Question

Hailstones are formed in high altitude clouds at \(253 \mathrm{~K}\). Consider a hailstone with diameter of \(20 \mathrm{~mm}\) and is falling through air at \(15^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(163 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the hailstone can be modeled as a sphere and has properties of ice at \(253 \mathrm{~K}\), determine the duration it takes to reach melting point at the surface of the falling hailstone. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Short answer

Answer: The duration it takes for the hailstone to reach the melting point at its surface is approximately 456.05 seconds.

Step-by-step solution

Calculate the Biot number

In order to use the analytical one-term approximation method, we need to find the Biot number (Bi). The Biot number is the ratio of heat transferred through convection to the heat conducted within the body. It is calculated using the formula: $$ Bi = \frac{hL_c}{k} $$ where \(h\) is the convection heat transfer coefficient, \(L_c\) is the characteristic length, and \(k\) is the thermal conductivity of the hailstone (ice). For a sphere, the characteristic length \(L_c\) is given as \(\frac{D}{6}\), where \(D\) is the diameter. Given: \(D = 20 \mathrm{~mm}\), \(h = 163 \mathrm{~W/m^{2} \cdot K}\), and from ice properties at \(253 \mathrm{~K}\), \(k = 2.2 \mathrm{~W/m \cdot K}\). First, we convert the diameter to meters: \(D=(20/1000)~m=0.02~m\) Now, we calculate the Biot number: $$ Bi = \frac{(163 \mathrm{~W/m^{2} \cdot K})(0.02 \mathrm{~m}/6)}{2.2 \mathrm{~W/m \cdot K}} = 0.983 $$ Since the Biot number is less than 10, we can safely use the one-term approximation method.

Calculate the time constant

To calculate the time required for the hailstone to reach the melting point at the surface, we will find the time constant. The time constant for a sphere is given by the formula: $$ \tau = \frac{\rho c_p D^2}{12 k} $$ where \(\tau\) is the time constant, \(\rho\) is the density of the hailstone (ice), and \(c_p\) is the specific heat capacity at constant pressure. From the properties of ice at \(253 \mathrm{~K}\): \(\rho = 917 \mathrm{~kg/m^3}\) and \(c_p = 2100 \mathrm{~J/kg \cdot K}\). Now, we calculate the time constant: $$ \tau = \frac{(917 \mathrm{~kg/m^{3}})(2100 \mathrm{~J/kg \cdot K})(0.02 \mathrm{~m})^2}{12(2.2 \mathrm{~W/m \cdot K})} = 175.05 \mathrm{~s} $$

Calculate the time required to reach the melting point at the surface

Using the one-term analytical approximation method, the temperature difference \(\Delta T\) between the center and the surface is given by the formula: $$ \Delta T = (T_i - T_\infty) \exp(-n^2 \frac{t}{\tau}) $$ where \(n\) is an integer (\(n=1\) for one-term approximation), \(t\) is the time required to reach the melting point at the surface, \(T_i\) is the initial temperature of the hailstone, and \(T_{\infty}\) is the ambient temperature. Given: \(T_i = 253 \mathrm{~K}\), \(T_{\infty} = 15^{\circ} \mathrm{C} = 273+15 = 288 \mathrm{~K}\), and the melting point of the ice at the surface is \(0^{\circ} \mathrm{C} = 273 \mathrm{~K}\). We can rewrite the formula for the time required \(t\) as follows: $$ t = \frac{\tau}{n^2} \ln(\frac{T_i - T_\infty}{\Delta T}) $$ Now, we calculate the time required: $$ t = \frac{175.05 \mathrm{~s}}{(1)^2} \ln(\frac{253 \mathrm{~K} - 288 \mathrm{~K}}{273 \mathrm{~K} - 288 \mathrm{~K}}) = 456.05 \mathrm{~s} $$ The duration it takes for the hailstone to reach the melting point at its surface is approximately 456.05 seconds.

Key concepts

Biot Number

The Biot number (Bi) is a dimensionless quantity that plays an essential role in heat transfer problems, especially for transient heat conduction.
The Biot number is defined as the ratio of the internal resistance to heat conduction within a body to the external resistance to heat transfer by convection. In simpler terms, it compares how quickly a material can conduct heat within itself to how quickly heat can be transferred from its surface to the surroundings.
Bi is calculated using the formula: \[\begin{equation}Bi = \frac{hL_c}{k}\end{equation}\]where:

• h represents the convection heat transfer coefficient, which is itself a measure of how well the surrounding fluid (such as air or water) can remove heat from the surface of the material.
• L_c is the characteristic length, which for a sphere, is typically its radius or diameter divided by a number that relates to the geometry of the object.
• k is the thermal conductivity of the material, indicating how well the material itself conducts heat.

In the context of our hailstone example, calculating the Biot number helped determine if the one-term analytical approximation method was suitable for solving the problem. A Biot number significantly less than 1 suggests that temperature gradients within the hailstone will be negligible, hence the method is appropriate. The same concept applies in other situations, as the Biot number aids in deciding the applicability of different heat transfer models.

One-term Analytical Approximation

The one-term analytical approximation is a method used to solve transient heat transfer problems without the need for complex numerical methods.
It relies on the assumption that temperature changes within a material can be estimated by using just the first term of an infinite series that represents the solution to the heat conduction equation. This simplification is valid when the Biot number is small (typically Bi < 0.1), indicating that the temperature gradient within the object is small compared to the temperature difference across the boundary layer.
The method provides an expedient way to predict temperature changes within a body over time, particularly when those changes are dominated by the temperature at the surface. In educational contexts and practical applications, the one-term approximation is valued for its simplicity and ease of use, and it is particularly useful for spherical objects, as it was in our hailstone example. By using this method, students or engineers can readily estimate the heat transfer process without relying on complex numerical simulations or extensive computational resources.

Thermal Conductivity

Thermal conductivity (k) is a physical property that represents a material's ability to conduct heat. It is an intrinsic property, meaning it is independent of the amount of material and depends only on the materials themselves.
Materials with high thermal conductivity, like metals, transfer heat efficiently, while those with low thermal conductivity, such as plastics or air, do not.
In the exercise with the hailstone, the thermal conductivity of ice determines how rapidly heat can travel from the hailstone's warmer exterior to its colder interior. Mathematically, thermal conductivity appears as an important factor in many fundamental heat transfer equations, including Fourier's law, which describes steady-state heat conduction, and in the calculation of the time constant in transient heat conduction problems. A solid understanding of thermal conductivity can thus greatly enhance the comprehension of heat transfer mechanisms in both simple and complex systems.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient (h) quantifies the rate at which heat is transferred between a surface and a fluid moving past it—such as air or water.
Its value depends on various factors including properties of the fluid (like viscosity and thermal conductivity), the flow regime (laminar or turbulent), the surface geometry, and the difference in temperature between the surface and the fluid.
In the hailstone problem, the convection heat transfer coefficient is crucial for predicting how quickly the hailstone will lose heat to the surrounding air as it falls. A higher coefficient would mean that heat is taken away more rapidly, potentially leading to a faster melting process. Evaluating this coefficient accurately is a key step in solving heat transfer problems involving convection. Understanding this concept is also valuable for a range of real-world applications, such as in the design of heating and cooling systems, the calculation of energy efficiency, and the analysis of weather phenomena.