One-Dimensional Steady Conduction: Plane Wall

Understanding how heat moves through solid materials is fundamental to designing everything from building insulation to electronic heat sinks. For a simple, flat barrier—a plane wall—under steady-state conditions, the analysis becomes elegantly straightforward, providing the foundational principles for tackling more complex thermal systems. Mastering this concept allows you to predict temperature distributions and heat flow rates, which are critical for energy efficiency, material selection, and safety in engineering applications.

Fourier's Law and the Linear Temperature Profile

The analysis begins with Fourier's Law of Heat Conduction, which states that the rate of heat transfer through a material is proportional to the negative of the temperature gradient and the area perpendicular to the direction of heat flow. For one-dimensional, steady-state conduction through a plane wall of constant thermal conductivity $k$, this law simplifies significantly.

Mathematically, it is expressed as: $$q_x=-kA\frac{dT}{dx}$$ Under steady-state conditions, with no heat generation within the wall and constant $k$, the heat transfer rate $q_x$ is constant. Integrating Fourier's Law leads to the classic result: the temperature distribution $T(x)$ is linear. If the inner surface at $x=0$ is maintained at $T_{s,1}$ and the outer surface at $x=L$ is at $T_{s,2}$, the temperature at any point is: $$T(x)=T_{s,1}-\frac{T_{s,1}-T_{s,2}}{L}x$$ This linear drop is a direct consequence of energy conservation. The constant slope of the temperature line, $(T_{s,2}-T_{s,1})/L$, is the driving force for conduction. The heat transfer rate can then be calculated simply as: $$q=kA\frac{T_{s,1}-T_{s,2}}{L}$$

Thermal Resistance and Composite Walls

Real-world walls are rarely single materials. They are composite walls, consisting of multiple layers (e.g., brick, insulation, drywall). Analyzing each layer individually would be cumbersome. This is where the powerful concept of thermal resistance becomes indispensable. By analogy to electrical circuits, we define the conduction resistance for a plane wall as: $$R_{cond}=\frac{L}{kA}$$ This transforms the heat transfer equation into a form analogous to Ohm's Law: $q=\frac{\Delta T}{R_{cond}}$.

For multiple layers in perfect thermal contact and arranged in series—meaning heat flows sequentially through each—the total thermal resistance is simply the sum of the individual resistances: $$R_{tot}=R_1+R_2+...+R_n=\frac{L_1}{k_{1}A}+\frac{L_2}{k_{2}A}+...+\frac{L_n}{k_{n}A}$$ The total heat transfer rate through the composite wall is then: $$q=\frac{T_{\infty ,1}-T_{\infty ,2}}{R_{tot}}$$ Here, $T_{\infty ,1}$ and $T_{\infty ,2}$ are the fluid temperatures on either side of the wall. This approach dramatically simplifies analysis, allowing you to quickly assess the impact of adding or changing a layer's material or thickness.

The Overall Heat Transfer Coefficient (U)

While thermal resistance is excellent for analysis, design often uses the overall heat transfer coefficient, denoted as $U$. This coefficient condenses all resistances—including convective resistances at the wall surfaces—into a single, convenient number. It is defined such that: $$q=UA\Delta T_{overall}$$ where $\Delta T_{overall}$ is the total temperature difference between the two surrounding fluids. The relationship between $U$ and total resistance is inverse: $$U=\frac{1}{R_{tot}A}$$ For a plane wall with convection on both sides, the total resistance per unit area is: $$R_{tot}^{\prime }\prime =\frac{1}{h_1}+\frac{L}{k}+\frac{1}{h_2}$$ Therefore, $U=1/R_{tot}^{\prime }\prime$. A low $U$-value indicates high insulation (high resistance), while a high $U$-value indicates a good conductor (low resistance). This metric is paramount in building codes and appliance ratings (like for windows and refrigerators).

Contact Resistance at Interfaces

Our assumption of "perfect thermal contact" between layers in a composite wall is often inaccurate. In reality, surfaces are rough, and only peaks make contact; the gaps are filled with air or another fluid, which usually has a much lower thermal conductivity than the solids. This imperfect contact creates an additional resistance known as thermal contact resistance, $R_{t,c}$.

Contact resistance is defined experimentally and depends on surface roughness, contact pressure, interstitial material, and temperature. It is incorporated into the series resistance model just like a material layer: $$R_{tot}=\frac{1}{h_{1}A}+\frac{L_A}{k_{A}A}+\frac{R_{t,c}}{A}+\frac{L_B}{k_{B}A}+\frac{1}{h_{2}A}$$ Where $R_{t,c}$ is the thermal contact resistance for the interface area. Neglecting this resistance can lead to significant overestimations of heat transfer, especially in systems involving multiple mechanical joints or stacked materials, such as in electronics cooling.

Worked Example: Analyzing a Composite Wall

Consider a furnace wall comprised of a 10-cm layer of fireclay brick ($k=1.0$ W/m·K) and a 20-cm layer of common brick ($k=0.5$ W/m·K). The inner surface is exposed to gases at $1200^{\circ }$C with a convection coefficient $h=50$ W/m²·K, and the outer surface is exposed to air at $30^{\circ }$C with $h=20$ W/m²·K. Assume a contact resistance of $0.01$ K·m²/W between the bricks. Calculate the heat flux (q/A) and the temperature at the interface between the bricks.

Step 1: Calculate resistances per unit area.

• Convection (inside): $R_{conv,i}^{\prime }\prime =1/h_1=1/50=0.02$ K·m²/W
• Fireclay brick: $R_A^{\prime }\prime =L_A/k_A=0.10/1.0=0.10$ K·m²/W
• Contact: $R_{t,c}^{\prime }\prime =0.01$ K·m²/W
• Common brick: $R_B^{\prime }\prime =L_B/k_B=0.20/0.5=0.40$ K·m²/W
• Convection (outside): $R_{conv,o}^{\prime }\prime =1/h_2=1/20=0.05$ K·m²/W

Step 2: Find total resistance and heat flux. $$R_{tot}^{\prime }\prime =0.02+0.10+0.01+0.40+0.05=0.58\text{ Kcdotpm²/W}$$ $$q/A=\frac{T_{\infty ,1}-T_{\infty ,2}}{R_{tot}^{\prime }\prime }=\frac{1200-30}{0.58}\approx 2017\text{ W/m²}$$

Step 3: Find interface temperature. The temperature drop from the inside gas to the interface is: $$T_{\infty ,1}-T_{interface}=(q/A)\times (R_{conv,i}^{\prime }\prime +R_A^{\prime }\prime )=2017\times (0.02+0.10)\approx 242^{\circ }\text{C}$$ $$T_{interface}=1200-242=958^{\circ }\text{C}$$ This example shows how the systematic resistance method, including contact resistance, provides a clear path to finding any unknown temperature in the system.

Common Pitfalls

1. Assuming constant thermal conductivity ($k$): The value of $k$ often varies with temperature. For large temperature differences, using an average $k$ based on the mean temperature of the layer is necessary for accuracy. Applying the linear formula with a room-temperature $k$ value for a furnace wall will give incorrect results.
2. Ignoring contact resistance: As demonstrated, even a seemingly small contact resistance can be significant compared to conductive resistances, especially if materials are thin or highly conductive. Always include it in precision calculations unless specifically stated that contact is perfect.
3. Misapplying the series resistance model: The series addition $R_{tot}=R_1+R_2+...$ only holds if heat flows normally through each layer in sequence. If layers are arranged in parallel (e.g., studs in an insulated wall), you must use the parallel resistance formula. Carefully examine the physical heat flow path.
4. Confusing heat transfer rate ($q$) with heat flux ($q/A$ or $q^{\prime }\prime$): The formulas $q=\Delta T/R$ and $U=1/(R_{tot}A)$ require careful attention. Resistance $R$ has units of K/W, while resistance per unit area $R^{\prime }\prime$ has units of K·m²/W. Mixing these will lead to calculation errors by a factor of the area $A$.

Summary

• Under steady-state, one-dimensional conditions with constant properties, the temperature profile through a homogeneous plane wall is strictly linear, a direct result of Fourier's Law and energy conservation.
• The thermal resistance concept ($R=L/kA$)