Building Envelope Heat Transfer & Thermal Performance Analysis
Building Envelope Heat Transfer & Thermal Performance Analysis
Building envelope thermal performance determines heating and cooling loads, occupant comfort, and energy efficiency. This guide covers heat transfer mechanisms, U-value calculations, thermal bridging, fenestration performance, infiltration, and moisture control for residential and commercial building design.
Heat Transfer Mechanisms
Steady-State Conduction Through Walls
Fourier’s Law (one-dimensional conduction):
$$q = -k \cdot A \cdot \frac{dT}{dx}$$
Where:
- $q$ = heat transfer rate (Btu/hr)
- $k$ = thermal conductivity (Btu·in/hr·ft²·°F or Btu/hr·ft·°F)
- $A$ = area (ft²)
- $dT/dx$ = temperature gradient (°F/ft)
For composite wall (series resistance):
$$q = \frac{A \cdot \Delta T}{R_{total}} = U \cdot A \cdot \Delta T$$
Where:
- $R_{total}$ = total thermal resistance (ft²·°F·hr/Btu)
- $U$ = overall heat transfer coefficient (Btu/hr·ft²·°F)
- $\Delta T$ = temperature difference (°F)
Total resistance:
$$R_{total} = R_{out} + R_{1} + R_{2} + … + R_{n} + R_{in}$$
Where:
- $R_{out}$ = exterior surface film resistance
- $R_{1}, R_{2}, …$ = material layer resistances
- $R_{in}$ = interior surface film resistance
Surface film coefficients (still air):
| Surface | Winter (15 mph wind) | Summer (7.5 mph wind) |
|---|---|---|
| Exterior vertical wall | R = 0.17 | R = 0.25 |
| Interior vertical wall | R = 0.68 | R = 0.68 |
| Horizontal surface (heat flow up) | R = 0.61 | R = 0.61 |
| Horizontal surface (heat flow down) | R = 0.92 | R = 0.92 |
Material resistance:
$$R = \frac{L}{k}$$
Where:
- $L$ = thickness (ft or in)
- $k$ = thermal conductivity (Btu/hr·ft·°F or Btu·in/hr·ft²·°F)
U-Value Calculation
Overall heat transfer coefficient:
$$U = \frac{1}{R_{total}}$$
Example: Insulated wall assembly:
graph LR
A[Outdoor Air<br/>T_out] -->|R_out<br/>0.17| B[Brick Veneer<br/>4 in]
B -->|R_air<br/>Air space| C[Sheathing<br/>1/2 in]
C -->|R_insulation<br/>Cavity insulation| D[Drywall<br/>1/2 in]
D -->|R_in<br/>0.68| E[Indoor Air<br/>T_in]
Resistance calculation:
| Layer | Thickness | k or R | Resistance |
|---|---|---|---|
| Outside air film | — | — | 0.17 |
| Brick veneer | 4 in | k = 5.0 | 4/5.0 = 0.80 |
| Air space (3/4") | — | — | 1.01 |
| Sheathing (OSB) | 1/2 in | k = 0.62 | 0.5/0.62 = 0.81 |
| Fiberglass batt | 5.5 in | R-19 | 19.00 |
| Gypsum drywall | 1/2 in | k = 1.11 | 0.5/1.11 = 0.45 |
| Inside air film | — | — | 0.68 |
| Total | — | — | R-22.92 |
U-value: $U = 1/22.92 = 0.044$ Btu/hr·ft²·°F
Worked Example 1: Wall Heat Loss Calculation
Given:
- Wall area: 1,000 ft²
- Wall assembly: R-22.92 (from table above)
- Indoor temperature: 70°F
- Outdoor temperature: 10°F (winter design)
Find: Heat loss through wall
Solution:
$$q = U \cdot A \cdot \Delta T$$
$$q = 0.044 \times 1,000 \times (70 - 10) = 2,640 \text{ Btu/hr}$$
Annual heating energy (simplified, 5,500 HDD):
$$E_{annual} = U \cdot A \cdot HDD \times 24$$
$$E_{annual} = 0.044 \times 1,000 \times 5,500 \times 24 = 5,808,000 \text{ Btu} = 5.8 \text{ MMBtu}$$
At $3/therm natural gas (100,000 Btu, 80% furnace efficiency):
$$Cost = \frac{5,808,000}{100,000 \times 0.80} \times 3.00 = $218$$
Answer: Peak heat loss: 2,640 Btu/hr; Annual cost: ~$218
Thermal Bridging
Thermal bridge: Conductive path through insulation (studs, structural members)
Effect: Local U-value increase, condensation risk, reduced overall R-value
Parallel Path Method
For walls with framing:
$$U_{effective} = (U_{cavity} \times f_{cavity}) + (U_{framing} \times f_{framing})$$
Where:
- $f_{cavity}$ = fraction of wall area with cavity insulation
- $f_{framing}$ = fraction of wall area with framing (studs)
Typical framing fractions:
- 2×4 studs @ 16" o.c.: 15-20% framing
- 2×6 studs @ 24" o.c.: 10-15% framing
- Steel studs @ 16" o.c.: 25% framing (worse thermal bridging)
Example:
- Cavity R-value: R-19 (fiberglass between studs)
- Framing R-value: R-5 (wood studs through cavity)
- Framing fraction: 15%
$$U_{eff} = (1/19 \times 0.85) + (1/5 \times 0.15) = 0.045 + 0.030 = 0.075$$
$$R_{eff} = 1/0.075 = 13.3$$
Thermal bridging reduces R-19 wall to effective R-13.3 (30% reduction)
Mitigation Strategies
Continuous exterior insulation: Eliminates thermal bridges
- Rigid foam over sheathing (1-2 in typical)
- Adds R-5 to R-10 without bridging
Advanced framing: Reduce framing fraction
- 24" o.c. spacing instead of 16"
- Single top plate, eliminate headers on non-bearing walls
- Reduces framing to 10-12%
Structural insulated panels (SIPs): Minimal thermal bridging
- Effective R-value ≈ 95% of nominal
Thermal break in steel framing: Critical for metal buildings
- Z-girts with thermal spacers
- Exterior insulation over studs
Fenestration Performance
Window and Glazing U-Values
Window performance metrics:
- U-factor: Overall heat transfer coefficient (lower is better)
- SHGC (Solar Heat Gain Coefficient): Fraction of solar radiation transmitted (0-1)
- VT (Visible Transmittance): Fraction of visible light transmitted (0-1)
Typical residential window U-factors (NFRC rated):
| Glazing Type | Frame | U-factor (Btu/hr·ft²·°F) | SHGC |
|---|---|---|---|
| Single clear | Aluminum | 1.30 | 0.76 |
| Single clear | Vinyl | 1.00 | 0.75 |
| Double clear | Aluminum | 0.80 | 0.70 |
| Double clear | Vinyl | 0.50 | 0.70 |
| Double low-E (ε=0.10) | Vinyl | 0.30 | 0.40 |
| Triple low-E, argon | Vinyl | 0.20 | 0.35 |
Low-emissivity (low-E) coating: Reduces radiant heat transfer between panes
- $\epsilon$ = 0.84 (clear glass)
- $\epsilon$ = 0.10 (hard-coat low-E)
- $\epsilon$ = 0.04 (soft-coat low-E)
Gas fill: Argon or krypton between panes reduces conduction
- Air: k = 0.015 Btu/hr·ft·°F
- Argon: k = 0.010 Btu/hr·ft·°F (reduces U-factor ~10%)
- Krypton: k = 0.005 Btu/hr·ft·°F (expensive, thin gaps)
Solar Heat Gain
Solar heat gain through windows:
$$Q_{solar} = A_{window} \times SHGC \times I_{solar}$$
Where:
- $A_{window}$ = glazing area (ft²)
- $SHGC$ = solar heat gain coefficient (0-1)
- $I_{solar}$ = incident solar radiation (Btu/hr·ft²)
Peak solar radiation (clear day, perpendicular surface):
- South-facing vertical: 200-250 Btu/hr·ft² (winter)
- West-facing vertical: 180-220 Btu/hr·ft² (summer afternoon)
- Horizontal (roof): 250-300 Btu/hr·ft² (summer)
Shading coefficient (older metric):
$$SC = \frac{SHGC_{actual}}{SHGC_{reference}} \approx 1.15 \times SHGC$$
Where reference is single clear glass (SHGC ≈ 0.87)
Infiltration and Air Leakage
Air Leakage Measurement
Blower door test: Pressurize building to 50 Pa, measure airflow
Metrics:
- ACH50: Air changes per hour at 50 Pa pressure
- CFM50: Cubic feet per minute leakage at 50 Pa
Typical ACH50 values:
- Existing homes (no air sealing): 8-15 ACH50
- Standard new construction: 4-7 ACH50
- Energy Star homes: 3-5 ACH50
- Passive House: < 0.6 ACH50
Conversion to natural infiltration:
$$ACH_{natural} = \frac{ACH_{50}}{N}$$
Where $N$ = 20 typical (depends on climate, shielding, height)
Infiltration heat loss:
$$Q_{inf} = \rho \cdot c_p \cdot V \cdot ACH \cdot \Delta T$$
Simplified:
$$Q_{inf} = 1.08 \times CFM_{inf} \times \Delta T$$
Worked Example 2: Infiltration Load
Given:
- House volume: 12,000 ft³
- Blower door result: ACH50 = 5.0
- Indoor: 70°F
- Outdoor: 20°F (winter design)
Find: Infiltration heat loss
Solution:
Natural air changes: $$ACH_{natural} = \frac{5.0}{20} = 0.25 \text{ ACH}$$
Infiltration airflow: $$CFM_{inf} = \frac{12,000 \times 0.25}{60} = 50 \text{ CFM}$$
Sensible heat loss: $$Q_{inf,s} = 1.08 \times 50 \times (70 - 20) = 2,700 \text{ Btu/hr}$$
Latent heat loss (winter, small): $$Q_{inf,l} \approx 0$$
Answer: 2,700 Btu/hr infiltration heat loss
For tight house (ACH50 = 2.0): $$Q_{inf,tight} = 1.08 \times 20 \times 50 = 1,080 \text{ Btu/hr}$$
Savings: 1,620 Btu/hr (60% reduction)
Thermal Mass Effects
Thermal mass: Material capacity to store and release heat
Heat storage capacity:
$$Q_{stored} = m \cdot c_p \cdot \Delta T = \rho \cdot V \cdot c_p \cdot \Delta T$$
Where:
- $\rho$ = density (lb/ft³)
- $c_p$ = specific heat (Btu/lb·°F)
- $V$ = volume (ft³)
Volumetric heat capacity:
$$C_v = \rho \cdot c_p$$
Materials:
| Material | ρ (lb/ft³) | c_p (Btu/lb·°F) | C_v (Btu/ft³·°F) |
|---|---|---|---|
| Concrete | 140 | 0.20 | 28 |
| Brick | 120 | 0.20 | 24 |
| Gypsum | 50 | 0.26 | 13 |
| Wood | 35 | 0.45 | 16 |
| Water | 62.4 | 1.00 | 62 |
Benefit: Thermal mass dampens temperature swings, reduces peak loads
Example: Concrete floor slab (4 in thick, 1,000 ft²)
- Volume: $1,000 \times 4/12 = 333$ ft³
- Mass: $333 \times 140 = 46,667$ lb
- Heat capacity: $46,667 \times 0.20 = 9,333$ Btu/°F
Temperature swing reduction: 10°F swing → 5°F swing with thermal mass
Moisture Control and Condensation
Vapor Diffusion
Fick’s Law (vapor diffusion):
$$\dot{m}_v = -M \cdot A \cdot \frac{dP_v}{dx}$$
Where:
- $\dot{m}_v$ = vapor mass flow rate (lb/hr)
- $M$ = permeance (perm = grains/hr·ft²·in Hg)
- $dP_v/dx$ = vapor pressure gradient
Permeability: $M = \mu/L$ (material property / thickness)
Vapor retarder classifications:
- Class I (impermeable): < 0.1 perm (polyethylene sheet, foil)
- Class II (semi-impermeable): 0.1-1.0 perm (kraft paper, some paints)
- Class III (semi-permeable): 1-10 perm (latex paint)
Condensation Risk
Dewpoint temperature: Temperature at which moisture condenses
At each layer interface, check:
$$T_{interface} > T_{dewpoint}$$
If $T_{interface} < T_{dewpoint}$, condensation occurs.
Condensation prevention strategies:
- Warm-side vapor retarder: Prevent moisture entry into wall
- Exterior insulation: Keep sheathing warm (above dewpoint)
- Ventilation: Remove moisture from building
- Dehumidification: Control indoor humidity (< 40% RH winter)
Related Technical Guides:
- Material Thermal Properties
- Heat Transfer Fundamentals
- Heating Load Calculations
- Cooling Load Calculations
- Psychrometric Fundamentals
References:
- ASHRAE Fundamentals Handbook, Chapter 26: Heat, Air, and Moisture Control in Building Assemblies
- ASHRAE Fundamentals Handbook, Chapter 25: Thermal and Water Vapor Transmission Data
- Building Science Corporation: www.buildingscience.com
- Lstiburek, J., “Builder’s Guide to Cold Climates”
- Straube, J., “High Performance Enclosures”