Examples of U-value calculations using BS EN ISO 6946:1997 Prepared for: DETR/BR By: S M Doran and L Kosmina BRE East Kilbride Dec De cembe emberr 1999
Repo Re port rt No 78 7812 129 9
(revised June 2000) Final approval on behalf of BRE : Final approval on behalf of BRE: Signed ________________ ______________ __ Date
________________ ______________ __
Mrs H J Cuckow, Director, BRE East Kilbride
BRE East Kilbride Kelvin Road East Kilbride Glasgow G75 0RZ Tel : 01355 576200 Fax : 01355 576210 Email :
[email protected] [email protected]
© Building Research Establishment Ltd 1999
Examples of U-value calculations using BS EN ISO 6946:1997 S M Doran and L Kosmina, BRE East Kilbride December 1999 This document illustrates the procedures given in BS EN ISO 6946:1997 1 for calculating the U-value U-value of opaque elements. The procedures are explained explained using examples of U-value calculations for some typical wall, roof and floor designs which contain repeating thermal bridges.
Contents 1.
Introduction
1
2.
Outline of the procedure
2
3.
Cavity vity wall with with lig lightw htweight ight ma maso sonr nry y leaf leaf an and d insul nsula ated ted dry linin ining g
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.
Fl 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 joists between 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 the U-value. At present, Building Regulations Regulations specify that U-values U-values should be calculated using the Proportional Area Method, which is described in the CIBSE Guide, Section A32. Future regulations, regulations, however, are likely to be based upon upon the method for calculating U-values defined in BS EN ISO 6946:1997, which includes the Combined Method for repeating thermal bridges and correction procedures for the effects of metal fixings, air gaps and unheated unheated spaces. This paper illustrates the use of BS EN ISO 6946:1997 for some typical wall, roof and floor designs. Thermal conductivity values for common building materials can be obtained from the CIBSE Guide Section A3, 1999 Edition (especially for masonry) or from prEN 12524 3. For specific insulation products, however, data should be obtained from manufacturers’ declared declared values. A table is provided at the end of this document document giving typical conductivities for some common building materials.
Examples of U-value calculations using BS EN ISO 6946:1997 S M Doran and L Kosmina, BRE East Kilbride December 1999 This document illustrates the procedures given in BS EN ISO 6946:1997 1 for calculating the U-value U-value of opaque elements. The procedures are explained explained using examples of U-value calculations for some typical wall, roof and floor designs which contain repeating thermal bridges.
Contents 1.
Introduction
1
2.
Outline of the procedure
2
3.
Cavity vity wall with with lig lightw htweight ight ma maso sonr nry y leaf leaf an and d insul nsula ated ted dry linin ining g
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.
Fl 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 joists between 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 the U-value. At present, Building Regulations Regulations specify that U-values U-values should be calculated using the Proportional Area Method, which is described in the CIBSE Guide, Section A32. Future regulations, regulations, however, are likely to be based upon upon the method for calculating U-values defined in BS EN ISO 6946:1997, which includes the Combined Method for repeating thermal bridges and correction procedures for the effects of metal fixings, air gaps and unheated unheated spaces. This paper illustrates the use of BS EN ISO 6946:1997 for some typical wall, roof and floor designs. Thermal conductivity values for common building materials can be obtained from the CIBSE Guide Section A3, 1999 Edition (especially for masonry) or from prEN 12524 3. For specific insulation products, however, data should be obtained from manufacturers’ declared declared values. A table is provided at the end of this document document giving typical conductivities for some common building materials.
2. Outline of the procedure procedure The following is an outline of the calculation procedure: 1. Calculate the upper upper resistance limit (Rupper) by combining in parallel the total resistances of all possible heat-flow paths (i.e. sections) through the building element. 2. Calculate the lower resistance limit (Rlower) by combining in parallel the resistances of the heat flow paths of each layer separately and then summing the resistances of all layers of the building element. 3. Calculate the total thermal resistance (RT) from RT
=
R upper
+ Rlower 2
4. Calculate, where appropriate, appropriate, corrections for air gaps (∆Ug) and mechanical fasteners (∆Uf). Examples of corrections corrections for air gaps gaps are shown in sections sections 3, 4, 10 and 12 and examples of corrections for mechanical fasteners are shown in sections 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 amount to less than 3% of the U-value. This has been done in the examples examples that follow.
3. Cavity wall with lightweight lightweight masonry masonry leaf and insulated 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 mm AAC blockwork, blockwork, 38 89 mm timber studs studs with insulation between the studs and one one sheet of 12.5 mm plasterboard. 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 0.25 W/m·K) heat flow
Total thickness
354 mm
U-value
0.31 W/m·K
The thickness of each layer, together with the thermal conductivities of the t he 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 the bridging part in each case. case. For each homogeneous homogeneous layer and for each section (expressed in metres) by the thermal conductivity.
Layer
Thickness (mm) -
1
outer leaf brick air cavity
3(a) 3(b) 4(b) 5
mortar (6.6%) mineral wool (90.5%)
50 1 00 89 (89)
plasterboard -
a)
conductivity (W/m·K) 0 .7 7 0.11 0.88 0.13 0.25 -
resistance (m²K/W) 0.132 0.180 0.114 2.342 0.050
Due to requirements for sound insulation this wall construction may only be suitable for
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 R upper
=
1 F1 R1
+
F2 R2
+
F3 R3
+
F4 R4
where F1, F2, F3 and 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:-
F1 1
2
3(a)
4(a)
external 1 surface
2
3(b)
4(a)
5
1
2
3(a)
4(b)
5
1
2
3(b)
4(b)
5
5
F2 internal surface
F3 F4
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 Resistance of bricks Resistance of air cavity Resistance of AAC blocks (93.4%) Resistance of mineral wool (90.5%) Resistance of plasterboard Internal surface resistance Total thermal resistance (R1) Fractional area F1 = 0.845 (93.4% × 90.5%)
= 0.040 = 0.132 = 0.180 = 0.909 = 2.342 = 0.050 = 0.130 = 3.783 m²K/W
Resistance through section containing mortar and mineral wool External surface resistance Resistance of bricks Resistance of air cavity Resistance of mortar (6.6%) Resistance of mineral wool (90.5%) Resistance of plasterboard Internal surface resistance Total thermal resistance (R2)
= 0.040 = 0.132 = 0.180 = 0.114 = 2.342 = 0.050 = 0.130 = 2.988 m²K/W
Fractional area F2 = 0.060 (6.6% × 90.5%) Resistance through section containing AAC blocks and timber External surface resistance Resistance of bricks Resistance of air cavity Resistance of AAC blocks (93.4%) Resistance of timber (9.5%) Resistance of plasterboard Internal surface resistance Total thermal resistance (R3)
= 0.040 = 0.132 = 0.180 = 0.909 = 0.685 = 0.050 = 0.130 = 2.126 m²K/W
Fractional area F3 = 0.089 (93.4% × 9.5%) Resistance through section containing mortar and timber External surface resistance Resistance of bricks Resistance of air cavity Resistance of mortar (6.6%) Resistance of timber (9.5%) Resistance of plasterboard Internal surface resistance Total thermal resistance (R4)
= 0.040 = 0.132 = 0.180 = 0.114 = 0.685 = 0.050 = 0.130 = 1.331 m²K/W
Fractional area F4 = 0.006 (6.6% × 9.5%) Combining these resistances we obtain: R upper
=
1 F1 R1
F + 2 R2
F3
F + + 4 R3 R4
=
1 0.845 3.783
+
0.060 2.988
+
0.089 2.126
+
0.006 1.331
= 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:
3(a) external surface
1
4(a)
2
5 3(b)
4(b)
internal surface
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: R first
=
1 Fblocks R blocks
+
Fmortar R mortar
and the resistance of the bridged layer consisting of insulation and timber is calculated using: R second
=
1 Finsul R insul
+
Ftimber R timber
The lower limit of resistance is then obtained by adding together the resistances of the layers: External surface resistance Resistance of bricks Resistance of air cavity Resistance of first bridged layer 1 R first = 0.934 0.066 + 0.909 0.114
= 0.040 = 0.132 = 0.180 = 0.622
Resistance of second bridged layer 1 R second = 0.905 0.095 + 2.342 0.685
= 1.904
Resistance of plasterboard Internal surface resistance Total (Rlower)
= 0.050 = 0.130 = 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:
RT
=
R upper
+ R lower 2
=
3.450 + 3.058 2
= 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), RI is 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. 102 mm brick outer leaf (conductivity 0.77 W/m·K) 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 2 × 12.5 mm plasterboard (conductivity 0.25 W/m·K)
Total thickness
336 mm
U-value
0.31 W/m·K
heat flow
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. Layer
1 2 3 4(a¹) 4(a²) 4(b) 5
Material
external surface outer leaf brick air cavity plywood mineral wool between timber studs air space next to mineral wool 38 mm × 140 mm timber studs at 400 mm centres plasterboard internal surface
Thickness (mm) 102 50 19 120
Thermal conductivity (W/m·K) 0.77 0.13 0.038
Thermal resistance (m²K/W) 0.040 0.132 0.090 0.146 3.158
20 (140)
0.13
0.180 1.077
25 -
0.25 -
0.100 0.130
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
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: 1 R upper = F1 F2 + R1 R 2 where F1 and F2 are the fractional areas of the two sections and R 1 and R2 are the total resistances of the two sections. The method of calculating the upper resistance limit is illustrated conceptually below:F1 external
1
2
3
4(a)
5
surface
internal surface
F2 1
2
3
4(b)
5
Figure 4.2 : Conceptual illustration of how to calculate the upper limit of thermal resistance
Resistance through the section containing insulation External surface resistance Resistance of bricks Resistance of air cavity Resistance of plywood Resistance of mineral wool (90.5%) Resistance of air space next to mineral wool Resistance of plasterboard Internal surface resistance Total (R1) Fractional area F1 = 0.905 (90.5%)
= 0.040 = 0.132 = 0.090 = 0.146 = 3.158 = 0.180 = 0.100 = 0.130 = 3.976 m²K/W
Resistance through section containing timber stud External surface resistance Resistance of bricks Resistance of air cavity Resistance of plywood Resistance of timber studs (9.5%) Resistance of plasterboard Internal surface resistance Total (R2)
= 0.040 = 0.132 = 0.090 = 0.146 = 1.077 = 0.100 = 0.130 = 1.715 m²K/W
Fractional area F2 = 0.095 (9.5%)
The upper limit of resistance is then: R upper
1
=
F1 R1
+
F2
=
R2
1 0.905 3.976
0.095 + 1.715
=
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: R=
1 Finsul R insul
+
Ftimber R timber
The method of calculating the lower limit of resistance is illustrated conceptually below. F1 4(a) external surface
1
2
3
5 F2
internal surface
4(b)
Figure 4.3 : Conceptual illustration of how to calculate the lower limit of thermal resistance
The lower limit of resistance is then obtained by adding up the resistances of all the layers: External surface resistance Resistance of bricks Resistance of air cavity Resistance of plywood Resistance of bridged layer =
= 0.040 = 0.132 = 0.090 = 0.146 1 0.905 3.158 + 0.180
0.095 + 1.077
Resistance of plasterboard Internal surface resistance Total (Rlower)
= 2.783
= 0.100 = 0.130 = 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: RT
=
R upper
+ R lower 2
=
3.533 + 3.421 = 3.477 m²K/W 2
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 RI d) is the thermal resistance of the layer containing gaps, R T 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 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 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 In this example RI is the same as the resistance of the bridged layer used in the calculation of the lower resistance limit d)
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
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. 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. Air circulation possible on the warm side of the insulation. Air gaps may penetrate the insulation.
∆U’’
∆Ug
0
W/m²K 0.00
W/m²K 0.000
1
0.01
0.006
2
0.04
0.026
Correction level
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. 102 mm brick (conductivity 0.77 W/m·K)
75 mm cavity filled with mineral wool (conductivity 0.038 m²K/W) 100 mm AAC blocks (conductivity 0.11 m²K/W) bridged by mortar (conductivity 0.88 W/m·K)
Total thickness
290 mm
U-value
0.32 W/m·K
13 mm lightweight plaster, (conductivity 0.18 W/m·K)
heat flow
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
1 2 3(a) 3(b) 4
e)
Material
external surface outer leaf brick mineral wool batts AAC blocks (93.4%) mortar (6.6%) lightweight plaster internal surface
Thickness (mm) 102 75 100 (100) 13 -
Thermal conductivity (W/m·K) 0.77 0.038 0.11 0.88 0.18 -
Thermal resistance (m²K/W) 0.040 0.132 1.974 0.909 0.114 0.072 0.130
This construction provides satisfactory sound insulation from neighbouring dwellings
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: 1 R upper = F1 F2 + R1 R 2 where F1 and F2 are the fractional areas of the two sections and R 1 and R2 are the total resistances of the two sections. A conceptual diagram of the upper limit of resistance is shown immediately below F1 external
1
2
3(a)
4
surface
internal surface
F2 1
2
3(b)
4
Figure 5.2 : Conceptual illustration of how to calculate the upper limit of resistance
Resistance through section containing concrete blocks External surface resistance Resistance of bricks Resistance of mineral wool slabs Resistance of AAC blocks (93.4%) Resistance of lightweight plaster Internal surface resistance Total (R1)
0.040 0.132 1.974 0.909 0.072 0.130 3.257
Fractional area F1 = 0.934 (93.4%) Resistance through section containing mortar External surface resistance Resistance of bricks Resistance of mineral wool slabs Resistance of mortar (6.6%) Resistance of lightweight plaster Internal surface resistance Total (R2)
0.040 0.132 1.974 0.114 0.072 0.130 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: R upper
=
1 F1 R1
+
F2 R2
=
1 0.934 3.257
+
0.066 2.462
= 3.189 m²K/W
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:F1 3(a) external
1
2
surface
4 F2
internal surface
3(b)
Figure 5.3 : Conceptual illustration of how to calculate the lower limit of resistance
External surface resistance Resistance of bricks Resistance of mineral wool slabs Resistance of AAC blocks & mortar =
= 0.040 = 0.132 = 1.974 1 0.934 0.909
+
0.066 0.114
Resistance of lightweight plaster Internal surface resistance Total (Rlower)
= 0.622
= 0.072 = 0.130 = 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: RT
=
R upper
+ R lower 2
=
3.189 + 2.970 2
= 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
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 λf nf Af where α is the scaling factor for mechanical fixings, which is 6 for wall ties f), and λf is 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, n f, is calculated as follows using the above information on the wall tie spacing. nf
=
1000 000 900 × 450
= 2.47
Af is the cross-sectional area of the wall tie, expressed in m², which is 0.00001075 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 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 ∆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, no correction 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, where applicable, is incorporated into the overall U-value calculation:
U-value in absence of wall ties
effect of wall ties
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.
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. 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)
Total thickness
335 mm
U-value
0.30 W/m·K
heat flow
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
The wall construction may be summarised as follows: Layer
Material
external surface outer leaf brick mineral wool batts concrete blockwork dense plaster internal surface Total (RT)
1 2 3 4
Thickness (mm) 102 120 100 13 -
Thermal conductivity (W/m·K) 0.77 0.038 1.13 0.57 -
Thermal resistance (m²K/W) 0.040 0.132 3.158 0.088 0.023 0.130 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 λf nf Af where α is the scaling factor for mechanical fixings, which is 6 for wall ties h), and λf is 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, n f, 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 Uvalue 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. 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 0.1 m²K/W
7. Pitched roof with insulation between and over the joists
(loft space and pitched roof above) U-value
0.20 W/m·K
joist
insulation
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)
external surface roof space beneath tiled roof with felt or boardsi) continuous layer of mineral wool mineral wool between 48 × 100 mm timber joists with 600 mm between centres 48 × 100 mm timber joists between insulation plasterboard internal surface
1 2 3(a) 3(b) 4
-
Thermal conductivity (W/m·K) -
Thermal resistance (m²K/W) 0.040 0.200
100 100
0.042 0.042
2.381 2.381
(100) 12.5 -
0.13 0.25 -
0.769 0.050 0.100
Upper resistance limit:A conceptual illustration of how the upper limit of resistance is calculated is shown immediately below F1 external
1
2
3(a)
4
surface
internal surface
F2 1
2
3(b)
4
Figure 7.2 : Conceptual illustration of how to calculate the upper limit of resistance
i)
Using Table 3 of BS EN ISO 6946
Resistance through section containing both layers of insulation External surface resistance Resistance of roof space i) Resistance of mineral wool over joists Resistance of mineral wool between joists Resistance of plasterboard Inside surface resistance Total (R1)
= 0.040 = 0.200 = 2.381 = 2.381 = 0.050 = 0.100 = 5.152 m²K/W
Fractional area F1 = 0.92 (92%) Resistance through section containing timber joists External surface resistance Resistance of roof space Resistance of mineral wool over joists Resistance of timber joists Resistance of plasterboard Inside surface resistance Total (R2)
= 0.040 = 0.200 = 2.381 = 0.769 = 0.050 = 0.100 = 3.540 m²K/W
Fractional area F2 = 0.08 (8%) The upper resistance limit is given by R upper
=
1 F1 R1
+
F2 R2
=
1 0.92 5.152
+
0.08 3.540
= 4.971 m²K/W
Lower resistance limit:A conceptual illustration of the method of calculating the lower limit of resistance is shown below:F1 3(a) external
1
2
surface
4 F2
internal surface
3(b)
Figure 7.3 : Conceptual illustration of how to calculate the lower limit of resistance
External surface resistance Resistance of roof space Resistance of mineral wool over joists Resistance of bridged layer 1 1 = = Finsul Ftimber 0.92 0.08 + + 2.381 0.769 R insul R timber Resistance of plasterboard Inside surface resistance Total (Rlower)
= 0.040 = 0.200 = 2.381 = 2.039
= 0.050 = 0.100 = 4.810 m²K/W
Total resistance of roof RT
=
R upper
+ R lower 2
=
4.971 + 4.810 2
= 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.
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. U-value
A
0.27 W/m·K
Plan at A-A
A 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) 0.100
external surface* 1 tiles* 19 2 roofing felt* 1 3 ventilated airspace between rafters 50 and spacers* 4(a) insulation board j) occupying 88% of 100 0.025 4.000 face area (between rafters and spacers) 4(b) rafters (beneath ventilated area) (100) 0.13 0.769 occupying 12% of face area 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
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:-
F1
F2
external surface
external surface insulation
rafters
insulation
insulation
rafters
insulation
plasterboard
plasterboard
internal surface
internal surface
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 Resistance of insulation between rafters Resistance of insulation beneath rafters Resistance of plasterboard Internal surface resistance Total thermal resistance (R1) Fractional area F1 = 0.88 (88%)
= = = = = =
0.100 4.000 0.620 0.038 0.100 4.858 m²K/W
Resistance through the section through the rafters Effective external surface resistance Resistance of rafters Resistance of insulation beneath rafters Resistance of plasterboard Internal surface resistance Total thermal resistance (R2) Fractional area F2 = 0.12 (12%)
= = = = = =
0.100 0.769 0.620 0.038 0.100 1.627 m²K/W
The upper limit of resistance is then obtained from: R upper
=
1 F1 R1
+
F2 R2
=
1 0.88 4.858
0.12 + 1.627
= 3.923 m²K/W
Lower resistance limit Effective external surface resistance Resistance of bridged layer 1 = 0.88 0.12 + 4.000 0.769 Resistance of insulation beneath rafters Resistance of plasterboard Internal surface resistance Total thermal resistance (Rlower)
= 0.100 = 2.659 = = = =
0.620 0.038 0.100 3.517 m²K/W
Total resistance of roof The total resistance is the average of the upper and lower limits RT =
R upper
+ R lower 2
=
3.923 + 3.517 2
= 3.720 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.
A
Plan at A-A
A
Figure 9.1 : Roof construction U-value
0.33 W/m·K
The construction may be summarised as follows:Layer
Material
Thickness (mm)
Thermal conductivity (W/mK) -
Thermal resistance (m²K/W) 0.100
external surface* 1 tiles* 19 2 roofing felt* 1 3 ventilated airspace between rafters 50 and spacers* 4(a) insulation board occupying 88% of 25 0.025 1.000 face area (between rafters and spacers) 4(b) rafters (beneath ventilated area) (25) 0.13 0.192 occupying 12% of face area 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.
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:-
F1
F2
external surface
external surface
insulation
insulation
rafters
rafters
insulation
insulation
plaster board internal surface
plasterboard internal surface
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 Resistance of insulation between rafters Resistance of insulation beneath rafters Resistance of plasterboard Internal surface resistance Total thermal resistance (R1) Fractional area F1 = 0.88 (88%)
= = = = = =
0.100 1.000 2.300 0.050 0.100 3.550 m²K/W
Resistance through the section through the rafters Effective external surface resistance Resistance of rafters Resistance of insulation beneath rafters Resistance of plasterboard Internal surface resistance Total thermal resistance (R2) Fractional area F2 = 0.12 (12%)
= = = = = =
0.100 0.192 2.300 0.050 0.100 2.742 m²K/W
The upper limit of resistance is then obtained from: R upper
=
1 F1 R1
+
F2 R2
=
1 0.88 3.550
+
0.12 2.742
= 3.429
Lower resistance limit:Effective external surface resistance Resistance of bridged layer 1
=
0.88
+
0.12
1.000 0.192 Resistance of insulation beneath rafters Resistance of plasterboard Internal surface resistance Total thermal resistance
= 0.100 = 0.664
= = = =
2.300 0.050 0.100 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 =
R upper
+ R lower 2
=
3.429 + 3.214 2
= 3.322 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 n f where nf
=
1000 000 400 × 150
= 16.7 / m²
The nails are made of steel with a their thermal conductivity, λf, 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 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 insulation penetrated by the nails and RT is 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)
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 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 continuous 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 situated above an unheated space such as a garage or an unheated corridor. 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
U-value
0.25 W/m·K
12.5 mm plasterboard above unheated area
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
1 2(a) 2(b) 3 4
internal surface plywood glass mineral wool timber joists (occupying 12%) plasterboard external
Thickness (mm) 19 150 (150) 12.5 -
Thermal conductivity (W/m·K) 0.13 0.040 0.13 0.25 -
Thermal resistance (m²K/W) 0.170 0.146 3.750 1.154 0.050 0.040
A conceptual illustration of the calculation of the limits of resistance is shown below:
F1 internal surface
internal surface
plywood plywood 2(a)
2(a)
2(b)
2(b) plasterboard
plasterboard
Ru
Ru
external surface
external surface
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 Resistance of plywood Resistance of mineral wool insulation Resistance of plasterboard Ru External surface resistance Total thermal resistance (R1)
= 0.170 = 0.146 = 3.750 = 0.050 = 0.490 = 0.040 = 4.646 m²K/W
Resistance through the section containing joists: Internal surface resistance Resistance of plywood Resistance of timber joists Resistance of plasterboard Ru External surface resistance Total thermal resistance (R2)
= 0.170 = 0.146 = 1.154 = 0.050 = 0.490 = 0.040 = 2.050 m²K/W
The upper limit of resistance is then obtained from: R upper
=
1 F1 R2
+
F2 R2
=
1 0.88 4.646
+
0.12 2.050
= 4.033
m²K/W
Lower resistance limit:Internal surface resistance Resistance of plywood Resistance of bridged layer 1 = 0.88 0.12 + 3.750 1.154 Resistance of plasterboard Ru External surface resistance Total (Rlower)
= 0.170 = 0.146 = 2.953 = 0.050 = 0.490 = 0.040 = 3.849 m²K/W
Total resistance of floor The total resistance of the wall is the average of the upper and lower resistance limits RT
=
R upper
+ R lower 2
=
4.033 + 3.849 2
= 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), RI is 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 ), 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 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
1 2 3(a) 3(b) 4
Material
Thickness (mm)
internal surface screed polystyrene light concrete blocks, 440 mm wide concrete beams, 70 mm wide lower surface
65 100 100
Thermal conductivity (W/m·K) 0.41 0.040 0.18
Thermal resistance (m²K/W) 0.170 0.159 2.500 0.556
(100)
1.13
0.088
-
-
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. 70
440 U-value 100
Figure 11.1 : Beam and block suspended floor
0.24 W/m·K
internal surface
internal surface
screed
screed insulation
insulation beams
beams
blocks lower surface
blocks
lower surface
underfloorspace & soi underfloor space & soil
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 Resistance of screed Resistance of polystyrene Resistance of light concrete blocks Resistance of lower surface of floor deck Total thermal resistance (R1) Fractional area F1 = 0.863 (i.e. 86.3%)
= 0.170 = 0.159 = 2.500 = 0.556 = 0.170 = 3.555 m²K/W
Resistance through section containing concrete beams Internal surface resistance Resistance of screed Resistance of polystyrene Resistance of beams Resistance of lower surface of floor deck Total thermal resistance (R2) Fractional area F2 = 0.137 (i.e. 13.7%) The upper limit of resistance is then obtained from: R upper
=
1 F1 R1
+
F2 R2
=
1 0.863 3.555
+
0.137 3.087
= 3.483
= 0.170 = 0.159 = 2.500 = 0.088 = 0.170 = 3.087 m²K/W
Lower resistance limit (of floor deck):Internal surface resistance Resistance of screed Resistance of polystyrene Resistance of bridged layer 1 = 0.863 0.137 + 0.556 0.088 Resistance of lower surface of deck Total (Rlower)
= 0.170 = 0.159 = 2.500 = 0.322 = 0.170 = 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 R=
Rupper
+ R lower 2
=
3.483 + 3.321 = 3.402 m²K/W 2
Uf = 1 / R = 1 / 3.402 = 0.294 W/m²K This gives a U-value (U f) 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: U=
1 1 Uf
1 + − R si,upper − R deck,uninsulated − R si,lower U0
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), R deck,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,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 of the remaining part of the floor is combined with the U-value of the floor deck calculated above. Summary of floor details exposed perimeter (P) floor area (A) perimeter to area ratio (P/A) wall thickness (w) soil type ventilation parameter ( ε)
35.6 m 79.1 m² 0.45 m-1 0.3 m sandy 0.015 m²/m
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 U=
=
1 1 Uf
+
1 U0
− (R si,upper + R deck,uninsulated + R si,lower ) 1
1 0.294
+
1 0.76
− (0.17 + 0.2 + 0.17 )
= 0.24 W / m²K 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.
12. Suspended timber ground floor
U-value
0.22 W/m·K
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)
1 2(a) 2(b)
internal surface chipboard glass mineral wool on netting timber joists (occupying 12%) lower surface
19 150 (150) -
Thermal conductivity (W/m·K) 0.13 0.040 0.13 -
Thermal resistance (m²K/W) 0.170 0.146 3.750 1.154 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:-
upper surface
upper surface
chipboard
chipboard mineral wool
joists
mineral wool
lower surface underfloor space and soil
joists lower surface
underfloor space and soil
Figure 12.2 : Conceptual illustration of the methods of calculating the upper and lower limits of thermal resistance
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: R upper
=
1 F1 R1
F + 2 R2
=
1 0.88 4.236
0.12 + 1.640
= 3.560 m²K/W
Lower resistance limit (for floor deck) Internal surface resistance = 0.170 Resistance of chipboard = 0.146 Resistance of bridged layer 1 = = 2.953 0.88 0.12 + 3.750 1.154 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 RT
=
R upper
+ R lower 2
=
3.560 + 3.439 2
= 3.500 m²K/W
Correction for air gaps Since the insulation layer is entirely between joists a correction should be applied to the floor deck U-value in order to account for air gaps. The overall U-value of the floor deck should be adjusted by adding a term ∆Ug, where
∆Ug = ∆U’’ × (R /R I T )² and where ∆U’’ = 0.01 (referred to in BS EN ISO 6946 as correction level 1), RI is 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 / 3.500)² = 0.007 W/m²K U-value of the floor deck (Uf) 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%. 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 less than 1 / RT the U-value of the floor deck is Uf = 1 / 3.500 = 0.286 W/m²K U-value of the floor Since the floor perimeter is 40 m, the area 100 m, and the ground of clay soil, the U-value of the floor ignoring insulation is U0 = 0.65 W/m²K (using Table 4) U=
1 1 Uf
+
1 U0
− (0.17 + 0.2 + 0.17)
=
1 1 0.286
+
1 0.65
− (0.17 + 0.2 + 0.17)
= 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 notional uninsulated floor deck. The figures 0.17 and 0.17 represent the surface resistances of the upper and lower surface resistances for the same notional floor deck (taken from 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 flow Upwards Horizontal Downwards inside resistance 0.10 0.13 0.17 outside resistance 0.04 0.04 0.04 underfloor space* 0.13 0.17 *These values should be used for the upper and lower surfaces of the underfloor space 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 Thickness of air Direction of heat flow layer (mm) Upwards Horizontal Downwards 0 0.00 0.00 0.00 5 0.11 0.11 0.11 7 0.13 0.13 0.13 10 0.15 0.15 0.15 15 0.16 0.17 0.17 25 0.16 0.18 0.19 50 0.16 0.18 0.21 100 0.16 0.18 0.22 300 0.16 0.18 0.23 Table 3 Scaling factors for ceiling fixings and wall ties from BS EN ISO 6946 type of mechanical fastenings scaling factor (α) 5 roof fixings 6 wall ties between masonry leaves
The following table provides U-values for suspended floors without insulation. For a detailed calculation of the thermal resistance below the deck of a ground floor the reader is referred to the procedure in BS EN ISO 13370.
Table 4 U-values of uninsulated suspended floors (from CIBSE Guide A3)
perimeter/area 0.05 0.10 0.15 0.20
Soil type and ventilation opening area per unit perimeter of underfloor space (in m²/m) clay/silt sand/gravel homogeneous rock 0.0015 0.003 0.0015 0.003 0.0015 0.003 0.16 0.17 0.19 0.20 0.27 0.28 0.27 0.29 0.32 0.33 0.43 0.44 0.36 0.38 0.42 0.43 0.54 0.55 0.44 0.46 0.49 0.51 0.63 0.64
0.25 0.30 0.35 0.40
0.50 0.56 0.61 0.65
0.52 0.58 0.63 0.68
0.56 0.62 0.67 0.72
0.58 0.64 0.69 0.74
0.70 0.76 0.81 0.85
0.71 0.77 0.82 0.87
0.45 0.50 0.55 0.60
0.69 0.73 0.76 0.79
0.72 0.76 0.79 0.83
0.76 0.79 0.83 0.86
0.78 0.82 0.85 0.88
0.89 0.92 0.95 0.98
0.91 0.94 0.97 1.00
0.65 0.70 0.75 0.80
0.82 0.85 0.87 0.90
0.85 0.88 0.91 0.93
0.88 0.91 0.93 0.95
0.91 0.94 0.96 0.98
1.00 1.03 1.05 1.06
1.02 1.05 1.07 1.09
0.85 0.90 0.95 1.00
0.92 0.94 0.96 0.98
0.95 0.97 0.99 1.01
0.97 0.99 1.01 1.03
1.00 1.02 1.04 1.06
1.08 1.10 1.11 1.13
1.11 1.12 1.14 1.15
Table 5 Thermal conductivity of some common building materials Density (kg/m³)
Conductivity (W/m·K)
Walls Brickwork (outer leaf) Brickwork (inner leaf)
1700 1700
0.77 0.56
Concrete block (medium density) Concrete block (low density)
1400 600
0.57 0.18
Concrete (medium density) (inner leaf) 1800 2000 2200 Concrete (high density) : 2400
1.13 1.33 1.59 1.93
Reinforced concrete (1% steel) Reinforced concrete (2% steel)
2300 2400
2.3 2.5
Mortar (protected) Mortar (exposed)
1750 1750
0.88 0.94
Gypsum
600 900 1200
0.18 0.30 0.43
900
0.25
Sandstone Limestone, soft Limestone, hard
2600 1800 2200
2.3 1.1 1.7
Fibreboard Plasterboard Tiles ceramic Timber (softwood) Steel Stainless steel
400 900 2300 500 700 7800 7900
0.1 0.25 1.3 0.13 0.18 50.0 17.0
Surface finishes External rendering Plaster (dense) Plaster (lightweight)
1300 1300 600
0.57 0.57 0.18
Roofs Aerated concrete slab Asphalt Felt/bitumen layers Screed Stone chippings Tiles (clay) Tiles (concrete) Wood wool slab
500 2100 1100 1200 2000 2000 2100 500
0.16 0.70 0.23 0.41 2.0 1.0 1.5 0.10
Gypsum plasterboard
Floors Cast concrete Metal tray (steel) Screed Hardwood timber Softwod timber, plywood, chipboard Insulation Expanded polystyrene (EPS) board Mineral wool quilt Mineral wool batt Phenolic foam board Polyurethane board
2000 7800 1200 700 500 1000 15 12 25 30 30
1.35 50.0 0.41 0.18 0.13 0.24 0.040 0.042 0.038 0.025 0.025
Note: If available, certified test values should be used in preference to those in the table.
