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.
Understanding Conduction Through Composite Walls and Thermal Resistance
A resistance network does more than predict the total heat loss. It can show the temperature at every boundary between materials. In a series wall, the same heat rate passes through each layer at steady state.
The temperature drop is not shared equally. A layer with greater resistance takes a greater share of the drop. This is why insulation can have a large temperature change across a small thickness, while a metal sheet beside it has only a small change.
Within a uniform layer with constant properties, temperature changes in a straight line with distance. The line is steeper in materials that conduct heat poorly. Finding interface temperatures matters when checking whether moisture may condense inside a wall or whether a component could exceed a safe operating temperature.
Real walls rarely have one continuous path. A framed building wall has insulation in most of its area, yet wooden or steel studs create separate paths. These paths experience nearly the same indoor and outdoor temperatures, but they do not carry the same heat rate.
A steel stud can bypass much of the insulation because it conducts heat far more easily. This effect is called thermal bridging. It raises total heat loss and can create cold patches on an indoor surface.
Those patches may lead to condensation or mould when indoor air is humid. When calculating parallel paths, use the actual area of each path. A result based only on the insulation properties can look very good while missing the heat loss through framing, fasteners, window edges, or concrete connections.
Contact resistance comes from the fact that solid surfaces are rough at a microscopic scale. Two plates that seem to touch are actually supported by many tiny high points. Small gaps between those points often contain air, which resists heat flow.
Contact resistance becomes smaller when surfaces are smoother, pressure is higher, or a suitable thermal paste fills the gaps. It can be important in electronic devices, heat exchangers, bolted joints, and metal assemblies.
Students should distinguish between contact resistance for a whole interface and contact resistance given per unit area. Mixing these forms creates an area error that can change the answer by a large factor.
Resistance models depend on assumptions that must be checked before using them. The simplest model assumes steady conditions, heat moving mainly in one direction, constant thermal conductivity, and no heat generated inside the wall. Thin walls near corners, pipes, screws, or windows often violate the one dimensional assumption.
Surface effects deserve care too. Convection depends on airflow, while radiation can matter for hot surfaces or air gaps. Keep units consistent throughout the calculation.
Temperature differences in kelvin have the same numerical size as temperature differences in degrees Celsius. Finally, check physical sense. Heat must move from warmer to cooler regions, temperatures should fall in that direction, and adding insulation should reduce heat loss unless a major parallel bridge remains.