Radiation Network Simulator Back
Heat Transfer Simulator

Radiation Network Simulator — Surface & Space Resistances

Visualize radiative heat transfer between two gray surfaces with the same resistance network you would draw for an electric circuit. Vary temperatures and emissivities to see why heat flow drops on non-black surfaces.

Parameters
Surface 1 temperature T_1
K
Surface 2 temperature T_2
K
Emissivity ε_1
Emissivity ε_2

Assumes A_1 = A_2 = 1 m² and view factor F_12 = 1 (infinite parallel plates).

Results
Radiative heat flow Q_12
Blackbody limit Q_BB
Gray efficiency η = Q/Q_BB
Total resistance R_tot
Radiation Resistance Network

Left node = σT_1^4 / right node = σT_2^4 / surface resistances R_1, R_2 and space resistance R_12 connect in series

Theory & Key Formulas

The radiative heat flow is computed by treating σT^4 as a driving potential and connecting surface and space resistances in series, just like an electric circuit.

Surface resistance (reflection loss when ε < 1):

$$R_s = \frac{1-\varepsilon}{A\,\varepsilon}$$

Space resistance (geometric loss controlled by the view factor F_12):

$$R_{12} = \frac{1}{A_1\,F_{12}}$$

Heat flow between the two surfaces. σ = 5.670×10⁻⁸ W/(m²·K⁴):

$$Q_{12} = \frac{\sigma\,(T_1^4 - T_2^4)}{R_1 + R_{12} + R_2}$$

Blackbody limit (ε_1 = ε_2 = 1) and gray efficiency:

$$Q_{BB} = \sigma A (T_1^4 - T_2^4),\quad \eta = \frac{Q_{12}}{Q_{BB}}$$

For infinite parallel plates (F_12 = 1, A_1 = A_2 = A) this simplifies to η = 1/(1/ε_1 + 1/ε_2 - 1).

What is the Radiation Network Simulator

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I heard heat can travel through vacuum — that's radiation, right? But why do we call it a "network"?
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Yes, radiation is electromagnetic waves so it travels in vacuum too. Roughly speaking, when you map radiative heat transfer onto an electric circuit, the analysis becomes much clearer. "Voltage" is σT^4, "resistors" are surface and space resistances, "current" is the heat flow Q — that mapping is exactly what we call a "radiation network".
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So what's the difference between a "surface resistance" and a "space resistance"?
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The surface resistance R_s = (1−ε)/(Aε) represents the "reflection loss" caused by the surface not being a blackbody. A mirror-like surface with low ε reflects radiation back to itself. The space resistance R_12 = 1/(A_1·F_12) represents the "geometric loss" set by the view factor F_12. Try sliding ε down in the simulator — the surface resistance jumps and the heat flow drops sharply.
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I keep hearing the temperature shows up to the fourth power. When I move the slider, Q_12 takes off really quickly.
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That's the strength of the Stefan-Boltzmann law E_b = σT^4. Push T_1 from 800 K up to 1600 K and T^4 multiplies by 16. Even with the same "800 K gap", the radiation at 800 K vs 0 K is orders of magnitude weaker than 1600 K vs 800 K. That is why in furnaces and combustors radiation dominates over convection.
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So when the gray efficiency drops to 66.7%, where does the "lost" energy go?
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It does not vanish. The energy reflected back at the surface resistance simply does not contribute to the net heat flow. Thermos flasks and MLI (multi-layer insulation) actually use this on purpose. Stack many aluminum-coated foils with very low ε, deliberately make the surface resistance huge, and you block radiation. Try setting ε_1 = ε_2 = 0.05 in the simulator — Q drops to less than a tenth.

Frequently Asked Questions

It is set by the material and surface condition. Polished aluminum is 0.04 to 0.07, oxidized aluminum 0.20 to 0.30, oxidized steel 0.60 to 0.85, white paint 0.80 to 0.95, and a near-black coating above 0.95. Even the same material varies strongly with temperature and oxidation, so use measured values or trusted handbook data. When the wavelength dependence is strong, you have to consider spectral emissivity.
It is the approximation that emissivity does not depend on wavelength, and it makes the analysis very simple. For most industrial materials it holds well in the mid-infrared (the radiation peak of typical 300 to 2000 K industrial temperatures). Materials with strong spectral signatures such as glass and ceramics, or selective absorber coatings on solar collectors, must be treated band by band.
It depends strongly on temperature. Near room temperature the natural-convection coefficient is around 5 to 10 W/(m^2*K) and the equivalent radiative coefficient is comparable, so the two compete. Above 500 K radiation starts to dominate, and above 1000 K it becomes several to ten times larger than convection. At low temperatures, conduction and convection dominate and radiation is sometimes neglected.
Three are common. The surface-to-surface (S2S) method precomputes the view-factor matrix and solves the same network as this tool — fast, optimal when the medium is transparent. The discrete-ordinates (DO) method, a finite-volume approach that handles absorption and scattering, is used for participating media such as combustion chambers. Monte-Carlo ray tracing handles complex geometries and specular reflection well, and is used for verification and high-accuracy work.

Real-World Applications

Furnace and combustor design: In steel furnaces, glass-melting tanks and boiler fireboxes, radiation accounts for 60 to 90% of the heat transfer. Solving the radiation network from burner to wall and from wall to load with high fidelity directly drives fuel consumption, temperature uniformity and refractory life. The S2S and Monte-Carlo radiation models in CFD codes are essentially this network scaled up.

Spacecraft thermal design: Satellites in vacuum have no convection, so almost all heat rejection is radiative. The standard design wraps equipment in MLI (multi-layer insulation, a dozen aluminized films at ε ≈ 0.05) to inflate the surface resistance, and dumps heat to deep space through high-emissivity radiator panels. Choosing surface treatments — white paint, silver coatings, OSR — controls the entire heat balance.

Building windows and insulation: Low-E (low-emissivity) glass coats the interior surface with a thin film of tin oxide or similar to drop the emissivity from 0.85 to about 0.10, dramatically cutting radiative exchange between indoors and outdoors. In double-pane windows the argon gap suppresses convection while the Low-E coating suppresses radiation, lowering the U-value to less than a third of single glazing.

Infrared thermography calibration: A thermal camera converts surface radiative emission into temperature, but the result depends strongly on ε. At the same temperature, black paint (ε ≈ 0.95) emits nearly ten times the signal of polished stainless (ε ≈ 0.10), so practitioners apply emissivity tape or known-emissivity paint for calibration. The change you see when moving the ε sliders in this tool is exactly this source of error.

Common Misconceptions and Cautions

The most common misconception is to oversimplify as "a black object absorbs more heat, so it gets hotter". In the visible band, "black equals high absorptivity" is indeed true, but for heat transfer what matters is the emissivity in the mid-infrared (the radiation peak at typical industrial temperatures), and this does not match the visible color. White paint, for instance, reflects visible light but behaves nearly like a blackbody with ε ≈ 0.9 in the mid-infrared. Remember that the ε slider in the simulator is the "emissivity at the relevant temperature band", not the visible color.

The next common error is to treat the view factor F_12 as "1 if the partner is in sight". F_12 is the fraction of radiation leaving surface 1 that reaches surface 2, and it equals 1 only in fully enclosed configurations such as infinite parallel plates or the inside of a concentric cylinder. For a general convex shape F_12 < 1, and the missing fraction (1 − F_12) goes to other surfaces or the surroundings. This tool simplifies to infinite parallel plates (F_12 = 1) so the network solution is very simple, but in real machines accurate view factors govern the accuracy of the analysis.

Finally, do not divide the surface resistance by a temperature difference or an absolute temperature. The radiation network is an Ohm's-law analogy in which σT^4 is the potential and Q is the current, so the driving potential is σ(T_1^4 − T_2^4), not T_1 − T_2. If you import the convective intuition of "heat-transfer coefficient times ΔT" and rebuild R on a temperature-difference basis, the answer can look fine at low temperatures but go badly wrong at high temperatures. The essence of radiation is the fourth-power law, and the network is rigorously a linear circuit driven by that fourth-power difference.