Question

The thickness of insulator is desired to be determined to reduce heat loss from the room at 20°C to air outside at 4°C. The thickness of the wall is 20 cm. The thermal conductivities of the wall and insulator are 1,4 and 0,055 W/m°C, respectively. The convection heat transfer coefficient at the wall surface facing the room is 2,6 W/m2°C while the convection heat transfer coefficient at the insulator surface facing the outside air is 5,1 W/m2°C. Determine the minimum thickness of the insulator in [cm] that reduces heat loss by 50% compared to the case where there is no insulator. Assume the perfect contact between the insulator and the wall. (The tolerance of your answer is 2%) air at 4°C Room at 20°C insulation wall

Answer 1

The problem requires calculating the thickness of an insulator needed to reduce heat loss by 50%, using the concept of thermal resistance and conductivity. The solution involves determining the combined thermal resistance of the wall and insulator, with the goal of doubling the wall's solo resistance.

Step-by-step explanation:

To solve this physics problem concerning thermal conductivity and insulation, we need to use the concept of heat transfer through conduction. The original heat loss through the wall without insulation is calculated using Fourier's law of heat conduction. The formula for heat loss Q through a wall is Q = (kAΔT)/d, where k is the thermal conductivity, A is the area, ΔT is the temperature difference, and d is the thickness of the material.

The rate of heat loss must be reduced by 50% when the insulator is added, so we need to determine the new thickness 'x' that satisfies this new condition. Assuming the surface area and temperature difference remain constant, both the wall and the added insulator will have their respective thermal resistances in series. Thus, the combined resistance with the insulator should be twice as much as the initial resistance of the wall alone.

The equation for combined resistance R_total is 1/R_total = 1/R_wall + 1/R_insulator. By solving this equation, where R_insulator = x/k_insulator, and the original heat loss is known, we can find the thickness 'x' of the insulator that will reduce the heat loss by 50%. To solve for 'x', we can rearrange the equation and input the given values for k_wall, k_insulator, and the original thickness of the wall.