A stainless steel spherical container, with \(k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used for storing chemicals undergoing exothermic reaction. The reaction provides a uniform heat flux of \(60 \mathrm{~kW} / \mathrm{m}^{2}\) to the container's inner surface. The container has an inner radius of \(50 \mathrm{~cm}\) and a wall thickness of \(5 \mathrm{~cm}\) and is situated in a surrounding with an ambient temperature of \(23^{\circ} \mathrm{C}\). The container's outer surface is subjected to convection heat transfer with a coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For safety reasons to prevent thermal burn to individuals working around the container, it is necessary to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). Determine the variation of temperature in the container wall and the temperatures of the inner and outer surfaces of the container. Is the outer surface temperature of the container safe to prevent thermal burn?

Step by step solution

01

List given information

Here is a summary of the information given in the problem statement: - Thermal conductivity of the container (\(k\)): \(15 \dfrac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\) - Heat flux at inner surface (\(q''\)): \(60,000 \dfrac{\mathrm{W}}{\mathrm{m}^2}\) - Inner radius of the container (\(r_1\)): \(0.5 \mathrm{m}\) - Wall thickness (\(\delta\)): \(0.05 \mathrm{m}\) - Ambient temperature (\(T_\infty\)): \(23^\circ \mathrm{C}\) - Convection heat transfer coefficient (\(h\)): \(1000 \dfrac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}}\) - Safety temperature limit (\(T_{limit}\)): \(50^\circ \mathrm{C}\)

02

Calculate the heat transfer rate across the container radius

We can use Fourier's law to find the heat transfer rate through the wall of the container: \(q = k \cdot A \cdot \frac{\Delta T}{\Delta x} = k \cdot A \cdot \frac{T_{inner} - T_{outer}}{\delta}\) In this case, we are given the heat flux at the inner surface, so we can rewrite the formula as follows: \(q = q'' \cdot A_{inner} = k \cdot A_{outer} \cdot \frac{T_{inner} - T_{outer}}{\delta}\) Since \(A_{inner} = 4 \pi r_1^2\) and \(A_{outer} = 4 \pi (r_1 + \delta)^2\), we can now use the above equation to find an expression for the temperature difference between the inner and outer surfaces of the container: \(T_{inner} - T_{outer} = \frac{q'' \cdot 4 \pi r_1^2}{k \cdot 4 \pi (r_1 + \delta)^2} \cdot \delta\)

03

Calculate the inner and outer surface temperatures

For the outer surface, we can use Newton's law of cooling: \(q''_{outer} = h \cdot (T_{outer} - T_\infty)\) We already have the expression that relates \(q''\) with the temperature difference. So, we will substitute it for \(q''\) in the Newton's law of cooling to find the expression for the outer surface temperature. Now, we have: \(q''_{outer} = q'' - h \frac{4 \pi r_1^2 \cdot \delta}{(r_1 + \delta)^2} = h \cdot (T_{outer} - T_\infty)\) By solving the above equation, we get the value for \(T_{outer}\). Then, we can calculate \(T_{inner}\) using the expression derived in step 2: \(T_{inner} = T_{outer} + \frac{q'' \cdot 4 \pi r_1^2}{k \cdot 4 \pi (r_1 + \delta)^2} \cdot \delta\)

04

Check the safety condition

Once we have found the temperatures of the inner and outer surfaces, we need to check if the outer surface temperature is below the safety limit of \(50^{\circ} \mathrm{C}\) to prevent thermal burns: Safe if \(T_{outer} < T_{limit}\)

05

Determine the temperature distribution in the container wall

Finally, to find the temperature distribution in the container wall, we can use the following formula: \(T(r) = T_{inner} - \frac{q'' \cdot 4 \pi r_1^2}{k \cdot 4 \pi (r_1 + \delta)^2} \cdot (r - r_1)\) In this formula, \(r\) varies from \(r_1\) to \(r_1 + \delta\). By varying the value of \(r\), we can find the temperature distribution in the container wall.

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