6.2 Conduction

Heat conduction is a crucial aspect of heat transfer, focusing on how thermal energy moves through materials. It's all about understanding how different substances conduct heat and how we can use this knowledge in engineering applications.

Fourier's Law is the cornerstone of heat conduction, helping us calculate heat transfer rates in various scenarios. We'll explore how it applies to different geometries and how thermal resistance affects heat flow in composite systems.

Thermal Conductivity and Heat Conduction

Definition and Properties of Thermal Conductivity

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• Thermal conductivity represents the ability of a material to conduct heat, measured in W/(m·K)
• Materials with higher thermal conductivity allow heat to flow through them more readily than materials with lower thermal conductivity (copper vs. wood)
• The thermal conductivity of a material depends on its composition, microstructure, and temperature
  • Composition: pure metals generally have higher thermal conductivity than alloys or non-metals
  • Microstructure: crystalline materials often have higher thermal conductivity than amorphous materials
  • Temperature: thermal conductivity may increase or decrease with temperature, depending on the material
• In heat conduction, thermal conductivity determines the rate at which heat is transferred through a material when a temperature gradient is present

Role of Thermal Conductivity in Heat Conduction

• Thermal conductivity values are used in calculating heat transfer rates and temperature distributions in conduction problems
• Higher thermal conductivity leads to faster heat transfer and more uniform temperature distributions
• Lower thermal conductivity results in slower heat transfer and steeper temperature gradients
• Understanding thermal conductivity helps engineers select appropriate materials for heat transfer applications (insulation, heat exchangers, thermal management systems)

Fourier's Law for Heat Transfer

One-Dimensional Steady-State Heat Conduction

• Fourier's law states that the heat transfer rate through a material is proportional to the negative temperature gradient and the area perpendicular to the gradient
• The one-dimensional form of Fourier's law is expressed as $q = -kA(dT/dx)$, where:
  • $q$ is the heat transfer rate (W)
  • $k$ is the thermal conductivity (W/(m·K))
  • $A$ is the cross-sectional area (m²)
  • $dT/dx$ is the temperature gradient (K/m)
• In steady-state conditions, the temperature distribution is linear, and the temperature gradient is constant along the heat transfer direction

Applying Fourier's Law

• To apply Fourier's law, the thermal conductivity, cross-sectional area, and temperature gradient must be known or determined
• The negative sign in Fourier's law indicates that heat flows from regions of higher temperature to regions of lower temperature
• Example: Calculate the heat transfer rate through a 2 m × 1 m wall with a thickness of 0.2 m and a thermal conductivity of 0.5 W/(m·K), if the temperature difference across the wall is 20 K
  • Given: $A = 2 m², L = 0.2 m, k = 0.5 W/(m·K), ΔT = 20 K$
  • Temperature gradient: $dT/dx = ΔT/L = 20 K / 0.2 m = 100 K/m$
  • Heat transfer rate: $q = -kA(dT/dx) = -(0.5 W/(m·K))(2 m²)(100 K/m) = -100 W$

Conduction in Different Geometries

Planar (Cartesian) Geometry

• Planar geometry refers to heat conduction through a flat surface or wall, where the temperature gradient is perpendicular to the surface
• In planar geometry, the heat transfer rate is calculated using the one-dimensional form of Fourier's law: $q = -kA(dT/dx)$
• Example: Calculate the heat transfer rate through a 3 m × 2 m window with a thickness of 0.01 m and a thermal conductivity of 0.8 W/(m·K), if the inside and outside temperatures are 20°C and 5°C, respectively
  • Given: $A = 6 m², L = 0.01 m, k = 0.8 W/(m·K), T₁ = 20°C, T₂ = 5°C$
  • Temperature gradient: $dT/dx = (T₂ - T₁)/L = (5°C - 20°C) / 0.01 m = -1500 K/m$
  • Heat transfer rate: $q = -kA(dT/dx) = -(0.8 W/(m·K))(6 m²)(-1500 K/m) = 7200 W$

Cylindrical and Spherical Geometries

• Cylindrical geometry involves heat conduction through a hollow or solid cylinder, where the temperature gradient is in the radial direction
  • For cylindrical geometry, the heat transfer rate is given by $q = -2πkL(dT/dr)$, where $L$ is the length of the cylinder and $dT/dr$ is the radial temperature gradient
• Spherical geometry deals with heat conduction through a hollow or solid sphere, with the temperature gradient in the radial direction
  • In spherical geometry, the heat transfer rate is calculated using $q = -4πkr₁r₂(dT/dr) / (r₂ - r₁)$, where $r₁$ and $r₂$ are the inner and outer radii of the sphere, respectively
• Example: Calculate the heat transfer rate through a cylindrical pipe with an inner radius of 0.02 m, an outer radius of 0.03 m, a length of 5 m, and a thermal conductivity of 50 W/(m·K), if the inner and outer surface temperatures are 100°C and 80°C, respectively
  • Given: $r₁ = 0.02 m, r₂ = 0.03 m, L = 5 m, k = 50 W/(m·K), T₁ = 100°C, T₂ = 80°C$
  • Radial temperature gradient: $dT/dr = (T₂ - T₁) / (r₂ - r₁) = (80°C - 100°C) / (0.03 m - 0.02 m) = -2000 K/m$
  • Heat transfer rate: $q = -2πkL(dT/dr) = -2π(50 W/(m·K))(5 m)(-2000 K/m) = 3,141,592.65 W$

Thermal Resistance in Composite Systems

Definition and Calculation of Thermal Resistance

• Thermal resistance is a measure of a material's ability to resist heat flow, analogous to electrical resistance in an electrical circuit
• The thermal resistance of a material is defined as $R = L / (kA)$, where:
  • $L$ is the thickness of the material
  • $k$ is its thermal conductivity
  • $A$ is the cross-sectional area perpendicular to the heat flow
• In composite systems, where multiple materials with different thermal conductivities are layered together, the overall thermal resistance is the sum of the individual thermal resistances: $R_{total} = R₁ + R₂ + ... + Rₙ$

Effect of Thermal Resistance on Heat Transfer

• The presence of thermal resistance in composite systems reduces the overall heat transfer rate compared to a single material with the same total thickness
• To evaluate the effect of thermal resistance on heat transfer:
  1. Calculate the individual thermal resistances
  2. Sum them to find the total resistance
  3. Use the total resistance in the appropriate heat transfer rate equation (e.g., $q = ΔT / R_{total}$ for planar geometry)
• Example: Calculate the heat transfer rate through a composite wall consisting of a 0.1 m layer of concrete (k = 1.4 W/(m·K)) and a 0.05 m layer of insulation (k = 0.04 W/(m·K)). The wall has an area of 10 m² and the temperature difference across it is 30 K.
  • Given: $L₁ = 0.1 m, k₁ = 1.4 W/(m·K), L₂ = 0.05 m, k₂ = 0.04 W/(m·K), A = 10 m², ΔT = 30 K$
  • Thermal resistance of concrete: $R₁ = L₁ / (k₁A) = 0.1 m / (1.4 W/(m·K) × 10 m²) = 0.00714 K/W$
  • Thermal resistance of insulation: $R₂ = L₂ / (k₂A) = 0.05 m / (0.04 W/(m·K) × 10 m²) = 0.125 K/W$
  • Total thermal resistance: $R_{total} = R₁ + R₂ = 0.00714 K/W + 0.125 K/W = 0.13214 K/W$
  • Heat transfer rate: $q = ΔT / R_{total} = 30 K / 0.13214 K/W = 227 W$