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.

Step by step solution

01

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. $$T_{inner} = 120^{\circ} \mathrm{C} + 273.15 = 393.15\mathrm{K}$$ $$T_{ambient} = 15^{\circ} \mathrm{C} + 273.15 = 288.15\mathrm{K}$$ $$T_{outer} = 50^{\circ} \mathrm{C} + 273.15 = 323.15\mathrm{K}$$

02

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.

03

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: $$A = 4\pi (r_{inner} + \Delta x)^2$$

04

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 \(h\): $$R_{cond} = \frac{\Delta x}{k(T)A} = \frac{\Delta x}{k_0(1+\beta T_{outer})(4\pi (r_{inner} + \Delta x)^2)}$$ $$R_{conv} = \frac{1}{hA} = \frac{1}{80 (4\pi (r_{inner} + \Delta x)^2)}$$

05

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 \(0.0635\mathrm{m}\).

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