Question

Carbon steel balls $\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}, k=64 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, \)\left.c_{p}=434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( initially at \)200^{\circ} \mathrm{C}\( are quenched in an oil bath at \)20^{\circ} \mathrm{C}$ for a period of \(3 \mathrm{~min}\). If the balls have a diameter of \(5 \mathrm{~cm}\) and the convection heat transfer coefficient is $450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, the center temperature of the balls after quenching will be (Hint: Check the Biot number.) (a) \(30.3^{\circ} \mathrm{C}\) (b) \(46.1^{\circ} \mathrm{C}\) (c) \(55.4^{\circ} \mathrm{C}\) (d) \(68.9^{\circ} \mathrm{C}\) (e) \(79.4^{\circ} \mathrm{C}\)

Short answer

Answer: The center temperature of the balls after quenching is approximately \(46.1^{\circ}\mathrm{C}\).

Step-by-step solution

Determine the Biot number

The Biot number (Bi) is given by: Bi \( =\frac{hL_\text{c}}{k}\) \(where:\) \(h = 450 \,\text{W}/\text{m}^2\cdot\text{K}\) (convection heat transfer coefficient) \(L_\text{c} = \frac{r}{3} \text{(assuming the sphere shape)}\) \(r = 0.025\,\text{m}\) (radius of the balls) \(k = 64 \,\text{W/m}\cdot\text{K}\) (thermal conductivity of the steel balls) Calculating the Biot number: Bi \(=\frac{450\times0.025/3}{64} = 0.104\). Since Bi \(<1\), we can assume the lumped capacitance method is applicable.

Calculate the temperature at the center of the balls after quenching

The lumped capacitance method provides the transient temperature response of a solid, assuming uniform temperature throughout. The temperature change equation can be written as: \(\frac{t-T_\infty}{T_0-T_\infty} = \text{exp}\left(-\frac{hA}{\rho c_p V}t\right)\) \(where:\) \(T_0 = 200^{\circ}\mathrm{C}\) (the initial temperature of the balls) \(T_\infty = 20^{\circ}\mathrm{C}\) (the temperature of the oil bath) \(\rho = 7830\, \text{kg/m}^3\) (density of the steel balls) \(c_p = 434\, \text{J/kg} \cdot \text{K}\) (specific heat capacity of the steel balls) \(A = 4\pi r^2\) (surface area of the ball) \(V = \frac{4}{3}\pi r^3\) (volume of the ball) >t = 3 min = 180 s (time) Rearrange the temperature change equation for \(t\) and substitute the given values: \(t-T_\infty = (T_0-T_\infty) \cdot \text{exp}\left(-\frac{hA}{\rho c_p V}t\right)\) \(t-20 = (200-20) \cdot \text{exp}\left(-\frac{450\times(4\pi(0.025)^2)}{7830\times434\times\frac{4}{3}\pi(0.025)^3} \cdot 180\right)\) \(t-20 = 180 \cdot \text{exp}(-0.339)\) Solve for \(t\) to find the center temperature of the balls after quenching: \(t = 20 + 180 \cdot \text{exp}(-0.339) \approx 46.1^{\circ}\mathrm{C}\) The center temperature of the balls after quenching will be approximately \(46.1^{\circ}\mathrm{C}\). The correct answer is (b) \(46.1^{\circ}\mathrm{C}\).