Radiation Heat Transfer

Why it matters

Radiation heat transfer is crucial in many engineering applications, such as designing thermal systems in spacecraft, understanding heat loss in buildings, and improving energy efficiency in industrial processes. It allows engineers to predict and control heat transfer without direct contact between objects.

Key ideas

• Radiation: The transfer of energy through electromagnetic waves. Unlike conduction and convection, it does not require a medium.
• Black Body: An idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.
• Emissivity (ε): A measure of a material's ability to emit energy as radiation. It ranges from 0 to 1, with 1 being a perfect emitter (black body).
• Stefan-Boltzmann Law: States that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's absolute temperature.
• View Factor (F): The fraction of radiation leaving a surface that strikes another surface. It depends on the geometry of the surfaces.

Formulas

• Q = ε·σ·A·(T₁⁴ - T₂⁴)

  • Q: Heat transfer rate (W)
  • ε: Emissivity (dimensionless)
  • σ: Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²·K⁴)
  • A: Surface area (m²)
  • T₁, T₂: Absolute temperatures of the surfaces (K)

• Σ_j F_ij = 1 for all surfaces of a complete enclosure, including j = i when self-view is possible.

  • F₁₂, F₁₃, ..., F₁n: View factors from surface 1 to surfaces 2, 3, ..., n (dimensionless)

The net Q expression assumes a diffuse-gray surface fully viewing large isothermal surroundings through a nonparticipating medium. For two finite gray surfaces use the appropriate radiation-resistance network rather than assigning a single emissivity to the whole exchange. View factors satisfy reciprocity A_i F_ij = A_j F_ji. Flat/convex surfaces have F_ii = 0, but concave ones can view themselves.

Worked example

Given: A surface with an area of 2 m², emissivity 0.8, at a temperature of 500 K, is enclosed by large isothermal surroundings at 300 K; take view factor to the surroundings as one.

1. Identify the known values:

   • A = 2 m²
   • ε = 0.8
   • T₁ = 500 K
   • T₂ = 300 K
   • σ = 5.67 × 10⁻⁸ W/m²·K⁴

2. Apply the Stefan-Boltzmann Law: Q = ε·σ·A·(T₁⁴ - T₂⁴)

3. Substitute the values: Q = 0.8 × 5.67 × 10⁻⁸ × 2 × ((500)⁴ - (300)⁴)

4. Calculate the result:

   • 500^4 = 62,500,000,000 K^4.
   • 300^4 = 8,100,000,000 K^4.
   • Difference = 54,400,000,000 K^4.
   • Q = 0.8 × 5.67e-8 × 2 × 5.44e10 = 4935.168 W.

Final answer: 4.935 kW net radiation from the surface.

Common mistakes

• Confusing emissivity with absorptivity.
• Forgetting to convert temperatures to Kelvin.
• Incorrectly calculating view factors.
• Neglecting the fourth power in the Stefan-Boltzmann Law.

For GATE ME

Questions often involve calculating heat transfer rates using the Stefan-Boltzmann Law, determining view factors, and understanding the concept of emissivity. Practice problems involving different geometries and surface interactions.

Quick check

1. What is the emissivity of a perfect black body?
2. Does radiation require a medium for heat transfer?
3. What is the unit of the Stefan-Boltzmann constant?

Answers: 1. 1, 2. No, 3. W/m²·K⁴