Conduction Through a Furnace Wall: 22-Agric-A3 Heat Engineering, May 2016, Q1

Problem 1: Conduction Through a Furnace Wall (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, k	1.7 W/m·K
Wall thickness, L	0.15 m
Inner-surface temperature, T₁	1400 K
Outer-surface temperature, T₂	1150 K
Wall dimensions	0.5 m × 1.2 m

Find. The rate of heat loss Q through the wall.

Approach. Steady-state, one-dimensional conduction with no internal generation reduces to Fourier's law across the plane wall: compute the heat flux from the two measured face temperatures, then multiply by the given wall area.

One-dimensional steady conduction across the fireclay-brick furnace wall, from the measured hot face (1400 K) to the measured cold face (1150 K).

Heat flux from Fourier's law. For steady 1-D conduction across a plane wall with no internal generation, $$q'' = k\,\frac{T_1-T_2}{L} = 1.7\times\frac{1400-1150}{0.15} = \boxed{2833.3\ \text{W/m}^2}$$
Total heat-loss rate. Multiplying by the given wall area $A = 0.5\times1.2 = 0.60\ \text{m}^2$, $$Q = q''A = 2833.3\times0.60 = \boxed{1700\ \text{W}}$$

The furnace wall carries 1.7 kW through a modest 0.6 m² opening because the 250 K temperature drop is large even though fireclay brick is a comparatively poor conductor — the same governing law used throughout the rest of this paper's conduction, convection and radiation problems, just applied here to a single homogeneous slab.

Quantity	Result
Heat flux, q″	2833.3 W/m²
Heat-loss rate, Q	1700 W

Concept & Theory

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

Key relations: $q'' = k\,\Delta T/L$ (flux, W/m²); $Q = q''A$ (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 through the wall, so the flux is constant and proportional to the temperature drop divided by the thickness. Because both face temperatures were measured directly, the unknown reduces to a single multiplication once the flux is found.

Common pitfall: Reporting the flux (per unit area) as the final answer instead of multiplying by the actual wall area — the question asks for the rate of heat loss through THIS wall, not the flux through an idealized unit area.

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

Question 1 of 4: Conduction Through a Furnace Wall

Problem 1: Conduction Through a Furnace Wall (25 points)