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Technology in ArchitectureTechnology in ArchitectureTechnology in ArchitectureTechnology in Architecture
Lecture 5Heat Flow in Opaque Materials
Thermal Mass
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)
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
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
Air FilmsAir Films
Film surface conductance coefficient
Outdoor air film: R= 1/hO
Indoor air film: R=1/hI
Finding hFinding hOO and h and hII – – EmittanceEmittance
Emittance(ε): absorption of radiant heat
S: p.1612, T.E.3B
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. 1612, T.E.3A
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.1612, T.E.3B
R-values for Enclosed Air R-values for Enclosed Air CavitiesCavities
Film surface conductance coefficient (S: p. 161, T4.4)
S: p. 1614, T.E.4
Emittance
Position of Air Space
Air Space Width
Air Space Temperatu
reDirection
of Heat Flow
R-values For Solid MaterialsR-values For Solid Materials
Table 4.2 Thermal Properties of Typical Building and Insulating MaterialsDensity
ConductanceConductivity
Resistance
S: p. 1593, T. E.1
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.)
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. 186, F.7.8
Example 1
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
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. 186, F.7.8
Table 4.2 Thermal Properties of Typical Building and Insulating Materials
Thermal Properties TableThermal Properties Table
S: p. 1603, T.E.1
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
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
Finding Indoor Air Film Coefficient Finding Indoor Air Film Coefficient –h–hII
Film surface conductance coefficient (S: p. 158, T4.3)
S: p. 1612, T.E. 3A
Indoor air filmVertical surfaceHorizontal heat flowNon-reflective surface
hI=1.46R=0.68
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
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. 1593, T.E.1
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
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. 1600-1, T.E.1
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
Table 4.2 Thermal Properties of Typical Building and Insulating Materials
Thermal Properties TableThermal Properties Table
S: p. 1594-5, T.E.1
3.5” InsulationMineral Fiber
R=13.00
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
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. 1593, T.E.1
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
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. 1600, T.E.1
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
Finding Outdoor Air Film Finding Outdoor Air Film Coefficient--hCoefficient--hOO
Film surface conductance coefficient (S: p. 158, T4.3)
S: p. 1612, T.E.3A
Outdoor air filmWinter WindHorizontal heat flowNon-reflective surface
hO=6.0R=0.17
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
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
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
Thermal Gradient Thermal Gradient CalculationCalculation
Temperature Conditions
Indoor Air Temperature (TI)
68ºF/50% RHOutdoor Air Temperature (TO)
13ºFTemperature Difference (TI-TO)
68-13=55ºFDewpoint Temperature (DPT)
58ºF
“WithInsulation”ΔT from Temp@
Component (RI) ΣR InteriorOuter Edge
Thermal Gradient Thermal Gradient CalculationCalculation
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
+
=
Vapor Barrier PlacementVapor Barrier Placement
Show Thermal Gradient on Wall Section
Select location forvapor barrier
T>DPT=58ºF
Section View
68ºF
13ºF
13.5ºF
17.4ºF
20.7ºF
65.8ºF
61.3ºF
Vapor Barrier
Thermal GradientsThermal Gradients
Observations: Interior surface temperature with
insulation(ΣRI=16.81) 65.8ºF
Interior surface temperature without insulation
(ΣRC=4.82) 60.2ºF
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
Insert Microclimate critiques hereInsert Exam results here