# Question 4.7: A two-layer furnace wall consists of a 60-mm ceramic wool ma......

A two-layer furnace wall consists of a 60-mm ceramic wool material on the hot side and an 80-mm rock wool layer on the cold (outer) side.

The initial temperature of the furnace and the atmosphere is 30°C. The heat is then turned on to obtain a temperature of 1250°C on the inside.

Find the heat absorbed per square meter by the wall in a heating period of 90 min and the distribution of the wall temperature at the end of heating.

For ceramic wool

Density $\rho_1$ = 300 kg/m³
Thermal conductivity $\lambda_1$ = 0.110 w/m°C
Specific heat $c_1$ = 1.07  kJ/kg°C

For rock wool

Density $\rho_2$ = 100 kg/m³

Thermal conductivity $\lambda_2$ = 0.032 w/m°C

Specific heat $c_2$ = 0.75 kJ/kg°C

Heat transfer coefficient inside the furnace = 0.25 kw/m²°C.
Heat transfer coefficient between outer wall and atmosphere = 10 w/m²°C.

Step-by-Step
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For ceramic wool

Thermal diffusivity $\alpha_1=\frac{\lambda_1}{c_1 \rho_1}$

$=\frac{0.110}{1.07 \times 10^3 \times 300} \\ =0.342 \times 10^{-6} \ \mathrm{~m}^2 / \mathrm{sec}$

For rock wool

Thermal diffusivity $\alpha_2=\frac{\lambda_2}{c_2 \rho_2}$

\begin{aligned}& =\frac{0.032}{0.75 \times 10^3 \times 10^2} \\& =0.4 \times 10^{-6} \ \mathrm{~m}^2 / \mathrm{sec}\end{aligned}

Let us divide the ceramic wool layer into three layers of 20 mm each. Then

\begin{aligned}\Delta \tau & =\frac{0.02^2}{2 \times 0.342 \times 10^{-6}} \\& =9.6 \ \mathrm{mins} \simeq 10 \ \mathrm{~min} \\\text { and } 2 \Delta x & =0.02 \sqrt{\frac{0.4}{0.342}} \\& =21 \ \mathrm{~mm} \simeq 20 \ \mathrm{~mm}\end{aligned}

Similarly, the thermal resistances of the 1∆x and 2∆x segments are

\begin{aligned}& r_1=\frac{0.1020}{0.110}=0.18 \frac{\mathrm{m}^2{ }^{\circ} \mathrm{C}}{\mathrm{w}} \\& r_2=\frac{0.020}{0.32}=0.625 \frac{\mathrm{m}^2{ }^{\circ} \mathrm{C}}{\mathrm{w}}\end{aligned}

The temperature distribution is shown in Table 4.2. The first temperature at 0∆τ, 1∆$x_o$ is shown as the average of 1250 and 30°C. The inner face then attains 1250°C and retains it throughout the 90-min period.
Except for, this, the whole lining, inner and outer, is at 30°C at the beginning and then gradually gets warmed up.

The temperatures are calculated by the usual formula

$t_n=\frac{t_{n-1 \Delta n-1}+t_{n+1 \Delta n+1}}{2}$

The temperature at the contact layer, i.e., 1∆$x_3$ is calculated by the formula

$t_1 \Delta x_3=\frac{t_{1 \Delta x_2} X r_2+t_{2 \Delta x_0} X r_1}{r_1+r_2}$

where $r_1$ and $r_2$ are the thermal resistances of the two materials.

$r_1=\frac{\Delta x_1}{\lambda_1} \quad \text { and } \quad r_2=\frac{\Delta x_2}{\lambda_2}$

After 90 min the average temperature of the outer wall is

$\frac{32+45+48}{3}=42^{\circ} \mathrm{C}$

Average temperature of the inner wall, i.e., ceramic fiber lining is

$\frac{1250+2 \times 778+2 \times 973+1 \times 606}{6}=893^{\circ} \mathrm{C}$

Average temperature of rock wool lining is

$\frac{378+2 \times 180+2 \times 106+48}{6}=166^{\circ} \mathrm{C}$

The amount of heat stored by 1 m² of the linings:
For ceramic wool

Thickness = 3 × 20 = 600 mm = 0.06 m

Specific heat = 1.07 kJ/kg°C

density = 300 kg/m³

Heat absorbed/m³ = 0.06 × 1 × 300 × 1.07 × 10³ × 893

= 17,200 kJ

For rockwool:

Thickness, 4 × 20 = 0.800 mm = 0.08 m

Specific heat = 0.75 kJ/kg°C

density = 100 kg/m³

Heat absorbed/m³  = 0.08 × 1 × 100 × 0.75 × 10³ × 166

= 1000 kJ

Total heat obsorbed by the lining in 90 min

= 17200 + 1000 = 18,200 kJ/m²

The temperature of the outer wall is > 30°C from 7∆t and has an average value of 42°C. Heat lost to the environment is

= 10 × (42 – 30) (90 – 60) × 60

= 216 kJ/m², 90 min

Suppose that for design purposes, the following restrictions apply:

1. Hot face temperature < 1400°C
2. Rock wool temperature < 750°C
3. Furnace outer wall temperature < 60°C

A glance at the table will show that restriction (1) is fixed by the data and selection of proper ceramic fiber.
The interface temperature (at 2∆$x_0$) is well below 750°C as required by restriction (2).

The outer wall temperature is below 60°C, satisfying restriction (3). However, if this restriction could be loosened, we can cut down the last outer layer, (2∆$x_3$), thus reducing the cost

Table 4.2

 Time Central Layer ∆τ 1∆$x_0$ 1∆$x_1$ 1∆$x_2$ 1∆$x_3$ 2∆$x_0$ 2∆$x_1$ 2∆$x_2$ 2∆$x_3$ 0∆τ 640 30 30 30 30 30 30 30 1∆τ 1250 335 30 30 30 30 30 30 2∆τ 1250 640 182 30 30 30 30 30 3∆τ 1250 716 335 60 30 30 30 30 4∆τ 1250 792 338 266 45 30 30 30 5∆τ 1250 830 549 312 148 38 30 30 6∆τ 1250 900 571 460 171 89 34 30 7∆τ 1250 910 680 482 274 102 60 32 8∆τ 1250 965 696 590 292 167 67 45 9∆τ 1250 973 778 606 378 180 106 48 60 mm 80 mm Totaling 140 mm

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