Heat Transfer: Conduction, Convection, and Radiation

Conduction — Fourier's Law

Conduction is heat transfer through a solid or stationary fluid by molecular interaction, with no bulk motion. The driving relation is Fourier's law; for steady, one-dimensional conduction through a plane wall it reduces to: $\dot{Q} = \frac{kA(T_1 - T_2)}{L}$ where k = thermal conductivity [W/(m·K)], A = cross-sectional area perpendicular to heat flow, L = wall thickness, and (T₁ − T₂) the temperature difference. The heat flux q″ = Q̇/A = kΔT/L has units W/m². The corresponding thermal resistance is R_cond = L/(kA).

For radial conduction through a cylindrical wall (pipe insulation is the classic case): $\dot{Q} = \frac{2\pi k L (T_1 - T_2)}{\ln(r_2/r_1)}, \qquad R_{cyl} = \frac{\ln(r_2/r_1)}{2\pi k L}$

Metals conduct ~10,000× better than insulation — which is exactly why fiberglass (low k) is used to resist heat flow. A common error is forgetting that ΔT in kelvin equals ΔT in °C (a difference of 150°C = a difference of 150 K), so conduction problems may use either as long as you take a difference.

Convection and the Resistance Network

Convection transfers heat between a surface and a moving fluid. Newton's law of cooling: $\dot{Q} = hA(T_s - T_\infty)$ where h = convection coefficient [W/(m²·K)], T_s = surface temperature, T∞ = bulk fluid temperature. The resistance is R_conv = 1/(hA). The coefficient h is not a property — it depends on geometry, fluid, and flow speed, ranging from ~5 W/(m²·K) for natural convection in still air to >10,000 for boiling or condensing. It is correlated through dimensionless groups: the Nusselt number Nu = hL/k (ratio of convection to conduction), the Prandtl number Pr = ν/α (momentum vs.

thermal diffusivity), and, for natural convection, the Grashof/Rayleigh numbers (buoyancy vs. viscous forces).

The electrical analogy. Heat flow is like current, temperature difference like voltage, and thermal resistance like electrical resistance, so Q̇ = ΔT/R. Resistances in the heat-flow path add in series: $\dot{Q} = \frac{\Delta T_{overall}}{R_{total}} = \frac{T_{\infty,1} - T_{\infty,2}}{\frac{1}{h_1 A} + \frac{L}{kA} + \frac{1}{h_2 A}}$ This lets you fold convection on each side and conduction through a wall into one calculation. Defining the overall heat-transfer coefficient U by 1/(UA) = R_total gives the compact form Q̇ = UA·ΔT.

Worked resistance example. A 0.1 m brick wall, k = 0.7 W/(m·K), area 10 m², has inside air at 22°C (h_i = 8) and outside air at −5°C (h_o = 25). Resistances: R_i = 1/(8×10) = 0.0125, R_wall = 0.1/(0.7×10) = 0.0143, R_o = 1/(25×10) = 0.004 K/W. R_total = 0.0308 K/W. Heat loss Q̇ = ΔT/R = (22 − (−5))/0.0308 = 27/0.0308 = 877 W. The wall and inside film dominate the resistance — adding insulation (low k) to the wall is the most effective fix. Resistances may also combine in parallel (two materials side by side carrying heat between the same two temperatures), exactly mirroring parallel electrical resistors: 1/R_total = Σ(1/R_i).

Free vs. forced convection. Free (natural) convection is driven only by buoyancy from density differences the heat itself creates, giving low h; forced convection uses a fan or pump to move the fluid, giving much higher h and faster cooling. Boiling and condensation involve phase change and produce the largest coefficients of all because latent heat is exchanged with very little temperature change.

Radiation and Heat Exchangers

Radiation is energy carried by electromagnetic waves and needs no medium. 67×10⁻⁸ W/(m²·K⁴), and temperatures MUST be absolute (kelvin) because they are raised to the fourth power — using °C is a guaranteed wrong answer. The T⁴ dependence means radiation is negligible at room temperature but dominates at high temperatures (furnaces, filaments): doubling absolute temperature multiplies radiated power by 2⁴ = 16. Kirchhoff's law states that at thermal equilibrium a surface's absorptivity equals its emissivity (α = ε).

Worked conduction example. A 0.2 m wall, k = 1.5 W/(m·K), has surfaces at 200°C and 50°C. Heat flux q″ = kΔT/L = 1.5 × (200 − 50)/0.2 = 1.5 × 150/0.2 = 225/0.2 = 1,125 W/m².

Heat exchangers transfer heat between two fluids separated by a wall. Because the temperature difference varies along the exchanger, use the log-mean temperature difference: $\dot{Q} = UA \cdot LMTD, \qquad LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}$ where ΔT₁ and ΔT₂ are the temperature differences at the two ends.

A counterflow exchanger maintains a more uniform ΔT and can let the cold outlet approach the hot inlet — impossible in parallel flow, where the outlets can at best converge to a common intermediate temperature. The alternative effectiveness–NTU method is used when outlet temperatures are unknown. On the FE, recognize which ΔT pairs go at each end and remember to take the natural log of their ratio.