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Environmental Controls I/IG Lecture 9 Heat Flow in Opaque Materials Thermal Mass.

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Presentation on theme: "Environmental Controls I/IG Lecture 9 Heat Flow in Opaque Materials Thermal Mass."— Presentation transcript:

1 Environmental Controls I/IG Lecture 9 Heat Flow in Opaque Materials Thermal Mass

2 Conductive Heat Flow Conductive Heat Flow through opaque materials: Q= U x A x ΔT Q: heat flow (Btuh) U: transmission coefficient (Btu/h-ºF-ft 2 ) A: area (ft 2 ) ΔT: temperature difference (T i -T o )

3 Transmission Coefficient Transmission Coefficient (U): U= 1/ΣR U: transmission coefficient (Btu/h-ºF-ft 2 ) ΣR: sum of resistance values (R-values) for layers of a construction assembly

4 Summing R-values Sum of R-values (ΣR): ΣR= 1/h O +R 1 +R 2 +R 3 +…+1/h I h O,h I : film surface conductance coefficients R 1,R 2,R 3,…:Resistance values (R-values) for each layer of a construction assembly

5 Air Films Film surface conductance coefficient Outdoor air film: R= 1/h O Indoor air film: R=1/h I

6 Finding h O and h I – Emittance Emittance(ε): absorption of radiant heat S: p.1570, T.E.3B

7 Finding h O and h I Film surface conductance coefficient (S: p. 158, T4.3) Position of Surface Direction of Heat Flow Emittance Air Motion S: p. 1570, T.E.3B

8 Finding h O and h I – Emittance Emittance(ε): absorption of radiant heat Effective Emittance (ε eff ): 1/ε eff =1/ε 1 +1/ε 2 -1 S: p.1570, T.E.3B

9 R-values for Enclosed Air Cavities Film surface conductance coefficient (S: p. 161, T4.4) S: p. 1571, T.E.1 Emittance Position of Air Space Air Space Width Air Space Temperature Direction of Heat Flow

10 R-values For Solid Materials Table 4.2 Thermal Properties of Typical Building and Insulating Materials Density Conductance Conductivity Resistance S: p. 1549, 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 Conductivity and Conductance Conductivity vs. Conductance ºF x Conductivity k=0.25 Btuh Say x=4 Conductance C=k/x=0.25/4= Btuh S: p. 182, F.7.8 Example 1

13 Converting to Resistance Resistance (R): measure of resistance to the passage of heat (h-ft 2 -ºF/Btu) R=1/C or R=x/k

14 Converting to Resistance Conductivity vs. Conductance ºF x Conductivity k=0.25 Btuh Resistance R=1/k=1/0.25= 4 Say x=4 Conductance C=k/x=0.25/4= Btuh Resistance R=x/k=4/0.25=16 Example 1 (cont.) S: p. 182, F.7.8

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

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

17 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F 0.123

18 Finding Indoor Air Film Coefficient –h I Film surface conductance coefficient (S: p. 158, T4.3) S: p. 1570, T.E. 3A Indoor air film Vertical surface Horizontal heat flow Non-reflective surface h I =1.46 R=0.68

19 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F 0.123

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. 1549, T.E.1

21 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F 0.123

22 Table 4.2 Thermal Properties of Typical Building and Insulating Materials Finding Framing R-value Nominal 2x4 Pine stud depth is 3.5 R avg =( )/2=1.23/inch R=3.5x1.23 =4.35 S: p. 1567, T.E.1

23 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F 0.123

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

25 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F 0.123

26 Finding Fiberboard R-value Table 4.2 Thermal Properties of Typical Building and Insulating Materials ½ Fiberboard R=1.32 S: p. 1549, T.E.1

27 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F 0.123

28 Table 4.2 Thermal Properties of Typical Building and Insulating Materials Finding Wood Shingle R-value Wood shingles (16, 12 exposure) R=1.19 S: p. 1567, T.E.1

29 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F 0.123

30 Finding Outdoor Air Film Coefficient--h O Film surface conductance coefficient (S: p. 158, T4.3) S: p. 1570, T.E.3A Outdoor air film Winter Wind Horizontal heat flow Non-reflective surface h O =6.0 R=0.17

31 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F 0.123

32 At At Insulation Frame Component (R I ) (R F ) Ref. U-Value Calculation indoor air film T.E.3A ½ gypsum board T.E.1 2x4 stud (3.5 pine)n.a.4.35T.E Insulation13.00n.a.T.E.1 ½ fiberboard T.E.1 wood shingles T.E.1 outdoor air film T.E.3A TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F U= 1/ΣR

33 At At Insulation Frame Component (R I ) (R F ) U-Value Overall Average TotalsΣR I 16.81ΣR F 8.16 U I 0.059U F % framing: U AVG =0.85(0.059)+0.15(0.123)=0.069

34 DensityWeight Component #/cf #/sf Thermal Mass indoor air film ½ gypsum board ½ insulation ½ fiberboard wood shingles outdoor air film #/sf Weight (#/sf)=Density (#/cf) x Thickness (ft.) ½ Gyp. Bd. =50#/cf x = 2.08 #/sf

35 Insert Microclimate critiques here Insert Exam results here


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