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

Based on the given properties of carbon steel balls and the annealing process conditions, the lumped capacitance method can be used to determine the required time and heat transfer rate. It takes approximately 1276.5 seconds for the annealing process, and the total heat transfer rate from the balls to the ambient air is about 39.2 kW.

Step-by-step solution

Determine if the lumped capacitance method is appropriate

First, we need to find the Biot number to check if the lumped capacitance method can be applied. The Biot number is given by the formula: Bi = \(\frac{h \cdot L_{c}}{k}\), where \(h\) is the heat transfer coefficient, \(L_{c}\) is the characteristic length, and \(k\) is the thermal conductivity. For a sphere, the characteristic length is given by: \(L_{c}=\frac{V}{A}=\frac{4}{3} \pi r^{3} / 4 \pi r^{2}=\frac{r}{3}\). The given values are \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(k=54 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(r=0.004\,\text{m}\). So, we can calculate the Biot number as follows: Bi = \(\frac{75 \cdot (0.004 / 3)}{54}\). **Step 2: Check if the lumped capacitance method is valid**

Verify the Biot number's value

For the lumped capacitance method to be valid, the Biot number should be less than 0.1. Calculate the Biot number: Bi = \(1.85 \times 10^{-3} < 0.1\). Since the Biot number is less than 0.1, the lumped capacitance method can be applied. **Step 3: Calculate the required time for annealing process**

Apply the lumped capacitance method

Use the formula for the lumped capacitance method: \(t=\frac{(T_{initial}-T_{final}) m c_{p}}{hA(T_{initial}-T_{ambient})}\), where \(t\) is the time required for the annealing process, \(T_{initial}\) and \(T_{final}\) are the initial and final temperatures of the balls, \(m\) is the mass of the ball, \(c_{p}\) is the specific heat capacity, \(A\) is the surface area of the ball, and \(T_{ambient}\) is the ambient temperature. The mass of the ball can be found using the density and volume: \(m=\rho V=\rho \frac{4}{3} \pi r^{3} = 7833 \cdot \frac{4}{3} \pi (0.004)^{3}\). The surface area of the ball is given by: \(A=4 \pi r^{2} = 4 \pi (0.004)^{2}\). Now substitute the values in the formula: \(t=\frac{(900-100) \cdot 7833 \cdot \frac{4}{3} \pi (0.004)^{3} \cdot 0.465 \times 10^{3}}{75 \cdot 4 \pi (0.004)^{2} \cdot (900-35)}\). **Step 4: Calculate the total heat transfer rate**

Find the heat transfer rate

The total rate of heat transfer from the balls to the ambient air can be found using the given number of balls to be annealed per hour and the heat transfer rate for a single ball: \(Q_{total} = \frac{2500 \cdot (T_{initial}-T_{final}) m c_{p}}{t}\). Substitute the values to find the total heat transfer rate: \(Q_{total} = \frac{2500 \cdot (900-100) \cdot 7833 \cdot \frac{4}{3} \pi (0.004)^{3} \cdot 0.465 \times 10^{3}}{t}\). **Step 5: Calculate the time and total heat transfer rate**

Solve for the required quantities

Solve for the time required for the annealing process and the total heat transfer rate: \(t \approx 1276.5 \,\text{s}\) \(Q_{total} \approx 39.2 \,\text{kW}\) The annealing process will take approximately 1276.5 seconds, and the total rate of heat transfer from the balls to the ambient air during the process will be approximately 39.2 kW.