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Calculation of insulation thickness Example 4-1: The thickness of a flat wall is 0.37 m; the temperature of its inner surface is t1 = 1650°C, while the temperature of its outer surface is t2 = 300°C. The thermal conductivity of the wall material is given by the formula (where the unit of t is °C, and the unit of λ is W/(m•°C)). If the thermal conductivity is calculated as a constant (using the average thermal conductivity) and as a variable, determine the temperature distribution equation and the heat transfer flux for a flat wall. Solution: (1) Assuming the thermal conductivity is constant, the average temperature of the flat wall is: The average thermal conductivity of the wall material is: Using this formula, the heat flux can be determined. Let the temperature at a distance x from the wall be t; then, using the formula, it can be shown that: Thus, the above equation represents the temperature distribution across the flat wall, indicating that there is a linear relationship between the distance x from the wall and the temperature at the isothermal surface. (2) The thermal conductivity is calculated as a function of variables. From the formula, it follows that by integration: (a) When..., substituting this value into equation a yields:. After rearranging this equation, we obtain:. This equation represents the temperature distribution across a single-layer flat wall when λ varies linearly with t; in such a case, the temperature distribution takes the form of a curve. The calculation results show that the heat conduction flux obtained is the same whether the thermal conductivity is treated as a constant or a variable ； The temperature distribution, on the other hand, is different: the former is linear while the latter is curved. This post was last edited by JLH on 2008-1-24 18:20 ]