Question

The both sides of a large plane wall are exposed to air. The difference, (TA – TB), of the two air temperatures, is 39ºC. The wall conducts 74 W/m2 when there is no insulation. Find the minimum thickness of insulation in [mm] with kins = 0.053 W/mºC in order that the heat conduction rate is less than 42 W/m2. Hint: You can solve this problem without knowing the thicknesses of the wall and insulator. (The tolerance of your answer is 2%.)

Answer 1

To calculate the minimum insulation thickness for reduced heat conduction, apply thermal resistance and Fourier's Law, using provided heat flux and temperatures, solve for thermal resistance of insulation, and finally calculate insulation thickness with the given thermal conductivity value.

Step-by-step explanation:

In this physics problem, we're challenged with calculating the minimum thickness of insulation required to reduce the heat conduction rate through a wall. In order to accomplish this, we need to apply the concept of thermal resistance and Fourier's Law of Heat Conduction. Initially, the wall conducts 74 W/m2 with a temperature difference of 39°C. The goal is to reduce the heat conduction rate to less than 42 W/m2. Given the thermal conductivity (kins) of the insulation material, we can use the following relationship:

Q = (TA - TB) / (Rwall + Rins)

where Q is the heat flux, TA - TB is the temperature difference, and Rwall and Rins are the thermal resistances of the wall and insulation, respectively. Since we're looking to find the insulation thickness, we rearrange to solve for Rins and then calculate the minimum thickness by thickness = Rins * kins. After working through the algebra, we reach the minimum insulation thickness required to achieve the objective in millimeters.