The outer surface of an engine is situated in a place where oil leakage can occur. Some oils have autoignition temperatures of approximately above \(250^{\circ} \mathrm{C}\). When oil comes in contact with a hot engine surface that has a higher temperature than its autoignition temperature, the oil can ignite spontaneously. Treating the engine housing as a plane wall, the inner surface \((x=0)\) is subjected to \(6 \mathrm{~kW} / \mathrm{m}^{2}\) of heat. The engine housing \((k=13.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a thickness of \(1 \mathrm{~cm}\), and the outer surface \((x=L)\) is exposed to an environment where the ambient air is \(35^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent fire hazard in the event the leaked oil comes in contact with the hot engine surface, the temperature of the engine surface should be kept below \(200^{\circ} \mathrm{C}\). Determine the variation of temperature in the engine housing and the temperatures of the inner and outer surfaces. Is the outer surface temperature of the engine below the safe temperature?

Step by step solution

01

Establish knowns and unknowns

We are given: - Heat generation rate: \(q = 6 \mathrm{~kW} / \mathrm{m}^{2}\) - Thermal conductivity: \(k = 13.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Wall thickness: \(L = 1 \mathrm{~cm} = 0.01 \mathrm{~m}\) - Ambient air temperature: \(T_{\infty} = 35 ^{\circ} \mathrm{C}\) - Convection heat transfer coefficient: \(h = 20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Maximum safe outer surface temperature: \(T_{\text{max}} = 200 ^{\circ} \mathrm{C}\) We will be finding: - Temperature distribution within the housing - Inner surface temperature - Outer surface temperature - A check for outer surface temperature safety

02

Apply the heat conduction equation

Since the problem involves heat transfer through the steady-state plane wall, we use the heat conduction equation: $$q = -kA\frac{dT}{dx}$$ Here \(A\) is the surface area, \(T\) is the temperature, and \(x\) is the distance from the inner surface (\(0 \le x \le L\)). We can rewrite the equation as: $$\frac{dT}{dx} = -\frac{q}{kA}$$

03

Integrate the equation to find the temperature distribution

Integrating the equation, we obtain: $$T(x) = -\frac{q}{kA}x + T_0$$ We need to find the constant \(T_0\). To do this, we need to use the condition at the inner surface \((x=0)\): $$T(0) = -\frac{q}{kA}(0) + T_0$$ This gives us \(T_{\text{inner}} = T_0\). Now, let's find the temperature at the outer surface \((x=L)\): $$T(L) = -\frac{q}{kA}(L) + T_0$$ $$T_{\text{outer}} = T_{\text{inner}} - \frac{qL}{kA}$$ We can see that \(Q = qA\). Thus, \(T_{\text{outer}} = T_{\text{inner}} - \frac{QL}{k}\)

04

Use convective heat transfer expression to find the temperature at the outer surface

Since we know the thermal conductivity and the heat transfer coefficient, we can relate the convection heat transfer rate to the temperature at the outer surface: $$q_{conv} = hA(T_{\text{outer}} - T_{\infty})$$ Substitute \(q_{conv}\) with \(Q\): $$Q = hA(T_{\text{outer}} - T_{\infty})$$ This equation helps us to eliminate an unknown A. Rearrange the equation to find \(T_{\text{outer}}\): $$T_{\text{outer}} = \frac{Q}{hA} + T_{\infty}$$

05

Use both expressions for the outer surface temperature to find the inner surface temperature

Now, we can equate the two expressions for \(T_{\text{outer}}\): $$T_{\text{inner}} - \frac{QL}{k} = \frac{Q}{hA} + T_{\infty}$$ Rearrange the equation to find \(T_{\text{inner}}\): $$T_{\text{inner}} = \frac{QL}{k} + \frac{Q}{hA} + T_{\infty}$$ Plug in the given values, we have \(T_{\text{inner}} = 200.904 ^\circ C\).

06

Use the inner surface temperature to find the outer surface temperature

Now, we know the inner surface temperature, and we can calculate the outer surface temperature using the expression: $$T_{\text{outer}} = T_{\text{inner}} - \frac{qL}{kA}$$ Plug in the given values, we get \(T_{\text{outer}} = 114.3515 ^\circ C\). Since, the outer surface temperature is below the safe limit of \(200 ^\circ C\). The system is safe.

07

Find the temperature distribution within the housing

Now we know the inner surface temperature. Thus, we can write the full temperature distribution equation: $$T(x) = -\frac{q}{kA}x + T_{\text{inner}}$$ This equation represents the temperature distribution within the housing.

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