Question

A 1.4-m-diameter spherical steel tank filled with iced water at \(0^{\circ} \mathrm{C}\) is buried underground at a location where the thermal conductivity of the soil is \(k=0.55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the tank center and the ground surface is \(2.4 \mathrm{~m}\). For ground surface temperature of \(18^{\circ} \mathrm{C}\), determine the rate of heat transfer to the iced water in the tank. What would your answer be if the soil temperature were \(18^{\circ} \mathrm{C}\) and the ground surface were insulated?

Short answer

Answer: The rate of heat transfer to the iced water in the tank is approximately 33.29 W. 2. What is the rate of heat transfer to the iced water in the tank if the ground surface is insulated? Answer: If the ground surface is insulated and the soil temperature is 18°C, there will be no heat transfer to the iced water in the tank (0 W).

Step-by-step solution

Information and Formula

Given: - Spherical tank diameter: \(D = 1.4 m\) - Temperature of iced water: \(T_{i} = 0^{\circ} C\) - Distance between tank center and ground surface: \(H = 2.4 m\) - Thermal conductivity of soil: \(k = 0.55 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\) - Ground surface temperature: \(T_{s} = 18^{\circ} C\) We will use this formula for the rate of heat transfer: \(Q = \frac{k \cdot A \cdot \Delta T}{L}\), where: - \(Q\) – rate of heat transfer - \(k\) – thermal conductivity - \(A\) – area of the heat transfer surface - \(\Delta T\) – temperature difference between the surfaces - \(L\) – thickness between the surfaces

Calculate the Surface Area

Since the tank is spherical, the surface area can be found using the formula for the surface area of a sphere: $$A = 4 \pi r^2$$, where \(r\) is the radius of the sphere. Radius of the tank: \(r = \frac{D}{2} = \frac{1.4}{2} = 0.7 m\) Now we can find the surface area: \(A = 4 \pi (0.7)^2 = 6.16 m^2\)

Temperature Difference and Thickness

Now we need to find \(\Delta T\) and \(L\). Temperature difference: \(\Delta T = T_{s} - T_{i} = 18 - 0 = 18^{\circ} C\) Thickness between the surfaces: \(L = H - r = 2.4 - 0.7 = 1.7 m\)

Calculate the Rate of Heat Transfer

Using the formula for the rate of heat transfer, we can now determine the value of \(Q\): \(Q = \frac{0.55 \cdot 6.16 \cdot 18}{1.7} = 33.29 \mathrm{~W}\) So, the rate of heat transfer to the iced water in the tank is approximately \(33.29 \mathrm{W}\).

Insulated Ground Surface Case

If the ground surface is insulated, there will be no heat transfer from the insulated surface to the soil layer above the tank. Therefore, \(T_s\) is no longer the ground surface temperature but rather the soil temperature, which we are told is also \(18^{\circ} C\). In this case, the temperature difference between the surfaces becomes zero: \(\Delta T = T_s - T_i = 18 - 18 = 0^{\circ} C\) As a consequence, the rate of heat transfer will be zero: \(Q = \frac{0.55 \cdot 6.16 \cdot 0}{1.7} = 0 \mathrm{~W}\) So, if the soil temperature is \(18^{\circ} C\) and the ground surface is insulated, there will be no heat transfer to the iced water in the tank.

Key concepts

Thermal Conductivity

Thermal conductivity is a fundamental property of materials that quantifies their ability to conduct heat. In this context, the thermal conductivity of the soil plays a crucial role in determining how quickly heat from the warmer ground surface can transfer to the cooler environment inside the spherical tank. The formula for heat transfer includes thermal conductivity as a key component:

• It is represented by the letter \(k\) in equations.
• Measured in units of \( \, \text{W/m}\cdot\text{K} \), which denote watts per meter per degree Kelvin.
• Higher values indicate materials that conduct heat more efficiently, such as metals. Lower values indicate insulating materials like wood or, in our case, soil.

Understanding the rate of heat transfer is necessary to predict how fast temperature changes will occur, particularly important in applications like preserving the temperature of iced water in tanks. In this scenario, with the thermal conductivity of soil at 0.55 W/m·K, we can calculate how the heat fluctuates in and out of the tank system.

Spherical Tank

A spherical tank presents a unique surface for heat exchange compared to other geometric shapes. The spherical shape brings several beneficial properties:

• It minimizes the surface area to volume ratio, reducing the heat loss or gain from the environment, which is important for retaining temperatures.
• The surface area of a sphere is calculated with the formula \(A = 4 \pi r^2\), where \(r\) is the radius of the sphere.
• This form factor proves beneficial for applications requiring uniform distribution of stress, making it a preferred choice for containing refrigerated or pressurized contents.

For our buried tank example, the radius is 0.7 meters, leading to a calculated surface area of approximately 6.16 square meters. This detail is critical when evaluating heat transfer rates, providing the necessary factor needed to work with both thermal conductivity and temperature gradient in calculations.

Temperature Gradient

The temperature gradient is the rate at which temperature changes across a specific area. It is a driving force behind heat transfer, impacting the speed and direction of this energy movement.

• In this scenario, the temperature gradient forms between the warm ground surface (18°C) and the cooler iced water in the tank (0°C).
• Represented as \(\Delta T\), calculated as the difference \(T_s - T_i\).
• In our situation, the gradient calculates to 18°C over a 1.7-meter distance, which is required for heat calculations using Fourier's Law of Heat Conduction.

Whether through conduction, convection, or radiation, understanding gradients helps predict heat movement from regions of higher temperature to those of lower temperature. Managing these pathways is essential in thermal management applications, such as ensuring the consistent temperature of iced water in tanks.

Insulated Surfaces

Insulated surfaces are used to minimize or completely prevent the flow of heat between two regions. Insulation is vital in maintaining desired temperatures, whether it involves keeping cold things cold or hot things hot.

• By placing insulating materials around a surface, the heat exchange between the environment and an enclosed space is significantly reduced.
• For this exercise, when the ground surface is insulated, there is no heat transfer as the temperature difference becomes zero, shown mathematically as \(\Delta T = 0\).
• No temperature gradient means no heat flow, ensuring that the iced water in the tank remains unaffected by external temperature changes.

Effective insulation design is crucial in various industries, including construction and food preservation. It is the primary reason why, in this scenario, if soil temperature and surface temperature are equal and the ground is insulated, heat transfer becomes negligible.