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 \(15^{\circ} \mathrm{C}\), and the vessel's outer surface experiences convection heat transfer with a coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2}\). K. To prevent thermal burns to workers who touch the vessel, 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

Based on the given conditions and solving the associated heat transfer equations, the minimum wall thickness of the spherical vessel can be determined to ensure that the outer surface temperature remains at or below \(50^{\circ} \mathrm{C}\). This is achieved by converting temperatures to Kelvin, writing expressions for temperature in the wall, using convective heat transfer equations, and integrating the differential equation considering boundary conditions. The final expression represents the minimum wall thickness required to meet the desired outer surface temperature requirement.

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert the temperatures given in Celsius to Kelvin, as the thermal conductivity equation uses Kelvin. \(T_\text{inner} = 120 + 273.15 = 393.15 K\) \(T_\text{outer,max=} 50 + 273.15 = 323.15 K\) \(T_\text{ambient} = 15 + 273.15 = 288.15 K\)

Write an expression for temperature in the vessel wall

We can write the expression for temperature in the vessel wall (\(T_\text{wall}\)) using the thermal conductivity equation: \(k(T) = k_0 (1+\beta T)\), where \(k_0=1.01 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\beta=0.0018 \mathrm{~K}^{-1}\) This will help us in relating the heat flux through the wall to the temperature difference between the inner and outer surface of the wall.

Convective heat transfer equation

Now let's write the convective heat transfer equation for the vessel's outer surface: \(q_\text{conv} = hA(T_\text{outer}-T_\text{ambient})\) where \(h = 80 \mathrm{~W} / \mathrm{m}^{2}\cdot \mathrm{K}\) (heat transfer coefficient) and \(T_\text{outer}\) is the outer surface temperature. Since the reaction provides a uniform heat flux on the inner surface, the same amount of heat must be transferred from the outer surface to the environment. Therefore, the heat flux through the wall is equal to the convective heat transfer from the outer surface. So, we can write: \(q_\text{cond} = q_\text{conv}\). For spherically symmetrical heat conduction, the heat flux through the wall can be written as: \(q_\text{cond} = -k(T) r \frac{dT}{dr}\), where \(r\) is the radial position from the center of the vessel, and \(T\) is the temperature as a function of \(r\). Now we can write the differential equation for the temperature distribution in the wall: \(r \frac{dT}{dr} = -\frac{k(T)}{hA}\) Let's solve this equation to find the temperature distribution T(r).

Find the temperature distribution and wall thickness

To solve the problem, we integrate the differential equation and apply boundary conditions using the known inner surface temperature \(T_\text{inn =}393.15 K\) and the upper limit of wall_thickness for the maximum allowed outer temperature to be \(T_\text{outer,max =} 323.15 K\). This will allow us to find the minimum wall thickness for meeting the requirement of outer surface temperature below \(50^{\circ} \mathrm{C}\). Upon integrating and applying boundary conditions, we obtain \(T(r) = A\ln(\frac{1+\beta T(r)R}{1+\beta T(r)r})\), where \(A\) is a constant. Let \(T_\text{outer,max} = A\ln(\frac{1+\beta T_\text{outer,max}R}{1+\beta T_\text{outer,max}(r+wall_\text{thickness})})\) Finally, we can solve for the minimum \(wall_\text{thickness}\), which will ensure that \(T_\text{outer,max}\) is below or equal to \(50^{\circ} \mathrm{C}\).