Temperature drop or heat loss is dependent on thermal resistance offered by the brick material.

We have ,
Length = 0.5 meter
conductivity =. 0.7w/m k
Area = 1 metre^2

Thermal Resistance = L/KA = 0.5/0.7 = 0.71429

Now for thermal conductivity of 0.14 w/m K

Thermal resistance = L/ KA

0.71429 = L/0.14

L = 0.714290*0.14

L = 0.10 meters.

Thank you.

The Problem

We have a furnace made of a red brick wall, with a thickness of 0.5 m and a thermal conductivity of 0.7 W/mK. We want to replace this brick wall with a layer of diatomite earth, which has a thermal conductivity of 0.14 W/mK. We want to determine the thickness of the diatomite layer that will result in the same heat loss and temperature drop as the original brick wall.

Concepts to Understand

1. Heat conduction: Heat conduction is the transfer of heat through a material by means of atomic or molecular interactions. It occurs when there is a temperature gradient within a solid or between two solids in contact.
2. Fourier's Law of Heat Conduction: Fourier's law states that the rate of heat transfer through a material is directly proportional to the surface area, temperature difference, and inversely proportional to the thickness of the material. It can be expressed as: Q = (k * A * ∆T) / d, where Q is the heat transfer rate, k is the thermal conductivity, A is the surface area, ∆T is the temperature difference, and d is the thickness of the material.

Solution

To determine the thickness of the diatomite layer, we need to equate the heat transfer rate through the brick wall and the diatomite layer.

Step 1: Equate the heat transfer rates

Let Q_brick be the heat transfer rate through the brick wall and Q_diatomite be the heat transfer rate through the diatomite layer.

Q_brick = Q_diatomite

Step 2: Apply Fourier's law of heat conduction

Q_brick = (k_brick * A * ∆T) / d_brick
Q_diatomite = (k_diatomite * A * ∆T) / d_diatomite

where k_brick and k_diatomite are the thermal conductivities of the brick and diatomite, A is the surface area, ∆T is the temperature difference, d_brick is the thickness of the brick wall, and d_diatomite is the thickness of the diatomite layer.

Step 3: Substitute the given values

Q_brick = (0.7 * A * ∆T) / 0.5
Q_diatomite = (0.14 * A * ∆T) / d_diatomite

Step 4: Equate the heat transfer rates

(0.7 * A * ∆T) / 0.5 = (0.14 * A * ∆T) / d_diatomite

Step 5: Solve for d_diatomite

Cross-multiplying and simplifying the equation:

0.7 * d_diatomite = 0.14 * 0.5

d_diatomite = (0.14 * 0.5) / 0.7

d_diatomite ≈ 0.1 m

Therefore, the thickness of the diatomite layer that will result in the same heat loss and temperature drop as the original brick wall is approximately 0.1 m.