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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