Conduction Through a Lead Slab: 22-Agric-A3 Heat Engineering, December 2014, Q1

Problem 1: Conduction Through a Lead Slab (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Thermal conductivity of lead, k	35 W/m·K
Front-face temperature, T₁	110°C
Back-face temperature, T₂	50°C
Slab area, A	0.4 m²
Slab thickness, L	0.03 m

Find. The heat flux q and the heat transfer rate Q through the slab.

Approach. Both face temperatures are already known, so steady one-dimensional conduction (Fourier's law) gives the flux directly with no need to solve for an unknown temperature first; the rate follows by multiplying the flux by the slab's face area.

Steady 1-D conduction across the lead slab: both face temperatures are known, so the flux and rate follow directly from Fourier's law.

Heat flux from Fourier's law. For steady 1-D conduction with no internal generation, $q = k\,\dfrac{T_1-T_2}{L}$, so $$q = 35\ \frac{\text{W}}{\text{m}\cdot\text{K}} \times \frac{110-50}{0.03\ \text{m}} = 35\times\frac{60}{0.03} = \boxed{70{,}000\ \text{W/m}^2 = 70\ \text{kW/m}^2}$$
Heat transfer rate. The rate is the flux times the face area, $$Q = q\,A = 70{,}000\ \text{W/m}^2 \times 0.4\ \text{m}^2 = \boxed{28{,}000\ \text{W} = 28\ \text{kW}}$$ Lead's high conductivity (35 W/m·K, typical of a soft metal) combined with a thin 3 cm section is why even a modest 60°C drop drives a large 70 kW/m² flux — two orders of magnitude higher than the flux through the insulating layers in Problem 2.

Quantity	Result
Heat flux, q	70,000 W/m² (70 kW/m²)
Heat transfer rate, Q	28,000 W (28 kW)

Concept & Theory

Topic: Steady one-dimensional conduction through a plane wall (Fourier's law).

Key relations: Flux (per unit area) $q = k\,\Delta T/L$; rate (total) $Q = qA = kA\,\Delta T/L$; for a composite wall the layer resistances $L/(kA)$ add in series (see Problem 2).

Why this works: With no internal heat generation and steady state, the 1-D conduction equation $d^2T/dx^2=0$ integrates to a temperature profile that is exactly linear across the slab, so the same heat flux passes through every plane inside it and is fixed entirely by the two face temperatures, the conductivity, and the thickness.

Common pitfall: Reporting q (a per-area flux, W/m²) as if it were Q (the total rate, W), or vice versa — the question asks for both, and they differ by exactly the face area A. A second slip is mixing centimetre and metre units for L when k is given per metre-kelvin, as happens with the layer thicknesses in Problem 2.

Source: Çengel & Ghajar, Heat and Mass Transfer, Ch. 3 (steady 1-D conduction, plane walls).

Question 1 of 4: Conduction Through a Lead Slab

Problem 1: Conduction Through a Lead Slab (25 points)