6.2 Envelope Heat Transmission & Infiltration/Exfiltration Load Methods

6.2 Envelope Heat Transmission & Infiltration/Exfiltration Load Methods

Building envelope heat transmission and unconditioned air infiltration represent the primary heating loads during winter design conditions and substantial components of summer sensible cooling loads. Calculating envelope loads requires evaluating the steady-state thermal resistance ($R$-value) and overall heat transfer coefficient ($U$-factor) of multi-layer assemblies, accounting for two-dimensional thermal bridging across structural framing members, sizing below-grade and slab-on-grade boundary losses, and quantifying infiltration air volume driven by stack effect, wind velocity, and mechanical room pressurization.

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1. Steady-State Conduction & Multilayer Thermal Resistance

Under steady-state conditions with one-dimensional heat flow, the thermal transmission rate ($q$) through an opaque building component of surface area $A$ is expressed by Fourier's law:

$q = U \cdot A \cdot (T_{\text{outdoor}} - T_{\text{indoor}}) = \frac{A \cdot \Delta T}{R_{\text{total}}}$

Where:

• $q = \text{Heat transfer rate } (\text{Btu/hr})$
• $U = \text{Overall heat transfer coefficient } (\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)})$
• $A = \text{Net surface area of assembly } (\text{ft}^2)$
• $\Delta T = \text{Temperature difference between ambient outdoor air and indoor air } ({}^\circ\text{F})$
• $R_{\text{total}} = \text{Total thermal resistance of the composite assembly } (\text{hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu})$

+-----------------------------------------------------------------------------------------+
| COMPOSITE MULTI-LAYER WALL THERMAL CIRCUIT                                              |
+-----------------------------------------------------------------------------------------+
| Outdoor Air (To)                                                        Indoor Air (Ti) |
|      |                                                                         |        |
|     [R_o]  ---  [R_ext_finish]  ---  [R_sheathing]  ---  [R_insul]  ---  [R_gyp] --- [R_i]  |
|  (Film Out)       (Brick/Siding)       (Plywood/OSB)     (Cavity)      (Drywall) (Film In)
+-----------------------------------------------------------------------------------------+

Overall Thermal Resistance ($R_{\text{total}}$) & Series Layer Summation

The total thermal resistance of an unbridged, homogeneous multi-layer assembly is the direct series sum of the individual layer resistances, including the interior ($R_i$) and exterior ($R_o$) surface air film resistances:

$R_{\text{total}} = R_i + \sum_{k=1}^n R_k + R_o = \frac{1}{h_i} + \sum_{k=1}^n \frac{x_k}{k_k} + \frac{1}{h_o}$

$U = \frac{1}{R_{\text{total}}}$

Where:

• $x_k = \text{Thickness of material layer } k\text{ (inches or feet)}$
• $k_k = \text{Thermal conductivity of material layer } k\text{ (Btu}\cdot\text{in./(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F) or Btu/(hr}\cdot\text{ft}\cdot{}^\circ\text{F)})$
• $h_i, h_o = \text{Inside and outside convective/radiative surface film heat transfer coefficients } (\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)})$

Standard Surface Air Film Resistances ($R_i$ and $R_o$)

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2. Thermal Bridging & Parallel Heat Flow in Framed Assemblies

Framed envelope assemblies (wood studs, light-gauge steel studs, metal curtain wall mullions) contain high-conductivity structural framing that bridges the low-conductivity cavity insulation. This requires calculating an area-weighted parallel heat flow path.

+-----------------------------------------------------------------------------------------+
| PARALLEL HEAT PATHS THROUGH FRAMED WALL CAVITY                                         |
+-----------------------------------------------------------------------------------------+
| Path 1 (Through Stud / Framing):      R_path1 = R_i + R_gyp + R_stud + R_sheath + R_o   |
| Path 2 (Through Insulated Cavity):    R_path2 = R_i + R_gyp + R_cavity + R_sheath + R_o |
| Area-Weighted Overall U-factor:       U_overall = (f_frame * U_path1) + (f_cav * U_path2)|
+-----------------------------------------------------------------------------------------+

Area-Weighted Parallel Path Equation

$U_{\text{overall}} = f_{\text{framing}} \cdot U_{\text{framing}} + (1 - f_{\text{framing}}) \cdot U_{\text{cavity}} = \frac{f_{\text{framing}}}{R_{\text{framing}}} + \frac{1 - f_{\text{framing}}}{R_{\text{cavity}}}$

Where $f_{\text{framing}}$ is the framing fraction (typically $0.15\text{ to }0.25$ for wood framing at $16\text{ in.}$ or $24\text{ in.}$ on center, and $0.10\text{ to }0.18$ for steel stud construction).

Steel Stud Thermal Derating (Modified Zone Method / ASHRAE 90.1)

Steel has a thermal conductivity ($k \approx 314\text{ Btu}\cdot\text{in./(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$) over 300 times greater than fiberglass batt insulation ($k \approx 0.27$). Steel studs act as massive thermal shorts, derating cavity insulation effectiveness by 40% to 60%. ASHRAE Standard 90.1 provides effective cavity $R$-values ($R_{\text{effective}}$):

Design Rule: To meet modern energy codes (ASHRAE Standard 90.1 / IECC), steel-framed walls require continuous exterior insulation (ci) (such as rigid polyisocyanurate or extruded polystyrene) to break the framing thermal bridge.

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3. Below-Grade Envelopes & Slab-on-Grade Heat Losses

Heat transfer from basements and on-grade concrete slabs does not follow simple 1D one-layer equations because ground temperature varies with depth and thermal path length through the earth increases radially.

+-----------------------------------------------------------------------------------------+
| SLAB-ON-GRADE PERIMETER HEAT FLOW DYNAMICS                                              |
+-----------------------------------------------------------------------------------------+
|                 Indoor Space (Ti)                                                       |
|                 =============================                                           |
|                 | Concrete Slab Floor       |                                           |
|     Outside     |                           | Perimeter Length P (ft)                   |
|     Air (To)    +---------------------------+                                           |
|     ~~~~~~~~~   | Perimeter Insulation (R)  |                                           |
|        |        +---------------------------+ ===> q = F_p * P * (T_in - T_out)         |
|        |              | Ground Heat Paths                                               |
|        +--------------+ (Curvilinear Earth Flow)                                        |
+-----------------------------------------------------------------------------------------+

1. Slab-on-Grade Perimeter Heat Loss Formulation

Concrete slab-on-grade heat loss occurs almost entirely around the exposed perimeter. Sizing is based on the perimeter heat loss coefficient ($F_p$) and slab perimeter length ($P$):

$q_{\text{slab}} = F_p \cdot P \cdot (T_{\text{indoor}} - T_{\text{outdoor}})$

Where:

• $q_{\text{slab}} = \text{Slab heating transmission rate } (\text{Btu/hr})$
• $F_p = \text{Perimeter heat loss coefficient } (\text{Btu/(hr}\cdot\text{ft}\cdot{}^\circ\text{F)})$
• $P = \text{Exposed perimeter length of the slab floor } (\text{ft})$
• $T_{\text{indoor}} - T_{\text{outdoor}} = \text{Design dry-bulb temperature difference } ({}^\circ\text{F})$

Standard $F_p$ Perimeter Loss Coefficients (ASHRAE 90.1)

2. Below-Grade Basement Walls & Floors

Below-grade heat transfer is evaluated using depth-averaged $U$-factors ($U_{\text{avg}}$) applied to below-grade wall area ($A_{\text{bg}}$) against ground temperature ($T_{\text{ground}}$):

$q_{\text{basement, wall}} = U_{\text{bg, wall}} \cdot A_{\text{bg, wall}} \cdot (T_{\text{indoor}} - T_{\text{ground}})$

$q_{\text{basement, floor}} = U_{\text{bg, floor}} \cdot A_{\text{floor}} \cdot (T_{\text{indoor}} - T_{\text{ground}})$

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4. Infiltration & Exfiltration Load Methodologies

Infiltration is the uncontrolled ingress of outdoor air through cracks, gaps, building envelope joints, and door openings. Exfiltration is the corresponding egress of conditioned air.

Driving Physical Forces for Infiltration

1. Wind Pressure (Velocity Head): $\Delta P_{\text{wind}} = C_p \cdot \frac{\rho_{\text{air}} \cdot V_{\text{wind}}^2}{2 g_c}$ Wind creates positive pressure on the windward facade (causing infiltration) and negative suction pressure on the leeward and sidewall facades (causing exfiltration).
2. Stack Effect (Thermal Buoyancy): $\Delta P_{\text{stack}} = \rho_0 \left(\frac{g}{g_c}\right) h \left(\frac{T_{\text{indoor}} - T_{\text{outdoor}}}{T_{\text{outdoor}}}\right)$ During winter ($T_{\text{indoor}} > T_{\text{outdoor}}$), warm air rises inside the building, creating negative pressure and infiltration at the ground level, and positive pressure and exfiltration at upper levels. The height where internal and external pressures equalize is the Neutral Pressure Level (NPL).
3. Mechanical Imbalance: Exhaust air flow in excess of supply/ventilation air creates net negative building pressure, accelerating infiltration.

+-----------------------------------------------------------------------------------------+
| WINTER STACK EFFECT & INFILTRATION PRESSURE PROFILE                                     |
+-----------------------------------------------------------------------------------------+
| Top of Building (Z = H):      P_in > P_out  -->  EXFILTRATION (Positive Pressure)       |
|                                     |                                                   |
| Neutral Pressure Level (NPL): P_in = P_out  -->  ZERO FLOW                              |
|                                     |                                                   |
| Base of Building (Z = 0):     P_in < P_out  -->  INFILTRATION (Negative Suction)        |
+-----------------------------------------------------------------------------------------+

Infiltration Volumetric Airflow Calculation Methods

1. Air Change Method

Used widely for general commercial and residential room load estimation:

$\text{CFM}_{\text{inf}} = \frac{\text{ACH} \times \text{Volume }(\text{ft}^3)}{60\text{ min/hr}}$

Where $\text{ACH}$ is Air Changes per Hour ($0.1\text{ to }0.3\text{ ACH}$ for tight commercial construction; $0.5\text{ to }1.0\text{ ACH}$ for loose/older construction).

2. Crack Length Method

Based on fluid flow through window and door perimeter gaps:

$\text{CFM}_{\text{inf}} = K \cdot L \cdot (\Delta P)^n$

Where $L$ is total crack length in feet, $K$ is crack flow coefficient, and $n$ is flow exponent ($0.5 \le n \le 0.7$).

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5. Infiltration Heating & Cooling Load Synthesis

Infiltration air enters directly into the occupied space without passing through the central cooling/heating coil. It must be treated as a direct room sensible and latent load.

+-----------------------------------------------------------------------------------------+
| INFILTRATION PSYCHROMETRIC LOAD EQUATIONS (STANDARD SEA-LEVEL AIR)                      |
+-----------------------------------------------------------------------------------------+
| Sensible Load:   q_s = 1.08 * CFM_inf * (T_outdoor - T_indoor)                          |
| Latent Load:     q_l = 4840 * CFM_inf * (W_outdoor - W_indoor)                          |
| Total Load:      q_t = 4.5  * CFM_inf * (h_outdoor - h_indoor)                          |
+-----------------------------------------------------------------------------------------+

• Winter Heating Infiltration: Both $\Delta T = (T_o - T_i)$ and $\Delta W = (W_o - W_i)$ are negative, imposing direct sensible and humidification (latent) heating loads on the zone terminal units.
• Summer Cooling Infiltration: Hot, humid air entering the room adds to the zone Sensible Heat Ratio (RSHR) and requires coil dehumidification.

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6. NCEES Reference Handbook Navigation Tactics

• Envelope Transmission Formulas: Search "Thermal Resistance" or "Heat Transmission" to find steady-state conduction formulas ($U = 1/R_{\text{total}}$) and typical material thermal conductivity ($k$) tables.
• Slab-on-Grade Coefficients: Search "Slab-on-Grade" or "F_p" to locate the perimeter heat loss formula and coefficient table.
• Air Change Method: Search "Infiltration" or "Air Changes" to locate air change rate tables and conversion formulas.

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7. Worked Computational Examples

Example 1: Multilayer Steel-Stud Wall U-Factor with Thermal Bridging

An exterior wall assembly consists of the following series layers from outside to inside:

1. Exterior air film ($R_o = 0.17\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$)
2. 4-inch face brick ($k = 9.0\text{ Btu}\cdot\text{in./(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$, $R_1 = 4 / 9.0 = 0.44\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$)
3. 1-inch continuous extruded polystyrene (XPS) rigid board ($R_2 = 5.00\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$)
4. 5/8-inch exterior gypsum sheathing ($R_3 = 0.56\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$)
5. $2\times 6$ light-gauge steel studs at $16\text{ in.}$ on center with nominal R-19 fiberglass batt cavity insulation. Per ASHRAE Standard 90.1, the effective cavity resistance is $R_{\text{effective}} = 7.10\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$.
6. 5/8-inch interior gypsum wallboard ($R_4 = 0.56\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$)
7. Interior air film ($R_i = 0.68\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$)

Calculate: (a) The overall assembly $U$-factor with thermal bridging, and (b) The total winter heat loss through a $3,000\text{ ft}^2$ wall if $T_{\text{indoor}} = 70^\circ\text{F}$ and $T_{\text{outdoor}} = 10^\circ\text{F}$.

Solution:

1. Sum all series resistances using the derated effective cavity resistance: $R_{\text{total}} = R_o + R_1 + R_2 + R_3 + R_{\text{effective}} + R_4 + R_i$ $R_{\text{total}} = 0.17 + 0.44 + 5.00 + 0.56 + 7.10 + 0.56 + 0.68 = 14.51\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$
2. Calculate overall $U$-factor: $U = \frac{1}{R_{\text{total}}} = \frac{1}{14.51} = 0.0689\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$
3. Calculate total winter steady-state transmission heat loss: $\Delta T = 70 - 10 = 60^\circ\text{F}$ $q = U \cdot A \cdot \Delta T = 0.0689 \times 3,000\text{ ft}^2 \times 60^\circ\text{F} = 12,402\text{ Btu/hr}$

(Note: If thermal bridging were ignored and nominal R-19 were used without derating, $R_{\text{total}} = 26.41$, giving $q = 6,815\text{ Btu/hr}$—an underestimation of 45%!)

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Example 2: Infiltration & Slab-on-Grade Combined Heating Load

A standalone commercial building in Columbus, OH has the following characteristics:

• Building floor plan: $100\text{ ft} \times 60\text{ ft}$ rectangle ($A_{\text{floor}} = 6,000\text{ ft}^2$, ceiling height $12\text{ ft}$).
• Unheated concrete slab-on-grade floor with R-10 perimeter insulation ($F_p = 0.54\text{ Btu/(hr}\cdot\text{ft}\cdot{}^\circ\text{F)}$).
• Infiltration rate: $0.35\text{ ACH}$.
• Indoor winter design conditions: $70^\circ\text{F}$ DB, $30%\text{ RH}$ ($W_i = 0.00470\text{ lb}w/\text{lb}{da}$).
• Outdoor winter design conditions (99.6%): $2^\circ\text{F}$ DB, $80%\text{ RH}$ ($W_o = 0.00085\text{ lb}w/\text{lb}{da}$).

Calculate: (a) The perimeter slab-on-grade heat loss, (b) Infiltration volumetric airflow, (c) Infiltration sensible heating load, and (d) Infiltration humidification (latent) heating load.

Solution:

1. Slab-on-grade perimeter heat loss: $P = 2 \times (100 + 60) = 320\text{ ft}$ $\Delta T = 70 - 2 = 68^\circ\text{F}$ $q_{\text{slab}} = F_p \cdot P \cdot \Delta T = 0.54 \times 320 \times 68 = 11,750.4\text{ Btu/hr}$
2. Infiltration volumetric airflow: $\text{Volume } = 6,000\text{ ft}^2 \times 12\text{ ft} = 72,000\text{ ft}^3$ $\text{CFM}_{\text{inf}} = \frac{0.35 \times 72,000\text{ ft}^3}{60\text{ min/hr}} = 420\text{ CFM}$
3. Infiltration sensible heating load: $q_{s,\text{inf}} = 1.08 \times \text{CFM}_{\text{inf}} \times (T_i - T_o) = 1.08 \times 420 \times 68 = 30,844.8\text{ Btu/hr}$
4. Infiltration latent (humidification) heating load: $\Delta W = 0.00470 - 0.00085 = 0.00385\text{ lb}_w/\text{lb}_{da}$ $q_{l,\text{inf}} = 4840 \times \text{CFM}_{\text{inf}} \times \Delta W = 4840 \times 420 \times 0.00385 = 7,826.3\text{ Btu/hr}$
5. Total combined zone heating load from slab edge and infiltration: $q_{\text{total}} = 11,750.4 + 30,844.8 + 7,826.3 = 50,421.5\text{ Btu/hr} = 50.42\text{ MBH}$