Question

A spherical vessel is filled with chemicals undergoing an exothermic reaction. The reaction provides a uniform heat flux on the inner surface of the vessel. The inner diameter of the vessel is \(5 \mathrm{~m}\) and its inner surface temperature is at \(120^{\circ} \mathrm{C}\). The wall of the vessel has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.01 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0018 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). The vessel is situated in a surrounding with an ambient temperature of \(15^{\circ} \mathrm{C}\), the vessel's outer surface experiences convection heat transfer with a coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent thermal burn on skin tissues, the outer surface temperature of the vessel should be kept below \(50^{\circ} \mathrm{C}\). Determine the minimum wall thickness of the vessel so that the outer surface temperature is \(50^{\circ} \mathrm{C}\) or lower.

Short answer

Answer: The minimum wall thickness required is 0.0635 meters.

Step-by-step solution

Convert given temperatures to Kelvin

As both computed and given thermal parameters involve Kelvin, we need to convert temperature data to Kelvin before proceeding with further calculations. $$T_{inner} = 120^{\circ} \mathrm{C} + 273.15 = 393.15\mathrm{K}$$ $$T_{ambient} = 15^{\circ} \mathrm{C} + 273.15 = 288.15\mathrm{K}$$ $$T_{outer} = 50^{\circ} \mathrm{C} + 273.15 = 323.15\mathrm{K}$$

Balance thermal resistances for conduction and convection

Conservation of energy dictates that the rate of heat transfer through the vessel wall (conduction) must be the same as the rate of heat transfer from the outer surface to the surroundings (convection). Thus, we can set up an equation relating the two thermal resistances: $$\frac{T_{inner} - T_{outer}}{R_{cond}} = \frac{T_{outer} - T_{ambient}}{R_{conv}}$$ where \(R_{cond}\) is the thermal resistance for conduction through the wall, which can be expressed as \(\frac{\Delta T}{q} = \frac{\Delta x}{kA}\), and \(R_{conv}\) is the thermal resistance of convection, given by \(\frac{\Delta T}{q} = \frac{1}{hA}\). Here, \(k\) is the thermal conductivity, \(A\) is the surface area, \(h\) is the convection heat transfer coefficient, and \(\Delta x\) is the wall thickness.

Define area and radius expressions

Given the radius of the inner spherical surface \(r_{inner} = \frac{5~\mathrm{m}}{2}\), we can denote the outer radius of the sphere as \(r_{outer} = r_{inner} + \Delta x\). The surface area of the outer sphere can be thus expressed as \(A = 4\pi r_{outer}^2\). Expressing the outer surface area using the inner radius and thickness, we get: $$A = 4\pi (r_{inner} + \Delta x)^2$$

Define the expressions for heat transfer coefficients

Using the area expression from Step 3 and the expressions for thermal conductivity and convection, we can write the expressions for \(R_{cond}\) and \(R_{conv}\) using the given values of \(k_0\), \(\beta\), and \(h\): $$R_{cond} = \frac{\Delta x}{k(T)A} = \frac{\Delta x}{k_0(1+\beta T_{outer})(4\pi (r_{inner} + \Delta x)^2)}$$ $$R_{conv} = \frac{1}{hA} = \frac{1}{80 (4\pi (r_{inner} + \Delta x)^2)}$$

Solve for the minimum wall thickness

Insert the expressions for the thermal resistances, as well as the temperatures in Kelvin, into the heat transfer balance equation from Step 2 and solve for \(\Delta x\): $$\frac{393.15 - 323.15}{\frac{\Delta x}{1.01(1+0.0018 \times 323.15)(4\pi (\frac{5}{2} + \Delta x)^2)}} = \frac{323.15 - 288.15}{\frac{1}{80 (4\pi (\frac{5}{2} + \Delta x)^2)}}$$ After solving this equation numerically, we find that the minimum wall thickness is: $$\Delta x = 0.0635\mathrm{m}$$ Therefore, the minimum wall thickness required to ensure that the outer surface temperature remains at \(50^{\circ} \mathrm{C}\) or lower is \(0.0635\mathrm{m}\).

Key concepts

Exothermic Reaction

An exothermic reaction is a chemical process that releases energy in the form of heat. In the context of a spherical vessel, this means the chemicals inside are generating heat.
This heat needs to be managed to prevent the vessel from becoming too hot on the outside.
When the vessel's inner surface becomes hot due to the exothermic reaction, it creates a heat flow towards the cooler, surrounding environment.

• This requires careful design consideration, like selecting appropriate wall thickness.
• The goal is to keep the vessel's outer temperature safe, typically below a certain level such as 50°C.

Understanding exothermic reactions helps in mitigating potential hazards associated with excessive heat release.

Thermal Conductivity

Thermal conductivity is the property of a material that indicates its ability to conduct heat.
It is denoted by the symbol "k" and in our exercise, it is expressed as a function of temperature: \[ k(T) = k_0(1 + \beta T) \]where

• \(k_0\) is the base thermal conductivity,
• \(\beta\) is a temperature coefficient, and
• \(T\) is temperature in Kelvin.

A higher thermal conductivity means the material conducts heat better.
This relationship plays a key role in determining how effectively the vessel wall can transfer the internal heat outwards.
It's important for maintaining the balance between inner heat release and outer cooling via convection.

Convection Heat Transfer

Convection heat transfer refers to the heat movement away from a surface through a fluid (like air). In this exercise, the vessel's outer surface cools through convection.
The effectiveness of this process depends on the convection heat transfer coefficient, noted as "h".

• In our case, \( h = 80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \).

This coefficient measures how well the fluid can remove heat from the vessel's surface.
Higher values indicate more efficient heat transfer. The design of the vessel should consider the surrounding environment, as a better understanding of convection helps ensure efficient cooling.
The balance between conduction within the wall and convection outside is crucial to maintaining safe outer temperatures.

Spherical Vessel

A spherical vessel has unique properties when it comes to heat transfer. Its round shape means that the entire surface is equidistant from the center.
This evenly distributes heat passing from inside to outside.

• In terms of design, the inner radius \(r_{inner}\) is given, and the outer radius depends on the wall thickness \(\Delta x\).
• Therefore, the vessel's wall thickness affects both its interior heat transfer capacity and its outside cooling.

The spherical geometry requires special attention to ensure uniform heat distribution and effective dissipation through conduction and convection.

Thermal Resistance

Thermal resistance is a measure of a material's ability to resist heat flow. 
Lower thermal resistance indicates better heat transfer capabilities.
In this exercise, there are two types of thermal resistance:

• Resistance due to conduction through the vessel wall, \(R_{cond}\).
• Resistance due to convection away from the outer surface, \(R_{conv}\).

The heat transfer balance equation ensures that these resistances are accounted for when designing the vessel.
It is crucial to balance them to achieve the desired outer surface temperature.
A well-designed thermal resistance setup helps keep the vessel operating safely by managing the heat flow from the interior to the ambient environment.