Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The equivalent thickness of air layers in millimeters is to be calculated.

Concept Introduction:

The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow.

  $Q=-kA\frac{\left(T_2-T_1\right)}{x}$..... (1)

Here, $x$ is the thickness of the material, $T_1$ is the temperature of the hot side of the wall, $T_2$ is the temperature of the cold side of the wall, $A$ is the area of the heat conduction, and $k$ is the thermal conductivity of the material.

For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become:

  $Q=-A\frac{\left(T_2-T_1\right)}{{\displaystyle \sum _i^N\frac{x_i}{k_i}}}$..... (2)

Answer to Problem 10.1P

The air layer thickness is calculated as, $x_{air}=298\text{ mm}$ .

Explanation of Solution

Given information:

Thickness of the refractory fireclay brick, $x_A=200\text{ mm}$

Thickness of the kaolin brick, $x_B=100\text{ mm}$

Thickness of the steel pipe, $x_C=6\text{ mm}$

Thin layers of air are between brick and steel.

Fire side refractory temperature, $T_1={1150}^{\circ }\text{C}$

Temperature of the outside of steel side, $T_2={30}^{\circ }\text{C}$

Heat loss from the wall per meter square of the wall, $Q=300\frac{\text{W}}{{\text{m}}^2}$

For the appendix 11, take the thermal conductivity of fireclay brick, kaolin brick, steel plate, and air as:

  $\begin{array}{c}{k}_A=1.7\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}\\ k_B=0.11\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}\\ k_C=45.34\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}\\ k_{air}=0.11\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}\end{array}$

Let the thickness of air layer be $x_{air}$ . Use equation (2) for 1 m² of wall area to calculate the air layer thickness as:

  $\begin{array}{c}Q=-\frac{\left(T_2-T_1\right)}{\frac{x_A}{k_A}+\frac{x_B}{k_B}+\frac{x_C}{k_C}+\frac{x_{air}}{k_{air}}}\\ 300\frac{\text{W}}{{\text{m}}^2}=-\frac{\left({30}^{\circ }\text{C}-{1150}^{\circ }\text{C}\right)}{\frac{200\text{ mm}\times \frac{\text{m}}{1000\text{ mm}}}{1.7\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}}+\frac{100\text{ mm}\times \frac{\text{m}}{1000\text{ mm}}}{0.11\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}}+\frac{6\text{ mm}\times \frac{\text{m}}{1000\text{ mm}}}{45.34\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}}+\frac{x_{air}}{0.11\frac{\text{W}}{\text{m}{\cdot }^{\circ }\text{C}}}}\\ x_{air}=0.298\text{ m}\\ x_{air}=298\text{ mm}\end{array}$