1. Thermal Bridges peter warm AECB Conference Norwich June 2008 peter warm AECB Conference Norwich June 2008

3. Importance of thermal bridges

4. Repeating linear thermal bridges Repeating thermal bridges Included in U-value calculation

5. Non-repeating linear thermal bridges Extra Heat loss = ∑ψ x length Add as Correction to heat loss

8. Lintels (box) Insulation has no effect!

9. Lintels Provide separate lintels in each leaf. Timber box fixes window. Provide separate lintels in each leaf. Timber box fixes window. Brenda and Robert Vale

11. Foundation Insulation Thermal bridges at joins with floors Lightweight blocks being used to prevent thermal bridge. Screed Beam and block

12. Hillcrest Park, Exeter (2004)

13. Taking Account of Thermal Bridging TAKE “HEATING ENGINEER’S” HEAT LOSS = U x A ADD CORRECTION TERM FOR EACH BRIDGE = ψ x L ψ = linear thermal transmittance W/mK L = length m TAKE “HEATING ENGINEER’S” HEAT LOSS = U x A ADD CORRECTION TERM FOR EACH BRIDGE = ψ x L ψ = linear thermal transmittance W/mK L = length m

14. Taking Account of Thermal Bridging ψ ψ values for Accredited Details (SAP 2005) ψ ψ values can also be calculated – see BRE IP 1/06

15. Taking Account of Thermal Bridging For unknown thermal bridges –Default rate of additional heat loss y W/m 2 K If accredited construction details are used y = 0.08 W/m 2 K If accredited details are not used y = 0.15 W/m 2 K –Total overall additional heat loss is given by H TB = y x A HL W/K where A HL is the total heat loss area m 2 For unknown thermal bridges –Default rate of additional heat loss y W/m 2 K If accredited construction details are used y = 0.08 W/m 2 K If accredited details are not used y = 0.15 W/m 2 K –Total overall additional heat loss is given by H TB = y x A HL W/K where A HL is the total heat loss area m 2

16. Non repeating thermal bridges not at joins

17. Pictures courtesy Malcolm Bell

18. Thermal Bridges Geometrical and Material Conventions – internal and external Calculation methods –Therm Examples Geometrical and Material Conventions – internal and external Calculation methods –Therm Examples

19. Exercise 1 Building Plan 9m 8m If Uwall=0.1 Height = 5 m What is “Engineers”heat loss a) Using internal (UK) b) Using external (D) dimensions?

20. Internal (UK) 8m Perim = 8 * 4 = 32m Area = 32 * 5 = 160m2 Heat loss = 160 * 0.1 = 16 W/C

21. External (D) 9m Perim = 9 * 4 = 36m Area = 36 * 5 = 180m2 Heat loss = 180 * 0.1 = 18 W/C 12.5% larger

22. Exercise 2 : calculate actual corner heat loss 1m 0.5m 1.5m

23. Therm Setup load and get Therm running, then: LIBARYS –Pw10lib and Pw10bc on stick or on server –Libraries / material lib / load lib / pw10lib.lib –Libraries / boundary library / load lib / pw10bc.lib SNAP –Set to 5mm H and V and set on –Options / Preferences / grid and H,V 5mm UFACTOR names –Libraries / check Ufactor names / check internal and external available, if not add names DISPLAY –Can only input once simulation calculated on screen –Calcs / DispOpts / advanced / colourIR 0 20 no auto prog default / colour flux 0 35 no auto prog default ACCURACY –recommended settings in options/prefs/therm settings –Quad tree mesh = 8; max error = 2%; max iteration = 10 –Thanks to Kersttin Rosemeier ( NZ) LIBARYS –Pw10lib and Pw10bc on stick or on server –Libraries / material lib / load lib / pw10lib.lib –Libraries / boundary library / load lib / pw10bc.lib SNAP –Set to 5mm H and V and set on –Options / Preferences / grid and H,V 5mm UFACTOR names –Libraries / check Ufactor names / check internal and external available, if not add names DISPLAY –Can only input once simulation calculated on screen –Calcs / DispOpts / advanced / colourIR 0 20 no auto prog default / colour flux 0 35 no auto prog default ACCURACY –recommended settings in options/prefs/therm settings –Quad tree mesh = 8; max error = 2%; max iteration = 10 –Thanks to Kersttin Rosemeier ( NZ)

24. Exercise 2 : Step 1 draw shape 1m 0.5m

25. Exercise 2 : Step 1 draw shape 1m 0.5m 1.5m

26. Exercise 2 : Step 2 define material conductivities Material k= 0.05085 W/m2K (chosen to give U=0.1 with standard boundary conditions)

30. Exercise 2 summary internal 8m Perim = 8 * 4 = 32m Area = 32 * 5 = 160m2 Simple Heat loss = 160 * 0.1 = 16 W/C Complex heat loss = Simple heat loss + Psi internal * length ( length?) (psi * length?) (Complex heat loss = ?) ( % change over simple heat loss?)

32. Exercise 3 – thinner insulation Building Plan 8.5m 8m If Uwall=0.1 Height = 5 m What is heat loss a)Using internal b)Using external dims?

34. Exercise 4 summary internal 8m Perim = 8 * 4 = 32m Area = 32 * 5 = 160m2 Simple Heat loss = 160 * 0.1 = 16 W/C Complex heat loss = Simple heat loss + Psi internal * length ( length?) (psi * length?) (Complex heat loss = ?) ( % change over simple heat loss?)

35. Exercise 4 summary external 8.5m Perim = 8.5 * 4 = 34m Area = 34 * 5 = 170m2 Simple Heat loss = 170 * 0.1 = 17 W/C Complex heat loss = Simple heat loss + Psi external * length ( length?) (psi * length?) (Complex heat loss = ?) ( % change over simple heat loss?) 8.5m

36. Exercise 5 : Poor material choice 1m 0.5m 1.5m K=0.13

37. Exercise 5 : Poor material choice 1m 0.5m 1.5m K=0.13

38. Exercise 5 : Poor material choice 1m 0.5m 1.5m K=0.13 Heat flux

39. Passivhaus convention For an overestimate of heat loss: Use external measurements To outside of construction, or to outside of insulation if large rainscreen voids ( eg lofts) Make insulation continuous – no material bridges, only geometrical (strictly psi external <0.01) As thermal bridges are calculated, psi values tend to be negative, ie more accurate calculation lowers heat loss. Hence external measurements good for first estimate.

40. psi internal and psi external Both give “correct” L2D and hence total heat loss Can be calculated from each other: Psi ext = Psi int - Uwall times difference in length Check this relations ship out for walls. Note critical to define plane of measurements Note can have psi for “mid insulation” plane, or any other pane is possible, but value different Similarly there generally exists a plane for which psi=zero, but this is of theoretical interest only. Both give “correct” L2D and hence total heat loss Can be calculated from each other: Psi ext = Psi int - Uwall times difference in length Check this relations ship out for walls. Note critical to define plane of measurements Note can have psi for “mid insulation” plane, or any other pane is possible, but value different Similarly there generally exists a plane for which psi=zero, but this is of theoretical interest only.

41. Wall to roof junction Note difference in external and internal lengths on slope Note have wall and roof U*A to subtract from L2D heat flow Note difference in external and internal lengths on slope Note have wall and roof U*A to subtract from L2D heat flow

42. Window and Internal floor details No difference in internal and external measurements Hence psi int = psi ext Psi is material only, no geometrical component No difference in internal and external measurements Hence psi int = psi ext Psi is material only, no geometrical component window

43. Internal floor between flats & party wall to external wall junctions Significant difference in internal and external measurements Hence psi int not equal to psi ext Significant difference in internal and external measurements Hence psi int not equal to psi ext

44. Ground floor Have to take into account ground Usually calculated fro default breadth of 8 metres Requires 20 by 20m lump of earth Some different conventions – see BRE 497 Have to take into account ground Usually calculated fro default breadth of 8 metres Requires 20 by 20m lump of earth Some different conventions – see BRE 497 4m 1m

45. Ground floor Have to take into account ground Usually calculated for default breadth of 8 metres Requires 20 by 20m lump of earth Some different conventions – see BRE 497 Have to take into account ground Usually calculated for default breadth of 8 metres Requires 20 by 20m lump of earth Some different conventions – see BRE 497 20m 4m

47. PH window conventions Need to calculate psi spacer for each window type and orientation Then need psi installation – at present single value for each type Need to calculate psi spacer for each window type and orientation Then need psi installation – at present single value for each type window

48. Summary – we have looked at Difference between Geometrical and material psi Have used Therm Link between psi and measurement plane PHPP conventions Difference between Geometrical and material psi Have used Therm Link between psi and measurement plane PHPP conventions