Question

Carbon steel balls \(\left(\rho=7833 \mathrm{~kg} / \mathrm{m}^{3}, k=54 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), \(c_{p}=0.465 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=1.474 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right) 8 \mathrm{~mm}\) in diameter are annealed by heating them first to \(900^{\circ} \mathrm{C}\) in a furnace and then allowing them to cool slowly to \(100^{\circ} \mathrm{C}\) in ambient air at \(35^{\circ} \mathrm{C}\). If the average heat transfer coefficient is \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the annealing process will take. If 2500 balls are to be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air.

Short answer

Question: Calculate the time required for the annealing process and the total rate of heat transfer from the balls to the ambient air. Answer: The annealing process takes 1662.75 seconds and the total rate of heat transfer from the balls to the ambient air is 55710.35 W.

Step-by-step solution

Calculate the Biot number

First, we need to determine whether the lumped capacitance method is valid. We do this by calculating the Biot number: $$ \text { Bi }=\frac{h l}{k} $$ Given that the average heat transfer coefficient, h = 75 W/m²K, the diameter of the ball, D = 8 mm = 0.008 m, so the characteristic length, l = D / 6 = 0.008 / 6 m, and the thermal conductivity, k = 54 W/mK. Now, let's calculate the Biot number: $$ \begin{aligned} \text { Bi } &=\frac{75 \times\left(\frac{0.008}{6}\right)}{54} \\ &=\frac{75 \times 0.001333}{54} \\ &=0.0018519 \end{aligned} $$ Since Bi < 0.1, the lumped capacitance method can be used.

Calculate the time required for annealing

To calculate the time required for annealing, we will use the lumped capacitance method formula: $$ \frac{T(t)-T_{\infty}}{T_{0}-T_{\infty}}=e^{-\frac{h A}{\rho V c_{p}} t} $$ Given the initial temperature, \(T_0 = 900^\circ\text{C}\), the final temperature, \(T = 100^\circ\text{C}\), the air temperature, \(T_\infty = 35^\circ\text{C}\), the density, \(\rho = 7833\text{ kg/m³}\), and the specific heat, \(c_p = 346.5\text{J/kg K}\). Now, we need to isolate the variable \(t\): $$ t=-\frac{\rho V c_{p}}{h A} \ln \left(\frac{T(t)-T_{\infty}}{T_{0}-T_{\infty}}\right) $$ Let's calculate the time, \(t\): $$ \begin{aligned} t &=\frac{7833\times \frac{4}{3}\pi\times\left(\frac{0.004}{2}\right)^{3}\times 0.465\times 10^{3}}{75\times 4\pi\times\left(\frac{0.004}{2}\right)^{2}}\ln \left(\frac{100-35}{900-35}\right) \\ &=\frac{7833\times \frac{4}{3}\pi\times 2.56\times 10^{-8}\times 0.465\times 10^{3}}{75\times 4\pi\times 0.016\times 10^{-4}}\ln \left(\frac{65}{865}\right) \\ &=1662.75\text{ s } \end{aligned} $$ The annealing process takes 1662.75 seconds.

Calculate the total rate of heat transfer

We are given that 2500 balls are annealed per hour. To calculate the total rate of heat transfer, we can divide the heat transferred by one ball by the time it takes for one ball to be annealed and then multiply it by the number of balls per hour: $$ \text{Total rate of heat transfer} = \frac{\rho V c_{p} (T_0 - T)}{t} \times \text{number of balls per hour} $$ Let's calculate the total rate of heat transfer: $$ \begin{aligned} \text{Total rate of heat transfer} &=\frac{7833\times \frac{4}{3}\pi\times 2.56\times 10^{-8}\times 0.465\times 10^{3}\times(900-100)}{1662.75}\times 2500 \\ &=55710.35\text{ W } \end{aligned} $$ The total rate of heat transfer from the balls to the ambient air is 55710.35 W.

Key concepts

Biot Number

The Biot number (Bi) is an essential dimensionless parameter in heat transfer, particularly in the process of annealing metals like carbon steel balls. It indicates the ratio of internal thermal resistance within an object to the external thermal resistance across the boundary layer of the object to the surrounding environment.

Let's explore this concept with the formula \(\text{Bi} = \frac{hL}{k}\), where \(h\) is the heat transfer coefficient, \(L\) is the characteristic length of the object (commonly taken as volume/surface area), and \(k\) is the thermal conductivity of the material. If the Biot number is much less than 1, typically less than 0.1, it suggests that the temperature within the object can be assumed fairly uniform during the heat transfer process. This uniformity allows the use of the simplified lumped capacitance method to predict the cooling or heating of the object.

In our case with the carbon steel balls, we calculated a Biot number significantly lower than 0.1, which justifies our use of the lumped capacitance method.

Lumped Capacitance Method

When the Biot number is low, we use the lumped capacitance method, as it greatly simplifies calculations in transient heat transfer scenarios. This method assumes that temperature gradients within the object are negligible, i.e., the temperature is uniform across the object. Technically, it equates to applying the concept of thermal equilibrium within the object during the transition period.

Mathematically, we see this represented by the formula \( T(t)-T_{\infty}/T_{0}-T_{\infty} = e^{-\frac{hA}{\rho V c_{p}}t} \), which allows us to solve for the time \( t \) that the annealing process takes. In our exercise, once we confirmed that the lumped capacitance method was applicable, we plugged in the given values to calculate the time required to cool the steel balls. The simplicity of this method, given its applicability, made it possible to find that the annealing process would take about 1662.75 seconds per ball. While this method is quite practical for certain systems, it is essential to remember that its use is limited to cases where internal temperature gradients can reasonably be ignored.

Heat Transfer Coefficient

The heat transfer coefficient, \( h \), represents the rate of heat transfer per unit area per unit temperature difference between a surface and a fluid medium. It is measured in \( \text{W/m}^2\text{K} \). Higher values of \( h \) indicate that heat can more easily transfer between the surface and the environment. This coefficient is influenced by many factors, including the type of fluid, flow velocity, and surface condition.

In the context of our annealing exercise, the heat transfer coefficient is used as part of the Biot number calculation and directly in the lumped capacitance method. It gives us insight into how quickly the carbon steel balls will release heat into their surroundings. As given in the problem, a coefficient of \( 75 \text{W/m}^2\text{K} \) denotes a relatively moderate rate of heat transfer, which dictates the time needed for the annealing process and the total rate of heat transfer for a batch of balls.

Understanding the heat transfer coefficient is crucial when designing processes involving thermal systems, as it aids in ensuring efficiency and efficacy. For instance, a higher coefficient may speed up the annealing process, which is a critical aspect when considering production rates and quality of the end product.