Conduction, Convection & Radiation

Conduction moves no matter; it passes energy from one jostling particle to the next, and in a metal through the conduction electrons as well. Fourier's law sets the rate, P = −k A dT/dx, where the minus sign makes P positive when T falls as x grows. The material property is k, and its range is enormous: copper about 400 W m⁻¹ K⁻¹, stainless steel 15, glass 0.80, liquid water 0.60, wood 0.13, glass-fibre quilt 0.040, still air 0.026. In steady state, with nothing lost through the sides, the same P crosses every slice of a uniform slab, so dT/dx is constant and the temperature profile is a straight line. Separating the variables and integrating across the thickness, P ∫₀^L dx = −k A ∫ dT from T₁ at x = 0 to T₂ at x = L, and both integrands are constants, so P L = k A (T₁ − T₂) and P = k A (T₁ − T₂)/L. That straight profile is the whole content of steady state: while the slab is still warming it is curved, the two faces pass different powers, and this formula does not apply.

Write the slab result as P = ΔT/R with R = L/(k A) in K W⁻¹ and conduction looks exactly like Ohm's law. The point of the rearrangement is that layers crossed one after another carry the same P, so their resistances add. Per unit area it is easier still, each layer contributing L/k in m² K W⁻¹. Take an insulated cavity wall, with every k in W m⁻¹ K⁻¹ and every resistance in m² K W⁻¹: 13 mm plaster at k = 0.50 gives 0.026, 100 mm dense block at 1.13 gives 0.0885, 80 mm mineral wool at 0.040 gives 2.00, 102 mm brick at 0.77 gives 0.132, and the two surface films add 0.125 inside and 0.0435 outside, for 2.42 in all. The wool alone is 83% of that. Double its thickness and the loss falls by 45%; double the brickwork instead and it falls by 5%. Only the largest resistance is worth attacking. Parallel paths need the opposite treatment: a timber stud bridging the insulation is a second route across the same wall, and there the conductances add, not the resistances.

Move the fluid and you no longer have to pass energy from particle to particle; you carry it away. Right at the surface the fluid is still at rest, so the first fraction of a millimetre conducts anyway, and the rate is set by how thin the flow keeps that boundary layer. Because the answer depends on speed, geometry and the fluid, it is bundled into one empirical coefficient: P = h A (Tₛ − T_f), with h in W m⁻² K⁻¹. Typical values run from 2 to 10 for air moving only under its own buoyancy, 20 to 100 for air driven by a fan, and hundreds to thousands for flowing water. As a resistance, R_conv = 1/(h A). An indoor film of 0.125 m² K W⁻¹, which is what hᵢ = 8.0 W m⁻² K⁻¹ means, is worth 0.125 × 0.026 = 3.3 mm of still air — an invisible layer that resists more than two 4.0 mm panes of glass put together. Wind outside collapses that film, which is why an exposed face is given hₒ = 23 rather than 8 W m⁻² K⁻¹.

Every surface above absolute zero radiates electromagnetic energy, and no medium is required. A perfect emitter sends out σ T⁴ per unit area, with σ = 5.670 × 10⁻⁸ W m⁻² K⁻⁴; a real surface sends out a fraction ε of that, between 0 and 1. Kirchhoff's result ties the two ends together: at a given wavelength a surface absorbs the same fraction it emits, so a good emitter is a good absorber and ε appears once in each term. Facing walls at Tₛᵤᵣᵣ the surface is absorbing ε σ Tₛᵤᵣᵣ⁴ per unit area at the same time, which leaves the net loss P = ε σ A (Tₛ⁴ − Tₛᵤᵣᵣ⁴). The temperatures must be absolute. A black surface at 293 K emits 418 W m⁻² and at 313 K emits 544 W m⁻², so 20 K buys 126 W m⁻². Feeding 20 and 40 into the same expression returns 0.14 W m⁻², which is not a small error but a different universe.

Radiation and convection act on the same surface at the same moment, but one is quartic and the other linear, so they cannot be compared until they share a form. Factor the difference: Tₛ⁴ − Tₛᵤᵣᵣ⁴ = (Tₛ − Tₛᵤᵣᵣ)(Tₛ + Tₛᵤᵣᵣ)(Tₛ² + Tₛᵤᵣᵣ²). When ΔT is small beside the absolute temperatures the last two brackets are close to 2Tₘ and 2Tₘ², where Tₘ is the mean temperature, so the product is about 4Tₘ³ ΔT. The radiative loss therefore reads hᵣ A ΔT with hᵣ ≈ 4 ε σ Tₘ³. Both routes now leave the one surface side by side, which makes them parallel: the surface coefficient is the sum h = h_conv + hᵣ, and the powers add, while putting 1/(h_conv A) and 1/(hᵣ A) in series as though the energy had to cross them in turn would undercount the loss by a factor of several. Put numbers in: a dull surface with ε = 0.90 near 293 K has hᵣ = 5.1 W m⁻² K⁻¹, squarely inside the natural-convection range of 2 to 10, so indoors radiation is never the small correction it is assumed to be. The linearisation is sturdy too: at ε = 0.85, over the 48 K span of the hard example below, it gives 6.1 against an exact 6.2 W m⁻² K⁻¹.

Rank the resistances first, then change one thing that only one mechanism can feel. Each mechanism has its own handle: h responds to the fluid's motion, ε to the surface finish, k and L to the solid. A vacuum flask pulls all three levers — evacuating the gap removes conduction and convection together, and silvering the walls to ε ≈ 0.03 removes most of what remains, which is why the silvering does the heavy lifting. Geometry matters too: a narrow sealed gap leaves the air little room to organise a circulation, so a first calculation treats it as conducting with resistance L/k — an idealisation that a wider gap, or radiation straight across it, will eat into. The deciding experiment follows the same logic. Switch on a fan and the convective loss climbs while the radiative loss does not; wrap the surface in bright foil and the radiative loss collapses while h barely moves. Whichever change moves the total is the mechanism that was carrying it.