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.
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 | ||||||