The Three Modes of Heat Transfer
Heat flows from hot to cold through three mechanisms. Engineers calculate each to design insulation, heat exchangers, and cooling systems.
Conduction (Fourier's Law)
Q/t = k × A × ΔT / d
k = thermal conductivity (W/m·K)
A = area (m²)
ΔT = temperature difference (K or °C)
d = thickness (m)
Steel wall 0.01m, ΔT=50°C, A=1m²:
Q/t = 50 × 1 × 50 / 0.01 = 250,000 WThermal Resistance and R-Value
R = d / (k × A) (thermal resistance, K/W)
R-value (building) = d / k (m²·K/W)
100mm fibreglass (k=0.04): R = 0.1/0.04 = 2.5 m²·K/W
Layers in series: R_total = R1 + R2 + R3Convection (Newton's Law of Cooling)
Q/t = h × A × (T_surface - T_fluid)
h = convection coefficient (W/m²·K)
Natural convection: h ≈ 5-25 W/m²·K
Forced (fan): h ≈ 25-250 W/m²·KRadiation (Stefan-Boltzmann)
Q/t = ε × σ × A × (T₁⁴ - T₂⁴)
ε = emissivity (0-1)
σ = 5.67×10⁻⁸ W/m²·K⁴
T in Kelvin (K = °C + 273.15)Calculate heat transfer: Free Heat Transfer Calculator
The Three Heat Transfer Modes
- Conduction: Q = kA(ΔT)/L. Heat flows through solids. k = thermal conductivity (W/mK). Steel: 50, concrete: 1.7, timber: 0.12, mineral wool: 0.04, air: 0.026 W/mK.
- Convection: Q = hAΔT. Heat transferred to/from a moving fluid. h = convective heat transfer coefficient (W/m²K). Natural convection (air): 2–25 W/m²K. Forced convection (water): 500–10,000 W/m²K.
- Radiation: Q = εσA(T₁⁴ - T₂⁴). All objects emit thermal radiation. σ = Stefan-Boltzmann constant (5.67×10⁻⁸ W/m²K⁴). ε = emissivity (0–1).
Applications in Building and Engineering
Thermal insulation in buildings resists conductive heat loss through walls, roofs, and floors. U-value (W/m²K) = 1/R-value = inverse of total thermal resistance; lower U-value = better insulation. UK Building Regulations require external walls U ≤ 0.18–0.30 W/m²K. Heat exchangers (boilers, radiators, condensers, coolers) transfer heat between fluids through conduction and convection. Electronic cooling (heatsinks, fans, thermal paste) manages conductive and convective heat removal from chips. Industrial processes use heat transfer calculations to size furnaces, dryers, and heat recovery systems.
Frequently Asked Questions
What is thermal resistance and how is it used?
Thermal resistance R_th = L/(kA) for conduction or 1/(hA) for convection (K/W). Like electrical resistance, thermal resistances in series add directly. A composite wall with brick (R₁), insulation (R₂), and plasterboard (R₃) has total R = R₁+R₂+R₃. Total heat flow Q = ΔT/R_total. This analogy makes heat transfer calculations tractable by circuit analysis methods.
Why does doubling insulation thickness not halve heat loss?
It depends on the dominant resistance. If insulation already dominates wall thermal resistance, doubling thickness approximately halves conductive loss. But surface convective resistances (inner and outer air films) are fixed regardless of insulation thickness. Above a certain insulation level, further increases give diminishing returns because convective surface resistance becomes the limiting factor, not the insulation itself.
What is the Biot number?
Bi = hL/k compares convective heat transfer at the surface to conductive heat transfer within the solid. Bi << 0.1 means the solid is thermally uniform (lumped capacitance model is valid). Bi >> 1 means surface temperature responds quickly but interior lags — relevant for thick parts in quenching, casting, and fire engineering.