Question

A metal spherical tank is filled with chemicals undergoing an exothermic reaction. The reaction provides a uniform heat flux on the inner surface of the tank. The tank has an inner diameter of \(5 \mathrm{~m}\), and its wall thickness is \(10 \mathrm{~mm}\). The tank wall has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where $k_{0}=9.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.0018 \mathrm{~K}^{-1}\(, and \)T$ is in \(\mathrm{K}\). The area surrounding the tank has an ambient temperature of \(15^{\circ} \mathrm{C}\), the tank's outer surface experiences convection heat transfer with a coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\). Determine the heat flux on the tank's inner surface if the inner surface temperature is \(120^{\circ} \mathrm{C}\).

Short answer

Answer: The heat flux on the inner surface of the metal spherical tank is approximately \(1.33 \times 10^5 \mathrm{~W}\).

Step-by-step solution

Find the inner and outer surface areas of the tank

First, we need to find the radii of the inner and outer surfaces of the tank. The given inner diameter is 5 m, so the inner radius, \(r_i\), is half that value: \(r_i = \frac{5}{2} = 2.5 \mathrm{~m}\) The wall thickness is given as 10 mm, which is equivalent to 0.01 m. To find the outer radius, \(r_o\), we add the wall thickness to the inner radius: \(r_o = r_i + 0.01 = 2.5 + 0.01 = 2.51 \mathrm{~m}\) Now we can calculate the surface areas of the inner and outer surfaces of the tank using the formula for the surface area of a sphere, \(A=4\pi r^2\): \(A_i = 4\pi r_i^2 = 4\pi (2.5)^2 = 78.54 \mathrm{~m}^2\) \(A_o = 4\pi r_o^2 = 4\pi (2.51)^2 = 78.75 \mathrm{~m}^2\)

Calculate the temperature difference between the inner and outer surfaces

Given the inner surface temperature, \(T_i = 120^{\circ} \mathrm{C}\), and the ambient temperature, \(T_{\infty} = 15^{\circ} \mathrm{C}\), we can calculate the temperature difference, \(\Delta T\), between the inner and outer surfaces: \(\Delta T = T_i - T_{\infty} = 120 - 15 = 105^{\circ} \mathrm{C}\). Since the given parameters \(k_0\) and \(\beta\) require temperatures in Kelvin, we need to convert \(\Delta T\) to Kelvin: \(\Delta T = 105 \mathrm{~K}\)

Determine the effective thermal conductivity of the tank wall

Using the given formula for the variable thermal conductivity, \(k(T)=k_{0}(1+\beta T)\), we can find the effective thermal conductivity at the average temperature of the wall. To do so, we first need to calculate the average temperature of the wall, \(T_{avg}\): \(T_{avg} = \frac{T_i + T_{\infty}}{2} = \frac{120 + 15}{2} = 67.5^{\circ} \mathrm{C}\) Convert to Kelvin: \(T_{avg} = 67.5 + 273.15 = 340.65 \mathrm{~K}\) Now, we can find the effective thermal conductivity, \(k_{eff}\): \(k_{eff} = k_0(1+\beta T_{avg}) = 9.1(1+0.0018 \times 340.65) = 15.97 \mathrm{~W/m\cdot K}\)

Apply the conduction-convection heat transfer equation to find the heat flux

We can now use the conduction-convection heat transfer equation to determine the heat flux, \(q\), on the inner surface of the tank: \(q = \frac{k_{eff}\Delta T}{\frac{1}{h_oA_o}}\) Where \(h_o\) is the convection heat transfer coefficient given as \(80 \mathrm{~W/ m}^2 \mathrm{~K}\), and \(A_o\) is the outer surface area we calculated previously (78.75 \(m^2\)). Plugging in the values, we get: \(q = \frac{15.97 \times 105}{\frac{1}{80 \times 78.75}} = 133439.84 \mathrm{~W}\) Thus, the heat flux on the tank's inner surface is approximately \(1.33 \times 10^5 \mathrm{~W}\).