HVAC Systems Encyclopedia

A comprehensive encyclopedia of heating, ventilation, and air conditioning systems

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):

SurfaceWinter (15 mph wind)Summer (7.5 mph wind)
Exterior vertical wallR = 0.17R = 0.25
Interior vertical wallR = 0.68R = 0.68
Horizontal surface (heat flow up)R = 0.61R = 0.61
Horizontal surface (heat flow down)R = 0.92R = 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:

LayerThicknessk or RResistance
Outside air film0.17
Brick veneer4 ink = 5.04/5.0 = 0.80
Air space (3/4")1.01
Sheathing (OSB)1/2 ink = 0.620.5/0.62 = 0.81
Fiberglass batt5.5 inR-1919.00
Gypsum drywall1/2 ink = 1.110.5/1.11 = 0.45
Inside air film0.68
TotalR-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

  1. Continuous exterior insulation: Eliminates thermal bridges

    • Rigid foam over sheathing (1-2 in typical)
    • Adds R-5 to R-10 without bridging
  2. 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%
  3. Structural insulated panels (SIPs): Minimal thermal bridging

    • Effective R-value ≈ 95% of nominal
  4. 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 TypeFrameU-factor (Btu/hr·ft²·°F)SHGC
Single clearAluminum1.300.76
Single clearVinyl1.000.75
Double clearAluminum0.800.70
Double clearVinyl0.500.70
Double low-E (ε=0.10)Vinyl0.300.40
Triple low-E, argonVinyl0.200.35

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:

Materialρ (lb/ft³)c_p (Btu/lb·°F)C_v (Btu/ft³·°F)
Concrete1400.2028
Brick1200.2024
Gypsum500.2613
Wood350.4516
Water62.41.0062

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:

  1. Warm-side vapor retarder: Prevent moisture entry into wall
  2. Exterior insulation: Keep sheathing warm (above dewpoint)
  3. Ventilation: Remove moisture from building
  4. Dehumidification: Control indoor humidity (< 40% RH winter)

Related Technical Guides:

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”