environmental controls i/ig lecture 9 heat flow in opaque materials thermal mass
TRANSCRIPT
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Environmental Controls I/IGEnvironmental Controls I/IGEnvironmental Controls I/IGEnvironmental Controls I/IG
Lecture 9Heat Flow in Opaque Materials
Thermal Mass
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Conductive Heat FlowConductive Heat Flow
Conductive Heat Flow through opaque materials:
Q= U x A x ΔT
Q: heat flow (Btuh)U: transmission coefficient (Btu/h-ºF-ft2)A: area (ft2)ΔT: temperature difference (Ti-To)
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Transmission CoefficientTransmission 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
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Summing R-valuesSumming R-valuesSum 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
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Air FilmsAir Films
Film surface conductance coefficient
Outdoor air film: R= 1/hO
Indoor air film: R=1/hI
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Finding hFinding hOO and h and hII – – EmittanceEmittance
Emittance(ε): absorption of radiant heat
S: p.1570, T.E.3B
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Finding hFinding hOO and h and hII
Film surface conductance coefficient (S: p. 158, T4.3)
Position of Surface
Direction of Heat FlowEmittance
Air Motion
S: p. 1570, T.E.3B
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Finding hFinding hOO and h and hII – – EmittanceEmittance
Emittance(ε): absorption of radiant heat
Effective Emittance (εeff):
1/εeff=1/ε1+1/ε2-1
S: p.1570, T.E.3B
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R-values for Enclosed Air R-values for Enclosed Air CavitiesCavities
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 Temperatu
reDirection
of Heat Flow
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R-values For Solid MaterialsR-values For Solid Materials
Table 4.2 Thermal Properties of Typical Building and Insulating MaterialsDensity
ConductanceConductivity
Resistance
S: p. 1549, T. E.1
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Conductivity and Conductivity and ConductanceConductance
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.)
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Conductivity and Conductivity and ConductanceConductance
Conductivity vs. Conductance
1’
1’
1”
1ºF
x”
1ºF
Conductivity k=0.25 Btuh
Say x=4”
Conductance C=k/x=0.25/4”=0.0625 Btuh
S: p. 182, F.7.8
Example 1
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Converting to ResistanceConverting to Resistance
Resistance (R): measure of resistance to the passage of heat (h-ft2-ºF/Btu)
R=1/C or R=x/k
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Converting to ResistanceConverting to Resistance
Conductivity vs. Conductance
1’
1’
1”
1ºF
x”
1ºF
Conductivity k=0.25 BtuhResistance R=1/k=1/0.25= 4
Say x=4”
Conductance C=k/x=0.25/4”=0.0625 BtuhResistance R=x/k=4/0.25=16
Example 1 (cont.)
S: p. 182, F.7.8
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Table 4.2 Thermal Properties of Typical Building and Insulating Materials
Thermal Properties TableThermal Properties Table
S: p. 1522-3, T.E.1
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U-Value CalculationU-Value Calculation
Wall 1indoor air film½” gypsum board2”x4” nominal stud (pine) w/3.5” Ins.½” fiberboardwood shingles (16” long, 12” exposure)outdoor air filmSection View
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123
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Finding Indoor Air Film Coefficient Finding Indoor Air Film Coefficient –h–hII
Film surface conductance coefficient (S: p. 158, T4.3)
S: p. 1570, T.E. 3A
Indoor air filmVertical surfaceHorizontal heat flowNon-reflective surface
hI=1.46R=0.68
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123
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Finding Gypsum Board R-Finding Gypsum Board R-valuevalue
Table 4.2 Thermal Properties of Typical Building and Insulating Materials
½” Gypsum Board
R=0.45
S: p. 1549, T.E.1
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123
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Table 4.2 Thermal Properties of Typical Building and Insulating Materials
Finding Framing R-valueFinding Framing R-valueNominal 2x4 Pine stud depth is 3.5”Ravg=(1.35+1.11)/2=1.23/inch
R=3.5x1.23 =4.35
S: p. 1567, T.E.1
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123
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Table 4.2 Thermal Properties of Typical Building and Insulating Materials
Thermal Properties TableThermal Properties Table
S: p. 1522-3, T.E.1
3.5” InsulationMineral Fiber
R=13.00
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123
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Finding Fiberboard R-valueFinding 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
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123
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Table 4.2 Thermal Properties of Typical Building and Insulating Materials
Finding Wood Shingle R-Finding Wood Shingle R-valuevalue
Wood shingles (16”, 12” exposure)
R=1.19
S: p. 1567, T.E.1
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123
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Finding Outdoor Air Film Finding Outdoor Air Film Coefficient--hCoefficient--hOO
Film surface conductance coefficient (S: p. 158, T4.3)
S: p. 1570, T.E.3A
Outdoor air filmWinter WindHorizontal heat flowNon-reflective surface
hO=6.0R=0.17
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123
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At AtInsulation Frame
Component (RI) (RF) Ref.
U-Value CalculationU-Value Calculation
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 ΣRI 16.81 ΣRF 8.16
UI 0.059 UF 0.123U= 1/ΣR
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At AtInsulation Frame
Component (RI) (RF)
U-Value — Overall Average U-Value — Overall Average
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
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Density WeightComponent #/cf #/sf
Thermal MassThermal Mass
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 0.004.29 #/sf
Weight (#/sf)=Density (#/cf) x Thickness (ft.)
½” Gyp. Bd. =50#/cf x 0.0416’= 2.08 #/sf
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Insert Microclimate critiques hereInsert Exam results here