View factors and radiation exchange between surfaces

View factor definition and algebra, radiosity and the resistance network, two-surface enclosures, concentric cylinders and radiation shields.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

In a furnace the tubes, the refractory walls and the flame exchange radiation with each other; in a cryogenic vessel, the inner tank exchanges radiation with the outer shell across a vacuum; in a reflective insulation system, thin shields cut radiation by orders of magnitude. All of these are calculated with view factors and a radiation resistance network, which turn a geometric problem into an electrical-circuit problem.

Key ideas

View factor F_ij. The fraction of the radiation leaving surface i that strikes surface j directly. It is purely geometric (sizes, distances, orientations); it does not depend on temperatures or emissivities, provided surfaces emit and reflect diffusely and radiosity is uniform over each surface. Also called shape or configuration factor. 0 ≤ F_ij ≤ 1.

View factor rules.

  • Reciprocity: A_i·F_ij = A_j·F_ji.
  • Summation: in an enclosure of N surfaces, Σ_j F_ij = 1 for each i (all radiation leaving i must land somewhere, including on i itself).
  • Self-viewing: F_ii = 0 for a flat or convex surface; F_ii > 0 for a concave one (e.g. the inside of a cylinder or hemisphere).
  • Superposition: the view factor to a composite surface is the sum of the view factors to its parts (F_i(j+k) = F_ij + F_ik); the reverse is not true without reciprocity.
  • Symmetry: identical geometric arrangements have equal view factors.
  • Crossed-strings (Hottel) for long two-dimensional geometries: F₁₂ = [(sum of crossed strings) − (sum of uncrossed strings)]/(2 × L₁). More complex cases use charts or analytical formulas (aligned parallel rectangles, perpendicular rectangles with a common edge, coaxial discs).

Useful results.

  • Small (convex) body 1 inside a large enclosure 2: F₁₂ = 1.
  • Infinite parallel plates: F₁₂ = F₂₁ = 1.
  • Long concentric cylinders or concentric spheres: F₁₂ = 1, F₂₁ = A₁/A₂, F₂₂ = 1 − A₁/A₂.
  • Two long plates of equal width w meeting at a right angle: F₁₂ = 1 − sin 45° = 0.293.

Radiosity and the network method (diffuse gray surfaces). Radiosity J is all radiation leaving a surface (emitted + reflected). Each surface has a surface resistance (1 − ε)/(εA) between its black-body potential E_b = σT⁴ and J; each pair of surfaces has a space resistance 1/(A_i·F_ij) between J_i and J_j. A black surface has zero surface resistance. A reradiating (adiabatic, e.g. refractory) surface has zero net heat, so its radiosity floats; it is a node with no connection to E_b.

Two-surface enclosure. Combining the three series resistances gives q₁₂ = σ(T₁⁴ − T₂⁴)/[(1 − ε₁)/(ε₁A₁) + 1/(A₁F₁₂) + (1 − ε₂)/(ε₂A₂)]. Special cases follow directly (parallel plates, concentric cylinders, small body in a large room).

Radiation shields. A thin, low-emissivity sheet placed between two surfaces adds two surface resistances and one space resistance. With N shields between infinite parallel plates of equal emissivity, the heat flux falls to 1/(N + 1) of the unshielded value; low-ε shields cut it far more.

Formulas

A_i·F_ij = A_j·F_ji; Σ_j F_ij = 1 Reciprocity and summation; A (m²), F (–).

J = ε·E_b + (1 − ε)·G Radiosity (W/m²); G irradiation (W/m²); opaque gray diffuse surface.

q_i = (E_bi − J_i)/[(1 − ε_i)/(ε_i·A_i)] Net radiation leaving surface i (W).

q_ij = (J_i − J_j)/[1/(A_i·F_ij)] Net exchange between i and j (W).

q₁₂ = σ·(T₁⁴ − T₂⁴)/[(1 − ε₁)/(ε₁·A₁) + 1/(A₁·F₁₂) + (1 − ε₂)/(ε₂·A₂)] Two-surface gray enclosure; T in K.

q″ = σ·(T₁⁴ − T₂⁴)/(1/ε₁ + 1/ε₂ − 1) Infinite parallel plates (W/m²).

q = σ·A₁·(T₁⁴ − T₂⁴)/[1/ε₁ + ((1 − ε₂)/ε₂)·(r₁/r₂)] Long concentric cylinders (r₁/r₂) or concentric spheres ((r₁/r₂)²).

q″_shielded = σ·(T₁⁴ − T₂⁴)/[(1/ε₁ + 1/ε₂ − 1) + Σ(1/ε_s1 + 1/ε_s2 − 1)] Parallel plates with shields; ε_s1, ε_s2 the two faces of each shield.

F₁₂ = [(crossed) − (uncrossed)]/(2·L₁) Hottel crossed-strings, 2-D geometries.

Worked examples

Example 1 (standard). A long pipe, 100 mm OD (ε₁ = 0.8, 500 K), runs concentrically inside a 300 mm ID duct (ε₂ = 0.6, 300 K). The annulus is evacuated (no convection). Find the radiation exchange per metre, and the view factors F₂₁ and F₂₂.

  1. A₁ = π × 0.1 × 1 = 0.3142 m²; r₁/r₂ = 1/3; F₁₂ = 1.
  2. Denominator: 1/ε₁ + ((1 − ε₂)/ε₂)·(r₁/r₂) = 1.25 + 0.6667 × 0.3333 = 1.4722.
  3. T₁⁴ − T₂⁴ = 6.25 × 10¹⁰ − 0.81 × 10¹⁰ = 5.44 × 10¹⁰ K⁴.
  4. q = 5.67 × 10⁻⁸ × 0.3142 × 5.44 × 10¹⁰/1.4722 = 658 W per metre.
  5. Reciprocity: F₂₁ = A₁F₁₂/A₂ = 0.1/0.3 = 0.333; summation: F₂₂ = 1 − 0.333 = 0.667.

Answer: q ≈ 658 W/m; F₂₁ = 1/3, F₂₂ = 2/3.

Example 2 (GATE level). Two large parallel plates at 800 K and 400 K both have ε = 0.8. (a) Find the net radiation flux. (b) A thin shield with ε = 0.05 on both faces is placed between them. Find the new flux, the percentage reduction and the shield temperature.

(a)

  1. q″ = σ·(T₁⁴ − T₂⁴)/(1/ε₁ + 1/ε₂ − 1); T₁⁴ − T₂⁴ = 4.096 × 10¹¹ − 0.256 × 10¹¹ = 3.840 × 10¹¹ K⁴.
  2. Denominator 1/0.8 + 1/0.8 − 1 = 1.5; q″ = 5.67 × 10⁻⁸ × 3.840 × 10¹¹/1.5 = 14 515 W/m². (b)
  3. Shield adds 1/0.05 + 1/0.05 − 1 = 39; total = 1.5 + 39 = 40.5.
  4. q″ = 5.67 × 10⁻⁸ × 3.840 × 10¹¹/40.5 = 537.6 W/m², i.e. 3.7 % of the original — a 96.3 % reduction.
  5. By symmetry (equal resistances on both sides of the shield), σT_s⁴ is midway: T_s⁴ = (800⁴ + 400⁴)/2 = 2.176 × 10¹¹, so T_s = 683 K.

Answer: (a) 14.5 kW/m²; (b) 538 W/m² (about 96 % reduction), shield at about 683 K.

Common mistakes

  • Assuming F₁₂ = F₂₁; they are equal only when A₁ = A₂.
  • Forgetting F_ii for concave surfaces (the outer cylinder sees itself).
  • Using view factors for finite plates as if they were infinite (F = 1 only for infinite plates or enclosed bodies).
  • Using °C in T⁴, or (T₁ − T₂)⁴.
  • Making emissivity part of the view factor; ε enters only through surface resistances.
  • Counting a shield as adding only one surface resistance; it adds two plus a space resistance.

For GATE CH

Expect view-factor algebra (reciprocity, summation, superposition) for enclosures such as a hemisphere over a disc, cylinders in ducts, cubes and triangular ducts; net exchange between parallel plates, concentric cylinders and spheres; shield problems; and occasionally a three-surface network with a reradiating wall. Practise drawing the network first, and checking that all view factors from each surface sum to 1.

Quick check

  1. A hemisphere (2) covers a flat disc (1). Find F₁₂ and F₂₁.
  2. Inside a cube, what is the view factor from one face to the opposite face if the face-to-adjacent-face value is 0.2?
  3. One shield between two plates, all surfaces with the same ε: by what fraction is the flux reduced?
  4. Is F₁₁ zero for the inside of a sphere?
  5. What is the surface resistance of a black surface?

Answers: 1. F₁₂ = 1; F₂₁ = πr²/(2πr²) = 0.5. 2. 1 − 4 × 0.2 = 0.2. 3. It is halved. 4. No, F₁₁ = 1 for a sphere's inner surface seeing itself. 5. Zero.

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is a view factor in the context of radiation heat transfer?Concept

    A view factor, also known as a configuration factor or shape factor, is a dimensionless quantity that represents the fraction of radiation leaving a surface that directly reaches another surface. It depends on the geometry of the surfaces and their relative orientation.

  2. 2.Explain the significance of view factors in radiation exchange between surfaces.Concept

    View factors are crucial in calculating the net radiative heat exchange between surfaces. They help determine how much radiation emitted by one surface is intercepted by another, which is essential for accurate thermal analysis in systems involving radiative heat transfer.

  3. 3.How are view factors used in the analysis of radiation heat transfer in enclosures?Application

    In enclosures, view factors are used to calculate the radiative heat exchange between surfaces. By knowing the view factors, one can set up and solve the radiative heat balance equations for each surface, allowing for the determination of heat transfer rates and surface temperatures.

  4. 4.Why is the reciprocity relation important in view factor calculations?Application

    The reciprocity relation states that the product of the view factor from surface A to surface B and the area of surface A is equal to the product of the view factor from surface B to surface A and the area of surface B. This relation simplifies the calculation of view factors and ensures consistency in radiative heat transfer analysis.

  5. 5.When is the view factor between two parallel plates equal to 1?Application

    Only when the plates are effectively infinite in both directions compared with their spacing, so that every ray leaving one plate must hit the other. Two long parallel strips of finite width w at spacing H are not enough: for equal strips F₁₂ = √(1 + (H/w)²) − H/w, which is less than 1 and falls as the spacing grows. In practice F ≈ 1 when the plate dimensions are much larger (say ten times or more) than the gap.

  6. 6.Describe how the summation rule is applied to view factors in an enclosure.Concept

    In a closed enclosure of N surfaces, all radiation leaving surface i (emitted plus reflected) must strike some surface of the enclosure, so Σ_j F_ij = 1, including the self-view factor F_ii (zero for flat or convex surfaces, non-zero for concave ones). Together with reciprocity A_iF_ij = A_jF_ji, it lets you find unknown view factors from a few known ones; an N-surface enclosure needs only N(N − 1)/2 factors to be found directly. For example, for a disc covered by a hemisphere, F₁₂ = 1 gives F₂₁ = A₁/A₂ = 0.5 and F₂₂ = 0.5.

  7. 7.How does surface emissivity affect radiation exchange between surfaces?Application

    Surface emissivity affects the amount of thermal radiation emitted by a surface. A higher emissivity means more radiation is emitted, which can increase the net radiation exchange between surfaces. Emissivity is a critical factor in determining the radiative heat transfer rate.

  8. 8.Two long plates, each 2 m wide, meet at a common edge at right angles. Find the view factor from one to the other.Numerical

    For long (two-dimensional) plates the crossed-strings method gives F₁₂ = (w₁ + w₂ − √(w₁² + w₂²))/(2w₁). With w₁ = w₂ = 2 m: F₁₂ = (4 − 2.828)/4 = 0.293. Equivalently F₁₂ = 1 − sin(α/2) for equal plates at angle α = 90°. By symmetry F₂₁ is also 0.293.

  9. 9.What is the effect of surface orientation on view factors?Application

    Surface orientation significantly affects view factors. If surfaces are oriented such that they face each other directly, the view factor is higher. Conversely, if they are oriented away from each other, the view factor decreases, as less radiation is exchanged directly between them.

  10. 10.Two long parallel strips, each 2 m wide, face each other 1 m apart. Calculate the view factor from one to the other.Numerical

    These are not infinite plates, so F₁₂ < 1. For two long, directly opposed strips of equal width w at spacing H, crossed strings give F₁₂ = √(1 + (H/w)²) − H/w. With H/w = 0.5: F₁₂ = √1.25 − 0.5 = 1.118 − 0.5 = 0.618. About 38 % of the radiation leaving one strip escapes through the open sides.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousRadiation laws: Stefan-Boltzmann, Kirchhoff and WienNext Overall heat transfer coefficient and fouling factors