Steady State Conduction

Steady-state conduction occurs when the temperature field within a material does not change with time. The temperature at any point remains constant, and the heat flux is time-invariant. This condition is fundamental to HVAC heat transfer analysis for building envelopes, insulation systems, and heat exchangers.

Fourier’s Law of Heat Conduction

Fourier’s law provides the fundamental relationship between heat flux and temperature gradient:

One-Dimensional Form:

q = -kA(dT/dx)

Where:

• q = heat transfer rate (W)
• k = thermal conductivity (W/m·K)
• A = cross-sectional area perpendicular to heat flow (m²)
• dT/dx = temperature gradient (K/m)
• Negative sign indicates heat flows from high to low temperature

Heat Flux Form:

q" = -k(dT/dx)

Where q" = heat flux (W/m²)

The thermal conductivity k is a material property. For most building materials, k varies weakly with temperature and can be treated as constant over typical HVAC temperature ranges.

Plane Wall Conduction

For a plane wall of thickness L with constant thermal conductivity and uniform surface temperatures T₁ and T₂:

Temperature Distribution:

T(x) = T₁ - (T₁ - T₂)(x/L)

The temperature varies linearly through the wall thickness.

Heat Transfer Rate:

q = kA(T₁ - T₂)/L

Thermal Resistance:

The thermal resistance concept simplifies analysis by analogy to Ohm’s law (V = IR):

R_cond = L/(kA)

Heat transfer rate becomes:

q = ΔT/R_cond = (T₁ - T₂)/(L/kA)

Units: R_cond in K/W or °C/W

Composite Walls - Series Configuration

For multiple layers in series (typical building wall construction), resistances add:

Total Resistance:

R_total = R₁ + R₂ + R₃ + … + R_n

For n layers:

R_total = Σ(L_i/k_i A)

Heat Transfer Rate:

q = ΔT_overall/R_total = (T_∞1 - T_∞2)/(R_conv,1 + R_cond,1 + R_cond,2 + … + R_conv,2)

Where:

• R_conv,1 = 1/(h₁A) = interior surface convection resistance
• R_conv,2 = 1/(h₂A) = exterior surface convection resistance
• R_cond,i = L_i/(k_i A) = conduction resistance of layer i

Interface Temperatures:

For any interface between layers i and i+1:

T_interface = T_upstream - q·R_layers_upstream

R-Values and U-Values

R-Value (Thermal Resistance per Unit Area):

R = L/k (units: m²·K/W or ft²·°F·hr/BTU)

The R-value is independent of area and commonly used in building construction:

• R-13 fiberglass batt: R = 2.29 m²·K/W (13 ft²·°F·hr/BTU)
• R-19 fiberglass batt: R = 3.35 m²·K/W (19 ft²·°F·hr/BTU)
• R-30 blown cellulose: R = 5.28 m²·K/W (30 ft²·°F·hr/BTU)

U-Value (Overall Heat Transfer Coefficient):

U = 1/(R_total) (units: W/m²·K or BTU/hr·ft²·°F)

Heat flux through assembly:

q" = U·ΔT

Lower U-values indicate better insulation performance.

Contact Resistance

At interfaces between solid materials, microscopic surface roughness creates air gaps that impede heat transfer. Contact resistance R_c depends on:

• Surface roughness and waviness
• Contact pressure
• Interstitial fluid (air, thermal grease)
• Material hardness

Typical Values:

Contact resistance is included in series with other resistances:

R_total = R₁ + R_c,1-2 + R₂ + R_c,2-3 + R₃

Cylindrical Coordinates

For radial conduction through cylindrical geometries (pipes, insulated ducts):

Temperature Distribution:

T(r) = T₁ - (T₁ - T₂)[ln(r/r₁)/ln(r₂/r₁)]

Logarithmic profile, not linear.

Heat Transfer Rate (per unit length):

q’ = 2πk(T₁ - T₂)/ln(r₂/r₁)

Where:

• q’ = heat transfer per unit length (W/m)
• r₁ = inner radius (m)
• r₂ = outer radius (m)

Thermal Resistance (per unit length):

R’_cyl = ln(r₂/r₁)/(2πk) (units: K·m/W)

For cylindrical pipe with insulation:

R’_total = ln(r₂/r₁)/(2πk_pipe) + ln(r₃/r₂)/(2πk_insulation)

Thermal Resistance Networks

Complex geometries are solved using resistance networks analogous to electrical circuits:

Series Resistances:

• Heat flows through each element sequentially
• Same heat flow through all elements
• R_total = R₁ + R₂ + R₃ + …

Parallel Resistances:

• Heat flow divides among parallel paths
• Same temperature difference across all paths
• 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + …

Example: Wall with thermal bridge

Wood stud (k = 0.12 W/m·K) and fiberglass insulation (k = 0.038 W/m·K) in parallel create a thermal bridge. Total heat flow is the sum of paths.

Worked Example 1: Composite Wall Heat Loss

Problem: Calculate heat loss through a wall assembly:

• Interior surface: T₁ = 22°C, h₁ = 8 W/m²·K
• Gypsum board: L = 12.7 mm, k = 0.17 W/m·K
• Fiberglass insulation: L = 140 mm, k = 0.038 W/m·K
• Wood siding: L = 19 mm, k = 0.12 W/m·K
• Exterior surface: T₂ = -5°C, h₂ = 30 W/m²·K

Solution:

Calculate resistances per unit area (R-values):

R_conv,i = 1/h₁ = 1/8 = 0.125 m²·K/W

R_gypsum = L/k = 0.0127/0.17 = 0.075 m²·K/W

R_insulation = L/k = 0.140/0.038 = 3.684 m²·K/W

R_siding = L/k = 0.019/0.12 = 0.158 m²·K/W

R_conv,o = 1/h₂ = 1/30 = 0.033 m²·K/W

R_total = 0.125 + 0.075 + 3.684 + 0.158 + 0.033 = 4.075 m²·K/W

Heat flux:

q" = ΔT/R_total = (22 - (-5))/4.075 = 27/4.075 = 6.63 W/m²

U-value:

U = 1/R_total = 1/4.075 = 0.245 W/m²·K

For a 20 m² wall:

q = q"·A = 6.63 × 20 = 133 W

Worked Example 2: Insulated Pipe Heat Loss

Problem: A copper steam pipe carries saturated steam at 150°C. Calculate heat loss per meter length:

• Pipe: inner radius r₁ = 50 mm, outer radius r₂ = 57 mm, k = 401 W/m·K
• Insulation: thickness = 50 mm (r₃ = 107 mm), k = 0.045 W/m·K
• Ambient air: T_∞ = 20°C, h = 10 W/m²·K
• Inner surface temperature = steam temperature (neglect steam-side resistance)

Solution:

Calculate thermal resistances per unit length:

R’_pipe = ln(r₂/r₁)/(2πk_pipe) = ln(57/50)/(2π × 401) R’_pipe = ln(1.14)/(2519) = 0.131/2519 = 0.000052 K·m/W

R’_insulation = ln(r₃/r₂)/(2πk_insulation) = ln(107/57)/(2π × 0.045) R’_insulation = ln(1.877)/(0.283) = 0.630/0.283 = 2.227 K·m/W

R’_conv = 1/(h × 2πr₃) = 1/(10 × 2π × 0.107) R’_conv = 1/6.72 = 0.149 K·m/W

R’_total = 0.000052 + 2.227 + 0.149 = 2.376 K·m/W

Heat loss per unit length:

q’ = (T_steam - T_∞)/R’_total = (150 - 20)/2.376 = 130/2.376 = 54.7 W/m

Note: The copper pipe resistance is negligible (0.002% of total). This is typical for metal pipes where insulation dominates.

Outer surface temperature of insulation:

T_surface = T_∞ + q’·R’_conv = 20 + 54.7 × 0.149 = 28.2°C

Worked Example 3: Critical Radius of Insulation

Problem: For the pipe in Example 2, does adding more insulation always reduce heat loss?

Solution:

The critical radius for cylindrical insulation is:

r_critical = k_insulation/h = 0.045/10 = 0.0045 m = 4.5 mm

Current insulation outer radius: r₃ = 107 mm » r_critical

Since r₃ > r_critical, adding more insulation will reduce heat loss. However, if the bare pipe radius exceeded r_critical, adding thin insulation would initially increase heat loss by increasing surface area more than it increases resistance.

For typical building applications with low h values and moderate insulation thickness, r > r_critical, so adding insulation always helps.

Temperature-Dependent Thermal Conductivity

When thermal conductivity varies significantly with temperature:

k(T) = k₀(1 + βT)

Use mean thermal conductivity:

k_mean = k₀[1 + β(T₁ + T₂)/2]

Or integrate Fourier’s law:

q = A(k₀ΔT + k₀β(T₁² - T₂²)/2)/L

For most HVAC applications over moderate temperature ranges, constant k is adequate.

Design Implications

Building Envelope Design:

• Maximize R-values in walls, roofs, and floors
• Minimize thermal bridges (studs, structural elements)
• Ensure continuous insulation where possible
• Account for contact resistance at material interfaces

Duct and Pipe Insulation:

• Use logarithmic equations for cylindrical geometry
• Verify insulation thickness exceeds critical radius
• Consider vapor barriers to prevent condensation
• Account for degradation of k due to moisture

Resistance Network Analysis:

• Identify series and parallel heat flow paths
• Calculate effective R-value for composite assemblies
• Locate interface temperatures to check for condensation risk
• Use U-values for assembly comparison and energy modeling

Components

• Fourier Law One Dimensional
• Thermal Conductivity Temperature Dependent
• Thermal Resistance Concept
• Composite Walls Series
• Composite Walls Parallel
• Multilayer Systems
• Cylindrical Coordinates
• Spherical Coordinates
• Critical Radius Insulation
• Contact Resistance
• Thermal Interface Materials