Conduction Through a Slab: 22-Agric-A3 Heat Engineering, December 2013, Q2

Problem 2: Conduction Through a 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, k	0.2 W/m·K
Front-face temperature, T₁	415°C
Conduction heat rate, Q	3 kW (3000 W)
Slab area, A	10 m²
Slab thickness, L	2.5 cm (0.025 m)

Find. The back-face temperature T₂.

Approach. One-dimensional steady conduction with no internal generation reduces to Fourier's law across the plane slab; with Q, k, A and L already known, invert the law to solve directly for the unknown back-face temperature.

One-dimensional conduction across the slab: a steady 3 kW flows from the known hot face (415°C) through the 2.5 cm thickness to the unknown back face.

Invert Fourier's law for the unknown back-face temperature. For steady 1-D conduction across a plane wall, $Q = kA(T_1-T_2)/L$, so $$T_2 = T_1 - \frac{Q\,L}{kA} = 415 - \frac{3000\ \text{W}\times0.025\ \text{m}}{0.2\ \dfrac{\text{W}}{\text{m}\cdot\text{K}}\times10\ \text{m}^2} = 415 - 37.5 = \boxed{377.5\,{}^{\circ}\text{C}}$$ The 37.5°C drop across just 2.5 cm reflects the slab's low thermal conductivity (0.2 W/m·K is typical of a firebrick or low-density refractory, not a metal), which is why even a modest 3 kW load through a fairly large 10 m² face produces a temperature difference this size.

This slab problem is a direct inversion of the same steady-state conduction law used throughout the rest of this paper's collector and radiation problems: whichever of Q, T₁, T₂, k, A or L is unknown, the other five pin it down uniquely because the temperature profile through a plane wall with no internal generation is always linear.

Quantity	Result
Back-face temperature, T₂	377.5°C

Concept & Theory

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

Key relations: $Q = kA\,\dfrac{\Delta T}{L}$ (rate, W); rearranged for an unknown face temperature, $T_2 = T_1 - QL/(kA)$; 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 heat rate is constant through the slab and proportional to the temperature drop divided by the thermal resistance $L/(kA)$. Because Q, k, A and L are all given here, the temperature drop $\Delta T = QL/(kA)$ is fixed and subtracts directly from the known front-face temperature.

Common pitfall: Forgetting to convert the thickness (2.5 cm) to metres before dividing by $kA$, which understates $\Delta T$ by a factor of 100 and gives an implausibly small temperature drop.

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

Question 2 of 4: Conduction Through a Slab

Problem 2: Conduction Through a Slab (25 points)