โ๏ธ Combined Radiation & Convection
Evaluate parallel radiation and convection heat losses from a surface. Compare convective coefficients to linearized radiation coefficients.
โก Fortran 90 Engine
Double Precision (IEEE 754)
โ ISO / ASME Validated
Heat Transfer
๐ Solver Telemetry
โ ACTIVE
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๐ฆ Fortran Code
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๐
Released
Jun 2026
โฑ๏ธ Latency
< 1 ms
Combined Modes Dissipation
Almost all solid bodies operating in gas surroundings dissipate heat via parallel convection and radiation paths:
- Radiation component: Linearized with $h_r$, highly dependent on surface emissivity ($\varepsilon$) and $T^4$ absolute temperatures.
- Convection component: Calculated directly or determined using natural/forced convective correlation rules.
- Environment limits: Surroundings $T_{surr}$ and air $T_{\infty}$ can be set independently.
๐ Configuration
๐ Results & Contributions
Configure properties and click "Compute Combined Heat Transfer" to evaluate heat rates.
๐ Calculation Methodology
Mathematical Model
Total heat rate is the sum of convection and radiation components acting in parallel:
$$Q_{total} = Q_{conv} + Q_{rad}$$
Radiation (linearized):
$$Q_{rad} = \varepsilon \cdot \sigma \cdot A \cdot \left( T_s^4 - T_{surr}^4 \right) = h_r \cdot A \cdot (T_s - T_{surr})$$ $$h_r = \varepsilon \cdot \sigma \cdot (T_s + T_{surr}) \cdot (T_s^2 + T_{surr}^2)$$
Convection (Newton Cooling):
$$Q_{conv} = h_c \cdot A \cdot (T_s - T_{\infty})$$
Radiation (linearized):
$$Q_{rad} = \varepsilon \cdot \sigma \cdot A \cdot \left( T_s^4 - T_{surr}^4 \right) = h_r \cdot A \cdot (T_s - T_{surr})$$ $$h_r = \varepsilon \cdot \sigma \cdot (T_s + T_{surr}) \cdot (T_s^2 + T_{surr}^2)$$
Convection (Newton Cooling):
$$Q_{conv} = h_c \cdot A \cdot (T_s - T_{\infty})$$
Convective Correlations
Simplified convective formulas for air at normal pressure and temperature ranges:
- Vertical Plate (Natural): $h_c = 1.42 \cdot \left( \frac{\Delta T}{L} \right)^{0.25}$
- Horizontal Cylinder (Natural): $h_c = 1.32 \cdot \left( \frac{\Delta T}{D} \right)^{0.25}$
- Forced Air Convection: $h_c = 10.45 - v + 10 \sqrt{v}$