Thermal Radiation Heat Transfer Back
Thermal Analysis

Thermal Radiation Heat Transfer Calculator

Compute Q = εσA(T₁⁴−T₂⁴) in real time using the Stefan-Boltzmann law. Visualize heat flux, radiation resistance, and radiation arrows scaled by intensity.

Design Parameters

T₁ =
K
(527°C)
T₂ =
K
(27°C)
ε =
A =
Results
Q (W)
Heat Flux (W/m²)
R_rad (K/W)
h_rad (W/m²K)
Visualization
Theory & Key Formulas

$$q = \\varepsilon \sigma T^4$$

ステファン-ボルツマン則:放射熱流束(W/m²)。$\sigma = 5.67 \times 10^{-8}$ W/m²K⁴、$\varepsilon$ は放射率。

$$q_{net} = \\varepsilon \sigma (T_1^4 - T_2^4)$$

2面間の正味放射熱流束(W/m²):黒体近似の場合。放射率 $\varepsilon \lt 1$ で実際の放熱量が減少。

$$\\lambda_{max} T = 2898 \\, \mu\text{m·K}$$

ウィーンの変位則:最大放射波長(μm)と絶対温度の積は定数。太陽≈0.5μm、人体≈10μm。

What is Thermal Radiation Heat Transfer?

🙋
What exactly is thermal radiation? I know things get hot and glow, but how do we calculate the energy being transferred that way?
🎓
Basically, it's energy emitted by all objects with a temperature above absolute zero, via electromagnetic waves. No physical contact is needed—it's how the Sun warms the Earth! The core equation is the Stefan-Boltzmann law. In this simulator, you control the key variables: the temperatures of two surfaces ($T_1$ and $T_2$), the emissivity ($\epsilon$), and the area ($A$). Try moving the $T_1$ slider above and watch how the calculated heat transfer rate changes dramatically.
🙋
Wait, really? The temperature is to the fourth power? That seems huge. What does emissivity do in practice?
🎓
Yes, the fourth-power dependence is why it's so non-linear! Double the absolute temperature, and the radiation increases by a factor of 16. Emissivity is a "efficiency factor" for a real surface compared to a perfect blackbody. For instance, black paint has an $\epsilon$ near 0.95, while polished aluminum is around 0.05. In the simulator, slide the emissivity control from 0.1 to 0.9 and you'll see the heat transfer multiply by nine, even with the same temperatures.
🙋
So the simulator shows heat transfer between two surfaces. What if $T_2$ is hotter than $T_1$? And what's the "Geometry" parameter for?
🎓
Great question! The equation calculates the net radiation exchange. If $T_2 \gt T_1$, the result ($Q$) becomes negative, meaning the net heat flows from surface 2 to surface 1. The "Geometry" factor, often called the view factor ($F$), accounts for how surfaces "see" each other. For two infinite parallel plates, they see each other fully ($F=1$). For complex shapes, like a small component inside a large enclosure, the view factor is different. Changing this in the tool adjusts the effective area for radiation exchange.

Physical Model & Key Equations

The fundamental law governing thermal radiation from a surface is the Stefan-Boltzmann Law, which states the total emissive power of a perfect blackbody is proportional to the fourth power of its absolute temperature.

$$E_b = \sigma T^4$$

Here, $E_b$ is the blackbody emissive power (W/m²), $\sigma = 5.67 \times 10^{-8}$ W/m²K⁴ is the Stefan-Boltzmann constant, and $T$ is the absolute temperature in Kelvin (K).

For real-world engineering between two surfaces, we use the net radiation heat transfer equation, which incorporates emissivity, area, geometry, and both temperatures.

$$Q = \epsilon \sigma A F (T_1^4 - T_2^4)$$

$Q$: Net radiation heat transfer rate (Watts).
$\epsilon$: Emissivity of the surface (0 to 1), a material property.
$A$: Surface area (m²).
$F$: View factor (geometry factor), accounting for the fraction of radiation leaving one surface that strikes the other.
The sign of $Q$ indicates the direction of net energy flow.

Frequently Asked Questions

This simulator assumes input in absolute temperature (K). If you wish to input in Celsius, please convert using K = ℃ + 273.15 before entering, as there is no automatic conversion function. Using the wrong unit can cause significant errors, especially due to the fourth-power difference.
Emissivity varies between 0 and 1 depending on the surface material and condition. For example, a blackbody (ideal radiator) is 1.0, polished aluminum is about 0.04, and oxidized iron is about 0.7. Practically, it is recommended to refer to material data sheets or literature values, or if unknown, start with a value around 0.9.
The area A is proportional to the radiative heat transfer Q. For example, doubling the area will double Q. This is because larger surfaces radiate and absorb more heat. In the visualization feature for shape effects, changing the area allows you to intuitively see how radiation efficiency changes.
This tool is specialized for simple radiative calculations between two surfaces. For complex systems with three or more surfaces (e.g., radiation in an enclosure furnace), mutual reflections and view factors between each pair of surfaces must be considered, and this formula alone cannot provide accurate calculations. For multi-surface analysis, please use a separate tool based on the radiation network method.

Real-World Applications

Spacecraft Thermal Control: In the vacuum of space, radiation is the only way to reject heat. Satellites use special radiators with high-emissivity coatings to dump excess heat into space, while low-emissivity surfaces (like multi-layer insulation) prevent heat loss to the cold environment.

Building & HVAC Design: Radiant barriers in attics, often made of foil-faced material with very low emissivity, reflect thermal radiation to reduce summer heat gain and winter heat loss, significantly impacting energy efficiency.

Industrial Furnaces & Processing: High-temperature furnaces for metal treatment or glass manufacturing rely heavily on radiation heat transfer. Engineers design heating elements and chamber geometry to ensure uniform radiation exposure on the product for consistent quality.

Electronics Cooling: Heat sinks often have black anodized finishes to increase emissivity ($\epsilon \approx 0.9$), enhancing the radiation of heat to the surroundings. This is a critical supplement to convective cooling in compact devices like gaming consoles and servers.

Common Misconceptions and Points to Note

Here are a few points where beginners often stumble when mastering this simulator. First is confusing absolute temperature (K) with Celsius (°C). This is a really common mistake. For example, when you want to set a surface temperature to "100°C", the input value should be "373K". If you input "100" as is, the calculation results will be wildly off. Before using the simulator, make a habit of mentally converting using "°C + 273 = K".

Next is the point that emissivity ε is not a fixed value. While approximate values are determined by material, it changes based on surface roughness, oxidation state, temperature, and even wavelength. For instance, for the same stainless steel, a polished surface might have ε≈0.1, but it can exceed 0.7 with significant oxidation. In practice, it's important to confirm "this value for this condition" through literature or experiments.

Finally, understand the limitations of this tool's model. What's calculated here is an ideal configuration: two parallel, infinitely wide surfaces, or where the view factor between them is 1 (they see each other completely). In reality, calculations involving multiple objects with complex shapes or within an enclosure (closed space) where reflection and absorption occur repeatedly cannot be done with this alone. Please use it strictly as a tool to "grasp the basic principles and intuition of radiation".

How to Use

  1. Enter the hot surface temperature (T1) in Kelvin—typical values range 300 K (room) to 1200 K (industrial furnace)
  2. Enter the cold surface or ambient temperature (T2) in Kelvin
  3. Input the emissivity (ε) between 0 and 1—use 0.95 for oxidized steel, 0.85 for aluminum, 0.05 for polished aluminum
  4. Specify the radiating surface area in m²
  5. The calculator applies Q = εσA(T₁⁴−T₂⁴) where σ = 5.67×10⁻⁸ W/(m²·K⁴) to output heat transfer rate in watts

Worked Example

A steel pipe at 800 K with emissivity 0.9 and surface area 2.5 m² radiates to surroundings at 300 K. Calculate: Q = 0.9 × 5.67×10⁻⁸ × 2.5 × (800⁴ − 300⁴) = 0.9 × 5.67×10⁻⁸ × 2.5 × (4.096×11 − 8.1×9) = approximately 41,850 W or 41.85 kW. This represents dominant heat loss in high-temperature equipment design.

Practical Notes