Question

Consider steady one-dimensional heat conduction in a plane wall in which the thermal conductivity varies linearly. The error involved in heat transfer calculations by assuming constant thermal conductivity at the average temperature is \((a)\) none, \((b)\) small, or \((c)\) significant.

Short answer

The error involved in heat transfer calculations by assuming constant thermal conductivity at the average temperature for a plane wall with linearly varying thermal conductivity can be either (b) small or (c) significant, depending on the specifics of the problem.

Step-by-step solution

Understand the concepts of heat conduction and thermal conductivity.

To solve this problem, we need to understand that heat conduction is the process in which heat energy transfers through a material due to the temperature gradient. Thermal conductivity is a property of the material and depends on its temperature, and it tells us how effective a material is at conducting heat. In this case, we have steady one-dimensional heat conduction in a plane wall, and the thermal conductivity varies linearly.

Examine the temperature profile and thermal conductivity variation.

Since the wall's heat conduction is one-dimensional and steady, the temperature profile depends only on the wall's width. The thermal conductivity varies linearly, meaning it increases or decreases at a constant rate with temperature. We can represent this linear variation as: k(T) = k_0 + β(T - T_0) where k(T) is the thermal conductivity at temperature T, k_0 is the thermal conductivity at a reference temperature T_0, and β is the linear coefficient of thermal expansion.

Compare the heat transfer calculations for constant and linearly varying thermal conductivity.

To find out the error involved in assuming constant thermal conductivity, we will compare the heat transfer calculations for both cases: constant and linearly varying thermal conductivity. For constant thermal conductivity, the heat transfer through the plane wall can be calculated using Fourier's law: q = -kA(ΔT/Δx) For linearly varying thermal conductivity, the heat flux equation becomes: q = -A∫(k_0 + β(T - T_0))(dT/dx)dx Now, we need to analyze how much difference there is between these two calculations.

Determine the significance of the error.

Generally, assuming constant thermal conductivity at the average temperature can lead to some error in heat transfer calculations, but the error depends on how much the thermal conductivity varies across the temperature range. If the variation is small, the error will also be small. However, if the variation is significant, the error could be significant as well. One can assume that the error involved in heat transfer calculations by assuming constant thermal conductivity at the average temperature is small if the linear variation's coefficient β is small. In cases where β is large, the error could be significant. Therefore, without specific temperature and thermal conductivity values, we can only determine that the error involved in heat transfer calculations by assuming constant thermal conductivity at the average temperature can be \((b)\) small or \((c)\) significant, depending on the specifics of the problem.