During a fire, the trunks of some dry oak trees (es) \(\left(k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(30^{\circ} \mathrm{C}\) are exposed to hot gases at \(520^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~h}\), with a heat transfer coefficient of \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. The ignition temperature of the trees is \(410^{\circ} \mathrm{C}\). Treating the trunks of the trees as long cylindrical rods of diameter \(20 \mathrm{~cm}\), determine if these dry trees will ignite as the fire sweeps through them. Solve this problem using analytical one-term approximation method (not the Heisler charts).

To determine if the trees will ignite, let's first find out the Biot number (Bi) and overall efficiency (η₁) for the given conditions. The Biot number (Bi) is given by: Bi = \(\frac{hL_c}{k}\) Where h is the heat transfer coefficient, L_c is the characteristic length, and k is the thermal conductivity of the material. For cylindrical coordinates, L_c is given by: L_c = \(\frac{d}{4}\) where d is the diameter of the cylinder. Let's calculate Bi and L_c. L_c = \(\frac{0.2 m}{4}\) = \(0.05 m\) Bi = \(\frac{65 \frac{W}{m^2.K} \cdot 0.05 m}{0.17 \frac{W}{m.K}}\) = \(19.12\) Next, let's find the overall efficiency, η₁: η₁ = \(\frac{1 - \xi}{1 + \xi}\) Where ξ is given by: ξ = \(\frac{Bi}{tanh(Bi)}\) Let's calculate ξ and η₁. ξ = \(\frac{19.12}{tanh(19.12)}\) = \(0.9999\) η₁ = \(\frac{1 - 0.9999}{1 + 0.9999}\) = \(5.20\times10^{-5}\)

The Fourier number (Fo) is a dimensionless parameter that represents the ratio of heat conduction within the material to the heat conduction on the surface. The Fourier number is given by: Fo = \(\frac{\alpha t}{L_c^2}\) We were given the thermal diffusivity of the material (α), the exposure time (t), and we calculated the characteristic length (L_c) in the step 1. Let's calculate the Fourier number (Fo). Fo = \(\frac{1.28 \times 10^{-7} \frac{m^2}{s} \cdot (5 h) \cdot \frac{3600 s}{h}}{(0.05 m)^2}\) = \(0.576\)

Now, we have η₁ and Fo, which will help us find the temperature at the outer surface of the tree. The one-term approximation for the cylindrical coordinates under transient conditions is given by: \(T(t,r) - T_i = \frac{2 \eta_1 (T_\infty - T_i)}{\pi} \cdot K_0(R)\) Where \(T(t,r)\) is the transient temperature, \(T_i\) is the initial temperature, \(T_\infty\) is the exposure temperature, and \(K_0(R)\) is the zeroth-order modified Bessel function of the second kind evaluated at R. Considering only the outer surface of the tree, r = L_c: \(T(t,L_c) - T_i = \frac{2 \eta_1 (T_\infty - T_i)}{\pi} \cdot K_0(Bi \sqrt{Fo})\) Let's find the temperature at the outer surface of the tree (T_outer). \(K_0(Bi \sqrt{Fo}) = K_0(19.12 \cdot \sqrt{0.576}) = K_0(13.65) \approx\) -5.6 T_outer = \((30^{\circ}C + 5.6\;\frac{5.2\times10^{-5}}{\pi} (520^{\circ}C - 30^{\circ}C)) = 30^{\circ}C + 0.0019 \cdot 490^{\circ}C = 30^{\circ}C + 0.93^{\circ}C\) T_outer = 30.93\(^{\circ}\)C

Now we know the temperature at the outer surface after the given exposure time, we can determine if the trees will ignite. The trees will ignite if the obtained temperature surpasses the ignition temperature (T_ignition). Since the calculated temperature at the outer surface (T_outer = 30.93\(^{\circ}\)C) is significantly lower than the ignition temperature (T_ignition = 410\(^{\circ}\)C), these dry oak trees will not ignite as the fire sweeps through them during the 5-hour period.