The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with a hot surface that has a temperature above its autoignition temperature, the oil can ignite spontaneously. Consider an engine cover that is made of a stainless steel plate with a thickness of \(1 \mathrm{~cm}\) and a thermal conductivity of \(14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the engine cover is exposed to hot air with a convection heat transfer coefficient of \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at \(333^{\circ} \mathrm{C}\). The outer surface is exposed to an environment where the ambient air is \(69^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent fire hazard in the event of oil leak on the engine cover, a layer of thermal barrier coating (TBC) with a thermal conductivity of \(1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is applied on the engine cover outer surface. Would a TBC layer of \(4 \mathrm{~mm}\) in thickness be sufficient to keep the engine cover surface below autoignition temperature of \(200^{\circ} \mathrm{C}\) to prevent fire hazard?

We will first find the heat transfer rate through the engine cover using the given values. The heat transfer rate is given by the formula: \(q = \frac{T_{1} - T_{2}}{\frac{1}{h_{1}A} + \frac{L_{1}}{k_{1}A} + \frac{L_{2}}{k_{2}A} + \frac{1}{h_{2}A}}\) Where: \(q\) is the heat transfer rate (W) \(T_{1}\) and \(T_{2}\) are the temperatures of the inner and outer surface, respectively (K) \(h_{1}\) and \(h_{2}\) are the convection heat transfer coefficients for the inner and outer surface, respectively \((\mathrm{W} / \mathrm{m}^{2}\cdot \mathrm{K})\) \(L_{1}\) and \(L_{2}\) are the thicknesses of the stainless steel plate and TBC layer, respectively (m) \(k_{1}\) and \(k_{2}\) are the thermal conductivities of the stainless steel plate and TBC layer, respectively \((\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) \(A\) is the surface area for heat transfer (m²) In this case, we are not given the surface area, thus we can rearrange the equation to find the heat transfer rate per unit area: \(q'= \frac{T_{1} - T_{2}}{\frac{1}{h_{1}} + \frac{L_{1}}{k_{1}} + \frac{L_{2}}{k_{2}} + \frac{1}{h_{2}}}\) Substitute the given values into the equation: \(T_{1} = 333^{\circ} \mathrm{C} + 273.15 = 606.15 \mathrm{K}\) \(T_{2} = 69^{\circ} \mathrm{C} + 273.15 = 342.15 \mathrm{K}\) \(h_{1} = h_{2} = 7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) \(L_{1} = 1 \mathrm{~cm} = 0.01 \mathrm{~m}\) \(k_{1} = 14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) \(L_{2} = 4 \mathrm{~mm} = 0.004 \mathrm{~m}\) \(k_{2} = 1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Plugging these values into the formula, we get: \(q'= \frac{606.15 - 342.15}{\frac{1}{7} + \frac{0.01}{14} + \frac{0.004}{1.1} + \frac{1}{7}}\) \(q'= \frac{264}{\frac{2}{7} + \frac{1}{1400} + \frac{4}{1100}}\) Calculate \(q'\): \(q'= 158.81 \frac{\mathrm{W}}{\mathrm{m}^{2}}\)

Now, we will find the outer surface temperature of the TBC layer (\(T_{out}\)) using the heat transfer rate per unit area and the convection heat transfer coefficient of the outer surface: \(T_{out} = T_{amb} + \frac{q'}{h_{2}}\) Where: \(T_{amb}\) is the ambient temperature (K) Substitute the given values into the equation: \(T_{out} = 342.15 + \frac{158.81}{7}\) Calculate \(T_{out}\): \(T_{out} = 365.7 \mathrm{~K}\) Convert to Celsius: \(T_{out} = 365.7 - 273.15 = 92.55^{\circ} \mathrm{C}\)

Now we must check if the calculated surface temperature of the outer surface of TBC layer (\(T_{out}\)) is below the autoignition temperature (\(T_{auto}\)) of \(200^{\circ} \mathrm{C}\): \(T_{out} < T_{auto}\) \(92.55^{\circ} \mathrm{C} < 200^{\circ} \mathrm{C}\) Since the outer surface temperature of the TBC layer is below the autoignition temperature, the \(4 \mathrm{~mm}\) thick TBC layer is sufficient to prevent fire hazard due to oil leakage.