Engineering · Heat transfer and thermodynamics

Net radiative heat transfer between gray surfaces

For two diffuse gray surfaces in a two-surface enclosure, net heat transfer follows the fourth-power temperature difference and a resistance network.

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Formula

\dot{Q}_{12} = \frac{σ ( T _{1}^{4} - T _{2}^{4} )}{\frac{1 - ε _{1}}{A _{1} ε _{1}} + \frac{1}{A _{1} F _{12}} + \frac{1 - ε _{2}}{A _{2} ε _{2}}}

Q̇ in W; σ in W/(m²·K⁴); area in m²; temperature in K; ε and F dimensionless

Variables & units

Variables & units
Symbol	Meaning	Unit
Q̇12	Net radiation heat rate from surface 1 to 2	W
$σ$	Stefan–Boltzmann constant	W/(m²·K⁴)
$T1,T2$	Absolute surface temperatures	K
$ε1,ε2$	Hemispherical emissivities	dimensionless
$A1,A2$	Exchange areas	m²
$F12$	View factor from surface 1 to 2	dimensionless

LaTeX

\dot Q_{12}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\varepsilon_1}{A_1\varepsilon_1}+\frac{1}{A_1F_{12}}+\frac{1-\varepsilon_2}{A_2\varepsilon_2}}

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Derivation & background

This resistance relation assumes diffuse gray surfaces with constant emissivity exchanging radiation in a two-surface enclosure; other radiative paths must be excluded or modeled as additional network branches. Temperatures are absolute kelvin. View factors are geometric and obey reciprocity A₁F₁₂=A₂F₂₁. Convection, contact conduction, spectral behavior and nongray surfaces are not included.

Exam blueprint

Validity range

Derivation steps

For two diffuse gray surfaces in a two-surface enclosure, net heat transfer follows the fourth-power temperature difference and a resistance network.

1State the model assumptions and list quantities with units.
2Substitute into Radiation network with view factor F₁₂; convert quantities to SI first.
3Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K. The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2. The resistance has units m⁻².

Rearrangements

Base equation for the target quantity

\dot Q_{12}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\varepsilon_1}{A_1\varepsilon_1}+\frac{1}{A_1F_{12}}+\frac{1-\varepsilon_2}{A_2\varepsilon_2}}

Preserve signs, boundary conditions and units when rearranging.

Task variant

Work through the numerical example for “Net radiative heat transfer between gray surfaces” and check its unit and magnitude.

Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K.
The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2.
The resistance has units m⁻².

Common mistakes

Substituting a related quantity that shares the same symbol.

Use variable definitions and check: Q̇ in W; σ in W/(m²·K⁴); area in m²; temperature in K; ε and F dimensionless.

Exam context

Estimate radiation exchange between furnace surfaces, shields and enclosures.

Formula cluster

Heat transfer and thermodynamics

For two diffuse gray surfaces in a two-surface enclosure, net heat transfer follows the fourth-power temperature difference and a resistance network.

P = σ·A·T⁴

Stefan-Boltzmann Law

Q̇ = hA(Ts − T∞)

Newton’s law of cooling for convection

ΔTlm = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂); Q̇=UAΔTlm

Logarithmic mean temperature difference in a heat exchanger

Worked example

Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K. The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2. The resistance has units m⁻².

Applications

Estimate radiation exchange between furnace surfaces, shields and enclosures.

How do you calculate with Net radiative heat transfer between gray surfaces?

Here is how to work through a typical Net radiative heat transfer between gray surfaces (Radiation network with view factor F₁₂) task step by step:

1
Task
Work through the numerical example for “Net radiative heat transfer between gray surfaces” and check its unit and magnitude.
Solution path
1. Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K.
2. The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2.
3. The resistance has units m⁻².

Frequently asked questions about Net radiative heat transfer between gray surfaces

Why is the temperature difference raised to a power?+

Each diffuse gray surface emits ideally in proportion to σT⁴. Net exchange is therefore driven by σ(T₁⁴−T₂⁴), not a linear temperature difference. Near a reference temperature, it can be linearized approximately, but only locally. Always use kelvin because radiation depends on absolute temperature. Assumes diffuse gray surfaces in a two-surface enclosure. Use kelvin and geometrically reciprocal view factors. Application: Estimate radiation exchange between furnace surfaces, shields and enclosures. The numerical example Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K. The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2. The resistance has units m⁻² anchors the idea in a concrete measurement case.

What does view factor F12 represent?+

F₁₂ is the fraction of diffuse radiation leaving surface 1 that directly reaches surface 2. It depends on geometry and orientation, lies from zero to one and obeys reciprocity A₁F₁₂=A₂F₂₁. Shields or third surfaces change exchange paths, so one F₁₂ term is then insufficient. Assumes diffuse gray surfaces in a two-surface enclosure. Use kelvin and geometrically reciprocal view factors. The corresponding units are Q̇ in W; σ in W/(m²·K⁴); area in m²; temperature in K; ε and F dimensionless. Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K. The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2. The resistance has units m⁻². This checks that the reference quantities are compatible.

Why are there three resistances in the denominator?+

The radiation network contains a surface resistance for the nonblack behavior of surface 1, a geometric space resistance from the view factor and a second surface resistance for surface 2. They act in series for this ideal exchange. Emissivity near one reduces surface resistance; poor view increases space resistance. All terms must refer to consistent areas. Assumes diffuse gray surfaces in a two-surface enclosure. Use kelvin and geometrically reciprocal view factors. Scope: This resistance relation assumes diffuse gray surfaces with constant emissivity exchanging radiation in a two-surface enclosure; other radiative paths must be excluded or modeled as additional network branches. Temperatures are absolute kelvin. View factors are geometric and obey reciprocity A₁F₁₂=A₂F₂₁. Convection, contact conduction, spectral behavior and nongray surfaces are not included. A practical application is Estimate radiation exchange between furnace surfaces, shields and enclosures.

How should I interpret the sign of Q̇12?+

If T₁>T₂, then T₁⁴−T₂⁴ is positive and heat flows from 1 to 2 under this sign convention. Swapping temperatures reverses the sign. The magnitude is net transfer, not gross radiation in both directions. Convection and conduction can add parallel heat paths and must be included separately in a total balance. Assumes diffuse gray surfaces in a two-surface enclosure. Use kelvin and geometrically reciprocal view factors. Check the calculation with this case: Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K. The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2. The resistance has units m⁻². Keep the model limits and units in view.

How can I check the numerical example?+

The example denominator is about 2.583 m⁻² and σ(500⁴−300⁴) about 3084.5 W/m². Dividing gives roughly 1194 W. The denominator has units m⁻², so the numerator W/m² yields watts. Heat flows from the hotter first surface to the cooler second surface. Assumes diffuse gray surfaces in a two-surface enclosure. Use kelvin and geometrically reciprocal view factors. The relation describes For two diffuse gray surfaces in a two-surface enclosure, net heat transfer follows the fourth-power temperature difference and a resistance network. This helps distinguish its target quantity from neighbouring quantities.

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Subject references and further reading

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More formulas: Thermodynamics and heat transfer

\vec q=-k\nabla T, \qquad \dot Q=-kA\frac{dT}{dx}

Fourier’s law of heat conduction

a=\frac{k}{\rho c_p}

Thermal diffusivity

\mathrm{Bi}=\frac{hL_c}{k}

Biot number for transient conduction

\mathrm{Fo}=\frac{a t}{L_c^2}

Fourier number for transient heat conduction

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Net radiative heat transfer between gray surfaces

For two diffuse gray surfaces in a two-surface enclosure, net heat transfer follows the fourth-power temperature difference and a resistance network.

AdvancedUniversity/collegeBachelor foundationsSubject specialization

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Variables & units

Symbol

Meaning

Unit

Q̇12

Net radiation heat rate from surface 1 to 2

σ

Stefan–Boltzmann constant

W/(m²·K⁴)

T1,T2

Absolute surface temperatures

ε1,ε2

Hemispherical emissivities

dimensionless

A1,A2

Exchange areas

m²

F12

View factor from surface 1 to 2

dimensionless

Derivation & background

Exam blueprint

Validity range

Derivation steps

For two diffuse gray surfaces in a two-surface enclosure, net heat transfer follows the fourth-power temperature difference and a resistance network.

1State the model assumptions and list quantities with units.
2Substitute into Radiation network with view factor F₁₂; convert quantities to SI first.
3Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K. The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2. The resistance has units m⁻².

How do you calculate with Net radiative heat transfer between gray surfaces?

Here is how to work through a typical Net radiative heat transfer between gray surfaces (Radiation network with view factor F₁₂) task step by step:

Task

Work through the numerical example for “Net radiative heat transfer between gray surfaces” and check its unit and magnitude.

Solution path

Let A₁=1 m², A₂=2 m², ε₁=0.8, ε₂=0.6, F₁₂=0.5, T₁=500 K and T₂=300 K.
The denominator is 0.25+2+0.3333=2.5833 m⁻²; the numerator is about 3084.5 W/m², giving Q̇₁₂≈1194 W from 1 to 2.
The resistance has units m⁻².

Frequently asked questions about Net radiative heat transfer between gray surfaces

Why is the temperature difference raised to a power?+

What does view factor F12 represent?+

Why are there three resistances in the denominator?+

How should I interpret the sign of Q̇12?+

How can I check the numerical example?+

Retain Net radiative heat transfer between gray surfaces for exams

Create a curated FSRS exam set for Radiation network with view factor F₁₂: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential