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Technology in Architecture

Technology in Architecture. Lecture 5 Heat Flow in Opaque Materials Thermal Mass. Conductive Heat Flow. Conductive Heat Flow through opaque materials: Q= U x A x Δ T Q: heat flow (Btuh) U: transmission coefficient (Btu/h-ºF-ft 2 ) A: area (ft 2 )

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Technology in Architecture

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  1. Technology in Architecture Lecture 5 Heat Flow in Opaque Materials Thermal Mass

  2. Conductive Heat Flow Conductive Heat Flow through opaque materials: Q= U x A xΔT Q: heat flow (Btuh) U: transmission coefficient (Btu/h-ºF-ft2) A: area (ft2) ΔT: temperature difference (Ti-To)

  3. Transmission Coefficient Transmission Coefficient (U): U= 1/ΣR U: transmission coefficient (Btu/h-ºF-ft2) ΣR: sum of resistance values (R-values) for layers of a construction assembly

  4. Summing R-values Sum of R-values (ΣR): ΣR= 1/hO+R1+R2+R3+…+1/hI hO,hI: film surface conductance coefficients R1,R2,R3,…: Resistance values (R-values) for each layer of a construction assembly

  5. Air Films Film surface conductance coefficient Outdoor air film: R= 1/hO Indoor air film: R=1/hI

  6. Finding hO and hI – Emittance Emittance(ε): absorption of radiant heat S: p.1612, T.E.3B

  7. Direction of Heat Flow Emittance Position of Surface Air Motion Finding hO and hI Film surface conductance coefficient (S: p. 158, T4.3) S: p. 1612, T.E.3A

  8. Finding hO and hI – Emittance Emittance(ε): absorption of radiant heat Effective Emittance (εeff): 1/εeff=1/ε1+1/ε2-1 S: p.1612, T.E.3B

  9. Air Space Width Air Space Temperature Direction of Heat Flow Emittance Position of Air Space R-values for Enclosed Air Cavities Film surface conductance coefficient (S: p. 161, T4.4) S: p. 1614, T.E.4

  10. Conductance Resistance Conductivity Density R-values For Solid Materials Table 4.2 Thermal Properties of Typical Building and Insulating Materials S: p. 1593, T. E.1

  11. Conductivity and Conductance Conductivity (k) heat flow through a material per unit thickness Conductance (C): heat flow through a material of stated thickness C=k/x where x= unit thickness (in.)

  12. x” 1” 1’ 1ºF 1’ 1ºF Conductivity and Conductance Example 1 Say x=4” Conductivity vs. Conductance Conductance C=k/x=0.25/4”=0.0625 Btuh Conductivity k=0.25 Btuh S: p. 186, F.7.8

  13. Converting to Resistance Resistance (R): measure of resistance to the passage of heat (h-ft2-ºF/Btu) R=1/C or R=x/k

  14. x” 1” 1’ 1ºF 1’ 1ºF Converting to Resistance Example 1 (cont.) Say x=4” Conductivity vs. Conductance Conductance C=k/x=0.25/4”=0.0625 Btuh Resistance R=x/k=4/0.25=16 Conductivity k=0.25 Btuh Resistance R=1/k=1/0.25= 4 S: p. 186, F.7.8

  15. Thermal Properties Table Table 4.2 Thermal Properties of Typical Building and Insulating Materials S: p. 1603, T.E.1

  16. Section View U-Value Calculation Wall 1 indoor air film ½” gypsum board 2”x4” nominal stud (pine) w/3.5” Ins. ½” fiberboard wood shingles (16” long, 12” exposure) outdoor air film

  17. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard 1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123

  18. Finding Indoor Air Film Coefficient –hI Indoor air film Vertical surface Horizontal heat flow Non-reflective surface hI=1.46 R=0.68 Film surface conductance coefficient (S: p. 158, T4.3) S: p. 1612, T.E. 3A

  19. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard 1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123

  20. Finding Gypsum Board R-value Table 4.2 Thermal Properties of Typical Building and Insulating Materials ½” Gypsum Board R=0.45 S: p. 1593, T.E.1

  21. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard 1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123

  22. Finding Framing R-value Nominal 2x4 Pine stud depth is 3.5” Ravg=(1.35+1.11)/2=1.23/inch R=3.5x1.23 =4.35 Table 4.2 Thermal Properties of Typical Building and Insulating Materials S: p. 1600-1, T.E.1

  23. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard 1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123

  24. Thermal Properties Table 3.5” Insulation Mineral Fiber R=13.00 Table 4.2 Thermal Properties of Typical Building and Insulating Materials S: p. 1594-5, T.E.1

  25. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard 1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123

  26. Finding Fiberboard R-value ½” Fiberboard R=1.32 Table 4.2 Thermal Properties of Typical Building and Insulating Materials S: p. 1593, T.E.1

  27. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123

  28. Finding Wood Shingle R-value Wood shingles (16”, 12” exposure) R=1.19 Table 4.2 Thermal Properties of Typical Building and Insulating Materials S: p. 1600, T.E.1

  29. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard 1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123

  30. Finding Outdoor Air Film Coefficient--hO Outdoor air film Winter Wind Horizontal heat flow Non-reflective surface hO=6.0 R=0.17 Film surface conductance coefficient (S: p. 158, T4.3) S: p. 1612, T.E.3A

  31. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard 1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123

  32. U-Value Calculation At At Insulation Frame Component (RI) (RF) Ref. indoor air film 0.68 0.68 T.E.3A ½” gypsum board 0.45 0.45 T.E.1 2x4 stud (3.5” pine) n.a. 4.35 T.E.1 3.5” Insulation 13.00 n.a. T.E.1 ½” fiberboard 1.32 1.32 T.E.1 wood shingles 1.19 1.19 T.E.1 outdoor air film 0.17 0.17 T.E.3A Totals ΣRI16.81ΣRF8.16 UI0.059 UF0.123 U= 1/ΣR

  33. U-Value — Overall Average At At Insulation Frame Component (RI) (RF) Totals ΣRI 16.81 ΣRF 8.16 UI 0.059 UF 0.123 15% framing: UAVG=0.85(0.059)+0.15(0.123)=0.069

  34. Thermal Gradient Calculation Temperature Conditions Indoor Air Temperature (TI) 68ºF/50% RH Outdoor Air Temperature (TO) 13ºF Temperature Difference (TI-TO) 68-13=55ºF Dewpoint Temperature (DPT) 58ºF

  35. + = Thermal Gradient Calculation “With Insulation” ΔT from Temp@ Component (RI) ΣR Interior Outer Edge indoor air film 0.68 0.68 2.2º 65.8º ½” gypsum board 0.45 1.13 3.7º 61.3º 3-½” insulation 13.00 14.13 46.2º 21.8º ½” fiberboard 1.32 15.45 50.6º 17.4º wood shingles 1.19 16.64 54.5º 13.5º outdoor air film 0.17 16.81 55.0º 13.0º ΔT from Interior =ΣR x (TI-TO)/ΣRI =0.68x55/16.81=2.2º Temp@Outer Edge= TI- ΔT from Interior =68-2.2=65.8º

  36. Vapor Barrier 68ºF 65.8ºF 61.3ºF 20.7ºF 17.4ºF 13.5ºF 13ºF Section View Vapor Barrier Placement Show Thermal Gradient on Wall Section Select location forvapor barrier T>DPT=58ºF

  37. Thermal Gradients Observations: Interior surface temperature with insulation (ΣRI=16.81) 65.8ºF Interior surface temperature without insulation (ΣRC=4.82) 60.2ºF

  38. Thermal Mass Density Weight Component #/cf #/sf indoor air film 0.0 0.00 ½” gypsum board 50.0 2.08 3-½” insulation 1.2 0.35 ½” fiberboard 18.0 0.75 wood shingles 26.6 1.11 outdoor air film 0.0 0.00 4.29 #/sf Weight (#/sf)=Density (#/cf) x Thickness (ft.) ½” Gyp. Bd.=50#/cf x 0.0416’= 2.08 #/sf

  39. Insert Microclimate critiques here Insert Exam results here

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