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 to the ground is approximately 665.7 W.

Step-by-step solution

Identify the dimensions of the tank and ground

The diameter of the spherical tank is given as 3 meters. The radius of the tank can be calculated as half of the diameter, which is 1.5 meters. The distance between the top surface of the tank and the ground surface is given as 4 meters.

Calculate the conductive thermal resistance

The heat transfer through the ground can be modeled as a conductive thermal resistance. The conductive thermal resistance R_cond can be calculated as: \(R_\mathrm{cond} = \dfrac{r_\mathrm{outer}}{4 \pi \cdot k \cdot r_\mathrm{inner}}\) where r_inner is the inner radius of the tank, r_outer is the outer radius of the ground covering the tank, and k is the thermal conductivity of the ground. In this case, r_inner = 1.5 m, r_outer = 1.5 m + 4 m = 5.5 m, and k = 1.4 W/(m·K). Substituting these values, we get: \(R_\mathrm{cond} = \dfrac{5.5}{4 \pi \cdot 1.4 \cdot 1.5} = 0.1879 \, \mathrm{K} / \mathrm{W}\)

Determine temperature difference

The temperature difference between the surface of the tank (T_tank) and the ground (T_ground) is given as: \(\Delta T = T_\mathrm{tank} - T_\mathrm{ground}\) Substituting the given values, we get: \(\Delta T = 140 - 15 = 125^{\circ} \mathrm{C}\)

Calculate the rate of heat transfer

The rate of heat transfer (Q) can be determined using the overall thermal resistance (R_cond) and the temperature difference. The formula is as follows: \(Q = \dfrac{\Delta T}{R_\mathrm{cond}}\) Substituting the values, we get: \(Q = \dfrac{125}{0.1879} = 665.7 \, \mathrm{W}\) Therefore, the rate of heat transfer from the tank to the ground is approximately 665.7 W.

Key concepts

Conductive Thermal Resistance

Conductive thermal resistance is a key concept when analyzing heat transfer, especially through solid materials. It describes how resistant a material is to the flow of heat. Simply put, it's how hard it is for heat to pass through. The lower the thermal resistance, the better the material conducts heat.
To calculate the conductive thermal resistance, we use the formula:

• \(R_\mathrm{cond} = \dfrac{r_\mathrm{outer}}{4 \pi \cdot k \cdot r_\mathrm{inner}}\)

Here, \(r_\mathrm{outer}\) and \(r_\mathrm{inner}\) are the outer and inner radii, respectively, and \(k\) represents the thermal conductivity. The units of this measure are typically \(\mathrm{K/W}\).
This resistance plays a crucial role in determining how efficiently heat can be transferred from the warmer body to the cooler surroundings. Understanding this concept is essential for designing systems that need to manage heat flow efficiently.

Spherical Tank

When discussing heat transfer, the geometry of the object, such as a spherical tank, is significant. A spherical shape is efficient for containing materials like gases and liquids. It evenly distributes stress and minimizes the surface area for a given volume, which can affect heat transfer dynamics.
The spherical tank in our problem has a defined diameter of 3 meters, which is straightforward to calculate the radius from. The radius is key in determining the overall thermal resistance, as seen in the previous section. For this particular example, the spherical tank is buried underground to safeguard its contents and manage the thermal interactions with the environment. This configuration affects how heat is conducted through the earth, influencing the rate of heat transfer.

Temperature Difference

Understanding temperature difference, often denoted as \(\Delta T\), is essential in calculating heat transfer. It reflects how much warmer one body is compared to another. This difference is a driving force behind the heat flow. Heat naturally moves from a high temperature to a low temperature region, aiming to reach thermal equilibrium.
For our spherical tank example, the temperature difference is quite significant. The tank's surface is much hotter at \(140^\circ \mathrm{C}\), whereas the ground surface is cooler at \(15^\circ \mathrm{C}\). Thus, the calculated \(\Delta T\) is \(125^\circ \mathrm{C}\).
This large temperature gradient facilitates a higher rate of heat transfer, and hence, it is important in determining the amount of heat exchanged with the surroundings efficiently.

Thermal Conductivity

Thermal conductivity is a material's inherent ability to conduct heat. A higher thermal conductivity means the material can transfer heat more efficiently. It is a constant that varies among materials and influences how thermal energy moves through substances.
Defined by the symbol \(k\), its units are \(\mathrm{W/m \cdot K}\). In the context of our problem, the soil surrounding the spherical tank has a thermal conductivity \(k = 1.4\, \mathrm{W/m \cdot K}\), meaning it's not as conductive as metals, but adequate enough for typical ground conditions.
Thermal conductivity, combined with the specific geometric orientation of materials, governs how quickly heat can pass through a layer, such as the earth around our buried tank. By understanding and calculating thermal conductivity, we can predict heat transfer rates more accurately and design better for thermal management.