Heat Transfer — Conduction

In steady state, thermal energy flows from hot to cold through a material; Fourier's law quantifies the rate as proportional to area and temperature gradient.

Concept

Fourier's law (1-D steady state): $\frac{dQ}{dt} = -kA\frac{dT}{dx}$

where $k$ is thermal conductivity (W/m·K). For a uniform slab of thickness $L$: $H = \frac{kA(T_1 - T_2)}{L}$

By analogy with electrical resistance, thermal resistance is $R_{th} = L/(kA)$. Series resistances add; parallel conductances add.

Derivation

Two slabs in series, thicknesses $L_1, L_2$ and conductivities $k_1, k_2$. In steady state, the same heat current $H$ passes through both: $H = \frac{k_1 A(T_1-T_m)}{L_1} = \frac{k_2 A(T_m-T_2)}{L_2}$

Eliminating $T_m$: $H = \frac{A(T_1-T_2)}{L_1/k_1 + L_2/k_2}$

so the effective conductivity for equal-area slabs: $\frac{L_1+L_2}{k_{eff}} = \frac{L_1}{k_1} + \frac{L_2}{k_2}$

For parallel slabs of equal length and different areas, the effective $k$ is area-weighted.

JEE Worked Example

Problem: A compound wall is made of two layers: 10 cm thick brick ($k=0.8$ W/mK) and 5 cm thick wood ($k=0.2$ W/mK). Outside is at $-10$°C, inside at $25$°C. Find the heat current per m² and the interface temperature.

Solution: $R_{br}=\frac{0.10}{0.8}=0.125,\quad R_{wd}=\frac{0.05}{0.2}=0.25$ $H/A = \frac{35}{0.375}=93.3\;\text{W/m}^2$

Interface from inside: $T_m = 25 - H R_{wd} = 25 - 93.3\times 0.25 = 1.7$°C (wood between inside and brick).

Traps

• Fourier's law assumes steady state; transient problems need the heat equation.
• For radial conduction in a cylinder, $H = 2\pi k L (T_1-T_2)/\ln(r_2/r_1)$ — not linear.
• In series, the highest-resistance layer drops most of the temperature.
• Don't confuse conductivity $k$ (intensive) with conductance $kA/L$ (extensive).
• For composite rods with different areas, area cancellation does not apply directly.

Key Takeaways

• $H = kA\Delta T / L$ in 1-D steady state.
• $R_{th} = L/(kA)$; series adds, parallel like resistors.
• The largest thermal-resistance layer controls heat flow.
• Use radial form for cylindrical and spherical geometries.