Question

Stainless steel ball bearings $\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, \)k=15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, \quad c_{p}=0.480 \mathrm{KJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, \quad\( and \)\quad \alpha=3.91 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}\( ) having a diameter of \)1.2 \mathrm{~cm}$ are to be quenched in water. The balls leave the oven at a uniform temperature of $900^{\circ} \mathrm{C}\( and are exposed to air at \)30^{\circ} \mathrm{C}$ for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is $125 \mathrm{~W} / \mathrm{m}^{2},{ }^{\circ} \mathrm{C}$, determine how long they can stand in the air before being dropped into the water.

Short answer

Answer: It takes 4.46 seconds for the stainless steel ball bearings to cool down from 900°C to 850°C.

Step-by-step solution

Calculate surface area and volume of ball bearings

First, we'll need to calculate the surface area and volume of the stainless steel ball bearing to later calculate the heat transfer. The formula for surface area (A) of a sphere is \(A = 4 \pi r^2\), and the formula for the volume (V) of a sphere is \(V = \frac{4}{3} \pi r^3\) , where r is the radius of the sphere. Given the diameter of the ball bearing is \(1.2 \thinspace cm\), we can calculate the radius as: \(r = \frac{1.2}{2} = 0.6 \thinspace cm = 0.006 \thinspace m\) Now, we can calculate the surface area and volume as: \(A = 4 \pi (0.006)^2 = 4.52 \times 10^{-4} \thinspace m^2\) \(V = \frac{4}{3} \pi (0.006)^3 = 9.05 \times 10^{-7} \thinspace m^3\)

Set up the heat transfer equation

Next, we'll set up the heat transfer equation to calculate the heat transfer rate: \(q = hA(T - T_\infty)\) Where \(q\) is the heat transfer rate, \(h\) is the heat transfer coefficient, \(A\) is the surface area, \(T\) is the temperature of the ball bearing, \(T_\infty\) is the surrounding temperature. Given: \(h = 125 \thinspace W/m^2 K\) \(T = 900^\circ C\) \(T_\infty = 30^\circ C\) Now, we can substitute the values into the heat transfer equation: \(q = 125 \times 4.52 \times 10^{-4} (900 - 30) = 39.35 \thinspace W\)

Calculate the heat loss required to drop the temperature

To find the heat loss required to drop the temperature from \(900^\circ C\) to \(850^\circ C\), we will use the formula: \(Q = mc_p\Delta{T}\) Where \(Q\) is the heat loss, \(m\) is the mass of the ball bearing, \(c_p\) is the specific heat capacity at constant pressure, and \(\Delta{T}\) is the change in temperature. Given: \(\rho = 8085 \thinspace kg/m^3\) \(c_p = 0.480 \thinspace kJ/kg \cdot ^{\circ}C = 480 \thinspace J/kg \cdot ^{\circ}C\) (converting from kJ to J) \(\Delta{T} = 50^\circ C\) First, we need to calculate the mass of the ball bearing using density: \(m = \rho V = 8085 \times 9.05 \times 10^{-7} = 0.00732 \thinspace kg\) Now, we substitute the values into the heat loss formula: \(Q = 0.00732 \times 480 \times 50 = 175.6 \thinspace J\)

Calculate the time required for the temperature to drop

Now that we have the heat transfer rate and the heat loss required to drop the temperature, we can calculate the time required for the temperature drop using: \(t = \frac{Q}{q} = \frac{175.6 \thinspace J}{39.35 \thinspace W} = 4.46 \thinspace s\) So, the stainless steel ball bearings can stand in the air for 4.46 seconds before they need to be dropped into the water to maintain a temperature above \(850^\circ C\).