Heat Transfer: Conduction, Convection, Radiation, and Heat Exchangers

Conduction and Thermal Resistance

The first move in any heat-transfer problem is to identify the mode: conduction (through a solid or stationary fluid), convection (between a surface and a moving fluid), or radiation (electromagnetic exchange between surfaces). Many FE items combine modes in series, which is exactly what the resistance network handles.

Conduction through a solid follows Fourier's law. For a plane wall at steady state:

q = kA(T₁ − T₂)/L = ΔT / R_cond, where R_cond = L/(kA)

k is thermal conductivity (W/m·K), L the thickness, A the cross-sectional area. The thermal-resistance analogy is the FE workhorse: heat flow q is like electric current, ΔT like voltage, and R like resistance, so Ohm's-law intuition transfers directly. Resistances in series add (R_total = ΣR_i); resistances in parallel combine reciprocally. For a composite wall, q = ΔT_overall / R_total.

Convection resistance at a fluid boundary is R_conv = 1/(hA). Cylindrical (radial) conduction through a pipe wall or insulation uses R = ln(r₂/r₁)/(2πkL). Memorizing the three resistance forms (plane wall, cylinder, convection film) lets you assemble a network for almost any composite-wall or insulated-pipe item without re-deriving anything.

Worked Example — Composite Wall

A wall has two layers: brick (k = 0.7 W/m·K, L = 0.1 m) and insulation (k = 0.04 W/m·K, L = 0.05 m), area A = 10 m². Inside surface 25 °C, outside surface 0 °C.

R_brick = 0.1/(0.7 × 10) = 0.0143 K/W. R_ins = 0.05/(0.04 × 10) = 0.125 K/W. R_total = 0.139 K/W.

q = ΔT/R_total = 25/0.139 = 180 W.

Notice the insulation (lower k) carries ~90% of the resistance — the low-conductivity layer dominates. If convection films were included, R_conv = 1/(hA) would be added in series at each surface.

Convection and Radiation

Convection between a surface and a moving fluid obeys Newton's law of cooling:

q = hA(T_s − T_∞)

The convection coefficient h is not a property of the solid — it depends on flow regime, geometry, and fluid properties, captured through the Nusselt number Nu = hL/k (here k is the fluid's conductivity); for forced internal flow the Dittus-Boelter correlation Nu = 0.023 Re^0.8 Pr^n uses n = 0.4 for heating and n = 0.3 for cooling. Free (natural) convection uses the Rayleigh or Grashof number instead, since buoyancy drives the flow rather than an external pump.

Order-of-magnitude h values worth carrying: free convection in air 5–25 W/m²·K, forced convection in air 25–250, forced convection in water 100–15,000, and boiling/condensation 2,500–100,000 — boiling dwarfs single-phase convection, which is why phase-change heat exchangers are so compact.

Radiation between a surface and surroundings:

q = εσA(T_s⁴ − T_surr⁴), with σ = 5.67 × 10⁻⁸ W/m²·K⁴ and emissivity ε (0–1).

Temperatures must be absolute (K) because of the fourth-power dependence — using °C is catastrophically wrong here.

Worked example. A 2 m² surface at 500 K (ε = 0.8) radiates to surroundings at 300 K. q = 0.8 × 5.67e-8 × 2 × (500⁴ − 300⁴) = 0.8 × 5.67e-8 × 2 × (6.25e10 − 8.1e9) = 4934 W.

Combined modes. A real surface often loses heat by convection and radiation in parallel; the totals add: q_total = q_conv + q_rad. For small temperature differences, radiation can be linearized into an equivalent radiation coefficient h_rad so it slots into a resistance network beside the convection film.

Fins and Heat Exchangers

Fins extend a surface to boost convective heat transfer; fin efficiency η_f compares actual fin heat to the heat an isothermal fin (entire fin at base temperature) would transfer. Real heat = η_f × h × A_fin × (T_base − T_∞).

Heat exchangers transfer heat between two streams. The total rate is q = UA · ΔT_lm, the Log-Mean Temperature Difference method:

ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)

where ΔT₁ and ΔT₂ are the temperature differences at the two ends (counterflow vs. parallel-flow changes how you pair them). Use LMTD when all four terminal temperatures are known.

When outlet temperatures are unknown but UA and inlets are given, switch to the effectiveness-NTU method: NTU = UA/C_min, effectiveness ε = q/q_max, with q_max = C_min(T_h,in − T_c,in) and heat-capacity rate C = ṁc_p.

An energy balance ties the two streams: q = ṁ_h c_p,h (T_h,in − T_h,out) = ṁ_c c_p,c (T_c,out − T_c,in). The smaller heat-capacity rate C_min = (ṁc_p)_min limits the exchanger and experiences the largest temperature change. Counterflow exchangers achieve a larger ΔT_lm than parallel-flow for the same terminal temperatures, so they are more effective; counterflow can also raise the cold outlet above the hot outlet, which parallel-flow can never do.

Worked example. A counterflow exchanger: hot 100→60 °C, cold 20→50 °C. Pair the ends correctly: ΔT₁ = T_h,in − T_c,out = 100 − 50 = 50, ΔT₂ = T_h,out − T_c,in = 60 − 20 = 40. ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (50 − 40)/ln(50/40) = 10/0.223 = 44.8 °C. If UA = 2000 W/°C, then q = UA·ΔT_lm = 2000 × 44.8 = 89.6 kW.