🔲 Gray Surfaces & Shields Calculator

Model gray surface radiative exchange using radiosity networks, or design multi-layer radiation shields to inhibit thermal transfer.

📝 Configuration

⚙️ Calculation Mode
📐 Enclosure Geometry
🛡️ Shield Screen Configuration
🔴 Surface Properties & Temps

📊 Results

Configure properties and click Calculate to view results.

📘 Calculation Methodology & Theory

Mathematical Model & Theory

For grey, diffuse, and opaque surfaces, radiation exchange is solved using a network of thermal resistances. The surface resistance represents the emissivity barrier:

$$R_{\text{surface}} = \frac{1 - \varepsilon}{\varepsilon A}$$

The space resistance accounts for geometric view factor limitations:

$$R_{\text{space}} = \frac{1}{A_i F_{ij}}$$

For a parallel plate geometry with $N$ radiation shields, the net heat transfer rate is expressed as:

$$Q_{\text{shield}} = \frac{A \sigma (T_1^4 - T_2^4)}{\left( \frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} - 1 \right) + \sum_{i=1}^N \left( \frac{1}{\varepsilon_{s,i,1}} + \frac{1}{\varepsilon_{s,i,2}} - 1 \right)}$$

Worked Engineering Example

Problem Statement:
Two parallel plates of area $A = 2.0\text{ m}^2$ are at $T_1 = 500\text{°C}$ ($773.15\text{ K}$) and $T_2 = 50\text{°C}$ ($323.15\text{ K}$) with $\varepsilon_1 = \varepsilon_2 = 0.8$. A single radiation shield with emissivity $\varepsilon_s = 0.05$ on both sides is placed between them. Compute the unshielded heat rate, shielded heat rate, and reduction percentage.

Step-by-step Solution:
1. Unshielded heat transfer rate $Q_{\text{no shield}}$:
$$Q_{\text{no shield}} = \frac{A \sigma (T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} - 1}$$ $$Q_{\text{no shield}} = \frac{2.0 \times (5.67037 \times 10^{-8}) \times (773.15^4 - 323.15^4)}{\frac{1}{0.8} + \frac{1}{0.8} - 1} = 26,209.6 \text{ W}$$ 2. Shielded heat transfer rate $Q_{\text{shield}}$ with 1 shield:
$$Q_{\text{shield}} = \frac{A \sigma (T_1^4 - T_2^4)}{\left(\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} - 1\right) + \left(\frac{1}{\varepsilon_{s,1}} + \frac{1}{\varepsilon_{s,2}} - 1\right)}$$ $$Q_{\text{shield}} = \frac{39,314.4}{1.5 + (20 + 20 - 1)} = \frac{39,314.4}{40.5} = 970.7 \text{ W}$$ 3. Compute reduction percentage:
$$\text{Reduction} = \left( 1 - \frac{970.7}{26,209.6} \right) \times 100\% = 96.30\%$$
Final Result:
The single shield reduces heat flow from $26.21\text{ kW}$ to $970.7\text{ W}$, a $96.30\%$ reduction.

Standard Assumptions & References

Assumptions: Surfaces are diffuse, gray, and opaque. The temperature of each surface is uniform. The space between the plates is evacuated or filled with a non-participating gas.

References:

  • Siegel, R., & Howell, J. R. Thermal Radiation Heat Transfer, Taylor & Francis Group.
  • Incropera, F. P., & DeWitt, D. P. Fundamentals of Heat and Mass Transfer, John Wiley & Sons.
  • Cengel, Y. A. Heat and Mass Transfer: A Practical Approach, McGraw-Hill.