Heat conducts through a solid wall at a rate set by its thermal conductivity, thickness and the temperature difference across it: q = k·A·ΔT/L. Writing L/(k·A) as a thermal resistance turns walls with several layers, and the air films on each side, into resistors in series that add like an electrical circuit.
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Fourier’s law
Conduction is heat moving through a material from hot to cold by molecular contact. Fourier’s law says the heat flux is proportional to the temperature gradient: q″ = −k·dT/dx. The thermal conductivity k, in W/(m·K), is a material property. Metals conduct well, with k in the tens or hundreds; insulation such as mineral wool or foam has k of a few hundredths.
In a plane wall at steady state with constant k and no heat generated inside, the same heat passes every plane, so the temperature falls in a straight line from one face to the other. The heat rate is then q = k·A·(T₁ − T₂)/L.
Thermal resistance networks
Rearranged, q = ΔT/R with R = L/(k·A), in K/W. Temperature difference plays the role of voltage and heat rate the role of current. Layers in series add their resistances; parallel paths, such as studs and insulation side by side, combine like parallel resistors.
Convection at a surface has its own resistance, R = 1/(h·A), using the convection heat transfer coefficient. The total resistance from one fluid to the other gives the overall heat transfer coefficient U = 1/(R_total·A), the U-value quoted for wall and window panels.
The largest resistance controls the heat rate and takes most of the temperature drop. In an insulated metal wall, the metal skins hardly matter. Contact between layers adds an interface resistance that can be significant for metal-to-metal joints but is small next to insulation.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Fourier’s law
q″ = −k·dT/dx
q″ = heat flux (W/m²), k = thermal conductivity (W/(m·K)), x = position through the wall (m)
Incropera et al., Fundamentals of Heat and Mass Transfer
Plane wall heat rate
q = k·A·(T₁ − T₂) / L
A = area (m²), L = thickness (m), T in °C or K
Incropera et al., Fundamentals of Heat and Mass Transfer
Conduction and convection resistance
R_cond = L / (k·A), R_conv = 1 / (h·A)
R in K/W, h = convection coefficient (W/(m²·K))
Çengel & Ghajar, Heat and Mass Transfer
Series network and U-value
q = (T∞,1 − T∞,2) / ΣR, U = 1 / (ΣR·A)
U in W/(m²·K)
Çengel & Ghajar, Heat and Mass Transfer
Worked examples
Example: Heat loss through an oven wall panel
Given
An industrial curing oven wall panel is 3 mm steel, 75 mm mineral wool, and 1.5 mm steel. Its area is 2.4 m².
The inside surface is at 220 °C and the outside surface at 40 °C. Take k = 50 W/(m·K) for the steel and 0.040 W/(m·K) for the mineral wool, typical values.
Find
The resistance of each layer per square metre.
The heat flux and the heat loss through the panel.
The temperature drop across each layer.
Solution
Resistance per unit area, R″ = L/k, for each layer.R″_steel,in = 0.003/50 = 0.00006, R″_wool = 0.075/0.040 = 1.875, R″_steel,out = 0.0015/50 = 0.00003 m²·K/W
The layers are in series, so the resistances add.R″_total = 1.87509 m²·K/W
Heat flux and total heat rate.q″ = (220 − 40) / 1.875 = 96.0 W/m², q = 96.0 × 2.4 = 230 W
Temperature drop across each layer is q″ × R″.ΔT_wool = 96.0 × 1.875 = 180.0 °C, ΔT_steel,in = 96.0 × 0.00006 = 0.006 °C
Answer: The panel loses about 96 W/m², or 230 W over 2.4 m². The mineral wool carries effectively the whole 180 °C drop; the steel skins could be left out of the calculation with no visible change.
Example: Heat gain through a freezer panel, with air films
Given
A walk-in freezer has 30 m² of wall panel: 100 mm polyurethane foam between thin steel skins.
Inside air is −20 °C and room air is 24 °C. Take k = 0.024 W/(m·K) for the foam and h = 8 W/(m²·K) on each side, typical values for still air; the skins are neglected.
Heat flux and total heat gain.q″ = 0.226 × (24 − (−20)) = 9.96 W/m², q = 9.96 × 30 = 299 W
Outside surface temperature is the room air minus the drop across the outside film.T_s,o = 24 − 9.96 × 0.125 = 22.8 °C
Answer: U ≈ 0.23 W/(m²·K), and the walls let in about 300 W, roughly 7.2 kWh a day for the refrigeration system to remove. The outside face stays at about 22.8 °C, close to room air, because the foam takes almost all the temperature drop.
Common mistakes
Adding conductivities instead of resistances. Layers in series add resistances L/k. Their conductivities do not add.
Mixing per-area and total resistances. Work either in R″ (m²·K/W) and q″, or in R (K/W) and q. Do not combine the two.
Forgetting the convection films. If the problem gives fluid temperatures rather than surface temperatures, include 1/h on each side.
Using thickness in millimetres with k in W/(m·K). Convert L to metres before dividing.
Assuming the temperature profile is linear when it is not. A straight-line profile needs steady state, constant k and no internal heat generation. Cylinders and spheres have log and 1/r profiles.
In industry
Frequently asked questions
What is the difference between thermal conductivity and thermal resistance?
Conductivity k is a property of a material. Resistance R = L/(k·A) is a property of a particular layer, combining the material with its thickness and area.
What is an R-value?
The building trade’s name for resistance per unit area, R″ = L/k. In the US it is quoted in ft²·°F·h/Btu; one unit is about 0.176 m²·K/W.
What is contact resistance?
An extra resistance at the joint between two solids, caused by air in the gaps between surface peaks. It matters most between metals and can be reduced with pressure or thermal paste.
Does the same method work for pipes?
Yes, with a different conduction resistance. For a cylindrical layer R = ln(r₂/r₁)/(2π·k·L), so insulation on a small pipe behaves differently from a flat wall.