91Ó°ÊÓ

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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Conductivity

Thermal conductivity is a property of a material that measures its ability to conduct heat. In simple terms, it tells us how easily heat can flow through a material. If a material has a high thermal conductivity, heat can move through it quickly, whereas a low thermal conductivity means it acts more like an insulator, slowing down the flow of heat.

• This property is crucial when analyzing how heat transfers through different parts of an object.
• The unit of thermal conductivity is watts per meter-kelvin (W/m·°­).
• Different materials have different thermal conductivities, which affects how they behave in thermal applications.

In our example with the stainless steel sphere, the thermal conductivity helps to determine how heat from the internal reaction moves through the steel wall to the outside environment.

Convection Heat Transfer

Convection is a mode of heat transfer where fluids, such as gases or liquids, move and carry heat from one place to another. This movement can occur naturally, due to temperature-induced density differences, or it can be forced, by a fan or pump, for instance.

• Convection heat transfer is described by the heat transfer coefficient, denoted as \( h \).
• The unit for the convection heat transfer coefficient is watts per square meter-kelvin (W/m²Â·°­).
• Convection can significantly impact the thermal conditions of objects in contact with moving fluids.

In our stainless steel container problem, convection is crucial as it encompasses how the heat moves from the outside surface of the container to the surrounding air, impacting the container's surface temperature.

Fourier's Law

Fourier's Law is a foundational principle in heat transfer analysis, applied primarily to understand how heat conduction works through a material. According to Fourier's Law, the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which the heat flows. Mathematically, it is expressed as:
\[ q = -k \frac{dT}{dx} \]

• Here, \( q \) is the heat flux (rate of heat transfer per unit area), measured in watts per square meter (W/m²).
• \( k \) is the thermal conductivity of the material.
• \( \frac{dT}{dx} \) is the temperature gradient, or the rate of change of temperature with respect to distance."

In our exercise, Fourier's Law allows us to calculate how the heat produced from the chemical reaction transmits through the steel wall of the container, ultimately impacting the inner and outer surface temperatures.

Newton's Law of Cooling

Newton's Law of Cooling describes how an object's temperature changes over time as it interacts with its environment. It states that the rate of heat transfer from an object is proportional to the difference in temperature between the object and the surrounding fluid. This can be represented by the formula:
\[ q = h \cdot (T_{surface} - T_{ambient}) \]

• \( q \) is the rate of heat transfer, usually measured in watts (W).
• \( h \) is the convection heat transfer coefficient, representing the efficiency of heat transfer between the object and its environment.
• \( T_{surface} \) and \( T_{ambient} \) are the temperatures of the object's surface and the ambient fluid, respectively.

In the context of the exercise, Newton's Law helps determine the safe outer surface temperature of the container, ensuring it does not pose a burn hazard for individuals nearby. It combines with the information provided to check if the thermal safety criteria are met.