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Heat transfer and thermal systems

Exchange between large diffuse grey plates

Net radiant flux between two large parallel opaque diffuse grey plates with opposing faces at T1 and T2 and a transparent gap.

Ref
HTS-31
Rev
v1
Field
Heat transfer and thermal systems
Result
q″ [W/m²]

Between two large parallel opaque diffuse grey plates, the net radiant flux is q″ = σ(T₁⁴ − T₂⁴)/(1/ε₁ + 1/ε₂ − 1). The denominator is the sum of the surface resistances. Black plates (ε = 1) give the largest exchange, σ(T₁⁴ − T₂⁴). Positive flux goes from plate 1 to plate 2, and equal temperatures give none.

Net radiant flux between two large parallel opaque diffuse grey plates with opposing faces at T1 and T2 and a transparent gap. Positive flux goes from plate 1 to plate 2.

Calculator

Solve for
W/m²
K
1002,000
K
1002,000
0.051
0.051
  • Flux between black plates, σ(T₁⁴ − T₂⁴)992.3 W/m²
  • Ratio to the black-plate exchange, 1/(1/ε₁ + 1/ε₂ − 1)0.6667
  • Equivalent radiation coefficient, σ(T₁ + T₂)(T₁² + T₂²)/(1/ε₁ + 1/ε₂ − 1)6.615 W/(m² K)

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Net radiant flux (W/m²) current value 661.5 W/m²
T1Temperature of plate 1Key driver
−302605,000
T2Temperature of plate 2
−604,000964
ε1Emissivity of plate 1
49794
ε2Emissivity of plate 2
49794
  • Between plates at 400 K and 300 K the net exchange is 661.5 W/m², 66.7 % of the black-plate value.
  • The surface resistances add: for the emissivities used, they cut radiative exchange to 0.667 of the black-body case.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (W/m²)Local stepChange in result
T1 Temperature of plate 1100 to 2,000 K−302.4604,500604,800+10%+67.9 %
T2 Temperature of plate 2100 to 2,000 K−603,900964604,800+10%−21.5 %
ε1 Emissivity of plate 10.05 to 1 49793.9744.8+10%+8.2 %
ε2 Emissivity of plate 20.05 to 1 49793.9744.8+10%+8.2 %

Sweep of Emissivity of plate 1

Emissivity of plate 1Net radiant flux (W/m²)
0.0549
0.1688160.7
0.2875266.2
0.4062366
0.525460.5
0.6438550.2
0.7625635.5
0.8812716.6
1793.9

Sweep one variable

Net radiant flux (W/m²)Hover, touch, or use the arrow keys on the chart
0200400600800 0.20.40.60.81 Emissivity of plate 1 now 661.5

Variables and how each one contributes

q″

Net radiant flux

Result · W/m²

What the equation gives you. Everything else feeds it.

T1

Temperature of plate 1

K · typical 100 to 2,000

The temperature of plate 1, in K. Raises the net flux as the fourth power. Raising it by 10 % (to 440) moves the result by +67.9 %, from 661.5 to 1,111 W/m².

T2

Temperature of plate 2

K · typical 100 to 2,000

The temperature of plate 2, in K. Raises the opposing term and lowers the net flux. Raising it by 10 % (to 330) moves the result by −21.5 %, from 661.5 to 519.4 W/m².

ε1

Emissivity of plate 1

· typical 0.05 to 1

The emissivity of plate 1. A higher emissivity raises the flux. Raising it by 10 % (to 0.88) moves the result by +8.2 %, from 661.5 to 715.8 W/m².

ε2

Emissivity of plate 2

· typical 0.05 to 1

The emissivity of plate 2. A higher emissivity raises the flux. Raising it by 10 % (to 0.88) moves the result by +8.2 %, from 661.5 to 715.8 W/m².

Worked example with the values above

  1. Fourth-power difference: 400⁴ − 300⁴ = 1.75×1010 K⁴; × σ = 992.32 W/m².
  2. Resistance term: 1/ε₁ + 1/ε₂ − 1 = 1.5.
  3. q″ = 992.32 / 1.5 = 661.54 W/m².

Assumptions and limits

  • Opaque, diffuse and grey surfaces, isothermal with uniform properties.
  • A transparent, non-participating gap, and absolute temperatures in kelvin.

Common mistakes

  • Using °C in the fourth-power terms.
  • Using the plate formula when the surfaces are not large compared with the gap.
  • Forgetting that an emissivity of 1 means a perfect black surface and 0 a perfect reflector.

Statistics and uncertainty

Emissivity values are uncertain by 10 % to 30 % for real surfaces, and temperature errors are multiplied by four in the fourth-power law, so a 2 % error in absolute temperature gives roughly 8 % in flux.

Where it is used

  • Furnaces, cryostats, ovens and spacecraft thermal design.
  • Choosing shield layers and estimating radiant exposure.

Maths you need

  • Stefan–Boltzmann law: E = σT⁴, with σ = 5.670×10⁻⁸ W/(m² K⁴)
  • Emissivity: ε, the ratio of emission to that of a blackbody
  • View factor: the fraction of radiation leaving one surface that reaches another

Related formulas

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