Heat Transfer Calculator.

The heat crossing a slab is proportional to its area and to the temperature difference, and inversely proportional to its thickness. The constant k is a property of the material, which is what makes it tabulable.

The instrument is now set to a carbon steel wall 100 mm thick over 2 m², between 300 °C and 50 °C. That gives 302,500 W, with a resistance of 0.0008 K/W and a gradient of 2,500 K/m.

Change nothing but the material. The same 100 mm wall over the same 2 m² with the same 250 K across it passes 302,500 W in carbon steel and 190.0 W in fibreglass — a factor of about 1,592.

No geometry changed. Nothing was made thicker. This is why insulation is a material problem long before it is a thickness problem, and why the material list here deliberately runs from copper all the way down to still air.

Rewrite Fourier’s Law as Q = ΔT/R with R = L/(kA) and conduction becomes an Ohm’s law. Resistances in series add, which is how a real wall — plaster, brick, insulation, render — is actually solved.

The reason this matters here is that all three modes take the same form. Convection gives R = 1/(hA) and radiation gives R = 1/(h_rA). The readout shows R in every mode for exactly that reason, and it is the bridge to the overall coefficient U in a heat exchanger, where 1/U = ΣR.

Take the same steel wall down from 100 mm to 5 mm and the heat rises from 302,500 W to 6,050,000 W — twenty times, exactly as R ∝ L says it must.

This is why the metal wall of a heat exchanger tube contributes almost nothing to its total resistance, and why a heat exchanger is really a contest between the two convection films on either side of it. The metal is there for strength and corrosion, not for insulation.

The heat leaving a surface into a fluid is hA times the temperature difference. Simple to write — and the entire difficulty is hidden inside h, which is not a material property and cannot be looked up.

The stage is now a 1.5 m vertical panel at 60 °C in still 20 °C air. The correlation gives h = 4.7 W/m²·K from Ra = 1.09e+10 and Pr = 0.71. Watch the working: no number was typed in.

Same panel, same temperatures, now with 8 m/s of air across it. h goes from 4.7 to 15.1 W/m²·K — about 3.2 times — for nothing but moving the air.

The honest test of which regime you are in is the ratio Gr/Re². Well below one, forced flow dominates and buoyancy can be ignored. Well above one, it is really free convection. Near one it is mixed and neither correlation is trustworthy — and this tool says so instead of quietly picking a side.

A 50 mm pipe at 1.5 m/s. In water the correlation gives h = 6,240 W/m²·K; in air at the same speed it gives 18.9. A factor of about 330.

Water is denser, has a far higher specific heat and conducts better, and all three feed the Reynolds and Prandtl numbers. It is why engines are water-cooled and why an air-cooled heat exchanger needs enormous fins to keep up.

Fluid touching a wall does not move — the no-slip condition — so heat must conduct across a thin film before anything can carry it away. The thickness of that film essentially is 1/h, and every trick for improving convection works by thinning it.

The Prandtl number sets how the thermal layer compares with the velocity layer. In air (Pr ≈ 0.71) the thermal layer is the thicker of the two. In engine oil (Pr ≈ 1,214) it is dramatically thinner — 19.0 mm of velocity layer against 1.78 mm of thermal layer.

Everything above absolute zero radiates, and the emissive power goes as the fourth power of absolute temperature. σ is 5.670374419 × 10⁻⁸ W/m²·K⁴ — since 2019 an exactly defined constant rather than a measured one.

Note what the formula is not. It is not εσA(ΔT)⁴. With T₁ = 800 K and T₂ = 300 K the difference of fourth powers is about six times the fourth power of the difference, and the reference tool this page was built against prints the wrong one on its readout chip.

Against a 20 °C background at ε = 0.8, one square metre at 350 K radiates about 360 W — a light breeze would beat that easily. Take the same surface to 1200 K and it radiates about 93,744 W, which no practical air flow can match.

The temperature rose by a factor of 3.4; the heat rose by a factor of about 260. The curve in the right-hand pane is that fourth power, with your operating point marked on it — which is the fastest way to see where the crossover sits.

Same geometry, same temperatures, only the finish changes. At ε = 0.04 — polished aluminium — the surface radiates 4,687 W. At ε = 0.98 — matt black paint — it radiates 114,836 W. A factor of 25, from a coating.

Polished aluminium is 0.04 and anodised aluminium is 0.82: the same metal, twenty times apart, separated by a few microns of oxide. Low-emissivity window coatings, loft radiant barriers and the silvering in a vacuum flask are all this idea used deliberately.

Two surfaces exchange radiation only insofar as they can see each other. The view factor F is the fraction of radiation leaving one that lands on the other, and setting it to 1 assumes each sees only the other and nothing else.

That assumption is exactly right for a small body inside a large enclosure — a component in a furnace, a thermometer in a duct — and wrong for almost everything else. Halving F halves the net exchange, because the geometry simply removes that fraction of the paths.