Steady Conduction Through a Lead Slab: 22-Agric-A3 Heat Engineering, May 2013, Q1

Problem 1: Steady 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 (W/m²) and the total heat transfer rate Q (W) through the slab.

Approach. One-dimensional steady conduction with no internal generation reduces to Fourier's law across the plane slab; the flux is uniform, so the rate is simply the flux times the face area.

One-dimensional conduction across the lead slab: heat flux q flows from the hot face (110°C) to the cold face (50°C) over the 0.03 m thickness.

Heat flux from Fourier's law. For steady 1-D conduction across a plane wall, $$q = k\,\frac{T_1-T_2}{L} = 35\ \frac{\text{W}}{\text{m}\cdot\text{K}}\times \frac{110-50\ \text{K}}{0.03\ \text{m}} = \boxed{70{,}000\ \text{W/m}^2 = 70\ \text{kW/m}^2}$$
Heat transfer rate. The flux is uniform over the face, so the rate is $$Q = q\,A = 70{,}000\ \frac{\text{W}}{\text{m}^2}\times 0.4\ \text{m}^2 = \boxed{28{,}000\ \text{W} = 28\ \text{kW}}$$

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: $q = k\,\dfrac{\Delta T}{L}$ (flux, W/m²); $Q = qA$ (rate, W); for a composite wall the resistances $L/(kA)$ add in series.

Why this works: With no internal heat generation and steady state, the 1-D conduction equation $d^2T/dx^2 = 0$ integrates to a linear temperature profile, so the flux is constant through the slab and equal to $k$ times the temperature gradient. Because the geometry is planar (not radial), the area does not change with position, which is why $Q = qA$ needs no logarithmic or spherical correction.

Common pitfall: Forgetting to convert the thickness to metres (a bare "0.03" is often mis-read as already being in the same units as $k$), and reporting only q or only Q when the question asks for both.

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

Question 1 of 4: Steady Conduction Through a Lead Slab

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