Question

Copper balls $\left(\rho=8933 \mathrm{~kg} / \mathrm{m}^{3}, \quad k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, \)c_{p}=385 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, \alpha=1.166 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\( ) initially at \)180^{\circ} \mathrm{C}$ are allowed to cool in air at \(30^{\circ} \mathrm{C}\) for a period of $2 \mathrm{~min}\(. If the balls have a diameter of \)2 \mathrm{~cm}$ and the heat transfer coefficient is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the center temperature of the balls at the end of cooling is (a) \(78^{\circ} \mathrm{C}\) (b) \(95^{\circ} \mathrm{C}\) (c) \(118^{\circ} \mathrm{C}\) (d) \(134^{\circ} \mathrm{C}\) (e) \(151^{\circ} \mathrm{C}\)

Short answer

Answer: The center temperature of the copper balls after cooling for 2 minutes is approximately \(95^\circ\,\text{C}\).

Step-by-step solution

Write down the given information

We are given the following information: - Density of copper, \(\rho = 8933\,\text{kg}/\text{m}^3\) - Thermal conductivity of copper, \(k = 401\,\text{W}/\text{m}\cdot\text{K}\) - Specific heat at constant pressure, \(c_p = 385\,\text{J}/\text{kg}\cdot{ }^\circ\text{C}\) - Heat transfer coefficient, \(h = 80\,\text{W}/\text{m}^2\cdot\text{K}\) - Initial temperature, \(T_1 = 180^\circ\,\text{C}\) - Air temperature, \(T_\infty = 30^\circ\,\text{C}\) - Cooling time, \(t = 2\,\text{min}\) - Diameter of copper balls, \(d = 2 \,\text{cm}\)

Apply the Lumped Capacitance method

First we need to calculate the Biot number of the copper balls to determine if the lumped capacitance method is appropriate. Biot number, \(Bi = \frac{hL}{k}\), where L is the characteristic dimension of the geometry (radius in this case). $$L = \frac{d}{2} = \frac{2\,\text{cm}}{2} = 1\,\text{cm} = 0.01\,\text{m}$$ $$Bi = \frac{hL}{k} = \frac{80\,\text{W}/\text{m}^2\cdot\text{K} \cdot 0.01\,\text{m}}{401\,\text{W}/\text{m}\cdot\text{K}} \approx 0.02$$ Since \(Bi < 0.1\), the lumped capacitance method can be applied.

Calculate the final temperature

Using the lumped capacitance method, we can now calculate the final center temperature \(T\) of the copper balls by applying Newton's law of cooling: $$T = T_\infty + (T_1 - T_\infty) e^{-\frac{h A}{\rho V c_p}t}$$ - Surface area, \(A = 4\pi r^2 = 4\pi (0.01\,\text{m})^2 \approx 1.2566 \times 10^{-3}\,\text{m}^2\) - Volume, \(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.01\,\text{m})^3 \approx 4.1888 \times 10^{-6}\,\text{m}^3\) Now, plug in the values and calculate the final temperature. $$T \approx 30^\circ\,\text{C} + (180^\circ\,\text{C} - 30^\circ\,\text{C})e^{-\frac{80\,\text{W}/\text{m}^2\cdot\text{K} \cdot 1.2566 \times 10^{-3}\,\text{m}^2}{8933\,\text{kg}/\text{m}^3 \cdot 4.1888 \times 10^{-6}\,\text{m}^3 \cdot 385\,\text{J}/\text{kg}\cdot{ }^\circ\text{C} \cdot 120\,\text{s}}} \approx 95^\circ\,\text{C}$$ Thus, the center temperature of the copper balls at the end of cooling is about \(95^\circ\,\text{C}\), which corresponds to option (b).