W/m²K


Since the floor perimeter is 40 m, the area 100 m, and the ground of clay soil, theU-value of the floor ignoring insulation is

U0 = 0.65 W/m²K (using Table 4)

( ) ( )17.02.017.065.01

286.01

1

17.02.017.0U1

U1

1U

0f

++−+=

++−+=

= 0.22 W/m²K

Note

The value of 0.2 used in the above equation is based on the CIBSE Guide A3(part 3.5.5.2 of the 1999 edition) and represents the thermal resistance of a notionaluninsulated floor deck. The figures 0.17 and 0.17 represent the surface resistancesof the upper and lower surface resistances for the same notional floor deck (takenfrom Table 1 of this document).

Appendix : Data tables

Table 1 Surface resistances for roofs, walls and exposed floors (m²K/W)from BS EN ISO 6946

Direction of heat flowUpwards Horizontal Downwards

inside resistance 0.10 0.13 0.17outside resistance 0.04 0.04 0.04underfloor space* - 0.13 0.17*These values should be used for the upper and lower surfaces of the underfloorspace according to BS EN ISO 13370:1998

Table 2 Air space resistances for roofs, walls and exposed floors (m²K/W)from BS EN ISO 6946

Direction of heat flowThickness of airlayer (mm) Upwards Horizontal Downwards

0 0.00 0.00 0.005 0.11 0.11 0.117 0.13 0.13 0.13

10 0.15 0.15 0.1515 0.16 0.17 0.1725 0.16 0.18 0.1950 0.16 0.18 0.21100 0.16 0.18 0.22300 0.16 0.18 0.23

Table 3 Scaling factors for ceiling fixings and wall tiesfrom BS EN ISO 6946scaling factor (α) type of mechanical fastenings

5 roof fixings6 wall ties between masonry leaves

The following table provides U-values for suspended floors without insulation. For adetailed calculation of the thermal resistance below the deck of a ground floor thereader is referred to the procedure in BS EN ISO 13370.

Table 4 U-values of uninsulated suspended floors(from CIBSE Guide A3)

Soil type and ventilation opening area per unit perimeter ofunderfloor space (in m²/m)

clay/silt sand/gravel homogeneousrock

perimeter/area 0.0015 0.003 0.0015 0.003 0.0015 0.0030.05 0.16 0.17 0.19 0.20 0.27 0.280.10 0.27 0.29 0.32 0.33 0.43 0.440.15 0.36 0.38 0.42 0.43 0.54 0.550.20 0.44 0.46 0.49 0.51 0.63 0.64

0.25 0.50 0.52 0.56 0.58 0.70 0.710.30 0.56 0.58 0.62 0.64 0.76 0.770.35 0.61 0.63 0.67 0.69 0.81 0.820.40 0.65 0.68 0.72 0.74 0.85 0.87

0.45 0.69 0.72 0.76 0.78 0.89 0.910.50 0.73 0.76 0.79 0.82 0.92 0.940.55 0.76 0.79 0.83 0.85 0.95 0.970.60 0.79 0.83 0.86 0.88 0.98 1.00

0.65 0.82 0.85 0.88 0.91 1.00 1.020.70 0.85 0.88 0.91 0.94 1.03 1.050.75 0.87 0.91 0.93 0.96 1.05 1.070.80 0.90 0.93 0.95 0.98 1.06 1.09

0.85 0.92 0.95 0.97 1.00 1.08 1.110.90 0.94 0.97 0.99 1.02 1.10 1.120.95 0.96 0.99 1.01 1.04 1.11 1.141.00 0.98 1.01 1.03 1.06 1.13 1.15

Table 5 Thermal conductivity of some common building materials

Density Conductivity(kg/m³) (W/m·K)

WallsBrickwork (outer leaf) 1700 0.77Brickwork (inner leaf) 1700 0.56

Concrete block (medium density) 1400 0.57Concrete block (low density) 600 0.18

Concrete (medium density) (inner leaf) 1800 1.132000 1.332200 1.59

Concrete (high density) : 2400 1.93

Reinforced concrete (1% steel) 2300 2.3Reinforced concrete (2% steel) 2400 2.5

Mortar (protected) 1750 0.88Mortar (exposed) 1750 0.94

Gypsum 600 0.18900 0.30

1200 0.43

Gypsum plasterboard 900 0.25

Sandstone 2600 2.3Limestone, soft 1800 1.1Limestone, hard 2200 1.7

Fibreboard 400 0.1Plasterboard 900 0.25Tiles ceramic 2300 1.3Timber (softwood) 500 0.13

700 0.18Steel 7800 50.0Stainless steel 7900 17.0

Surface finishesExternal rendering 1300 0.57Plaster (dense) 1300 0.57Plaster (lightweight) 600 0.18

RoofsAerated concrete slab 500 0.16Asphalt 2100 0.70Felt/bitumen layers 1100 0.23Screed 1200 0.41Stone chippings 2000 2.0Tiles (clay) 2000 1.0Tiles (concrete) 2100 1.5Wood wool slab 500 0.10

FloorsCast concrete 2000 1.35Metal tray (steel) 7800 50.0Screed 1200 0.41Hardwood timber 700 0.18Softwod timber, plywood, chipboard 500 0.13

1000 0.24InsulationExpanded polystyrene (EPS) board 15 0.040Mineral wool quilt 12 0.042Mineral wool batt 25 0.038Phenolic foam board 30 0.025Polyurethane board 30 0.025

Note: If available, certified test values should be used in preference to those in thetable.

References

1. BS EN ISO 6946:1997; Building components and building elements – Thermalresistance and thermal transmittance – Calculation method. British StandardsInstitution, London, 1997

2. CIBSE Guide A3, Environmental design : Thermal properties of buildingstructures; Chartered Institution of Building Services Engineers, London, 1999

3. prEN 12524 Building materials and products – Hygrothermal properties –Tabulated design values

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Thermal conductivity