Heat Transfer · Chapter 12
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Heat Transfer · Chapter Twelve

Radiation Heat Transfer

Radiative exchange between surfaces depends not just on temperatures and properties, but on geometry — how much of one surface "sees" another. The view factor captures that geometry.

This chapter develops the view factor and its algebra, computes radiation exchange between black and between gray diffuse surfaces using the network method, analyzes radiation shields, and closes with exchange involving emitting and absorbing gases.

Learning objectives

  • Interpret view factor as a geometric fraction and identify self-viewing surfaces.
  • Determine enclosure view factors using summation, reciprocity, symmetry, and superposition.
  • Calculate black-surface exchange with consistent area and sign conventions.
  • Build diffuse-gray radiation networks from surface and space resistances.
  • Assess radiation shields and recognize participating-gas model requirements.
  • Audit view-factor bounds, network units, energy conservation, and limiting cases.
12–1

The View Factor

The view factor \(F_{i\to j}\) is the fraction of diffuse radiation leaving \(i\) that strikes \(j\) directly. It depends on geometry and orientation, not temperature or emissivity. The first index is the emitting area, which is essential because \(F_{ij}\ne F_{ji}\) when areas differ.

View Factor (Shape Factor)

\(0\le F_{ij}\le1\). A convex or planar surface cannot see itself, so \(F_{ii}=0\); a concave surface generally has \(F_{ii}>0\). Tabulated/chart values require the exact dimension ratios and alignment stated.

Physical meaning

The product \(A_iF_{ij}\) is the geometric exchange area. Reciprocity says the same ray paths connect the pair in either direction even though each directional fraction can differ.

Problem-solving strategy

  1. Sketch and number every surface that closes the enclosure, including openings.
  2. Mark convex surfaces and obvious zero or unity factors.
  3. Use a chart/integration only for the minimum independent factors, with correct dimension ratios.
  4. Complete the matrix using summation, reciprocity, symmetry, and superposition.
  5. Audit every factor against \([0,1]\), each row sum against one, and each reciprocal pair.
12–2

View Factor Relations

View-factor algebra often eliminates most integrations. Summation applies only after the modeled enclosure is closed; an aperture must be represented as a surface. Superposition may add adjacent target surfaces or subtract a blocking subdivision.

Reciprocity and Summation Rules

The reciprocity rule relates the two directions of exchange between a pair of surfaces through their areas. The summation rule states that the view factors from one surface of an enclosure to all surfaces (including itself) must total one, since all radiation must land somewhere.

\[ A_i F_{i\to j} = A_j F_{j\to i}, \qquad \sum_{j=1}^{N} F_{i\to j} = 1 \]

Superposition and crossed-strings (2-D)
\[ F_{i\to(j+k)} = F_{i\to j} + F_{i\to k}, \qquad F_{i\to j} = \frac{\sum(\text{crossed strings}) - \sum(\text{uncrossed strings})}{2\,L_i} \]
EXAMPLE 1Complete a three-surface enclosure matrix
Problem. A closed three-surface enclosure has convex surfaces 1 and 2 with \(A_1=1.00\) m², \(A_2=3.00\) m², and a concave remainder \(A_3=4.00\) m². A geometric calculation gives \(F_{12}=0.600\). Determine all nine view factors.

Strategy

Use convexity for diagonal zeros, then summation and reciprocity; no further geometric integration is needed.

Givens

  • Closed enclosure; \(A_1=1\), \(A_2=3\), \(A_3=4\) m²; \(F_{12}=0.600\); surfaces 1 and 2 convex.

Physical/model interpretation

Surface 3 can see itself because it wraps around the convex inserts. Factors are fractions leaving the row surface.

Numbered derivation
1. Convexity gives \(F_{11}=F_{22}=0\); row 1 gives \(F_{13}=1-0.600=0.400\).
2. Reciprocity gives \(F_{21}=A_1F_{12}/A_2=0.200\).
3. Row 2 gives \(F_{23}=1-0.200=0.800\).
4. Reciprocity gives \(F_{31}=A_1F_{13}/A_3=0.100\), \(F_{32}=A_2F_{23}/A_3=0.600\), then \(F_{33}=0.300\).

Unit/sign/limit checks

Units: Each view factor is a ratio of radiant power to radiant power and is dimensionless. Reciprocity also balances units explicitly: \(A_iF_{ij}\) and \(A_jF_{ji}\) both have units of m².

Signs: Every computed entry is nonnegative and no entry exceeds one, as required for fractions of leaving radiation. In particular, the positive \(F_{33}=0.300\) is physically allowed because surface 3 is concave, whereas negative self-view would be impossible.

Plausibility, closure, and reciprocity: All magnitudes remain within the physical bounds \(0\le F_{ij}\le1\). The row sums are \(0+0.600+0.400=1.000\), \(0.200+0+0.800=1.000\), and \(0.100+0.600+0.300=1.000\). Independent pair checks give \(A_1F_{12}=0.600=A_2F_{21}\), \(A_1F_{13}=0.400=A_3F_{31}\), and \(A_2F_{23}=2.400=A_3F_{32}\), all in m².

Final answer: \(\boxed{[F_{ij}]=\begin{bmatrix}0&0.6&0.4\\0.2&0&0.8\\0.1&0.6&0.3\end{bmatrix}}\).

Engineering insight

The nonzero \(F_{33}\) is not an error; it is the signature of a concave enclosing surface.

12–3

Radiation Heat Transfer: Black Surfaces

Black surfaces have \(J_i=E_{bi}=\sigma T_i^4\), so there is no reflected component. For a pair, positive \(\dot Q_{1\to2}\) means net transfer from 1 to 2; absolute kelvin temperatures are mandatory.

Net exchange between two black surfaces
\[ \dot{Q}_{1\to 2} = A_1 F_{12}\,\sigma\left(T_1^4 - T_2^4\right) \]
For a many-surface black enclosure, sum \(A_iF_{ij}\sigma(T_i^4-T_j^4)\) over all \(j\); do not double-count pair exchange.
EXAMPLE 2Black refractory panel viewing a cooler opening
Problem. A black \(0.500\) m² refractory panel at \(700\) K has view factor \(0.350\) to a black furnace opening represented at \(400\) K. Find the net pair exchange from panel to opening.

Strategy

Both surfaces are black and the pair view factor is known, so direct pair exchange is the minimum valid model.

Givens

  • \(A_1=0.500\) m², \(F_{12}=0.350\), \(T_1=700\) K, \(T_2=400\) K.

Physical/model interpretation

The opening is an equivalent radiative surface. Positive direction is from hot panel to opening.

Numbered derivation
1. \(\dot Q_{12}=A_1F_{12}\sigma(T_1^4-T_2^4)\).
2. \(A_1F_{12}=0.175\) m² and \(T_1^4-T_2^4=2.145\times10^{11}\) K&sup4;.
3. The blackbody driving flux is \((5.670\times10^{-8})(2.145\times10^{11})=12.16\) kW/m².
4. \(\dot Q_{12}=(0.175)(12.16)=2.13\) kW.

Unit/sign/limit checks

Units: \(A_1F_{12}\) has units m², and \(\sigma(T_1^4-T_2^4)\) has units W/m²; their product is W. Numerically, \((0.175\ {\rm m^2})(12.16\ {\rm kW/m^2})=2.13\ {\rm kW}\).

Signs: \(700>400\), so the positive result is outward from panel.

Limit: \(F_{12}\to0\) or \(T_1\to T_2\) makes pair exchange vanish.

Final answer: \(\boxed{\dot Q_{1\to2}=2.13\ {\rm kW}}\).

Engineering insight

This is only one radiative branch; the panel's total load requires exchange with every visible surface plus convection and conduction.

12–4

Radiation Heat Transfer: Diffuse, Gray Surfaces

For an opaque diffuse-gray surface, \(J_i=\varepsilon_iE_{bi}+(1-\varepsilon_i)G_i\). The network analogy uses blackbody potential \(E_{bi}\), radiosity nodes \(J_i\), surface resistance \((1-\varepsilon_i)/(A_i\varepsilon_i)\), and space resistance \(1/(A_iF_{ij})\). These resistances have units m\(^{-2}\), so dividing a W/m² potential difference by resistance gives W.

Surface and space resistances
\[ R_{\text{surface},i} = \frac{1-\varepsilon_i}{A_i\varepsilon_i}, \qquad R_{\text{space},ij} = \frac{1}{A_i F_{ij}} \]
Two-surface enclosure
\[ \dot{Q}_{12} = \frac{\sigma(T_1^4 - T_2^4)}{\dfrac{1-\varepsilon_1}{A_1\varepsilon_1} + \dfrac{1}{A_1 F_{12}} + \dfrac{1-\varepsilon_2}{A_2\varepsilon_2}} \]
Large parallel plates
\[ \dot{Q}_{12} = \frac{A\,\sigma(T_1^4 - T_2^4)}{\dfrac{1}{\varepsilon_1} + \dfrac{1}{\varepsilon_2} - 1} \]
EXAMPLE 3Gray panels across a vacuum dryer
Problem. Large parallel panels in a vacuum dryer are at \(650\) K and \(320\) K with emissivities \(0.75\) and \(0.25\). Find net radiative heat flux and the fraction of blackbody exchange transmitted by the gray-surface network.

Strategy

Large parallel diffuse-gray plates have \(F_{12}=1\); use the closed two-surface relation rather than a blackbody shortcut.

Givens

  • \(T_1=650\) K, \(T_2=320\) K, \(\varepsilon_1=0.75\), \(\varepsilon_2=0.25\).

Physical/model interpretation

Edge effects and participating gas are absent. Positive flux is from hot panel 1 to cool panel 2.

Numbered derivation
1. \(q''=\sigma(T_1^4-T_2^4)/(1/\varepsilon_1+1/\varepsilon_2-1)\).
2. Denominator \(D=1/0.75+1/0.25-1=4.333\).
3. Blackbody driving flux is \(\sigma(650^4-320^4)=9.527\) kW/m².
4. \(q''=9.527/4.333=2.20\) kW/m², and \(q''/q''_b=1/D=0.231\).

Unit/sign/limit checks

Units: the emissivity denominator is dimensionless, retaining W/m².

Signs: Defining \(q''_{1\to2}\) positive from panel 1 to panel 2, \(650^4-320^4>0\) produces \(+2.20\) kW/m², consistent with heat leaving the hotter panel. Reversing the surface order would reverse the temperature difference and give the same magnitude with a negative sign.

Limit: \(0.231<1\), and setting both emissivities to one would recover blackbody exchange.

Final answer: \(\boxed{q''_{1\to2}=2.20\ {\rm kW/m^2},\quad q''/q''_b=0.231}\).

Engineering insight

The cooler panel's low emissivity dominates the resistance; measured finish and oxidation state are critical input data.

12–5

Radiation Shields and the Radiation Effect

A thin opaque shield reaches a steady temperature at which net heat into it is zero. It creates two exchange gaps and two additional surface interactions. Low emissivity, not thickness, drives the ideal radiative benefit; supports, seams, gas conduction, and contact bridges can erase it.

One radiation shield between parallel plates
\[ \dot{Q}_{12,\text{shield}} = \frac{A\,\sigma(T_1^4 - T_2^4)}{\left(\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_3}-1\right) + \left(\dfrac{1}{\varepsilon_3}+\dfrac{1}{\varepsilon_2}-1\right)} \]
The \(N+1\) reduction applies only to ideal identical-emissivity, large parallel surfaces. Otherwise assemble each gap explicitly.
Course Note

The power of shields comes from low emissivity, not thickness — a sheet of aluminum foil (\(\varepsilon \approx 0.05\)) works far better than a thick black plate. This is why spacecraft are wrapped in shiny multilayer insulation rather than bulky blankets.

12–6

Radiation Exchange with Emitting and Absorbing Gases

Monatomic and symmetric diatomic gases are often nearly transparent, while CO₂, H₂O, CO, and hydrocarbons participate in bands. Gas emissivity depends on temperature, partial pressure, composition, and mean beam length; it is not a material constant.

Gas Emissivity

For chart methods, compute species partial pressures and \(pL\), choose mean beam length for the enclosure, apply total-pressure and band-overlap corrections, and evaluate gas absorptivity at the source temperature as directed. Strong gradients or soot may require spectral/radiative-transfer simulation.

Net radiation from a gas to its enclosure
\[ \dot{Q}_{\text{net}} = \varepsilon_g\,\sigma A_s\,T_g^4 - \alpha_g\,\sigma A_s\,T_s^4 \]
This compact form requires a compatible enclosure idealization and chart-derived \(\varepsilon_g,\alpha_g\); generally \(\alpha_g(T_s)\ne\varepsilon_g(T_g)\) when source temperatures differ.

Chapter recap

  • View factors encode geometry and must satisfy bounds, summation, and reciprocity.
  • Black exchange uses geometric exchange area and absolute-temperature potentials.
  • Diffuse-gray networks add surface reflection resistance to space resistance.
  • Shields add radiative resistance; participating gases require composition and path data.

Before moving on

You should be able to complete a view-factor matrix, build and audit a two-surface network, determine heat-flow direction, and explain when shields or gas-radiation charts are valid.

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