Conduction through composite walls is the study of heat flow across layered materials such as insulation, brick, metal, glass, and air gaps. Engineers use thermal resistance models to simplify one-dimensional steady conduction problems into circuits that look like electrical resistance networks. This cheat sheet helps students organize the main formulas, assumptions, and solution steps needed for walls with layers in series or parallel.
It is especially useful for heat transfer, HVAC, energy systems, and mechanical design courses.
The core idea is that heat transfer rate equals a temperature difference divided by total thermal resistance, q = ΔT/R_total. For a plane wall layer, the conduction resistance is R_cond = L/(kA), where L is thickness, k is thermal conductivity, and A is area normal to heat flow. Convection at a surface is modeled as R_conv = 1/(hA), while thermal contact resistance may be added between imperfectly touching surfaces.
For composite walls, resistances in series add directly, and parallel heat paths use reciprocal addition.
Key Facts
- Fourier’s law for one-dimensional steady conduction through a plane wall is q = -kA dT/dx.
- For a single plane wall with constant k, the heat transfer rate is q = kA(T_hot - T_cold)/L.
- The conduction thermal resistance of a plane layer is R_cond = L/(kA).
- The convection thermal resistance at a surface is R_conv = 1/(hA).
- For layers in series, total resistance is R_total = R_1 + R_2 + R_3 + ... and q = (T_hot - T_cold)/R_total.
- For parallel heat paths with the same temperature difference, the equivalent resistance satisfies 1/R_eq = 1/R_1 + 1/R_2 + 1/R_3 + ....
- Thermal contact resistance is included as R_contact = R''_contact/A, where R''_contact is contact resistance per unit area.
- The overall heat transfer coefficient is defined by q = UAΔT, so U = 1/(R_total A) when R_total is written for total area A.
Vocabulary
- Thermal conductivity
- Thermal conductivity k measures how easily a material conducts heat, with larger k giving lower conduction resistance.
- Thermal resistance
- Thermal resistance R is the opposition to heat flow, defined by R = ΔT/q for steady heat transfer.
- Composite wall
- A composite wall is a wall made of two or more layers or regions with different thermal properties.
- Contact resistance
- Contact resistance is the extra thermal resistance caused by imperfect contact, surface roughness, or small gaps between materials.
- Overall heat transfer coefficient
- The overall heat transfer coefficient U combines all conduction and convection resistances into q = UAΔT.
- Steady state
- Steady state means temperatures do not change with time, so the heat transfer rate is constant through each series layer.
Common Mistakes to Avoid
- Using area inconsistently in resistance formulas is wrong because R_cond = L/(kA) and R_conv = 1/(hA) both depend on the heat flow area.
- Adding parallel resistances directly is wrong because parallel paths share the same temperature difference, so use 1/R_eq = 1/R_1 + 1/R_2 + ....
- Forgetting surface convection resistances is wrong when the wall is exposed to fluids, because the inside and outside films can dominate the total resistance.
- Treating heat rate q and heat flux q'' as the same quantity is wrong because q is in watts, while q'' = q/A is in watts per square meter.
- Ignoring contact resistance is wrong when solid surfaces are bolted, pressed, or layered imperfectly, because microscopic air gaps can add significant resistance.
Practice Questions
- 1 A wall has area 12 m2, thickness 0.20 m, thermal conductivity 0.80 W/(m K), and surface temperatures 80°C and 25°C. Find the steady heat transfer rate.
- 2 A composite wall has two layers in series: L1 = 0.10 m, k1 = 0.20 W/(m K), L2 = 0.05 m, k2 = 1.5 W/(m K), and A = 5 m2. If the temperature difference is 40 K, find R_total and q.
- 3 A wall has inside convection h_i = 10 W/(m2 K), outside convection h_o = 25 W/(m2 K), area 8 m2, and a solid layer resistance of 0.30 K/W. Find the total thermal resistance and the heat transfer rate for ΔT = 60 K.
- 4 In a composite wall with insulation and a metal stud in parallel, explain why the metal path can strongly reduce the wall’s effective thermal resistance even if it covers a small fraction of the area.