1.2 Thermal Conduction and Fourier's Law

1.2 Thermal Conduction and Fourier's Law

Conduction is the primary mode of heat transfer within opaque solids, stationary liquids, and non-convecting gases. For infrared thermographers, conduction represents the bridge between hidden subsurface conditions—such as degraded refractory brick inside a furnace, a high-resistance electrical joint buried in a busway, or wet insulation inside a wall cavity—and the surface thermal signatures detected by an infrared camera.

Microscopic Mechanisms of Conduction

Conduction occurs at the atomic and molecular level without any bulk macroscopic displacement of the matter itself:

• In electrical conductors (metals), thermal conduction is dominated by the diffusion of free electrons combined with lattice vibrational waves (phonons). Because free electrons conduct both electricity and thermal energy, good electrical conductors (copper, aluminum, silver) are almost invariably excellent thermal conductors. This relationship is formalized by the Wiedemann-Franz law, which dictates that the ratio of thermal conductivity to electrical conductivity is directly proportional to absolute temperature.
• In non-metallic solids and insulators, free electron concentration is negligible. Heat transfers almost entirely via acoustic lattice vibrations (phonons). Disordered structures (such as fiberglass, cellular glass, and mineral wool) scatter phonons effectively, producing very low thermal conductivity.
• In liquids and gases, conduction occurs through direct molecular collisions and diffusion. Gas molecules transfer kinetic energy during random Brownian collisions. Because gases have large intermolecular distances, their thermal conductivity is exceptionally low unless bulk convection currents develop.

Fourier's Rate Equation of Heat Conduction

In 1822, French mathematician and physicist Joseph Fourier formulated the fundamental law governing conductive heat transfer. In one-dimensional form, Fourier's law states that the time rate of heat transfer through a homogeneous material is directly proportional to the surface area normal to the heat path and to the temperature gradient along that path:

q = -k · A · (dT/dx)

Where:

• q is the heat transfer rate in Watts (W, where 1 W = 1 J/s) or BTU/hr (1 W ≈ 3.412 BTU/hr).
• k is the thermal conductivity of the material (W/(m·K) or BTU/(hr·ft·°F)).
• A is the cross-sectional area perpendicular to the heat flow path (m² or ft²).
• dT/dx is the temperature gradient—the rate of temperature change with respect to distance along the conduction path (K/m or °F/ft).
• The negative sign is a thermodynamic necessity dictated by the Second Law of Thermodynamics: heat flows spontaneously down the temperature gradient, from higher temperature to lower temperature (i.e., if dT/dx is negative, q is positive in the direction of increasing x).

The heat flux (q''), defined as the heat transfer rate per unit surface area, is expressed as: q'' = q / A = -k · (dT/dx)

Thermal Conductivity (k) of Industrial Materials

Thermal conductivity (k) is an intensive physical property representing a material's inherent ability to conduct heat. Materials with high k values act as thermal conductors, rapidly equalizing temperature gradients. Materials with low k values act as thermal insulators, maintaining substantial temperature differentials across small physical dimensions.

Thermal Conductivity Reference Table

Thermal Resistance and Multi-Layer Systems

Just as electrical resistance opposes electrical current (R_e = ΔV / I), thermal resistance (R_th) opposes conductive heat flow:

R_th = ΔT / q = L / (k · A)

Where L is the thickness of the material layer (m), k is thermal conductivity (W/(m·K)), and A is heat transfer area (m²). The units of R_th are Kelvin per Watt (K/W).

In construction and building envelope thermography, thermal resistance is frequently normalized per unit area as the R-value (R_val = L / k, with SI units of m²·K/W and Imperial units of hr·ft²·°F/BTU).

Series Thermal Circuit Analysis

When heat conducts through multi-layer composite structures—such as an industrial furnace composed of inner refractory brick, an intermediate ceramic insulation blanket, and an outer structural steel shell—the layers are arranged in series. The total thermal resistance (R_total) is the algebraic sum of the individual layer resistances:

R_total = R_1 + R_2 + R_3 + ... + R_n = L_1 / (k_1 · A) + L_2 / (k_2 · A) + L_3 / (k_3 · A)

Because the steady-state heat flow rate q is constant through every planar layer: q = (T_hot,interior - T_cold,exterior) / R_total

If refractory brick erodes or cracks (reducing L_1), R_1 decreases, causing R_total to drop. As a direct mathematical consequence, the heat flow rate q increases and the outer steel shell temperature rises significantly, creating an external hot spot visible to an infrared camera.

Steady-State Versus Transient Heat Conduction

Thermal inspections encounter two distinct temporal regimes of conduction:

1. Steady-State Conduction:

   • System conditions are time-invariant: dT/dt = 0.
   • The temperature at every point within the component remains constant over time.
   • Heat entering any segment equals heat leaving that segment.
   • Common examples: an electrical busbar carrying continuous base load for four hours; an industrial boiler operating at stable production capacity; steady winter heating inside a commercial office building.
   • Steady-state conditions are ideal for quantitative thermography because temperature differences directly reflect thermal resistance values and heat generation rates.

2. Transient (Unsteady) Conduction:

   • Temperatures change as a function of time: dT/dt ≠ 0.
   • Heat transfer is governed by both thermal conductivity (k) and thermal capacitance (C_th = m · c = ρ · V · c).
   • The rate of temperature change is governed by thermal diffusivity (α = k / (ρ · c), in m²/s), which measures the rate at which heat propagates through a material relative to its ability to store thermal energy.
   • The thermal response time is characterized by the thermal time constant (τ = R_th · C_th).
   • Common examples: solar loading on a roof or masonry facade followed by nighttime cooling; a motor bearing heating up during startup; pulsed thermography where an active flash lamp excites a composite laminate to detect internal delaminations.

Worked Field Calculation: Industrial Furnace Refractory Degradation

Inspection Scenario

A thermographer conducts an infrared survey of a petrochemical process furnace wall. The nominal design consists of three layers:

1. Inner dense refractory firebrick: L_1 = 0.20 m, k_1 = 1.25 W/(m·K)
2. Intermediate calcium silicate insulation: L_2 = 0.10 m, k_2 = 0.08 W/(m·K)
3. Outer carbon steel casing: L_3 = 0.01 m, k_3 = 45.0 W/(m·K)

Operational data:

• Process internal hot face temperature: T_hot = 950 °C
• Outer casing baseline design temperature: T_shell,normal = 75.0 °C
• Evaluated surface area: A = 1.0 m²

During the thermographic scan, the thermographer spots an abnormal localized hot spot where the outer steel casing reaches T_shell,defect = 245.0 °C. The plant engineer suspects that internal refractory brick has spalled away, reducing the effective brick thickness to L_1,damaged = 0.04 m while the intermediate insulation has degraded to half thickness (L_2,damaged = 0.05 m).

Step-by-Step Solution

1. Calculate Normal Design Thermal Resistance (R_normal): R_1 = L_1 / (k_1 · A) = 0.20 / (1.25 · 1.0) = 0.160 K/W R_2 = L_2 / (k_2 · A) = 0.10 / (0.08 · 1.0) = 1.250 K/W R_3 = L_3 / (k_3 · A) = 0.01 / (45.0 · 1.0) = 0.00022 K/W R_total,normal = R_1 + R_2 + R_3 = 0.160 + 1.250 + 0.00022 ≈ 1.4102 K/W

2. Calculate Normal Heat Loss Rate (q_normal): Assuming an external surface temperature of 75 °C: q_normal = (T_hot - T_shell,normal) / R_total,normal = (950 - 75) / 1.4102 = 875 / 1.4102 = 620.5 W/m²

3. Calculate Degraded Area Thermal Resistance (R_damaged): R_1,damaged = 0.04 / (1.25 · 1.0) = 0.032 K/W R_2,damaged = 0.05 / (0.08 · 1.0) = 0.625 K/W R_3 = 0.00022 K/W R_total,damaged = 0.032 + 0.625 + 0.00022 ≈ 0.6572 K/W

4. Calculate Degraded Heat Loss Rate (q_damaged): With the outer casing measured at 245 °C: q_damaged = (T_hot - T_shell,defect) / R_total,damaged = (950 - 245) / 0.6572 = 705 / 0.6572 = 1,072.7 W/m²

5. Interpretation for Predictive Maintenance: The thermal resistance across the wall dropped by: (1.4102 - 0.6572) / 1.4102 × 100% = 53.4% This 53.4% loss of thermal resistance increased conductive heat flux through the wall by 72.9%, driving the outer shell temperature from 75 °C to 245 °C. This quantifies severe internal lining loss and indicates immediate risk of structural steel casing warping or metallurgical failure.