Question

A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is $4 \mathrm{~m}$. If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.

Short answer

Answer: The rate of heat transfer from the tank is approximately 475.87 W.

Step-by-step solution

Determine the radius of the tank and the temperature difference

We are given that the diameter of the tank is 3m. To find the radius (R1) of the tank, we can divide the diameter by two: \[R_1 = \frac{3}{2} = 1.5 \mathrm{~m}\] Next, we must find the temperature difference (ΔT) between the tank and the ground. We are given that the surface temperature of the tank is \(140^\circ \mathrm{C}\) and the ground is \(15^\circ \mathrm{C}\). This temperature difference can be calculated as follows: \[\Delta T = T_2 - T_1 = 15 - 140 = -125^\circ \mathrm{C}\] Note that the temperature difference is negative since the tank is hotter than the ground.

Compute the distance to the ground surface (R2)

We are given that the distance between the top surface of the tank and the ground surface is 4m. Since this is a diagonal distance from the top of the sphere to the ground, and our formula requires the radial distance from the center of the sphere to the ground surface, we can use the Pythagorean theorem to find R2. The theorem states: \[ R_2 = \sqrt{R_1^2 + d^2}\] Where d is the distance between the top surface of the tank and the ground surface. Plugging in the given values, we get: \[R_2 = \sqrt{(1.5)^2 + (4)^2} = \sqrt{2.25 + 16} = \sqrt{18.25}\] \[R_2 \approx 4.27 \mathrm{~m}\]

Calculate the heat transfer rate (Q)

Now that we have all the values needed, we can determine the rate of heat transfer (Q) using the given formula: \[Q = \frac{4 \pi k R_1 (T_2 - T_1)}{R_2 - R_1}\] We are given that the thermal conductivity of the ground (k) is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Plugging all the values into the formula, we get: \[Q = \frac{4 \pi (1.4)(1.5)(-125)}{4.27 - 1.5} = \frac{-1319.47}{2.77}\] Calculating the heat transfer rate, we get: \[Q \approx -475.87 \mathrm{~W}\] Since heat is transferred from the tank to the ground, the heat transfer rate is negative which indicates a transfer of heat from the tank to the ground. The rate of heat transfer from the tank is approximately 475.87 W.