General
A1.1 General guidance on the calculation of U-values is contained in Report BR 443 “Conventions for the Calculation of U-values” 2006. For building elements and components generally, the method of calculating U-values is specified in I.S. EN ISO 6946:2017. U-values of components involving heat transfer to the ground, e.g. ground floors with or without floor voids, basement walls, are calculated by the method specified in I.S. EN ISO 13370:2017. A soil thermal conductivity of 2.0 W/mK should be used, unless otherwise verified. U-values for windows, doors and shutters may be calculated using I.S. EN ISO 10077-1:2017 or I.S. EN ISO 10077-2:2017. U-values for curtain walling should be calculated using I.S. EN ISO 12631:2017 or tested in accordance to I.S. EN ISO 12567-1:2010. Information on U-values and guidance on calculation procedures contained in the 8th edition of CIBSE Guide Section A3: Thermal Properties of Building Structures are based on these standards and may be used to show compliance with this Part.
A method for assessing U-values of light steel-framed constructions is given in Digest 465 “U-values for light steel construction”, published BRE. Guidance in relation to the calculation of U-values for various forms of metal clad construction can be found in Technical Paper No. 14 “Guidance for the design of metal roofing and cladding to comply with Approved Document L2: 2001” published by MCRMA, Technical Information Sheet No. 312, “Metal cladding: U-value calculation assessing thermal performance of built-up metal roof and wall cladding systems using rail and bracket spacers” published by SCI and IP 10/02 “Metal cladding: assessing thermal performance of built-up systems which use ‘Z’ spacers” published by BRE.
Software packages to perform U-value calculations in accordance with the standards above are readily available.
A1.2 U-values derived by calculation should be rounded to two significant figures and relevant information on input data should be provided. When calculating U-values the effects of timber joists, structural and other framing, mortar bedding, window frames and other small areas where thermal bridging occurs must be taken into account. Similarly, account must be taken of the effect of small areas where the insulation level is reduced significantly relative to the general level for the component or structure element under consideration. Thermal bridging may be disregarded, however, where the general thermal resistance does not exceed that in the bridged area by more than 0.1 m2K/W. For example, normal mortar joints need not be taken into account in calculations for brickwork or concrete blockwork where the density of the brick or block material is in excess of 1500 kg/m3 . A ventilation opening in a wall or roof (other than a window, rooflight or door opening), may be considered as having the same U-value as the element in which it occurs.
A1.3 Examples of the application of the calculation method specified in I.S. EN 6946: 2017 are given below. An example of the calculation of ground floor U-values using I.S. EN ISO 13370:2017 is also given.
A1.4 Thermal conductivities of common building materials are given in Table A1. For the most part, these are taken from I.S. EN ISO 10456:2007 or CIBSE Guide A, Appendix 3.A7. Values for common insulation materials are also available in these documents. See paragraph 0.3.3 regarding application of these Tables.
Table HL8 - Thermal conductivity of some common building materials - Extract from TGD L
Simple structure without thermal bridging
A2.1 To calculate the U-value of a building element (wall or roof) using I.S.EN ISO 6946:2017, the thermal resistance of each component is calculated, and these thermal resistances, together with surface resistances as appropriate, are then combined to yield the total thermal resistance and U-value. The result is corrected to account for mechanical fixings (e.g. wall ties) or air gaps if required. For an element consisting of homogenous layers with no thermal bridging, the total resistance is simply the sum of individual thermal resistances and surface resistances. I.S. EN 6946:2017 provides for corrections to the calculated U-value. In the case of Example A1 (see Diagram A1), corrections for air gaps in the insulated layer and for mechanical fasteners may apply. However, if the total correction is less than 3 %of the calculated value, the correction may be ignored.
In this case no correction for air gaps applies as it is assumed that the insulation boards meet the dimensional standards set out in I.S. ENISO 6946:2017 and that they are installed without gaps greater than 5 mm. The construction involves the use of wall ties that penetrate fully through the insulation layer.
A potential correction factor applies which, assuming the use of 4 mm diameter stainless steel ties at 5 ties per m2 , is calculated as 0.006 W/m2K. This is equal to 3 % of the calculated U-value and the corrected U-value for this structure would be 0.19 W/m2K. It should be noted that, if galvanised steel wall ties were used, a correction of 0.02 W/m2K would apply, and the corrected U-value for this construction would be 0.20 W/m2K.
Example A1 - Masonry Cavity Wall
Diagram HL3 - Example A1: Masonry cavity wall - Extract from TGD L
Structure with Bridged Layer(s)
A2.2 For an element in which one or more layers are thermally bridged, the total thermal resistance is calculated in three steps as follows:
a) The upper thermal resistance is based on the assumption that heat flows through the component in straight lines perpendicular to the element's surfaces. To calculate it, all possible heat flow paths are identified, for each path the resistance of all layers are combined in series to give the total resistance for the path, and the resistances of all paths are then combined in parallel to give the upper resistance of the element.
b) The lower thermal resistance is based on the assumption that all planes parallel to the surfaces of the component are isothermal surfaces. To calculate it, the resistances of all components of each thermally bridged layer are combined in parallel to give the effective resistance for the layer, and the resistances of all layers are then combined in series to give the lower resistance of the element.
c) The total thermal resistance is the mean of the upper and lower resistances.
The percentage of timber bridging the insulation layer as a repeating thermal bridge can be calculated when the size and frequency of the timber members is known. Alternatively, the figures given in Table A2 can be used.
Diagram A2 Diagram HLA4 - Timber-frame wall - Extract from TGD L
Table A2 HL91 - Extract from TGD L - Timber fractions for bridged layers
The thermal resistance of each component is calculated (or, in the case of surface resistances, entered) as follows:
Upper resistance
Assuming that heat flows in straight lines perpendicular to the wall surfaces, there are four heat flow paths:
Path 1: through the internal timber battens and through the timber studs;
Path 2: through the internal timber battens and through insulation between the studs;
Path 3: not through the internal timber battens and through the timber studs;
Path 4: not through the internal timber battens and through the insulation between the timber studs.
The resistance of each of these paths is calculated as follows:
Resistance through path 1 [m2K/W]:
External surface resistance 0.040
Brick outer leaf 0.132
Air cavity 0.180
Sheathing ply 0.069
Timber studs 0.742
Insulation over studs 2.381
Timber battens 0.269
Plasterboard 0.006
Internal surface resistance 0.130
Total R1 = 3.949
Fraction of path 1: F1 = 0.12 x 0.15 = 1.8 %
Resistance through path 2 [m2K/W]:
External surface resistance 0.040
Brick outer leaf 0.132
Air cavity 0.180
Sheathing ply 0.069
Low-E air cavity 0.440
Insulation between studs 2.857
Insulation over studs 2.381
Timber battens 0.269
Plasterboard 0.006
Internal surface resistance 0.130
Total R2 = 6.504
Fraction of path 2: F2 = 0.12 x 0.85 = 10.2 %
Resistance through path 3 [m2K/W]:
External surface resistance 0.040
Brick outer leaf 0.132
Air cavity 0.180
Sheathing ply 0.069
Timber studs 0.742
Insulation over studs 2.381
Air cavity 0.180
Plasterboard 0.006
Internal surface resistance 0.130
Total R3 = 3.860
Fraction of path 3: F3 = 0.88 x 0.15 = 13.2 %
Resistance through path 4 [m2K/W]:
External surface resistance 0.040
Brick outer leaf 0.132
Air cavity 0.180
Sheathing ply 0.069
Low-E air cavity 0.440
Insulation between studs 2.857
Insulation over studs 2.381
Air cavity 0.180
Plasterboard 0.006
Internal surface resistance 0.130
*Total R4 = 6.415 *
Fraction of path 4: F4 = 0.88 x 0.85 = 74.8 %
The upper thermal resistance Ru is obtained from:
Ru = 1 / (F1 / R1 + F2 / R2+ F3 / R3 + F4 / R4)
where F1, F2, F3 and F4 are the fractional areas of heat flow paths 1, 2, 3 and 4 and R1, R2, R3 and R4 are the resistances of these paths.
Upper resistance Ru = 1 / (0.018 / 3.949 + 0.102 / 6.504 + 0.132 / 3.860 + 0.748 / 6.415) = 5.847 m2K/W
Lower resistance
Assuming an isothermal plane on each face of the layer of insulation which is bridged by timber studs, the thermal resistance of this bridged layer, Rb, is calculated from:
Rb = 1 / (Fins / Rins + Ft / Rt)
where Fins and Ft are the fractional areas of insulation and timber, and Rins and Rt are their resistances.
Rb = 1 / (0.85 / 2.857 + 0.15 / 0.742) = 2.001 m2K/W
Assuming an isothermal plane on each face of the service void layer which is bridged by timber battens, the thermal resistance of this bridged layer, Rsc, is calculated from:
Rsc = 1 / (Fal / Ral + Ftb / Rtb)
where Fal and Ftb are the fractional areas of air layer and timber battens, and Ral and Rtb are their resistances.
Rsc = 1 / (0.88 / 0.18 + 0.12 / 0.269) = 0.187 m2K/W
The resistances of all layers are then combined in series to give the lower resistance [m2K/W]:
External surface resistance 0.040
Brick outer leaf 0.132
Air cavity 0.180
Sheathing Ply 0.069
Bridged insulation layer 2.001
Insulation over studs 2.381
Bridged service void 0.187
Plasterboard 0.006
Internal surface resistance 0.130
Lower resistance (RL) = 5.126
Total resistance
The total resistance Rt is given by: Rt = (Ru + RL) / 2 = (5.847 + 5.126) / 2 = 5.487 m2K/W
The U-value is the reciprocal of the total resistance:
U-value = 1 / 5.487 = 0.18 W/m2K (to 2 decimal places). There is a potential correction for air gaps in the insulation layer. In this case no correction for air gaps applies as it is assumed that the insulation boards are factory fitted and meet the dimensional standards set out in I.S. EN ISO 6946:2017 and that they are installed without gaps greater than 5 mm.
Example A3: Domestic pitched roof with insulation at ceiling level (between and over joists)
A pitched roof has 100 mm of mineral wool tightly fitted between 44 mm by 100 mm timber joists spaced 600 mm apart (centres to centres) and 150 mm of mineral wool over the joists. The roof is slated or tiled with sarking felt under the slates or tiles. The ceiling consists of 13 mm of plasterboard. The fractional area of timber at ceiling level is taken as 9 %.
Diagram HLA5 - Domestic pitched roof - Extract from TGD L
Upper resistance (Ru)
Resistance through section containing both layers of insulation [m2K/W]:
External surface resistance 0.040
Resistance of roof space 0.200
Resistance of mineral wool over joists 3.750
Resistance of mineral wool between joists 2.500
Resistance of plasterboard 0.052
Inside surface resistance 0.100
Total = 6.642
Resistance through section containing timber joists:
External surface resistance 0.040
Resistance of roof space 0.200
Resistance of mineral wool over joists 3.750
Resistance of timber joists 0.769
Resistance of plasterboard 0.052
Inside surface resistance 0.100
Total = 4.911
The upper thermal resistance [Ru] is obtained from:
Ru = 1 / (F1 / R1 + F2 / R2)
where F1 and F2 are the fractional areas of heat flow paths 1 and 2, and R1 and R2 are the resistances of these paths.
Upper resistance Ru = 1 / (0.91 / 6.642 + 0.09 / 4.911) = 6.438 m2K/W
Lower resistance (RL)
Assuming an isothermal plane on each face of the layer of insulation which is bridged by timber studs, the thermal resistance of this bridged layer, Rb, is calculated from:
Rb = 1 / (Fins / Rins + Ft / Rt)
where Fins and Ft are the fractional areas of insulation and timber, and Rins and Rt are their resistances.
Rb = 1 / (0.91 / 2.500 + 0.09 / 0.769) = 2.079 m2K/W
The resistances of all layers are then combined in series to give the lower resistance [m2K/W]:
External surface resistance 0.040
Resistance of roof space 0.200
Resistance of mineral wool over joists 3.750
Resistance of bridged layer 2.079
Resistance of plasterboard 0.052
Inside surface resistance 0.100
Lower resistance (RL) = 6.221
Total resistance
The total resistance Rt is given by:
Rt = (Ru + RL) / 2 = (6.438 + 6.221) / 2 = 6.329 m2K/W
The U-value is the reciprocal of the total resistance:
U-value = 1 / 6.329 = 0.16 W/m2K (to 2 decimal places).
I.S. EN ISO 6946:2017 does not specify any potential correction for this construction.
Ground Floors and Basements
A3.1 The U-value of an uninsulated ground floor depends on a number of factors including floor shape and area and the nature of the soil beneath the floor. I.S. EN ISO 13370:2017 deals with the calculation of U-values of ground floors. Methods are specified for floors directly on the ground and for floors with vented and unvented sub-floor spaces. I.S. EN ISO 13370:2017 also covers heat loss from basement floors and walls.
A3.2 In the case of semi-detached or terraced premises, blocks of flats and similar buildings, the floor dimensions can be taken as either those of the individual premises or those of the whole building. Unheated spaces outside the insulated fabric, such as attached porches or garages, should be excluded when deriving floor dimensions but the length of the floor perimeter between the heated building and the unheated space should be included when determining the length of exposed perimeter. Where such ancillary areas have the potential to become part of the habitable area of the dwelling, floors should be insulated to the same level as the dwelling floors unless it is envisaged that a new insulated floor will be provided when converted.
Example A4: Slab-on-ground floor – full floor insulation
The slab-on-ground floor consists of a 150 mm dense concrete ground floor slab on 100 mm insulation.
Diagram HLA6 - Concrete slab-on-ground floor - Extract from TGD L
The insulation has a thermal conductivity of 0.020 W/mK. The floor dimensions are 8750 mm by 7250 mm with three sides exposed. One 8750 mm side abuts the floor of an adjoining semi-detached house.
In accordance with I.S. EN ISO 13370:2017, the following expression gives the U-value for well-insulated floors:
Rsi, Rf and Rse are internal surface resistance, floor construction (including insulation) resistance and external surface resistance respectively. Standard values of Rsi and Rse for floors are given as 0.17 m2K/W and 0.04 m2K/W respectively. The standard also states that the thermal resistance of dense concrete slabs and thin floor coverings may be ignored in the calculation and that the thermal conductivity of the ground should be taken as 2.0 W/mK unless otherwise known or specified.
Ignoring the thermal resistance of the dense concrete slab, the thermal resistance of the floor construction (Rf) is equal to the thermal resistance of the insulation alone, i.e. 0.1 / 0.020 or 5.00 m2K/W. Taking the wall thickness as 350 mm, this gives:
The edge insulation to the slab is provided to prevent thermal bridging at the edge of the slab. I.S. EN ISO 13370:2017 does not consider this edge insulation as contributing to the overall floor insulation and thus
reducing the floor U-value. However, edge insulation, which extends below the external ground level, is considered to contribute to a reduction in floor U-value and a method of taking this into account is
included in the standard. Foundation walls of insulating lightweight concrete may be taken as edge insulation for this purpose.
Elements Adjacent to Unheated Spaces
A4.1 As indicated in paragraph 0.3.5, the procedure for the calculation of U-values of elements adjacent to unheated spaces (previously referred to as semi-exposed elements) is given in I.S. EN ISO 6946:2017 and I.S. EN ISO 13789:2017.
The following formulae may be used to derive elemental U-values (taking the unheated space into account) for typical housing situations irrespective of the precise dimensions of the unheated space.
Uo = 1 / (1/U-Ru) or U = 1 / (1/Uo+Ru)
Where:
U – U-value of element adjacent to unheated space (W/m2K), taking the effect of the unheated space into account.
Uo – U-value of the element between heated and unheated spaces (W/m2K) calculated as if there was no unheated space adjacent to the element.
Ru – Effective thermal resistance of unheated space inclusive of all external elements (m2K/W).
This procedure can be used when the precise details on the structure providing an unheated space are not available, or not crucial.
Ru for typical unheated structures (including garages, access corridors to flats and unheated conservatories) are given in Tables A3, A4 and A5.
Table A5 applies only where a conservatory style sunroom is not treated as an integral part of the dwelling, i.e. is treated as an extension.
In the case of room-in-roof construction, the U-value of the walls of the room-in-roof construction and of the ceiling of the room below the space adjacent to these walls can be calculated using this procedure. See Diagram A5.
Alternatively Ru can be calculated using the following formula as detailed in BRE 443, to represent more accurately the resistance through the fabric and unheated space e.g. in apartment walls facing unheated corridors:
Table HL10 - Typical resistance (Rᵤ) for unheated space (a) integral and adjacent single garages or other similar unheated space - Extract from TGD L
Table HL11 - Typical resistance (Rᵤ) for unheated space (b) unheated stairwells and access corridors in flats - Extract from TGD L
Table HL12 - Typical resistance (Rᵤ) for unheated space (c) conservatory-type sunroom - Extract from TGD L
Table HL13 - Typical air change rates for unheated spaces (from BRE 443) - Extract from TGD L
Diagram HLA7 - Room in roof - Extract from TGD L
