Question

A thick wood slab \((k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.28 \times\) \(10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) ) that is initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) is exposed to hot gases at \(550^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~min}\). The heat transfer coefficient between the gases and the wood slab is $35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. If the ignition temperature of the wood is \(450^{\circ} \mathrm{C}\), determine if the wood will ignite.

Short answer

Answer: We can only provide an estimation, but based on the calculation (step 6), we have the expression for the temperature of the wood slab at distance Δx after 5 minutes: \(T_s = 823 - \frac{0.17\Delta T}{35 \times 1.19 \times 10^{-3}}\) Now, we need to compare the calculated temperature (T_s) with the ignition temperature (723 K). If T_s > 723 K, the wood will ignite. If T_s ≤ 723 K, the wood will not ignite. Please note that this is a rough estimation, and more detailed analysis is recommended for safety concerns.

Step-by-step solution

List the given values and convert them to SI units

Let's list the given values and convert them to SI units if required: k = 0.17 W/m·K (thermal conductivity) α = 1.28 x 10^{-7} m²/s (thermal diffusivity) T_initial = 25°C = 298 K (initial uniform temperature) T_gases = 550°C = 823 K (hot gases temperature) time = 5 min = 300 s (exposure time) h = 35 W/m²·K (heat transfer coefficient) T_ignition = 450°C = 723 K (ignition temperature)

Set up Fourier's law of heat conduction and Newton's law of cooling

Fourier's law of heat conduction: \(q = -k\frac{\Delta T}{\Delta x}\) Newton's law of cooling: \(q = h(T_g - T_s)\) (T_g - termperature of the hot gases, T_s - termperature of the wood surface)

Equate the heat flow

Equate the heat flow from conduction to convection: \(-k\frac{\Delta T}{\Delta x} = h(T_g - T_s)\)

Apply the boundary conditions

When the wood surface is in contact with the hot gases, the temperature of the wood surface will gradually increase. We need to find the distance from the surface (\(\Delta x\)) where the temperature increases. A good estimation for this distance can be calculated by multiplying the thermal diffusivity (α) with the exposure time (300 s), then taking the square root: \(\Delta x \approx \sqrt{2\alpha t}\)

Calculate the distance \(\Delta x\)

Plugging the values of \(\alpha = 1.28 x 10^{-7}\) m²/s and \(t = 300\) s, we get: \(\Delta x \approx \sqrt{2 \times 1.28 \times 10^{-7} \times 300} \approx 1.19 \times 10^{-3}\) m

Calculate temperature at x = \(\Delta x\)

Rearranging the equation mentioned in step 3 in order to find the temperature at distance \(\Delta x\). \(T_s = T_g - \frac{k\Delta T}{h\Delta x}\) Plugging the values \(k=0.17\:W/m·K\), \(T_g = 823\:K\), and \(h=35\:W/m^2·K\), we get: \(T_s = 823 - \frac{0.17\Delta T}{35 \times 1.19 \times 10^{-3}}\)

Evaluate if the temperature exceeds the ignition temperature

Now, we will evaluate if the temperature calculated in step 6 exceeds the ignition temperature of 723 K. Compare \(T_s\) with \(T_{ignition}\): If \(T_s > T_{ignition}\), the wood will ignite. If \(T_s \leq T_{ignition}\), the wood will not ignite. Keep in mind that this calculation is a rough estimation and, in practice, more detailed analysis is recommended for safety concerns. However, this approach provides an excellent starting point to understand if the wood slab is likely to ignite or not.