Inconel \(^{\circledast}\) refers to a class of nickel-chromium-based superalloys that are used in high-temperature applications, such as gas turbine blades. For further improvement in the performance of gas turbine engine, the outer blade surface is coated with ceramic-based thermal barrier coating (TBC). Consider a flat Inconel \({ }^{\circledR}\) plate, with a thickness of \(12 \mathrm{~mm}\), is coated with a layer of TBC, with a thickness of \(300 \mu \mathrm{m}\), on its surface. At the interface between the Inconel \({ }^{\circledR}\) and the TBC, the thermal contact conductance is \(10,500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermal conductivities of the Inconel \({ }^{\circledast}\) and the TBC are \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The plate is in a surrounding of hot combustion gasses at \(1500^{\circ} \mathrm{C}\), and the convection heat transfer coefficient is \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the mid-plane of the Inconel \({ }^{\oplus}\) plate, if the outer surface temperature is \(1200^{\circ} \mathrm{C}\).

Step by step solution

01

Write down the given parameters

The given parameters in the problem are as follows: Inconel thickness: \(L1 = 12 \times 10^{-3} \mathrm{~m}\) Inconel thermal conductivity: \(k1 = 25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) TBC thickness: \(L2 = 300 \times 10^{-6} \mathrm{~m}\) TBC thermal conductivity: \(k2 = 1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Thermal contact conductance at the interface: \(h_c = 10,500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Convection heat transfer coefficient: \(h = 750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Outer surface temperature: \(T_s = 1200^{\circ} \mathrm{C}\) Surrounding temperature: \(T_\infty = 1500^{\circ} \mathrm{C}\)

02

Calculate the thermal resistance of the materials

We can find the individual thermal resistances of Inconel and TBC layers using the formulas: \(R1 = \dfrac{L1}{k1}\space \), \(R2 = \dfrac{L2}{k2}\) Where \(R1\) and \(R2\) are the thermal resistances of Inconel and TBC, respectively. Calculating the thermal resistances: \( R1 = \dfrac{12 \times 10^{-3} \mathrm{~m}}{25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 4.8 \times 10^{-4} \mathrm{~m}^2 \cdot \mathrm{K}/ \mathrm{W} \) \( R2 = \dfrac{300 \times 10^{-6} \mathrm{~m}}{1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 2 \times 10^{-4} \mathrm{~m}^2 \cdot \mathrm{K}/ \mathrm{W} \)

03

Calculate the total thermal resistance

Now we will find the total thermal resistance by adding the individual resistances. Since these layers are in series, the total resistance is given by: \(R_{total} = R1 + \dfrac{1}{A h_c} + R2\) Where A is the surface area, and we have introduced the thermal contact conductance between the layers calculated as \(\dfrac{1}{A h_c}\). We can assume the surface area, A, to be 1 for simplicity. \(R_{total} = 4.8 \times 10^{-4} + \dfrac{1}{10,500} + 2 \times 10^{-4} = 5.695 \times 10^{-4} \mathrm{~m}^2 \cdot \mathrm{K} / \mathrm{W}\)

04

Calculate the temperature difference across Inconel and TBC layers

Now we need to find the temperature difference across the Inconel and TBC layers. We can determine this from the heat transfer process across the layers and by using the convection heat transfer coefficient, h. The convection heat transfer equation is given by: \(q = hA(T_s - T_{\infty})\) Where q is the heat transfer rate, and A is the surface area. We assume the surface area A to be 1. \(q = 750(1200 - 1500) = -225000 \mathrm{~W} / \mathrm{m}^2\) Since q is negative, it means the heat is being transferred from the surroundings to the layers.

05

Calculate the temperature at the mid-plane of the Inconel layer

Now, we can use the thermal resistance to find the temperature at the mid-plane of the Inconel layer. \(T_{mid} = T_s - q * \dfrac{L1}{2} * R1\) \(T_{mid} = 1200 - (-225000) * \dfrac{12 \times 10^{-3}}{2} \cdot 4.8 \times 10^{-4} \mathrm{~m}^2 \cdot \mathrm{K} / \mathrm{W}\) \(T_{mid} = 1200 + 648 \approx 1848^{\circ} \mathrm{C}\) Thus, the temperature at the mid-plane of the Inconel plate is approximately \(1848^{\circ} \mathrm{C}\).

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