Question 4.6: A furnace wall made of fire clay bricks has a thickness of 2......
A furnace wall made of fire clay bricks has a thickness of 200 mm and is initially at a temperature 60°C. The temperature of the furnace is suddenly increased to 1400°C. The outer surface of the wall is at an ambient temperature 30°C. Find the temperature distribution of the wall 2 h after the heating starts and the heat absorbed by the wall in this time
Density of bricks \rho \quad 1.9 \times 10^3 \ \mathrm{~kg} / \mathrm{m}^3
Thermal conductivity \lambda \quad 0.85 \ \mathrm{~kg} / \mathrm{m}^{\circ} \mathrm{C}
Specific heat \begin{array}{cc} c & 0.9 \ \mathrm{~kJ} / \mathrm{kg}^{\circ} \mathrm{C} \end{array}
Heat transfer coefficient inside the furnace h_i \quad 0.32 \ \mathrm{kw} / \mathrm{m}^{2{ }^{\circ} \mathrm{C}}
Heat transfer coefficient outside the wall h_o 12 \ \mathrm{w} / \mathrm{m}^2{ }^{\circ} \mathrm{C}
The thermal diffusivity will be
\begin{aligned}a & =\frac{\lambda}{c \rho}=\frac{0.85}{1.9 \times 10^3 \times 0.9 \times 10^3} \\& =0.5 \times 10^{-6} \ \mathrm{~m}^2 / \mathrm{sec}\end{aligned}Divide the wall into eight segments to obtain ∆x
\begin{aligned}\Delta x & =\frac{200}{8}=25 \ \mathrm{~mm} \\\Delta \tau & =\frac{\Delta x^2}{2 a} \\& =625 \ \mathrm{sec}=10.4 \ \mathrm{~min}\end{aligned}The total time is 2 h, hence,
time intervals are \frac{120}{10.4} = 11.6, say 12
The furnace at the beginning of heating, i.e., at ∆\tau < 0 is at 60°C. Hence, the whole wall, throughout its thickness is at 60°C
At ∆\tau = 0 when the heating is started the inner wall will attain 1400°C very quickly. The initial temperature of the inner side of the wall will be the mean of source and ambient temperatures
\begin{aligned}\therefore \Delta t \text { at } 0 \Delta x, & 0 \Delta \tau \\& =\frac{1400+60}{2}=730^{\circ} \mathrm{C}\end{aligned}All other segments are at 60°C at ∆\tau = 0.
In the next time increment, 1∆\tau, the inner temperature of the wall will attain 1400°C and will remain constant.
We assume that the temperatures of various segments of the wall, i.e., 1∆x, 2∆x, …, n∆x will increase progressively by one step at a time, i.e., at ∆\tau. Only the first segment will get heated. At the next instant the first and second segments will get heated, and so on.
For calculations the expression used will be
The results are best presented in a tabular form as shown in the Table 4.1. The time segments form rows and the wall segments form columns.
For example:
If the temperature at 4∆t and 2∆x is to be calculated, note that at
t_{3 \Delta \tau}, 1 \Delta x \text { is } 814^{\circ} \mathrm{C}and t_{3 \Delta \tau}, 3 \Delta x \text { is } 144^{\circ} \mathrm{C}
Hence , t_{4 \Delta \tau}, 2 \Delta x \\ =\frac{814+144}{2}=479^{\circ} \mathrm{C}
In a similar manner all the temperatures for various time and wall segments are calculated.
Note : The temperature of the outer wall is practically constant for all the heating time, i.e., 2 h.
The problem states that the outer temperature is held at 60°C and the atmospheric temperature is 30°C. Heat will be lost from the wall to the environment by convection and radiation at the given transfer rate of 12 w/m²°C. This loss is not calculated here.
The problem also states that the wall temperature (inside) is raised to 1400°C. The heat source (flame or electric element) will have to be at a slightly higher temperature (say 1450–1500°C) to have a positive transfer coefficient of 0.32 kW to the wall.
The average temperature of the wall at the end of the 2 h heating period is
\frac{\begin{aligned} & \frac{1400+1097}{2}+\frac{1097+817}{2}+\frac{817+577}{2}+\frac{577+390}{2}+\frac{390+253}{2} \\ & +\frac{253+165}{2}+\frac{165+111}{2}+\frac{111+82}{2}+\frac{82+68}{2}+\frac{68+60}{2} \\ & \end{aligned}}{10} \\ =\frac{4291}{10} \sim 430^{\circ} \mathrm{C}=t_aThe thickness of the wall is 200 mm. Let us calculate the heat stored for the 1 m² area
Volume of wall V = 1 × 0.2 = 0.2 m³
Specific heat c = 0.9 kJ/kg°C
Density \rho = 1.9 × 10³ kg/m
Time \tau = 2 h
Heat stored per 2 h period
\begin{aligned} & =V \rho c t_a \\ & =0.2 \times 1.9 \times 10^3 \times 0.9 \times 430 \\ & =147 \times 10^3 \ \mathrm{~kJ} / 2 \mathrm{~h} \end{aligned}Table 4.1
| Time | ← Furnace Inner Side | Segment Number | Furnace outer side → | ||||||||
| ∆τ | 0∆x | 1∆x | 2∆x | 3∆x | 4∆x | 5∆x | 6∆x | 7∆x | 8∆x | 9∆x | 10∆x |
| 0∆τ | 730 | 60 | 60 | 60 | 60 | 60 | 60 | 60 | 60 | 60 | 60 |
| 1∆τ | 1400 | 395 | 60 | 60 | 60 | 60 | 60 | 60 | 60 | 60 | 60 |
| 2∆τ | 1400 | 730 | 228 | 60 | 60 | 60 | 60 | 60 | 60 | 60 | 60 |
| 3∆τ | 1400 | 814 | 395 | 144 | 60 | 60 | 60 | 60 | 60 | 60 | 60 |
| 4∆τ | 1400 | 898 | 479 | 227 | 102 | 60 | 60 | 60 | 60 | 60 | 60 |
| 5∆τ | 1400 | 940 | 562 | 290 | 143 | 81 | 60 | 60 | 60 | 60 | 60 |
| 6∆τ | 1400 | 981 | 615 | 352 | 185 | 101 | 70 | 60 | 60 | 60 | 60 |
| 7∆τ | 1400 | 1008 | 666 | 400 | 226 | 127 | 80 | 65 | 60 | 60 | 60 |
| 8∆τ | 1400 | 1033 | 704 | 443 | 263 | 153 | 96 | 70 | 63 | 60 | 60 |
| 9∆τ | 1400 | 1052 | 738 | 484 | 298 | 180 | 111 | 80 | 65 | 61.5 | 60 |
| 10∆τ | 1400 | 1069 | 768 | 518 | 332 | 204 | 130 | 88 | 71 | 63 | 60 |
| 11∆τ | 1400 | 1084 | 794 | 550 | 361 | 231 | 146 | 100 | 76 | 65 | 60 |
| 12∆τ | 1400 | 1097 | 817 | 577 | 390 | 253 | 165 | 111 | 82 | 68 | 60 |