1. Conservation of Building Energy 3A9 Construction Technology Dr S Pavía Dept of Civil Engineering Trinity College Dublin

2. Lecture Overview Building Regulations. Technical guidance document L. Conservation of fuel and energy. Heat Loss in Buildings – Transmittance, Ventilation, Infiltration U Value Calculation – Proportional Area Method – Combined Area Method

3. Part L - Conservation of Fuel and Energy Aim: To limit the use of fossil fuel Energy and related CO2 emissions arising from the operation of buildings. Based on The EU Energy Performance of Buildings Directive EPBD (2002/91/EC). To demonstrate compliance, – Show that the calculated rate of CO 2 emissions associated to the operation of the building does not exceed the target value specified in this document. – Provide energy efficient measures including: Limit heat loss through fabric Limit heat loss from pipes, ducts and vessels used for transport and storage of heated water or air Control the demand and output of services

4. Heat flows from a hot to a colder body Newton’s Law of cooling: the rate of heat loss depends on the difference in temperature between the hot body and its surroundings the rate of cooling of a hot body is directly proportional to the difference in T between the body and its surroundings. Elliott 1977 Heat loss in buildings

5. Heat losses from buildings Heat flows from warm areas through the fabric of the buildings (floors, walls, roofs, windows and doors) to the colder areas. Most people like to live and work in room temperatures of approximately 20 °C. Heat loss is usually higher than expected/calculated (i.e. damp insulation, cold bridging)

6. Heat Loss in Buildings Heat Loss Transmission through walls, windows, doors, floors etc Ventilation/Infiltration – Ventilation is the deliberate process of replacing air in any space to provide high indoor air quality Infiltration is the uncontrolled entry of fresh air into a dwelling though air leakage paths in the building. Value depends on several variables - wind speed and strength, airtightness of the building construction etc.

7. Transmission Heat Loss A great amount of heat is lost by transmission through the building fabric The U-value is a measure of the rate at which a building element transmits heat The higher the U-value the more heat is transmitted, or lost, through the building element.

8. Thermal Properties Three related thermal properties – Thermal Conductivity (λ or k) – Thermal Resistance (R) – U Value

9. Thermal conductivity, λ Allows measurement of heat losses by conduction. DEF-number of joules of heat flowing each s. across the opposite faces of material 1 m long and of a cross sectional area 1m 2, the faces being maintained at a T difference of 1°C

10. Good conductors have high λ values. Most of the materials used in construction have low λ values. λ measures the heat flow through a material however building fabrics are not single materials. λ The denser a material is, the better it will conduct heat. Air has such low density that it is a very poor conductor and therefore makes a good insulator.

11. Thermal Resistance The amount of resistance that a material offers to the flow heat through it depends on the conductivity of the material (ʎ) and its thickness (d). R is the thermal resistance of the material (m 2 °C /W) d is the thickness of the material (m) λ is the thermal conductivity of the material (W/m °C )

12. Thermal transmittance or U value It applies to composite structures: heat travels at different rates through cement, brick and plaster (different λ values). A building element is composed of a number of materials, each of which has a resistance to the flow of heat. The U-value of a building element is calculated by adding together the thermal resistances of the components of the building element (ΣR) and dividing the result by one (taking its reciprocal). (W/m 2 °C)

13. Surface resistance: on each side of the structure, the surface has an influence on the heat transfer – surface properties such as emissivity- e.g. dark walls or North facing walls.

14. In a wall there are several materials, air gap resistances and surface resistances To determine the U Value – add the resistances to calculate the total resistance – consider the surface resistance on each side – divide the result by one (the U Value is the recyprocal). Calculation of U values

15. The higher the R the better a component’s insulation. The lower the U value the better the insulative properties of the structure.

16. U Value expressed in units of Watts per square metre per degree of air temperature difference (W/m 2 °C) measures the rate at which heat passes through a component or structure when a unit temperature difference is maintained between the ambient air temperatures on each side. U= Number of watts (or joules per second) of heat flowing across an area of 1 sq m, maintained at 1 degree temperature difference.

17. Calculating U Value Thermal Conductivity (λ) given by manufacturer – i.e. λ brick =1.15 (W/m °C) – Thermal Resistance R = d/ λ (W/m 2 °C) d= thickness of material i.e. R = 0.1/1.15 = 0.087 (brick thickness 100mm) U Value = 1/R – U = 1/0.087 = 11.5 (W/m 2 °C)

18. U Values - practical The thermal conductivity figures are based on measurements in controlled environments and make standard assumptions about, for example, moisture content. However, in practice, the moisture content of materials may be higher than assumed. E.g. if a material such as mineral wool insulation becomes wet, its thermal conductivity will be higher (because the thermal conductivity of water is greater than that of air). Therefore, on site, U values can be worse than expected.

19. Thermally homogeneous and inhomogeneous layers The effect of mortar joints (in solid brickwork walls and outer leaves of cavity walls) on U-values can be ignored as both components have similar density and insulative properties so fabric is considered unbridged: consisting of thermally homogeneous layers. If a wall comprises of materials with different thermal conductivity we have non homogeneous layers in the building fabric e.g. timber studs bridging insulation slabs in timber frame construction or aerated concrete blocks and mortar in inner wall leaves.

20. U Value of Window Calculate the U-value of both the single-glazed and the double-glazed windows in the figure. Calculate the % of saving in heat loss when introducing double glazing. λglass =1 W/m K λair gap=0.11

21. Single Glazed Pane

22. Double Glazed Pane

23. U Value of Window This means that in terms of a sq m surface maintained at 1 degree temperature difference, 4.92 joules are lost each second through the glass, however, on introducing double glazing, only 2.58 joules are lost. This represents a saving of 2.34 joules per second in heat loss with double glazing. 2.34/4.92 x 100= 47.56% saving in heat loss

24. THERMALLY HOMOGENEOUS LAYERS Calculate the thermal transmittance or U-value of the wall in the figure where thickness of the plaster is 15 mm and that of the brickwork 250 mm. λ p =0.50; λ b =1.15;R se =0.05;R si =0.06

25. Calculate the U-value of the wall in the figure. – λp=0.8 – λb=0.5

26. Calculate U Value Example from TG Document L dense concrete and mortar : similar density – unbridged layer

27. Calculate U Value in Wall D (m) ʎ R (D/ ʎ ) External Resistance0.04 External Render0.0191 Concrete Block0.11.330.075 Air0.18 Cavity Insulation0.10.0234.348 Concrete Block0.11.330.075 Plaster (lightweight)0.0130.180.072 Internal Resistance (wall)0.13 R(total) = 0.04+0.019+0.075+0.18+4.348+0.075+0.072+0.13 =4.939 U = 1/R = 1 / 4.939= 0.2

28. Thermal Bridging Thermal bridging occurs when a wall element is not homogeneous and comprises of materials with different thermal conductivity – Timber frame construction, the timber studs bridge the insulation layer in the framed wall. 2 Methods to calculate U values of bridged wall elements – Proportional Area Method – Combined Method

29. Bridged fabrics-Proportional area method Fabrics are bridged, materials act differently to heat transfer (THERMALLY INHOMOGENEOUS LAYERS) thus the surface areas of the different materials need to be accounted for. According to this method, R t is the sum of the resistances of the unbridged sections and the bridged sections. The method splits the component into unbridged and bridged sections. There will be as many bridged sections as thermal paths. unbridgedbridged

30. Calculate the U-value of the wall in the figure. Dimensions of a standard block with mortar=450x225mm. Standard block=440x215mm Standard block=440x215mm Block with mortar=450x225mm (10mm mortar surrounds each block)

31. Bridged fabrics-Proportional area method 1- Calculate the proportional surfaces of the bridged fabric: Area of block = 440x215 = 94600mm2 Area of Mortar and block = 450x225 = 101250mm2 Area of Mortar = 101250 - 94600 = 6650 Proportion of wall that is brick = 94600/101250 = 0.934 Proportion of wall that is mortar = 6650/101250 = 0.066 or Proportional area of mortar=6.65x100/101.25=6.57% Proportional area of block=100-6.57=93.43% Standard block=440x215mm block with mortar=450x225mm

32. 2 - Calculate R t

33. It considers the upper limit (R’t) and lower limit (R’’t) of thermal resistance. Bridged fabrics-Combined method Calculation of upper limit R’t All possible heat paths are identified. For each path, the resistance of all layers are combined in series to obtain the total resistance of the path. The resistances of all paths are then combined in parallel to give the upper limit of thermal resistance of the structure. R = Resistance of path F= Proportional Area

34. Proportional area of mortar=6.57% Proportional area of block=93.43% There are 2 possible heat paths: block and mortar

35. Calculation of lower limit R’’t Calculate the thermal resistance of the bridged layer (R bl ). Combine in series the R of all layers to obtain the lower limit of thermal resistance of the structure.

36. Proportional area of mortar=6.57% Proportional area of block=93.43%

37. Calculate the U value of the wall. Consider inner leaf bridged. heat flow perpendicular to wall surface. Dimensions of a standard block 440x215mm. Thickness of joints 20mm. Thermal resistance of the blockwork mortar is 0.114 m 2 K/W. Rse=0.06; R air cavity =0.18 Rsi=0.12 Thickness of brick outer leaf 102 mm. λ brick=0.77 λ woolbats=0.3035 λ plaster=0.18 λ block=0.18

38. Bridged fabrics-Proportional area method 1- Calculate the proportional surfaces of the bridged fabric: Area of block = 440x215 = 94600mm2 Area of Mortar and block = 450x225 = 101250mm2 Area of Mortar = 101250 - 94600 = 6650 Proportion of wall that is block = 94600/101250 = 0.934 Proportion of wall that is mortar = 6650/101250 = 0.066 Standard block=440x215mm Thickness of joint10mm block with mortar=450x225mm

39. 2 - Calculate R t λ brick =0.77 λ woolbats =0.035 λ plaster =0.18 λ block =0.18 R block mortar = 0.114 Rse=0.06 R air cavity =0.18 Rsi=0.12 102mm

40. Bridged fabrics-Combined method R = Resistance of path F= Proportional Area λ brick =0.77; λ woolbats =0.035; λ plaster =0.18; λ block =0.18; R block mortar = 0.114 ; R air cavity =0.18; Rsi=0.12; Rse=0.06 Proportion of wall that is brick = 0.934 Proportion of wall that is mortar = 0.066 There are 2 possible heat paths: block and mortar R brick =0.102/0.77=0.132 R insul =0.05/0.035=1.42 R block =0.1/0.18=0.55 R plaster =0.013/0.18=0.07

41. Calculation of lower limit R’’t Calculate the thermal resistance of the bridged layer (R bl ). Combine in series the R of all layers to obtain the lower limit of thermal resistance of the structure.

42. λ brick =0.77; λ woolbats=0.035; λ plaster=0.18; λ block=0.18; R block mortar = 0.114; Rsi=0.12; Rse=0.06; R aircavity=0.18; brick is 102mm thick. R brick =0.102/0.77=0.132 R insul =0.05/0.035=1.42 R block =0.1/0.18=0.55 R plaster =0.013/0.18=0.07

43. Calculate the U value of the platform-frame wall in the figure, with one insulating layer bridged by 150x50mm softwood studs at 400mm centres. Ignore the thermal conductivity of the vapour control layer. Height of studs 3m. Sheathing ply is 5 mm thick. Rse=0.06;Rsi=0.12; Rcavity=0.18 λ brick=0.77; λ ply=0.13; λ insul=0.04; λ plaster=0.25; λ studs=0.13 R brick =0.102/0.77= 0.13 R ply =0.005/0.13=0.03 R insul =0.150/0.04=3.75 R plaster =0.013/0.25=0.05 R studs =0.150/0.13=1.15

45. Calculation of fractional areas of heat paths (studs and insulation) Area of studs: 0.050 x 3= 0.15m 2 Area of insulation: 0.350 x 3= 1.05 m 2 Total area: 1.20 m 2 Proportional areas: Insulation:1.05x100/1.20=87.5 % Studs: 0.15x100/1.20=12.5% 150x50mm softwood studs at 400mm centres. Height of studs 3m.

46. Proportional area method (As many bridged sections as heat paths) Rse=0.06; Rsi=0.12; Rcavity=0.18 R brick =0.102/0.77= 0.13 R ply =0.005/0.13=0.03 R insul =0.150/0.04=3.75 R plaster =0.013/0.25=0.05 R studs =0.150/0.13=1.15 R unbridged section = 0.06+0.13+0.18+0.03=0.4 R studs = 1.15+0.05+0.12=1.32 R insulation = 3.75+0.05+0.12=3.92 Rt = 0.4+(1.32x0.125)+(3.92x0.875)=3.99 U= 1 / 3.99= 0.25 W/m 2 °C Proportional areas: Insulation:1.05x100/1.20=87.5 % Studs: 0.15x100/1.20=12.5%

47. Combined method R = Resistance of path F= Proportional Area There are 2 heat paths: insulation & studs Proportional areas: Insulation:1.05x100/1.20=87.5 % Studs: 0.15x100/1.20=12.5% Rse=0.06; Rsi=0.12; Rcavity=0.18; R brick = 0.13 R ply =0.03; R insul =3.75; R plaster =0.05; R studs =1.15

48. Rse=0.06; R brick = 0.13; Rcavity=0.18 R ply =0.03; R studs =1.15; R insul =3.75; R plaster =0.05; Rsi=0.12 Calculation of the lower limit of resistance W/m 2 °C

49. Alternative calculations Document L of BR (Conservation of fuel and Energy) contains tables of U-values calculated according to the methods studied. These tables allow to establish: A U-value for a given amount of insulation. The amount of insulation needed in order to achieve a specific U-value.

50. Building Regulations. Technical guidance document L. Conservation of fuel and energy.

52. Thermal conductivity of some common building materials- EN 12524:2000