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 xΔT Q: heat flow (Btuh) U: transmission coefficient (Btu/h-ºF-ft2) A: area (ft2) ΔT: temperature difference (Ti-To)

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

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

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

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

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

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

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

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

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.)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Insert Microclimate critiques here Insert Exam results here

Technology in Architecture