Question

A thin-walled spherical tank is buried in the ground at a depth of $3 \mathrm{~m}\(. The tank has a diameter of \)1.5 \mathrm{~m}$, and it contains chemicals undergoing exothermic reaction that provides a uniform heat flux of \(1 \mathrm{~kW} / \mathrm{m}^{2}\) to the tank's inner surface. From soil analysis, the ground has a thermal conductivity of $1.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and a temperature of \)10^{\circ} \mathrm{C}$. Determine the surface temperature of the tank. Discuss the effect of the ground depth on the surface temperature of the tank.

Short answer

Answer: The surface temperature of the spherical tank is approximately -14.0606°C.

Step-by-step solution

Calculate the radius of the tank

First, let's find the radius of the spherical tank. The diameter is given as \(1.5 \mathrm{~m}\), so the radius is half of that. $$ r = \frac{1.5}{2} = 0.75 \mathrm{~m} $$

Calculate the area of the tank

Next, we'll find the surface area of the spherical tank. The formula for the surface area of a sphere is: $$ A = 4 \pi r^2 $$ Plug in the value for the radius: $$ A = 4 \pi (0.75)^2 = 4 \pi (0.5625) \approx 7.0686 \mathrm{~m}^2 $$

Calculate the total heat transfer

Now, let's find the total heat transfer from the exothermic reaction. The provided heat flux is \(1 \mathrm{~kW/m^2}\), which we need to convert to \(\mathrm{W/m^2}\) and multiply by the surface area of the tank. $$ q = 1000 \cdot A = 1000 \cdot 7.0686 = 7068.6 \mathrm{~W} $$

Calculate the temperature gradient

Next, we'll calculate the temperature gradient in the soil. To do this, we will use the formula for heat transfer through conduction: $$ q = -kA\frac{dT}{dx} $$ We know the heat transfer, the thermal conductivity of the soil (\(k = 1.3 \mathrm{~W/m\cdot K}\)), and the surface area of the tank. We want to find the temperature gradient, \(\frac{dT}{dx}\). Rearrange the formula to solve for \(\frac{dT}{dx}\): $$ \frac{dT}{dx} = -\frac{q}{kA} $$ Plug in the values for \(q\), \(k\), and \(A\): $$ \frac{dT}{dx} = -\frac{7068.6 \mathrm{~W}}{1.3 \mathrm{~W/m\cdot K} \cdot 7.0686 \mathrm{~m}^2} \approx -8.0202 \mathrm{~K/m} $$

Calculate the surface temperature of the tank

Finally, we'll find the surface temperature of the tank. We know the temperature gradient and the depth of the tank (\(3 \mathrm{~m}\)). We can find the difference in temperature by multiplying the temperature gradient by the depth of the tank: $$ \Delta T = \frac{dT}{dx} \cdot d = -8.0202 \mathrm{~K/m} \cdot 3 \mathrm{~m} \approx -24.0606 \mathrm{~K} $$ The soil temperature at the surface is given as \(10^{\circ}\mathrm{C}\). To find the surface temperature of the tank, we subtract the temperature difference from the soil temperature: $$ T_{\text{surface}} = 10 - 24.0606 \approx -14.0606^{\circ}\mathrm{C} $$

Conclusion

The surface temperature of the spherical tank is approximately \(-14.0606^{\circ}\mathrm{C}\). As the depth of the tank increases, the temperature gradient becomes less steep, thus the surface temperature of the tank will decrease. This means that a greater depth will result in a lower surface temperature of the tank, as it is further away from the heat source and the soil acts as an insulator.