Heat conduction through a two-layer composite slab (equal thicknesses) Two layers of different materials with conductivities k1 and k2 are placed in series; each layer has the same thickness. What is the equivalent thermal conductivity of the composite slab?

Introduction / Context:
Composite walls are common in insulation and heat-exchanger design. Determining an equivalent thermal conductivity allows engineers to treat multiple layers as a single homogeneous slab for steady, one-dimensional conduction.

Given Data / Assumptions:

• Two layers in series, equal thickness L for each.
• Thermal conductivities: k1 and k2.
• Same cross-sectional area A; steady 1-D conduction; contact resistance neglected.

Concept / Approach:
The total thermal resistance in series adds: R_total = L/(k1A) + L/(k2A). Define an equivalent single slab of thickness 2L and conductivity k_eq such that R_total = 2L/(k_eqA).

Step-by-Step Solution:
R_total = L/(k1A) + L/(k2A) = (L/A) * (1/k1 + 1/k2).Set 2L/(k_eqA) = (L/A) * (1/k1 + 1/k2).Divide by L/A: 2/k_eq = 1/k1 + 1/k2.Solve for k_eq: k_eq = 2k1k2 / (k1 + k2).

Verification / Alternative check:
Check symmetry: if k1 = k2 = k, then k_eq = 2kk/(2k) = k, as expected.

Why Other Options Are Wrong:

• (k1 + k2)/2 and sqrt(k1k2) are for parallel or other averaging concepts, not series conduction with equal thickness.
• k1*k2 and (k1^2 + k2^2)/(k1 + k2) do not satisfy resistance addition.

Common Pitfalls:
Confusing series and parallel arrangements; forgetting that thickness doubles in the equivalent single-layer model.

Final Answer:
k_eq = 2 * k1 * k2 / (k1 + k2)