Reference ID: MET-A4B8 | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
Combined convection and radiation heat transfer calculations are essential for determining total heat loss from surfaces operating at moderate to high temperatures. In process engineering, this analysis is critical for designing insulation systems, sizing heat exchangers, predicting thermal losses from reactors, and ensuring safe surface temperatures on piping and equipment. The simultaneous treatment of convection—governed by fluid motion—and radiation—governed by electromagnetic emission—provides a realistic estimate of heat loads that individual mechanisms alone would misrepresent.
Rayleigh number for natural convection
The Rayleigh number quantifies the balance between buoyancy and viscous forces:
\[ Ra = \frac{g\,\beta\,\Delta T\,L^{3}}{\nu\,\alpha} \]
Regime
Range
Laminar on vertical plate
\(10^{4}\le Ra\le10^{7}\)
Nusselt number correlation
For laminar natural convection from a vertical surface:
\[ Nu = 0.54\,Ra^{0.25} \]
Pr Range
Validity
0.7–600
Correlation applicable
Convection coefficient
Convert the Nusselt number to the convection coefficient:
\[ h_{\text{conv}} = \frac{Nu\,k}{L} \]
Radiation heat-transfer coefficient
A linearized radiation coefficient simplifies mixed-mode calculations:
\[ h_{\text{rad}} = \frac{\varepsilon\,\sigma\,(T_{\text{s}}^{4}-T_{\text{sur}}^{4})}{T_{\text{s}}-T_{\text{sur}}} \]
where \( \sigma \) is the Stefan–Boltzmann constant.
Total heat-transfer coefficient and heat flux
Add convective and radiative contributions:
\[ h_{\text{total}} = h_{\text{conv}} + h_{\text{rad}} \]
The heat loss from area \( A \) is:
\[ q = h_{\text{total}}\,A\,\Delta T \]
Answer
Always include both mechanisms when surface temperatures exceed ~100 °C or when the surrounding environment is at a significantly lower temperature.
At 200 °C, radiation can contribute 30–50 % of the total heat loss from an uninsulated steel pipe in still air.
Below ~50 °C, radiation is usually < 10 % of the total and can often be neglected for quick estimates.
Answer
Calculate each coefficient separately: h_conv from appropriate Nusselt correlations and h_rad = εσ(T_s2 + T_∞2)(T_s + T_∞).
Add them directly: h_total = h_conv + h_rad.
Use h_total in Q = h_total A (T_s – T_∞) to obtain the combined heat-transfer rate.
Answer
For heavily oxidized carbon steel at 300–600 °C, use ε ≈ 0.85–0.90.
For refractory brick at 800–1200 °C, use ε ≈ 0.70–0.80.
When in doubt, measure with an IR thermometer or consult vendor data; a 5 % error in ε translates directly to a 5 % error in radiative heat loss.
Answer
Apply low-emissivity coatings (ε < 0.2) to cut radiation losses by up to 80 %.
Install radiation shields—thin stainless-steel or aluminum sheets spaced 5–10 mm away from the surface—to create multiple reflection barriers.
Enhance external airflow with fans to raise h_conv, then add a thin aerogel blanket (5–10 mm) to suppress both modes with minimal thickness.
Worked Example – Heat Loss from a Horizontal Reactor Roof
A 250 mm thick steel roof on a solvent reactor operates at 125 °C in a 25 °C ambient. The paint vendor claims the coating (ε = 0.79) will keep total heat loss below 1.4 kW m-2. Verify the claim assuming still air and surroundings at ambient temperature.
Knowns
g = 9.8 m s-2
σ = 5.67 × 10-8 W m-2 K-4
Pr = 0.7
k = 0.029 W m-1 K-1
ν = 2.04 × 10-5 m2 s-1
α = 2.91 × 10-5 m2 s-1
β = 0.00287 K-1
L = 0.25 m
A = 1.0 m2
ε = 0.79
T∞ = 25 °C
Ts = 125 °C
ΔT = 100 K
Ts,K = 398.15 K
Tsur,K = 298.15 K
Step-by-Step Calculation
Compute the Rayleigh number for the roof length:
\[
Ra = \frac{g \beta \Delta T L^3}{\nu \alpha}
= \frac{9.8 \times 0.00287 \times 100 \times 0.25^3}{2.04 \times 10^{-5} \times 2.91 \times 10^{-5}}
= 7.40 \times 10^7
\]
Obtain the average Nusselt number for a hot horizontal plate facing upward:
\[
Nu = 0.15 Ra^{1/3} = 0.15 (7.40 \times 10^7)^{1/3} = 50.1
\]
