A two-layer wall is made of two metal plates, with surface roughness of about \(25 \mu \mathrm{m}\), pressed together at an average pressure of \(10 \mathrm{MPa}\). The first layer is a stainless steel plate with a thickness of \(5 \mathrm{~mm}\) and a thermal conductivity of \(14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The second layer is an aluminum plate with a thickness of \(15 \mathrm{~mm}\) and a thermal conductivity of \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On the stainless steel side of the wall, the surface is subjected to a heat flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\). On the aluminum side of the wall, the surface experiences convection heat transfer at an ambient temperature of \(20^{\circ} \mathrm{C}\), where the convection coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the surface temperature of the stainless steel plate.

Step by step solution

01

Calculate the resistance of each layer

Start by calculating the thermal resistance of each layer. This can be found using the equation: \(R = \dfrac{L}{k A}\) where \(R\) is the thermal resistance, \(L\) is the thickness of the layer, \(k\) is the thermal conductivity, and \(A\) is the surface area. In this case, we will assume the surface area is the same for both layers. The resistance of stainless steel, \(R_1\), and aluminum, \(R_2\), can be calculated as follows: \(R_1 = \dfrac{5 \times 10^{-3}~\mathrm{m}}{14~\mathrm{W/m \cdot K} A}\) \(R_2 = \dfrac{15 \times 10^{-3}~\mathrm{m}}{237~\mathrm{W/m \cdot K} A}\)

02

Calculate total resistance of the system

To find the total thermal resistance of the system, add the resistance of the stainless steel layer, the resistance of the aluminum layer, and the convective resistance: \(R_{total} = R_1 + R_2 + R_{conv}\) The convective resistance, \(R_{conv}\), can be calculated as: \(R_{conv} = \dfrac{1}{h A}\) where \(h\) is the convection heat transfer coefficient. Thus, we get: \(R_{conv} = \dfrac{1}{12~\mathrm{W/m^2 \cdot K} A}\) Now, we can calculate the total resistance: \(R_{total} = \dfrac{5 \times 10^{-3}~\mathrm{m}}{14~\mathrm{W/m \cdot K} A} + \dfrac{15 \times 10^{-3}~\mathrm{m}}{237~\mathrm{W/m \cdot K} A} + \dfrac{1}{12~\mathrm{W/m^2 \cdot K} A}\)

03

Determine the heat flux through the system

The heat flux, \(q\), through the system is given, which is \(800 \mathrm{~W} / \mathrm{m}^{2}\). To find the temperature difference between the surface of the stainless steel plate and the ambient temperature, we can use the following equation: \(\Delta T = q \times R_{total}\) Substitute \(q\) and \(R_{total}\) in the equation to get the temperature difference: \(\Delta T = 800~\mathrm{W/m^2} \left(\dfrac{5 \times 10^{-3}~\mathrm{m}}{14~\mathrm{W/m \cdot K} A} + \dfrac{15 \times 10^{-3}~\mathrm{m}}{237~\mathrm{W/m \cdot K} A} + \dfrac{1}{12~\mathrm{W/m^2 \cdot K} A}\right)\) Solve for \(\Delta T\): \(\Delta T = 9.52~\mathrm{K}\)

04

Determine the surface temperature of the stainless steel plate

Since we know the temperature difference \(\Delta T\) between the stainless steel surface and the aluminum surface and the ambient temperature is given as \(20^{\circ} \mathrm{C}\), we can find the surface temperature of the stainless steel plate. The heat flux through the wall is in the direction of aluminum to stainless steel plate, so to find the surface temperature of the stainless steel plate, we will subtract the temperature difference from the ambient temperature: \(T_{stainless~steel} = 20^{\circ} \mathrm{C} - 9.52~\mathrm{K}\) \(T_{stainless~steel} = 10.48^{\circ} \mathrm{C}\) Thus, the surface temperature of the stainless steel plate is \(10.48^{\circ} \mathrm{C}\).

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