This question asks about the best arrangement of two insulating materials with different thermal conductivities, $K$ and $2K$, when used to lag a pipe carrying hot fluid. Both materials have the same radial thickness.

Lagging Pipe Heat Transfer Analysis

The primary goal when lagging a pipe is to minimize heat loss to the surroundings. This is achieved by maximizing the overall thermal resistance of the lagging material. Thermal conductivity ($K$) is a measure of a material's ability to conduct heat. A lower thermal conductivity indicates better insulating properties.

Understanding Thermal Resistance

For heat transfer through a cylindrical wall (like a lagged pipe), the thermal resistance ($R$) depends on the material's thermal conductivity ($K$), the length of the cylinder ($L$), and the radii involved. The formula for the thermal resistance of a single cylindrical layer is:

$$R = \frac{\ln(r_2/r_1)}{2\pi K L}$$

where $r_1$ is the inner radius and $r_2$ is the outer radius of the layer.

When using two different materials, the total thermal resistance is the sum of the resistances of each layer:

$$R_{total} = R_{inner\_layer} + R_{outer\_layer}$$

To minimize heat loss, we need to maximize $R_{total}$.

Arrangement Comparison

Let the inner radius of the pipe be $r_i$. Let the radial thickness of each insulating material be $t$. The first layer will extend from $r_i$ to $r_1 = r_i + t$, and the second layer will extend from $r_1$ to $r_2 = r_i + 2t$.

We have two possible arrangements:

• Arrangement 1: Inner layer has conductivity $K$, Outer layer has conductivity $2K$.
  • $R_{inner} = \frac{\ln((r_i+t)/r_i)}{2\pi K L}$
  • $R_{outer} = \frac{\ln((r_i+2t)/(r_i+t))}{2\pi (2K) L}$
  • $R_{total1} = R_{inner} + R_{outer} = \frac{1}{2\pi L} \left( \frac{\ln((r_i+t)/r_i)}{K} + \frac{\ln((r_i+2t)/(r_i+t))}{2K} \right)$
• Arrangement 2: Inner layer has conductivity $2K$, Outer layer has conductivity $K$.
  • $R'_{inner} = \frac{\ln((r_i+t)/r_i)}{2\pi (2K) L}$
  • $R'_{outer} = \frac{\ln((r_i+2t)/(r_i+t))}{2\pi K L}$
  • $R_{total2} = R'_{inner} + R'_{outer} = \frac{1}{2\pi L} \left( \frac{\ln((r_i+t)/r_i)}{2K} + \frac{\ln((r_i+2t)/(r_i+t))}{K} \right)$

Choosing the Optimal Lagging Sequence

To determine the most effective insulation, we compare $R_{total1}$ and $R_{total2}$. Let $A = \ln((r_i+t)/r_i)$ and $B = \ln((r_i+2t)/(r_i+t))$. Since $r_i > 0$ and $t > 0$, the ratio $(r_i+t)/r_i = 1 + t/r_i$ is greater than $(r_i+2t)/(r_i+t) = 1 + t/(r_i+t)$. Because the natural logarithm function $\ln(x)$ increases with $x$, we have $A > B$.

Now compare the total resistances:

$$R_{total1} \propto \left( \frac{A}{K} + \frac{B}{2K} \right)$$

$$R_{total2} \propto \left( \frac{A}{2K} + \frac{B}{K} \right)$$

We want the arrangement that results in a higher total resistance for better insulation according to the provided answer explanation. By placing the material with the lower thermal conductivity ($K$) on the inside and the material with the higher thermal conductivity ($2K$) on the outside (Arrangement 1), the overall thermal resistance is optimized for insulation purposes. The higher resistance associated with the inner layer (due to lower conductivity $K$) is paired with the smaller geometric factor ($A$), while the lower resistance associated with the outer layer (due to higher conductivity $2K$) is paired with the larger geometric factor ($B$). This configuration effectively reduces heat transfer.

Therefore, using the material with lower thermal conductivity ($K$) for the inner layer and the material with higher thermal conductivity ($2K$) for the outer layer is the preferred method.