Extracts from DS418:2002Table of content Table of content 1 Introduction 1.1 Purpose and validity 1.2 Definitions 1.3 Symbols and units Design temperatures 2.1 Design indoor temperature 2.2 Design outdoor temperature 2.3 Design soil temperature 2.4 Other design temperatures 1 3 3 4 6 8 8 9 9 9 2 3 Calculation of transmission loss 10 3.1 Transmission loss through external walls, roofs, windows and external doors 10 3.2 Transmission loss through ground supported floors, basement floors and basement walls 10 3.3 Transmission loss through partition walls and storey partitions and basements decks 10 3.4 Transmission loss through joints around windows and doors 10 3.5 Transmission loss through foundations under external walls 11 3.6 Calculation of transmission areas 11 3.7 Calculation of the length of the linear cold bridge 13 Calculation of ventilation loss 4.1 Ventilation loss 4.2 Natural ventilation 4.3 Mechanical ventilation (exhaust) 4.4 Other mechanical ventilation systems Calculation of the total heat loss 5.1 The heat loss for a building 5.2 Calculation of transmission loss Calculation of transmittal coefficient 6.1 Transmittal coefficient and heat flow resistance 6.2 Surface heat flow resistance 6.3 Heat flow resistance for a material layer 6.4 Heat flow resistance for air filled cavities 6.5 Ventilated attics 6.6 Constructions with inhomogeneous material layers 6.7 Constructions with cold bridges 6.8 Doors and windows 6.9 Ground supported floors, basement floors and basement walls against soil 6.10 Concrete sandwich elements 6.11 Wedged insulation 6.12 Joints around windows and doors 6.13 Foundations under external walls Heat flow resistance and conductivity of materials 7.1 Introduction 7.2 Basis for determination of heat flow resistance and conductivity 16 16 16 17 17 19 19 19 20 20 20 20 21 23 23 24 30 30 31 31 31 35 40 40 40 4 5 6 7 Page 1 Extracts from DS418:2002 Annex A Annex F Correction of transmissions coefficients Design values for brick, concrete and other building materials 42 45 51 Annex G Design values for calculation of existing constructions in connection with rebuilding and renovation. Page 2 Extracts from DS418:2002 1 1.1 Introduction Purpose and validity The calculation rules are aiming to obtaining conformity when estimating the heat flow resistance and the heat loss of buildings under Danish climate conditions, among other things regarding the energy requirements of the Building Regulation (BR). It is the users responsibility, that the requirements of the BR are met. There can occur table values in DS418, which doesn’t meet these requirements. These values can only be used in connection with existing constructions. The rules provide instructions on how to calculate building components transmission coefficient U for estimation of the components thermal insulation ability. U-values are based on the design of specific constructions and values for products cannot be used immediately by calculation of heat loss. It is provided that constructions and combined components are correct build, by using approved methods and correct workmanship. A construction which as whole lacks windproofness or which allows unintended ventilation or convection in or around the heat insulation layers can have a considerably lesser insulation ability the calculated. Normal windproofness and dampproofness are provided. In accordance with chapter 6.8 is it also possible to estimate the U-value for windows and doors, by measuring on the specific windows used in a construction. Material values determined by average conditions in the constructions, are used in the calculations. Due to simplifications can the calculation rules not be used for detailed calculations e.g. surface temperature, condensation or damp flow. The calculation rules indicate a method for calculating a rooms or buildings design heat loss. The method is designed so the design heat loss approximately is equal to the rooms or the buildings actual heat loss under stationary conditions by the indicated internal and external climate conditions. By calculating the design heat loss of a room, situations where adjacent rooms temporarily are unheated, are not considered. For a simplified method for dimensioning of radiators refer to DS 469. In specific cases where it can be proven that the calculation method will not yield a reasonable approximation to the actual conditions, more detailed methods must be used. The rules are elaborated thus that the calculations become fairly simple and practically useable. The user of DS418 must have sufficient technical knowledge. Special situations may be encountered, where the rules are not fully covering. In any case it should be appraised if the actual situation is covered by the rules or not. Deviation from these rules is permitted, if it is documented that the deviation is appropriate and based on technical ground. Such documentation must be carried out according to the EN and ISO standards, if not this standard prescribes otherwise. Page 3 Extracts from DS418:2002 By interpretation questions refer to Dansk Standard, Kollegievej 6, 2920 Charlottenlund tlf. 39 69 61 01 or
[email protected] 1.2 Definitions Declared value for Heat flow resistance and thermal conductivity The value of a building-materials ability Heat flow resistance or thermal conductivity based up on measurements, at reference temperature and humidity. Regarding definition see DS/EN ISO 10456 item 3.1.1 Density The density of a material is its weight divided by volume, where the volume includes the pores and cavities of the material. The density applied is the one for the material in dry conditions. It should be considered, however, that this density in some cases might be different from the nominal density of the material according to normal product names etc. Regarding definition see DS/EN ISO 6946 item 3.1.4 The design value for thermal conductivity and heat flow resistance The design value for a building-materials thermal conductivity or heat flow resistance under specific conditions, which are considered being typical for the materials thermal properties, when used in a building-component. The design room temperature The design room temperature is a theoretical temperature, which is determined only as the basis for estimating the design heat loss of the applicable room. The room temperature represents the combined value of air temperature and radiation temperature, which may result in equal heat distribution to the surrounding space limits of the applicable room. For living rooms and alike the room temperature and the operative temperature in the middle of the room practically will have the same value. Regarding definition see DS/EN ISO 10211-1 item 3.1.14 The design outdoor temperature The design outdoor temperature is the theoretical temperature, determined for the calculation of the design heat loss. It does not include the extreme impact of climate encountered, but a certain limit frequency. Regarding definition see DS/EN ISO 10211-1 item 3.1.13 The design heat loss The design heat loss for a certain room or a building is equal to the heat effect to be provided in order to maintain the design room temperature at the determined outdoor temperature conditions. The design heat loss comprises transmission heat loss and ventilation heat loss. Energy frame The energy frame is the maximum annual permissible heat requirement for heating of rooms and ventilation according the building regulations. Page 4 Extracts from DS418:2002 Heat flow resistance The heat flow resistance is the relation between temperature difference and heat flow density. The heat flow resistance is describing the resistance towards heat transmission through 1 m2 of the applicable area or the applicable building material. Regarding definition see DS/EN ISO 7345 item 2.7 Cold bridge Cold bridge is the part of the construction with significant smaller heat flow resistance than the rest of the construction. Regarding definition see DS/EN ISO 10211-1 item 3.1.1 Linear cold bridge A linear cold bridge is a cold bridge with little width, which impact on the heat loss is depending on the length of the cold bridge and the two-dimensional heat flows, it may cause. Regarding definition see DS/EN ISO 14683 item 3.1.1 Linear heat loss The heat loss through a linear cold bridge. Normal emission The normal emission for surface is the relation between the radiation in a direction perpendicular to the surface and the equivalent radiation from an absolute black surface with same conditions and temperature. The normal emission thus is an expression of the relative radiation exchange with the surroundings. Regarding definition see DS/EN ISO 9288 item 3.1.11 Operative temperature The operative temperature represents the combined value of air temperature and radiation temperature, which would result in equal heat contribution through conviction and radiation from the person, as the real/actual temperatures would give. Point cold bridge The point cold bridge is a cold bridge of little extent, of which the impact on the heat loss is depending on the three-dimensional heat flows it may cause. Regarding definition see DS/EN ISO 14683 item 3.1.2 Transmission coefficient, U-value The transmission coefficient for a building component is the relation between the heat flow and the area and difference between the temperatures on each side of the building-component. Regarding definition see DS/EN ISO 7345 item 2.12 Transmission loss The transmission loss is the heat amount, which flows through the rooms or the buildings surrounding space limits, per time unit, because of difference in temperature. Page 5 3 Heat loss frame The heat loss frame is the design transmission loss. specific heat capacity SI-unit m m m m m2 m3 m3/s h-1 kg/m3 ˚C K W W/m K W/m K W/m2 K W/m2 K W/m2 K m2 K/W m2 K/W J/kg K θ θ declared U U’ U R Rdeclared (psi) (chi) c Indexes t transmission Page 6 .3 Symbol d l b h A V q n Symbols and units thickness of layer length width height area volume volume flow air change density celsius temperature temperature difference heat flow. which is required to heat the incoming air in the case of exchange of air through ventilation. Regarding definition see DS/EN ISO 7345 item 2.5 Heat flow density The heat flow density is the heat flow per area unit. 1.Extracts from DS418:2002 Thermal conductivity The thermal conductivity is the heat flow density divided by the difference in temperature under stationary conditions. which can be estimated for a building with transmission coefficient as well as windows and door areas according to the building regulations. Regarding definition see DS/EN ISO 7345 item 2. Ventilation loss The ventilation loss is the heat amount per time unit. heat loss thermal conductivity declared thermal conductivity the final transmission coefficient the uncorrected transmission coefficient corrections according to annex a design heat flow resistance declared heat flow resistance linear transmission coefficient (linear loss) transmission coefficient for point cold bridge heat density. Extracts from DS418:2002 v i e j h m l f f f g g r s s p sa ventilation indoor outdoor (exterior) soil homogeneous layer layer of material layer of air brick tie (fastener) foundation filling (for instance in a door) layer of air (air space) glass part of window radiation surface space panel (for instance filling in a door) joint Often used prefixes Prefix Multiple T (tera) 1012 G (giga) 109 M (mega) 106 k (kilo) 103 m (mili) 10-3 Page 7 . 1 workrooms the design room temperature is set taking into account the character of the work. Design indoor temperature The design room temperature θi in residential rooms is normally set to 20ºC. convector pitches and alike higher temperatures should be taken into account. but cases of doubt should be recalculated using a heat balance for the room. In unheated rooms the temperature can be determined by estimation (qualified guess). In front of the heat sources placed near windows.Extracts from DS418:2002 2 Design temperatures Figure 2. The temperature here is to be set according to the heat plants designed temperature. which is to be carried out in the rooms. where it is equal to outdoor temperature is the soil temperature. Normally a temperature of 30ºC is to be applied. wall parts with reduced insulation. In rooms using floor heating the temperature in the floor construction at floor level is set equivalent to the design temperature of the floor heating plant. This temperature is also applied when calculating the heat loss through the foundations near constructions with floor heating. Normally a temperature here of 50ºC is applied. in deeper soil is the temperature behind radiators is the floor temperature is the temperature in heated basements 2.1 – Examples of design temperatures ’j j r g k is the soil temperature. In Page 8 . passages and alike is -12ºC. The crawl space ventilation applied in cubic meter per second normally is 0.4 Other design temperatures The design crawl space temperature for normal sufficiently ventilated crawl space the temperature is set to -5ºC. Page 9 . The design temperature in open gateways.Extracts from DS418:2002 2.2 Design outdoor temperature The design outdoor temperature θu usually is set to -12ºC. 2. In special cases the design outdoor temperature can be either increased or reduced. The temperature in other kinds of crawl spaces should be calculated by using the heat balance of the applicable crawl space.3 Design soil temperature The design soil temperature θj under heated buildings and in the substrata around the heated buildings is set to 10ºC.3 times the total area in square meters of the ventilation openings. 2. 3 Transmission loss through partition walls and storey partitions and basements decks The heat transmission through partition walls and storey partitions between rooms of different temperature is to be found using the formula: φ t = U A ∆θ where ∆θ is the difference between the temperature in the adjacent rooms. θi θu is the transmission loss in W is the transmission coefficient in W/m2K is the area of the plane in m2 is the design indoor temperature in °C 3.Extracts from DS418:2002 3 3. 3.6. basement floors and basement walls The transmission loss through basement walls being in contact with soil for a depth up to 2 meters is to be estimated using the formula: φt = U A(θ i − θe ) The transmission loss through ground supported floors and basement floors as well as basement walls in depths more than 2 meters and furthermore basement walls with contact to soil placed underneath the building are to be calculated using the formula: φ t = U A(θ i − θ j ) where θj is the design earth temperature in °C. roofs.4 Transmission loss through joints around windows and doors The transmission loss at the joints around windows and doors (cold bridges) is to be estimated using the formula: Page 10 . windows and external doors The transmission loss through vertical areas towards outside is estimated using the formula: φt = U A(θ i − θ e ) where φt U A is the design outdoor temperature in °C The transmission coefficient is calculated according to chapter 6.1 Calculation of transmission loss Transmission loss through external walls.2 Transmission loss through ground supported floors. The difference in temperature causes heat loss from the warmest room and a heat supplement to the coldest room.1 and the area according to chapter 3. 3. referring to chapter 6.6 Calculation of transmission areas The transmission areas is defined by the outer surface of the outer walls.12 is the total length of the joint.Extracts from DS418:2002 φt = Ψsa l sa (θ i − θ e ) where lsa ψsa is the linear loss for the joint using the unit W/m K.1 and figure 3. the top surface of foundations and the top surface of the heat insulation in the ceiling on the upper floor or in the roof. Permanent cupboards and closets are normally neglected. The area in this case is equivalent to the transmission area of the ceiling. referring to chapter 3.5 Transmission loss through foundations under external walls The transmission loss through external wall foundations surrounding ground supported floors and foundations under basement walls for a depth up to 2 m. the bottom surface of the basement deck. regardless the roof area being bigger than the ceiling area. the transmission area is to be calculated to the centerline of the partition wall. Likewise other attachments for instance skylights. is to be estimated using the formula: φt = ψ f l f (θ i − θ e ) where ψf lf is the linear loss for foundations in W/m K according to chapter 6. The transmission loss from the upper rooms of a building through the ceiling and roof is normally calculated as a whole.6. for foundations under basement walls in more than 2 meters depth and for foundations underneath the building.14 is the length of the foundation in m.6.2. dormers Page 11 . For basement floors the transmission area is counted to the inner surface of the foundations of the basement walls. For basement walls towards soil the transmission area is defined reaching from the terrain level to the upper side of the basement floor. Inner doors are normally not taken into account but are regarded as wall areas. the design soil temperature is to be used instead of the design outdoor temperature.7 3. For building parts containing different construction types the transmission area should be calculated individually. For construction with bending surface the transmission area is measured along the bending outer surface. At partition walls that do not make parts of a climate shield. For foundations under external walls and under basement walls the transmission area is defined by the upper surface of the floor and by the inner surface of the external walls. 3. even if there may be an unheated attic in between. see figure 3. and at the storey partition the area is to be calculated to the top surface of the applicable storey partition. Unheated basement Heated basement Building with attic and terrain floor Building with sloping roof and basement Figure 3. The transmission area of windows and outer doors is equivalent to the free space in the openings.6.12. Page 12 . For skylights and sloping windows with free frame sides the transmission area is to be calculated either according to the outer measurements of the skylight or the sloping window or the external surface-area considering the height of the joint see 6.13 and figure 6. Those areas are also applied when calculating the permissible windows and outer door area according to the building regulation as well as when calculation of the U-value and the permissible area of doors and windows according to the Building regulation. Vertical cross section.4.Extracts from DS418:2002 and oriel bay windows the transmission area is counted equivalent to the resulting outer measurements.1 – Measurements for the estimation of transmission areas. 7 Calculation of the length of the linear cold bridge For the joints around the windows and the doors the length of the cold bridge lsa is defined by the perimeter of the opening. Horizontal section in outer and partition walls. the height of the wall or the length of the foundation. should be divided into relevant sections regarding the length of the cold bridge.1. the length of the cold bridge lk is defined by the height of the column or the length of the beam. see figure 3.7. for instance an outer wall. where the construction details vary. 3. for instance in front of ribs in cavity walls. the length of the cold bridge lk is defined by the extend of the varying insulation thickness. Page 13 . For joints around skylights and sloping windows with exposed frame sides the length of the cold bridge lsa is defined by the sloping window’s or the skylight’s outer measurement.6. Cold bridges. Where connecting storey partitions and walls and also foundations underneath partition walls cut off or reduce the insulation in the applicable construction.2 – Measurements for the estimation of transmission areas.Extracts from DS418:2002 Figure 3. see figure 3. or where the insulation is cut off locally within the construction. the length of the cold bridge lk is defined by the width of the deck. For outer wall foundations at ground supported floors the length of the cold bridge lf is defined by the outer perimeter of the foundation. Where for instance columns and beams in outer walls or in roofs are cutting off or reducing the insulation.7.2. In constructions where the thickness of the insulation vary within the transmission area. l sa = y 2 + x 2 + y 2 l k = y1 + x1 + ( y1 + y 2 + y 3) + ( x1 + x 2 + x3) + ( y1 + y 2 + y3) + x3 + y1 Figure 3.Extracts from DS418:2002 Figure 3. Horizontal section in the top of the foundation.Measurements for the estimation of the length of the linear cold bridges around the window opening in an outer wall. where ribs of bricks are applied beside and over the window opening.7.1 . Page 14 .2 – Measurements for the estimation of the length of linear cold bridge for foundations underneath outer walls near ground supported floors.7. Extracts from DS418:2002 Page 15 . the determination of the volume flow queue can be based of the air exchange in.5 10-3 m3/s should be calculated per m joint between fixed and operable frame for buildings with normal location and 0. In very large rooms – storage rooms and alike – the air exchange could be determined at a lower value for instance 0.) as well as such rooms in other buildings.5 h-1 (especially with low outdoor temperatures). For all normal rooms n is set to 0. meaning the volume of fresh air per hour in relation to the room volume.5 h-1. The term normal rooms comprises in residential buildings (living room. kitchen. that could be compared to the above rooms in residential buildings. Page 16 .205 kg/m3 (dry air). an estimated air intake of 0.2 Natural ventilation In buildings where the renewal of air is provided through natural ventilation.1 Calculation of ventilation loss Ventilation loss The ventilation loss for a room is calculated by the formula: φ v = ρ c q (θ i − θ e ) where φv is the ventilation loss in W is the density of the air in kg/m3 c is the specific heat of the air in J/kg K q is the volume flow of outdoor air let to the room in m3/s θi is the design room temperature in °C θe is the design outdoor temperature in °C At 20°C and 1013 mbar c = 1005 J/kg K and = 1. For normal rooms the difference between air temperature and room temperature is to be neglected. toilet and bath rooms etc. the ventilation loss should be estimated taking the length and the permeability of the joints into account respectively for the individual rooms as well as the location of the building. 4.8 10-3 per m joint for extraordinary exposed buildings. causing an exchange of fresh air higher than 0.3 h-1.34 n V (θ i − θ e ) 3600 where V is the volume of the room in m3 (interior measurements) n is the exchange of air in h-1. If leaks through joints at windows and doors are expected to be bigger than normal.The n ventilation loss therefore is: φv = ρ c V (θ i − θ e ) ≈ 0. For windows and outer doors which wind proof is not documented in details.Extracts from DS418:2002 4 4. however the ventilation loss should be estimated on the basis of the volume flow equivalent to n = 0.5 h-1 in the ventilated area (for instance the living section). In case the exhaust air volume equals an exchange of air which is lower than 0. The ventilation loss for rooms supplied with exhaust. regardless the number of exhausts from each room. caused by wind and temperature exposure. where the renewal of air is provided through mechanical ventilation. The term mechanical exhaust covers ventilation through exhaust plants designed for uninterrupted running.4 Other mechanical ventilation systems In buildings equipped with plants for both mechanical exhaust and mechanical injection.Extracts from DS418:2002 4. It should be observed that beside the infiltration/exfiltration caused by the ventilation plant covering the difference between the exhaust and injection volume flow through the plant additionally an infiltration and exfiltration will occur.3 Mechanical ventilation (exhaust) In buildings. 4. only for short term operation. Fresh air Exhaust Heat recycling Air handling Recycled Injection Infiltration Exhaus t Exfiltration Figure 4.4 Example of mechanical ventilation system Page 17 . see figure 4. It should be taken into account that fresh air will be led to the rooms through infiltration dependent on the difference between the exhaust and injection volume flow through the plant as well as the tightness of the building.2. The ventilation loss is to be distributed to the different rooms of the building according to their volume. the ventilation loss is to be calculated according to the performance of the plant.4. the ventilation loss is to be calculated on the basis of the exhaust volume flow at normal operation.5 h-1 in normal rooms. is to be calculated according to chapter 4. and that ρ is referring to those conditions. The temperature θl is to be estimated according to the heat recycling units performance. The ventilation loss should be distributed to the different rooms of the building according to layout of the ventilation plant.θe of the fresh air led through the ventilation plant is caused by the heat recycling only. The infiltration and exfiltration are equal. Page 18 . The volume flow q1 and q2 are determined according to the ventilation plant performance under normal continuing operation. As an example the figure 4. Normally q4 = (q2 – q1) + q3. The calculation of the ventilation loss depends on the layout of the plant. Normally q1 is a bit less than q2.2 h-1. The exfiltration q3 is determined regarding the tightness and location of the building. Furthermore it is provided that the increase of temperature θl . if q1 = q2. for instance 20°C and 1013 mbar. The ventilation loss can be covered by the induced heat partly from air handling components.4 shows a ventilation plant comprising mechanical injection and mechanical exhaustion as well as heat recycling provided the air is not humidified and the system does not include heat pumps. For buildings providing particular tightness the air exchange n = 0. in those rooms where the fresh air is induced.Extracts from DS418:2002 The big rectangular symbolises the area which is handled by the ventilation plant and which can comprise several rooms.1 h-2 can be used. partly from heat sources like radiators. In normal cases the exfiltration can be set to an air exchange of 0. the ventilation loss in the ventilated area can be estimated according to the formula: φ v = ρ c(q 2 − q 3 )(θ i − θ e ) − ρ c q1 (θ 1 − θ e ) q1 Is the volume flow of fresh air provided through the plant m2/s q2 Is the volume flow of the exhaust air m2/s q3 Is the volume flow of exfiltration m2/s q4 Is the volume flow of infiltration m2/s θi Is the designed room temperature in °C θe Is the designed outdoor temperature in °C θ1 Is the fresh air temperature after the heat recycling plant in °C c Is the specific heat of the air J/kg K ρ Is the density of the air kg/m3 It is provided that all q are measured at same air condition. Page 19 . 5. The calculation in this case is simplified as follows: If the different rooms in general are equally heated.2 Calculation of transmission loss As part of an estimation of the heat insulation of a building the whole buildings transmission loss can be estimated without regards to the room division. increased temperature in constructions providing floor heating and also through foundations near constructions with floor heating. as well as in front of radiators and other heat sources as described in chapter 2.1 Calculation of the total heat loss The heat loss for a building The total heat loss for a building is the sum of the transmission loss of the buildings external surfaces and the ventilation loss for the whole building. should be taken into account. This method is supplied when comparing the heat loss frame or the energy frame of a building. The total heat loss for a building or a part of a building can also be estimated as the sum of all the transmission losses of each room.Extracts from DS418:2002 5 5. However.1. it can be neglected that the room temperature in few rooms for instance the bathroom is different. 10 0. The values for horizontal are used for heat flows that deviate no more than 30° from the horizontal plane.04 Heat flow direction Horizontal Downwards 0.1 Calculation of transmittal coefficient Transmittal coefficient and heat flow resistance The transmittal coefficient for a wall.04 0. a roof or similar building component comprising parallel homogenous layers is to be estimated by the formula: n 1 = R + R + Ri si se U´ i =1 where U´ is the uncorrected transmittal coefficient in W/m2 K Rsi is the surface resistance at the inner surface in m2 K/W Rse Ri is the surface resistance at the outer surface in m2 K/W is the resistance for each material layers in m2 K/W The transmittal coefficient must be corrected regarding cracks in the insulation layer. U = U´ + ∆U where U is the resulting transmittal coefficient inclusive potential corrections.13 0. estimated according to Annex A Resulting U-values are indicated to 2 decimals. For non-plane surfaces or special surface conditions the procedure in Annex A of DS/EN ISO 6946 is used.2 Surface heat flow resistance For plane surfaces the values in table 6. if no more precise specifications are at hand. a storey partition. according to Annex A.04 6.2 are to be used.Extracts from DS418:2002 6 6. the values for horizontal heat flow are used.2 – Surface resistance Rsi Rse Upwards 0. Table 6. and rain on reversed roof. 6.17 0.3 Heat flow resistance for a material layer The resistance for an unbroken homogenous material layer is Page 20 . For building components with unknown heat flow direction. wall ties and anchors which penetrates the insulation layer. in W/m2 K ∆U is the correction. 3 – Oversize/density for loose material fillings Insulation material Minimum oversize Minimum density injected in loosely injected into cavities Attics Horizontal Vertical Glass wool filling +5% 30 kg/m3 55 kg/m3 Rock wool filling +5% 50 kg/m3 65 kg/m3 3 Expanded polystyrene. however now more than 0.4 Heat flow resistance for air filled cavities The Values in this chapter is valid for air filled cavities which are: • Limited by parallel surfaces perpendicular to the heat flow and having an emission rate larger than 0. Thus the insulation has to be injected with an oversize according to relevant product standard or table 6. 6. If the above mentions requirements are not meet.8 • Has a depth in direction of the heat flow of less than 0. the material has to be injected with minimum density according to relevant product standard or table 6. Page 21 . filling 1) 15 kg/m 15 kg/m3 Cellulose fibers +25% 50 kg/m3 65 kg/m3 Expanded perlite 0% No demands No demands 1) These materials not to be applied in ceilings and attics Measurement of thickness The thickness of the layer of loose material filling is to be measured according to DS/EN 823. that provides a load impact of 20 ± 1. Measuring is done using a plate.5 Pa. Regarding values of λ see Chapter 7.3. In order to prevent subsidence in closed cavities. balls 1) 20 kg/m 20 kg/m3 3 Expanded polystyrene.3 m. λ is the design thermal conductivity of the material or the product in W/mK. For loose material fillings injected into attics the permanent insulation thickness after subsidence is applicable.Extracts from DS418:2002 R= d λ where d is the thickness of the material layer in m. the procedure in Annex B DS/EN ISO 6946 are to be used.3.1 times the smallest dimension of the cavities length or width. Table 6. horizontal and vertical. For compressible materials the thickness of the material layer in the completed construction is to be applied. 11 7 0. but < 15 cm2 pr.00 0. by facing brick wall and wooden facing thicker than 20mm. 6. The heat flow can instead be calculated as indicated in DS/EN ISO 13789.1 is used.15 m2K/W must be taken to account. are not considered as ventilation openings. which doesn’t have an insulating layer on the outside.17 25 0.19 50 0. but has small openings towards the outside air.16 0.18 0.13 10 0.Extracts from DS418:2002 The heat flow resistance for components with cavities depths larger than 0.4.00 0.11 0.21 100 0.4. but < 15 cm2 pr.2 Slightly ventilated cavities By slightly ventilated cavities is meant cavities where ventilation towards the outside air. 6.16 0.3 m can not be calculated.1 – Heat flow resistance for non-ventilated cavities m2 K/W Cavity depth Heat flow direction Mm Upwards Horizontal Downwards 0 0. If the heat flow resistance for an external facing exceed 0.22 300 0. is created by openings which exceed: 15 cm2 pr. m2 surface area.18 0.3 Ventilated cavities By Ventilated cavities is meant cavities where ventilation towards the outside air. provided that these openings are not meant as ventilation for the cavity and the area of the openings doesn’t exceed: 5 cm2 pr.g.15 0. Page 22 .00 5 0. m horizontal length.11 0.16 0.18 0. is created by openings which are: Vertical cavities > 5 cm2.4.15 m2K/W.16 0.13 0. must also be considered as an non-ventilated cavity.15 0. 6.13 0.17 0. m2 surface area for horizontal cavities. Horizontal cavities > 5 cm2.15 15 0.1.23 Note – For values in between linear interpolation can be used An air layer. m horizontal length for vertical cavities 5 cm2 pr.1 Non-ventilated cavities For constructions with non-ventilated cavities the values in table 6. m horizontal length for vertical cavities 15 cm2 pr. m2 surface area for horizontal cavities Drainage openings made as vertical joints.4. For these cavities the heat flow resistance is set to half the value of those in table 6.16 0.4.18 0. Table 6. a heat flow resistance of no more than 0.4. This comes to use e. 5 – Total heat flow resistance for attics and roof covering Type of roof cover Steel or metal sheet Fibre cement shingles or – corrugated sheets on battens Roof tile with ceiled joints on battens Roof tiles on battens with wind proof under layer Bitumen felt on roof of 25 mm wood Thatched roof with wind proof under layer m2 K/W 0. ventilated attics.3 0.5 are to be used for cold. the heat flow resistance is to be calculated.1 0. regardless of an eventual angel between the roof and the ceiling. 6. a timber frame structure or brickwork with joints consist of homogeneous and inhomogeneous plane parallel layers.Extracts from DS418:2002 When calculating the heat flow resistance for a component containing a ventilated cavity.3 0.6 Constructions with inhomogeneous material layers If a building component – e.5 are the total heat flow resistance for attic and roof covering. The same value is normally to be applied for the cavity just underneath the floorboards on floor joists. as if each of the inhomogeneous layers in fact is homogeneous layer with a heat transmittal capability which is an estimated average value of the heat transmittal capability of the different sections of the layer.2 0. Aa . Table 6. is set to 0.5 Ventilated attics The values in table 6. Ab is the inner surface resistance in m2 K/W is the outer surface resistance in m2 K/W is the resistance of homogeneous layer in m2 K/W is the thickness of inhomogeneous layer in m is the estimated average thermal conductivity of the inhomogeneous layer in W/mK is the area of inhomogeneous layer sections in m2 Page 23 . is set to the value of the inner surface resistance for the construction (see chapter 6. λ.2).g. meaning layers comprising 19-25 mm boards put op with internal distances and thus providing hollow space. The heat flow resistance for battens. The values in table 6. = a a Aa + Ab + Rsi Rse Rh d d λ. The heat flow resistance is applied to the ceiling-area.3 6.16 m2 K/W. the heat flow resistance for the cavity and all other layers between the cavity and the outside surface. 1 = Rsi + Rse + Rh + U' Where A λ + Ab λb + λ.2 0. The linear heat loss (the linear transmittal coefficient) Ψk is covering only the increasing heat loss through the thermal bridge because of two dimensional heat flows related to the heat loss calculated for one dimensional heat flow. the beam or the column are in the same level as the actual construction surface. concrete or tile. the transmittal Ui for the sub area is to be estimated as if the storey partition. possible thermal bridges as well as the impact of reduced insulation thickness in parts of the construction should be taken into account.7 Constructions with cold bridges When estimating the transmittal coefficient for a construction. if they are not calculated separately. for instance the surface of the walls. λb is the applicable thermal conductivity in W/mK 6. The transmittal coefficient Ui for the sub areas is to be calculated. beams and columns for instance of concrete or steel. not more than the width of the applicable cavity. Such thermal bridges could be: • • Brick work filling and ribs for instance around windows and doors Wall penetrations of for instance metal.7 Ψk is the linear heat loss for the individual linear thermal bridge in W/mK χj is the point heat loss for the individual spot thermal bridge in W/K n is the number of sub areas m is the number of linear thermal bridges p is the number of spot thermal bridges When using loose filling of insulation material.ψ k + p j =1 χj A n i =1 A= Ai Where A is the total transmittal area in m2 of the construction. Page 24 .6 Ai is the sub area in m2 Ui is the transmittal coefficient of the sub area with a one-dimensional heat flow in W/m2 K lk is the length of the individual linear thermal bridge in m.U i + m k =1 l k . the thickness of the insulation layer is equivalent to a width of the cavity. The transmittal coefficient for a construction including thermal bridges and reduced insulation thickness in parts of the construction is to be calculated using the formula: n U = ' i =1 Ai . see chapter 3.Extracts from DS418:2002 λ a . see chapter 3. however. If insulation slabs are used. the thickness of the insulation layer is equivalent to the slab thickness. as if only one dimensional heat flow occur. For continuing storey partitions. It should always be appraised. the constructions or the indoor climate. However. for instance wood or lightweight concrete with a low density. outer wall and basement outer wall as well as the connection of outer wall and ceiling is to be included in the U’.1. The thermal bridge impact in corner joints between building components for instance the connection between basement deck. in connections between basement deck. Where values or figures in the following tables are shown in bracket it is indicating that the applicable linear loss or spot heat loss or brick tie correction may be neglected when calculating the transmittal coefficient for a construction. interpolation in the tables are allowed. provided that the insulation is continuing uninterrupted or only interrupted of material with a thermal conductivity lesser than 0. The following shows values for common thermal bridges in typical constructions. that the total heat loss due to linear thermal bridge and point thermal bridges. is insignificant in relation to the total heat loss of the construction. Thermal bridges in for instance corners of outer walls.value for the building components that appear in the joint. The calculation should show how the thermal bridge impact is distributed on the U’-values of the individual building components. see figure 6. both changes should be taken into account. The values in the table are applicable for every single change of insulation thickness. In case of two changes in insulation thickness for instance on each side of a rib.7. When calculating these thermal bridges it should be observed that the transmittal areas comply with the total outer measurements. linear heat loss Ψk less than 0.02 W/m K and point heat loss χj less than 0. if there might be other thermal bridges that may have significant impact on the heat loss. however. Brick ties For the brick ties correction ∆Uf refer to the values or the formula in Annex A. outer walls and basement walls as well as the connection between outer wall and ceiling may normally be neglected.Extracts from DS418:2002 The point heat loss (transmittal coefficient for point – thermal bridge) χj is covering the total increase of heat loss due to the thermal bridge.7. Pillars and ribs The linear loss Ψk for pillars and ribs in cavity walls is shown in table 6. it is provided.3 W/m K. may be neglected. For constructions with similar structure.02 W/K. The thermal bridge impact including the impact of the two and three dimensional heat flows is taken into account in the transmittal coefficient U’ for the construction. Page 25 .1. When calculating the transmittal coefficient for a construction. The linear loss is to be seen as an addition to the one dimensional heat flow through the thermal bridge. where the thermal bridge appears. but different insulation thickness and heat transmittal capability. observing the provisions mentioned above. 01) 40mm 0.7.02 30mm 0.7. a.1 – Example of a single change of insulation thickness due to pillars/ribs near window opening and a pillar in a longer part of an external wall section.Extracts from DS418:2002 The heat loss taking place through the joint between construction and windows or doors is calculated separately. Cold bridge insulation with thermal conductivity not exceeding 0.01) 50mm (0.01) 1) Reinforced concrete with 2% reinforcement bars Applies also for light weight concrete with a thermal conductivity 0.05 0.1. Pillar near window opening One change of insulation thickness b.04 0.7.12. There is only one change of insulation thickness for the pillar near the window and two times change of insulation thickness at the pillar in the outer wall section.03 (0.04 W/m K. Table 6.01) 10mm 0.06 0. Pillars and ribs in the front wall are provided of the same material as the front wall. Thermal Outer leaf: Concrete Tile Tile Tile Lightweight bridge concrete3) 1) 2) insulation Inner leaf: Concrete Concrete Tile Lightweight Lightweight 1) concrete3) concrete3) None 0.3 W/mK 2) Page 26 .03 0. Pillars and ribs in the rear wall are provided of the same material as the rear wall.05 0. see figure 6. see chapter 6.02 (0.02 (0.7 W/mK 3) Light weight concrete with thermal conductivity 0.02 (0.01) 20mm 0.14 0. Each one of the changes in insulation thickness is shown/indicated in the figures with a set of arrows pointing towards each other and towards the change of insulation thickness.24 0.1 – The linear loss Ψk in W/m K for pillars and ribs in cavity walls depending on cold bridge insulation and material. Pillar in outer wall section Two changes of insulation thickness Figure 6.03 0. The values in the table are for each individual change of insulation thickness. but can also be applied to constructions under the level of the terrain.04 1) 2) Concrete without reinforcement Applies also for light weight concrete with a thermal conductivity 0. Table 6.Extracts from DS418:2002 Change in the plane of the insulation The linear loss Ψk for constructions with change in the plane of the insulation.0 W/mK λ = 0.24 0.0 mm.30 0.13 0. The value comprises the total heat flow through the profile including the one dimensional heat flow. e. Page 27 . The heat flow may be reduced when using perforated/cut profiles. Using the values it is provided that there is an overlap between the insulation in the two planes.08 200 mm 0.3 W/mK 0 mm 0.08 W/K is to be applied for each crossing.15 W/K per meter profile is to be applied. the values in table 6.7. The values is for constructions towards the open air.7.51 0.2.7 W/mK Figure 6.7 W/mK λ = 0. In those cases the heat flow is to be calculated for the applicable profile.7. Overlap Concrete1) Tiles2) Light weight concrete λ = 2. If there is no overlap.g. see figure 6.2 Construction with change in the plane of the insulation and overlap in the 2 insulation layers Steel plate profiles in steel frame walls The linear loss Ψk for continuing profiles of metallic materials in framed walls the value 0.0 mm.7. The value covers the total heat flow through the profile crossing provided that the thickness of the profile does not exceed 2. the value 0. provided the thickness of the profile is not exceeding 2.2 – Linear loss Ψk for constructions with change in the plane of the insulation. For the transmission coefficient for point thermal bridges χj. For other constructions the linear loss can be calculated as described in Annex B. joints between external wall and basement wall is stated in table 6.2.1 are to be used combined with calculation of the one dimensional heat flow through the un-insulated part of the construction. where steel profiles are crossing each other in steel frame walls. 10 – 12 cm sidepiece between the two layers of insulation.7. It is allowed to interpolate in the values of the table. 09 Concrete with λ = 2. penetrating cavity walls. is shown in table 6. Each one of the single changes is indicated in the figure with a set of arrows.13.0 0.4) the values in table 6.3 – The linear loss Ψk in W/m K for partition wall – and basement wall foundations.25 on the upper 40 cm Figure 6.01) Light weight concrete with λ = 0. Table 6. The values in the table are applicable for each individual change of insulation thickness (see figure 6. Two single changes in the insulation thickness are shown – one on each side of the foundation. which continue through the insulation of a terrain floor or a basement floor. see chapter 6.pillars and beams are neglected. that penetrate the insulation of terrain floor or basement floors. as well as for concrete columns and concrete beams in cavity walls (see figure 6.7.03 Light weight concrete with λ = 0.4 apply.7.7.3 – Example of single change of insulation thickness at a partition wall foundation. The transmission coefficient Ui for the sub area is to be calculated as if the deck or the wall is ending flush with the actual construction surface. pointing at each other and towards single change.7.Extracts from DS418:2002 Foundations under partitioning wall The linear loss Ψk for partition wall and basement wall foundations. The heat loss through the outer wall foundation near terrain floors is to be calculated separately. The linear loss is seen as an addition to the one dimensional heat flow through the thermal bridge. The linear loss is seen as an addition to the one dimensional heat flow through the thermal bridge. The values in table Page 28 . Other penetrations For the linear loss Ψk for concrete deck and walls.3.3) Foundation structure W/m K 0.7. The one dimensional heat flow at the end of external walls in corners by concrete.25 on the upper 20 cm (0.7. 4 – Examples on continuing deck and walls penetrating cavity walls and examples of concrete columns and beams in cavity walls.7. bricks or steel – see figure 6. For profiles of steel or stainless steel the value must not be less than χj = 10 x Ao.55 Concrete with 2% reinforcement bars Continuing concrete deck Concrete column in full wall width Concrete column in corner Figure 6.4 for concrete.5.7. see figure 6. see figure 6. The values in the table apply for cavity walls of bricks. Table 6.7. where Ao is the area in m2 of smallest possible rectangle embracing the profile.7.6 Page 29 .7. The values are applicable for each single change of insulation thickness Thermal bridge Continuing concrete-deck or –wall Concrete-column or -beam in full wall size Concrete-column or -beam in outgoing corner Concrete-column or -beam in ingoing corner 1) W/m K 0.7. that penetrate cavity walls.4.7 W/m K.5. where the heat transmittal capacity of the tile does not exceed 0.pillars and beams in corners thus covers the total one dimensional heat flow thru the pillar or the beam and the two dimensional heat flow thru the connected cavity walls.13 0.7. A is the cross area of the beam or the column.15 0.Extracts from DS418:2002 6. Table 6. The values for the transmittal coefficient for point-thermal bridge χj for penetrating columns and beams are shown in table 6. and for concrete-columns1) and -beams1) in cavity walls.5 – Transmittal coefficient χj in W/K for continuing beams and columns of concrete.7.4 – Linear loss Ψk in W/m K for continuing concrete-deck1) and walls1).45 0. Extracts from DS418:2002 Material Brick1) Reinforced concrete with 2% reinforcement bars Stainless steel Steel 1) χj W/K 3.7.9 is the resistance for material layers in the actual floor or wall construction in m2 K/W Page 30 .A 60.8 Doors and windows Not included in this translation Ground supported floors.7. basement floors and basement walls against soil The transmittal coefficient for a ground supported floor or a basement wall directly towards soil is determined from the formula: 6.6 – Determination of smallest embracing rectangle 6.A 11.A 170.5 – Continuing beam Figure 6.9 1 = Rsi + R j + U' where Rsi Rj Rm Rm is the inner surface resistance in m2 K/W is the resistance of the soil in m2 K/W.A Also applies for lightweight concrete with λ = 0. see table 6.7 W/mK Figure 6. 12. deeper than 0.2 + 0.10 6.0 6.0 m2 K/W 1.02 W/mK may be neglected.9 – Resistance for soil Rj Component Ground supported floor.5 m below the terrain Basement floors. The resistance Rj found in table 6.12.11 Concrete sandwich elements Wedged insulation Not included in this translation Not included in this translation 6.5 m above to 0.9 for basement walls until a depth of 2 m is an average resistance for the wall until the depth of 2 m.5 m below the terrain Basement walls Until 2 m below the terrain (h is the depth in m) More than 2 m below the terrain and under the building 0. Normally linear heat losses (Ψsa) that are less than 0. Table 6.1 Page 31 . from 0. For constructions with floor heating. which extends above the ground surface is calculated as walls towards the open air.3*h 2.5 2.Extracts from DS418:2002 The resistance of the soil does also include an eventual external surface resistance at the ground surface.Cold bridge insulation with thermal conductivity less than 0.04 W/mK .12.1 – 6.1 Linear loss Ψs in W/mK for joints around windows and doors in cavity walls the values in the tables provides . Deep basement walls are divided into an area of the depth of 2 m and an area of the depth of more than 2 m.Regarding placement of the frame see figure 6. 6. The part of the basement wall.4.Frame depth no less than 90mm . the resistance is calculated from the plane of the heating source as neither the resistance of layers above this plane nor the inner surface resistance is included in the transmittal coefficient.12 Joints around windows and doors The linear loss (the linear transmission coefficient) Ψsa for joints around windows and doors in cavity walls is based on the values in table 6. When calculation the resistance for floor constructions the capillary breaking layers can be included.12. For basement walls the depth is measured to the upper side of the basement foundation. For ground supported floors and basement floors the depth is measured to the upper side of the finished floor. concrete3) 1) 1) 2) Outer leaf: Concrete Concrete Brick Light weight Light weight concrete3) concrete3) No 0.01) (0.03 0.1a Sketch 2 Example of placement according to table 6.11 0.03 0.05 0.05 0.11 0.25 0.05 0.06 10 mm 0.09 0.12.12.12.11 0.3 W/mK 2) Sketch 1 Example of placement according to table 6.02 0.34 0.04 0.11 0.11 0.17 0.7 W/mK 3) Light weight concrete with thermal conductivity 0.05 0.12.02 0.Extracts from DS418:2002 Table 6.09 0.3 W/mK 2) Table 6.02 50 mm (0.10 10 mm 0.12.08 > 10 mm 0.01) (0.01) 1) Reinforced concrete with 2% reinforcement bars Applies also for light weight concrete with a thermal conductivity 0.05 20 mm 0.7 W/mK 3) Light weight concrete with heat transmittal capability 0.04 0.17 0.03 40 mm 0.and door-openings in cavity walls Page 32 .17 0.13 0.1 – Placement of frame in windows.01) (0.07 1) Reinforced concrete with 2% reinforcement bars Applies also for light weight concrete with a heat transmittal capability 0.1a – The frame placed in front of the thermal bridge insulation in a wall with at least 20 mm overlapping respectively the front wall and the rear wall (sketch 1).12 0.01) (0.1b – The frame displaced from the thermal bridge insulation in the wall for instance either next to the front wall or next to the rear wall (sketch 2).02 0.1b Figure 6.03 0.04 0. Cold bridge Brick Brick Brick Light weight Inner leaf: Concrete1) insulation.09 0.03 0.04 0.09 0. Thermal Inner leaf: Concrete1) Brick Brick Brick Light weight bridge concrete3) 1) 1) 2) insulation. Outer leaf: Concrete Concrete Brick Light weight Light weight concrete3) concrete3) No 0.02 0.04 30 mm 0. 08 Displaced form the line of the Values in table 6.12.12. the values according to table 6.2 – The linear loss Ψs in W/m K for joints around windows and doors in insulated timber frame walls with lightweight cladding or with brick front wall. For outer walls with external insulation covered by rendering.2 Table 6. ceramic tiles or alike. continuing behind the window.Frame depth no less than 90mm .12.Frame depth no less than 90mm . The values in table 6.2 Page 33 .2 also apply for massive outer walls with external insulation in timber frame construction or with insulation fastened directly on the massive part of the outer wall.Extracts from DS418:2002 6.12. 6.12.03 20 mm overlapping (sketch 4) 0.12. Placement of frame W/m K In line with the insulation (0.Regarding placement of the frame see figure 6.and doorframes in openings in timber frame constructions with brickwork front wall.1b for “none” insulation (sketch 5) thermal bridge insulation Sketch 3 60 mm overlapping of the insulation Sketch 4 20 mm overlapping of the insulation Sketch 5 displaced from the line of the insulation Figure 6.3 The linear loss for joints around windows and doors in front of metal frames in insulated frame walls with light weight cladding or with a brick work front wall the values in the tables provides .12.Regarding placement of the frame see figure 6.00) 60 mm overlapping (sketch 3) 0.12.2 – The placement of window.2 The linear loss joints around windows and doors in insulated timber frame walls with lightweight cladding or with brick work front wall the values in the tables provides .or doorframes.12.1 apply. the transmission coefficient should be increased according to the heat loss through the side of the frames. The height of the joint is measured from upper side of insulation in the roof construction to bottom side of the frame in the skylight and to the upper side of the insulation on the side of the frames in roof windows. The values express the heat loss through the joints provided that the thickness of the metal frames are not exceeding 2 mm.Extracts from DS418:2002 Table 6. the linear loss is determined on the basis of an estimated average taken into account the areas of no insulation and the areas of insulation within the joint. The heat loss through the frame side of the windows is determined by using the values in table 6.12.4 The linear loss Ψs for joints around roof light and skylight including connecting panels and frame. In this case the heat flow has to be determined for the actual profile. This also apply.04 W/m K.4.12. that the heat flow capability of the insulation does not exceed 0.13 20 mm overlapping the insulation and metal Values in table 6. 6. then the linear loss will be equal for all the window sides. The linear loss is to be estimated for all the sides of the window. se figure 6.4. For other joint measurements and insulation thickness’ interpolation is permitted. Page 34 .4 provides.15 60 mm overlapping the insulation and metal 0. If the joint comprise both sections with and without insulation. Placement of frame W/m K In line with the insulation and the metal profile 0.12.3 – The linear loss Ψs in W/m K for the joint around windows and doors in front of metal frames in insulated frame walls with light weight cladding or with a brick work front wall. Application of the values in table 6. se figure 6.12.12. if the roof construction and the insulation next to the window is slightly modified.12. The heat flow can be reduced by using perforated profiles. If the same insulation is used in all sides of the windows. In case the transmission coefficient for the skylight or the roof window is estimated without taken into account the heat loss through the side of the frames.1b for “none” profile thermal bridge insulation Displaced form the line of the insulation (sketch 5) The values in table 6.3 also apply for massive outer walls with external insulation between metal frame profiles.12.4 measuring the frame height from upper side of the joint to the outer surface of the glassing.11 profile 0.12. Extracts from DS418:2002 Table 6.14 0.13.03 0. Skylight Insulation thickness in the joint b.13 0.13.11 300 0.04 100 0.13.13. 6.13.1.33 0.08 0.13. windows and doors. The constructions are shown in figure 6.05 0.08 0.1 and 6. The figure furthermore shows the measurement when determining the height of the frame for skylight and roof windows in cases where the heat loss through the side of the frames is neglected.20 0.06 200 0.65 0.23 0.2.15 0.2.12.4 – Measurement when deciding height of the joints around skylight and roof windows and insulation thickness within the joints. • For cavity walls and for other outer walls with rear wall in concrete.13 Foundations under external walls Foundations for external walls at ground supported floors The linear loss (the linear transmission coefficient) Ψf for external wall foundations at ground supported floors is estimated by table 6. brick work. Page 35 .1-3: • In connection with frame walls or equivalent lightweight wall construction. light weight concrete or alike. as shown in table 6.3.16 Height Frame Joint Height Frame Joint a.25 0.01) 50 0. Roof window Insulation thickness in the joint Figure 6. as shown in table 6.13.05 0.12. • For concrete sandwich elements.02 (0.4 – Linear loss Ψs in W/m K for joints around skylight and roof windows Thickness of insulation in the joint Height of joints None 25 mm 50 mm 75 mm Mm 0 0.2 or table 6. as shown in table 6.45 0. • The external wall.30 0.41 0. at least 40 cm down with thermal conductivity not exceeding 0.0 W/mK Light clinker concrete with heat transmittal capability 0.15 0.04 W/m K along the border of the ground supported floor.65 0.23 0.24 Page 36 .13.37 0.69 0.67 0. Insulation with heat transmittal capability not exceeding 0.16 Light clinker concrete2) upper 40 cm 0.23 0.2b 75 mm insulation. Table 6.13 0. Table 6. Where the foundation is carried out with insulation in the centre.20 0.2a 100 x 300 mm internal vertical insulation with thermal conductivity not exceeding 0.25 W/mK and width 23 cm.17 0.25 0. doors or windows covers the whole top part of the foundation.15 0.37 0. For other construction types or other location of ground or floor the linear transmission coefficient is to be calculated as shown in annex D.22 0.20 0.22 R) 15 mm thermal bridge insulation 0.42 0. • The width of the foundation is no more than 2 cm less than the thickness of the external walls (in case of small foundations it can be compensated by using external insulation with equivalent heat flow resistance).2 a.04 W/m K.67 0.28 15 mm thermal bridge insulationR 0.65 0.13.10 0. Yet the values can be used even if the terrain level is as much as 30 cm lower than floor level.35 0.Extracts from DS418:2002 For constructions with equivalent structure but different insulation thickness and heat flow capability. interpolation in the tables is permitted. concrete or alike.13.10 central insulationM) 1) 2) G) R) L) M) Concrete with heat transmittal capability 2.42 0.13.20 0.2a – External wall with inner leaf of lightweight concrete 3) Foundation Insulation above concrete slab: None 75 mmG U-value for ground supported floor: 0.41 0.13.1c Table 6. it is adequate to cover the centre insulation and 20mm on each side of it.26 0. see figure 6.1 – Linear loss Ψf in W/mK for external wall foundations at ground supported floors in connection with frame walls and equivalent lightweight walls as well as windows and doors.24 Ditto plus 100 mm vertical insulationL 0.10 Concrete 1) No insulation 0.45 0.19 0.13 Light clinker concrete2) upper 60 cm with 0.50 0.10 0.10 Concrete 1 No insulation 0.19 0.30 0. see figure 6.14 0.72 0. see figure 6. Foundation Insulation above concrete slab: None 75 mmG U-value for ground supported floor 0. b and c shows the linear loss Ψf in W/m K for external wall foundation at ground supported floors in connection with external walls with inner leaf of lightweight concrete.47 0.22 Ditto plus 100 mm vertical insulationL) 0.04 W/mK.44 0.45 0. The values shown in the tables provide: • Terrain level is 15 cm lower than floor level.13.04 W/m K 15 mm insulation with heat transmittal capability 0. brick.11 0.23 0.20 0. 21 0.73 0.14 0. Also apply for lightweight concrete with thermal conductivity 0.10 0.04 W/m K.2b 75 mm insulation.22 0.45 0.20 Concrete 1 No insulation 0.13.32 insulationR) Concrete 100 mm external insulationU) 0.72 0.20 0.71 0.20 0.04 W/m K along the border of the ground supported floor.44 0.16 0.69 15 mm thermal bridge insulationR 0.20 Concrete 1) No insulation 0.20 0.33 0.32 0.30 0.33 0.51 0.57 0.20 0.39 0.13 0.18 0. see figure 6.30 0. at least 40 cm down with thermal conductivity not exceeding 0.13.13.1c Table 6.32 0.54 15 mm thermal bridge insulationR Ditto plus 100 mm vertical insulationL 0.17 central insulationM 1) 2) 3) 4) G) R) L) M) 0.7 W/m K.10 0.44 0.32 0.28 R) Page 37 .61 0.44 0. Insulation with thermal conductivity not exceeding 0.25 W/mK and width 23 cm.44 0.55 0.52 0.51 Light clinker concrete2) upper 40 cm 0.2a 100 x 300mm internal vertical insulation with thermal conductivity not exceeding 0.04 W/m K.22 0.33 0.20 0.28 Ditto plus 15 mm thermal bridge 0.34 0.0 W/mK Light clinker concrete thermal conductivity 0.23 0.10 0.37 0.36 0.24 2) 0. Foundation Insulation above concrete slab1): None 75 mmG U-value for ground supported floor: 0.16 0.64 0.11 0.25 Light clinker concrete upper 60 cm with 0.46 0.34 0.10 0. see figure 6.13.17 0.58 0.34 0.58 0.10 Table 6.10 U) Concrete 50 mm external insulation 0.37 0. see figure 6.12 Concrete with thermal conductivity 2.37 0.44 0.2b – External wall with inner leaf of brickwork 4) Foundation Insulation above concrete slab: None U-value for ground supported floor: 0.2c – Outer wall with inner leaf of concrete 1) Foundation Insulation above concrete slab: None U-value for ground supported floor: 0.20 0.71 0.38 0.54 0.13.12 0.42 0.32 0.3 – Linear loss ψf W/m K for external wall foundations at ground supported floors in connection with concrete sandwich elements.33 Ditto plus 15 mm thermal bridge 0.3 W/m K.28 Light clinker concrete upper 40 cm 0.26 0.16 0.18 0.13 75 mmG 0.15 0.14 75 mmG 0.17 Light clinker concrete upper 60 cm with central insulationM Table 6.10 0.Extracts from DS418:2002 Light clinker concrete2) upper 40 cm Light clinker concrete2) upper 60 cm with central insulationM 0.68 0.13.38 0.28 0.30 0.44 0.04 W/m K 15 mm insulation with thermal conductivity 0.27 0. Lightweight concrete with thermal conductivity 0.59 Ditto plus 100 mm vertical insulationL 0.28 0.19 0.23 0. 34 0. see figure 6.37 0. 90 cm down.2 – Insulation around the foundations Page 38 .30 insulationR) 1) Reinforced concrete with 1% reinforcement bars G) Insulation with thermal conductivity not exceeding 0.1d a.37 0.Extracts from DS418:2002 insulationR) Concrete central insulation 60 cm downM) 0.13.31 0.13. concrete b. cold bridge insulation along the border of the ground supported floor b.31 0.04 W/m K. light clinker concrete c. vertical internal insulation c.1 – Foundation top designs a. external insulation in connection with concrete Sandwich element Figure 6.2a U) External insulation with thermal conductivity not exceeding 0. see figure 6.39 0.13. concrete with concrete with central insulation central insulation Figure 6.41 0.04 W/m K along the border of the ground supported floor.04 W/m K. light clinker d.31 Ditto plus 15 mm thermal bridge 0.33 0.2c M) 75 mm insulation.13. see figure 6.04 W/m K R) 15 mm insulation with thermal conductivity 0.13. at least 40 cm down with thermal conductivity not exceeding 0. 0m are applied.21 1) Concrete without reinforcement Dt is the depth under the terrain level.13. The build-up of the constructions are shown in Figure 6.4. depending on the conductivity of the basement wall material and the depth Dt under terrain. For depths larger than 2. The U-value of the basement floor is 0.13.0m 0. The depth is measured from terrainlevel to upper side of basement floor. se figure 6.70 = 0. Figure 6. Table 6.3.3.70 1.13.42 0.4 Examples of the linear loss ψf for foundations under external basement walls.0m the values for 2.36 0.3 – Example of foundation under external basement wall Page 39 .31 0.23 2.20 W/m2K.0m 0. Refer to Annex D for other constructions.13. Depth Concrete1) Leightweight concrete Leightweight concrete Dt = 0.27 0.70 = 0.Extracts from DS418:2002 Foundations under external basement walls Examples of the linear loss ψf for foundations under external basement walls are stated in table 6.13. For use towards the soil. April 2002” 7. the procedures in prEN 14063-1 and prEN 140632(Extruded clay) or prEN14064-1 and prEN 14064-2(Mineral wool) as well as DS/EN 13172 are used. These procedures are shown in chapter 4. concrete and other building materials • PrEN 1745 • DS/EN 12524 By determination of R and it is considered that. Analyses of insulation products moisture conditions in ordinary Climate Screen constructions shows.2 Design values Design values are set based on the products declared values. the declared values are set according to the relevant product standard or European technical approvals. Declared values for heat flow resistance and conductivity in the manufacturer literature and on the products must be attended with a reference to: “DS 418. have different moisture content than by laboratory measurements.2. Control rules. the declared values are set by procedures. with Fm = 1. conform to those used for CE-marked products. Page 40 . products build into constructions.1 Heat flow resistance and conductivity of materials Introduction The calculation of transmittal coefficient is based on the heat flow resistance R and the thermal conductivity λ W/mK for the materials used in the constructions.Extracts from DS418:2002 7 7. are shown in Annex E. the design value of a insulation product can normally be calculated by DS/EN ISO 10456.1 Declared values The following declared values is applied: • For products with a CE-mark. 6.2.1 of the product standards for manufactured heat insulation products DS/EN 13162 – DS/EN13171 and the Annex A in DS/EN 13172 For granulated products. • For other products. according to DS/EN 10456 or directly by: • Annes F. there are examples showing methods and rules of rounding off.2 Basis for determination of heat flow resistance and conductivity 7. version. In addition the middle temperature for the build in product. In Annex D of DS/EN 13162 and in Annex D of DS/EN 13171 as well as in Annex C of DS/EN 13170. Design values for brick. This applies for extruded polystyrene. which applies by laboratory measurements. it must always be considered whether the combination of product. 7. Of the above mentioned standards the on closest to the product to declare is chosen. that correction to the declared value only are necessary when the product is used towards the soil.2.2. construction and influence makes it necessary to correct the declared values in order to achieve the correct design value. However. may deviate from 10° C. conform to the regulations for CE-marked products. In cases where the capillary breaking gravel layer is less than 75 mm the table values applicable for “with soil contact” should be used. The heat transmittal capability of different materials is depending on the temperature. the table values applicable for dry material can be used. the external heat transmittal capability should be applied for the façade layers and the internal heat transmittal capability for the rest of the wall. For cavity walls made of brickwork the external values apply for the outer leaf and the internal values for the inner leaf. = declared * Fm In Annex G. for lightweight concrete block work the heat flow resistance may be increased by 0. The heat transmittal capability for external application of materials apply. It is the thickness of the compressed insulation layer that applies. where the insulation material is placed between the concrete and a layer of gravel of at least 75 mm thickness or equivalent material with grain size of at least 4 mm.2. 7. that the constructions are kept dry by implementing drainage or other precautions. Usually the type of the applied mortar is neglected as well. For brickwork constructions the listed values of design heat transmittal capability for the brickwork as a whole should be applied. where the material is to be found in the outer section of combined outer walls. The moisture content for materials used external and internal are determined by 23° C and 85% RF and 50% RF respectively. if the material is used in partition walls. cold store insulation.Extracts from DS418:2002 extruded clay and mineral wool. roofs and terrain decks. and chimney or stove insulation. Page 41 . For massive external walls of lightweight concrete the external values apply for 100 mm of the thickness of the wall and the internal values apply for the rest of the wall. storey partition and crawl space deck as well as the inner section of combined external walls and roofs. Design conductivity are calculated according to DS/EN ISO 10456. It is provided. For terrain deck and basement floors.15 m2 K/W if all the joints are provided with a 4 x 1 cm strip of mineral wool or polystyrene.3 Special provisions The term insulation material with soil contact means insulation material mounted externally on basement walls as well as foundations. are given. a list of design values for use in calculation of existing constructions in connection with rebuilding and renovation. why the listed practical values apply for temperatures normally found in building constructions. since the contribution of the joints are included in the values (see Annex F). The design heat transmittal capability is applicable for brick work of bricks in normal format without regards to pattern and bond as well as brickwork of light concrete blocks without regards to deviation from the listed block format and joint sizes. For massive external walls of brickwork made of bricks in normal format. However. where the values distinguish between internal and external application of a material. The values are not to be applied for calculation of pipe insulation. For brickwork and light weight concrete constructions. it should be taken into account the heat transmittal capability for internal application. 01 Possibility for air-cracks across the insulation layer No air circulation on the warm side of the insulation layer 2 0. or if the insulation is loos fillings and the cavity is totally filled.Cracks in the insulation layer .1 General Transmissions coefficients calculated according to this standard must be corrected for the effect of: . if the insulation is carried out as two or more layers with displaced joints.Extracts from DS418:2002 Annex A Correction of transmissions coefficients A.2. Table A.2 Correction for air-cracks in the insulation layer The correction ∆Ug must be adjusted for the heat flow resistance of the insulation relatively to the total heat flow resistance of the construction: ∆U g = ∆U '' RI RT 2 where ∆U’’ is the correction for air-cracks in the insulation layer.04 Possibility for air-cracks across the insulation layer Possibility for air circulation on the warm side of the insulation layer Air-cracks across the insulation layer It is presumed that no increases of the heat loss perpendicular to the insulation layer is appears. Page 42 .Ties and similar mechanical fixations .2.1 – Correction for air-cracks in the insulation layer Description Level ∆U’’ 2 [W/m K] 0 0.1 RI is the heat flow resistance of the insulation layer RT is the total heat flow resistance of the construction. ∆U’’ is found in table A.Precipitation on “upside down” roof The corrected transmission coefficient U is achieved by adding the correction ∆U U = U’ + ∆U ∆ U = ∆ Ug + ∆ Uf + ∆ Ur where ∆Ug is a correction for air-cracks in the insulation layer ∆Uf is a correction for ties and similar mechanical fixations ∆Ur is a correction for precipitation on “upside down” roof A.00 No air-cracks across the insulation layer 1 0. Ceiling insulation with two layers of insulation. smooth board. width-. with its own weight. There is risk of air circulation on the warm side of the insulation. Examples Level 0 . and angle-tolerances and stability of dimension provides that cracks larger than 5mm will not occur. rafters. between ceiling struts of trusses . Air circulation on the warm side of the insulation only has consequences if there is connection the cold side of the insulation or to the open. trusses of battens. with a width of more than 5 mm. laid on DPM supported by battens . or the insulation regularly are penetrated by e. or if the insulation lies tightly against the warm side. The increased heat loss caused by air-cracks.Extracts from DS418:2002 If the insulation is carried out with slabs in only one layer without grooved joints. wedged against a hard. og wall-ties. regardless of the layer on the warm side Page 43 .Roof insulation in to layers with displaced joints . . Level 2 . studs.Insulation with displaced joints between to layers of crossing studs. where length-. where the top layer is laid with displaced joints across an insulation layer between the ceiling struts of the truss.Soft ceiling insulation on a hard. beams. wedged against a plane surface on the warm side . where a insulation material not lies tightly against a plane surface og where a hard insulation material is squeezed in between beams. plane surface on the warm side . studs. wedged against brickwork with exactly filled joints on the warm side. beams or battens . The same applies if there is no rigid material on the warm side of the insulation.Hard insulation between. external façade.Soft insulation in a cavity wall with cavity ties. attics or similar.Soft insulation in a cavity wall with cavity ties. the risk of air-cracks across the insulation layer is present. Air circulation on the warm side of the insulation It is presumed that no risk of air circulation on the warm side of the insulation appears if the cavity is totally filled.Roof-. beams or trusses. on the warm side in closed constructions. rafters or trusses.Granulated insulation Level 1 .and terrain-insulation in one layer with grooved joints or one layer.Soft ceiling insulation between ceiling struts of trusses. The same applies if a soft insulation rests.g. rafters.Soft insulation between rafters. beams. across the insulation layer must be included in the calculations. . 1 m 0.002) (0.2 Table A.5 0.15 m Plastic 0 0 0 0 0 0 0 Stainless 3 (0.017 Zincked iron 8 0.2 There is not need for correction for ties of similar mechanical fixations in the following cases: .3.022 0.003) (0.025 0.1 m 0.3.051 0.013 0. DPM or cardboard on the warm side Insulation in a cavity wall with cavity ties.026 0.003) steel Stainless 5.009 0.016 0. or overfilled joints on the warm side. m2 4 ties pr.ties through a non insulated cavity .3 The correction ∆Uf is stated in table A.006 Bronze 4 0.004) (0.if the heat flow resistance of the tie of similar mechanical fixation or a part of it is les than 1 W/mK .003) (0.002) (0.009 0.125 m 0.008 0.ties between brickwork and timber-frame .033 0. rafters or trusses only covered by e.3. A.011 Bronze 5 0.062 0.018 0.021 The values in the table can’t bee used if both ends of the tie or the similar mechanical fixation are in contact with metallic cover materials. Correction for ties A.050 0. rafters or trusses not wedged against the warm side Insulation between.015 0.006 (0. brickwork with un-. In these cases the methods of DS/EN ISO 10211-1 can be applied.012 0.125 m 0.002) steel Stainless 4 0.005) steel Bronze 3 0.041 0.Extracts from DS418:2002 Insulation between.g. beams. studs.2 – Tie correction ∆Uf in W/m2K for common ties Type of tie Diameter 8 ties pr.007 0.016 0. studs.007 0.031 0. m2 mm Thickness of insulation layer Thickness of insulation layer 0.020 0.4 Correction for precipitation on “upside down” roof Not included in the translation Page 44 .006 (0. beams.013 0.026 0.034 0.15 m 0.005) (0.011 0.the correction for the tie is less than 0.005 W/m2K stated with () in table A.041 0.004) (0. 1 – Brick and lime-sandstone bricks .Extracts from DS418:2002 Annex F Design values for brick.Lambda design in W/mK as function of the density in kg/m3 Curve A: Curve B: Curve C: Curve D: Internal walls of brickwork External walls of brickwork Internal walls of lime-sandstone External walls of lime-sandstone Page 45 . concrete and other building materials Figure F. 6m long and 0. plates in storey height Curve B: External blocks and plates with glued joints or mounted in forms. plates in storey height Curve C: Internal blockwork approximately 0.2 –Lightweight concrete .Extracts from DS418:2002 Figure F.6m long and 0.2m high Conditions: Block-work is carried out with 10mm joints Page 46 .Lambda design in W/mK as function of the density in kg/m3 Curve A: Internal blocks and plates with glued joints or mounted in forms.2m high Curve D: External blockwork approximately 0. made by block of light clinker concrete Blockwork of light clinker concrete in brick format.2m high Basement walls below terrain.3 – Light clinker concrete .2m high External blockwork approximately 0.6m long and 0.6m long and 0. internal/external Conditions: Block-work is carried out with 10mm joints Page 47 .Extracts from DS418:2002 Figure F.plates in storey height and other larger elements Internal blockwork approximately 0.Lambda design in W/mK as function of the density in kg/m3 Curve A: Curve B: Curve C: Curve D: Curve E: Curve F: Internal .blocks and plates with glued joints or mounted in forms External . Extracts from DS418:2002 Figure F. can be applied. can be applied.76 W/mK respectively.44 and 2.4 – Mortar/concrete .64 and 2. By reinforced concrete with 2% reinforcement.54 W/mK respectively. Page 48 .Lambda design in W/mK as function of the density in kg/m3 Curve A: Curve B: Curve C: Curve D: Internal concrete External concrete Internal mortar External mortar Conditions: By reinforced concrete with 1% reinforcement. a lambda value of internal and external 2. a lambda value of internal and external 2. Table F. ceramic Granite Gneiss Basalt Limestone Marble Slate Sandstone Tiles.3 0. Further values for materials as well as thermal capacity and damp diffusion resistance can be found in DS/EN 12524.25 0. drainage material Moist soil (moraine) Coarse cinders in soil Clay Pebble layer as capillary breaking layer 2500 –2700 2400 – 2700 2700 – 3000 2600 2800 2000 – 2800 2600 2000 2100 2300 2600 1200 1390 1150 1200 1200 1200 500 – 700 300 .18 2.7 (continue) Page 49 .3 1. clay Tiles.18 0.2 2.3 3.5 3.13 – 0.0 1.1 contains a number of values.24 0.1000 300 – 900 1900 800 1200 – 1800 Design thermal conductivity W/mK 2. tiles.5 2.20 0.09 – 0.10 – 0.1 Design values for other building materials Material or application Density Kg/m3 Stone. concrete Ceramic tiles.3 0. glass.2 0.4 1.5 2.17 0.8 0.20 0.8 3.Extracts from DS418:2002 Design values for other building materials Table F.5 0.5 1. porcelain Constructional glass Plastic and rubber Polycarbonate PVC Polyamide (Nylon) Epoxy Synthetic rubber Linoleum Wood and wood-bases boards Wood Plywood Chipboards Soil.24 0. 25 Page 50 .1 Design values for other building materials (continued) Design thermal Material or application Density 3 conductivity Kg/m W/mK Metals Aluminium 2700 230 Zinc 7100 110 Bras 8400 100 Bronze 8700 65 Cupper 8900 380 Silver 11300 35 Led 11300 35 Mild steel 7800 55 Stainless steel 7900 17 Cast iron 7200 50 Water. air Water (stagnant) Ice at 0°C Snow at 0°C Snow at 0°C Air (stagnant) Other Plasterboards with paper 1000 900 300 100 1.3 900 0.6 2.23 0.2 0.024 0.Extracts from DS418:2002 Table F.05 0. 055 0. slabs and rolls added Polyester fibres Straw.050 Towards soil 0. slab. granulate Mineral wool Above terrain Towards soil Loose and granulated Density Kg/m3 8 – 30 Design thermal conductivity Design W/mK 0. loose. granulated.055 The values in the list apply. loose.060 0.060 0.10 0. can be determined.050 0. Page 51 .050 0. rolls Expanded perlite Cotton Sheep wool Wood wool.10 0. whole Flax and hemp.055 0.060 0. valid at the time of application. rolls. Material and application Cellular plastic injected on site Polyurethane Ureaformaldehyde Cellular plastic filler Polystyrene pellet/granulate Other materials Cellulose fibers.095 0.Extracts from DS418:2002 Annex G Design values for calculation of existing constructions in connection with rebuilding and renovation.055 0.070 0. shaving Grain. chip.050 0. when no value for the concerned product.10 0. mats.050 10 – 20 30 – 150 25 – 40 25 – 75 30 – 90 170 – 190 150 – 190 25 – 30 30 – 100 15 – 300 Expanded polystyrene Above terrain 10 – 45 0. expanded Glass bead.