A 2-in-diameter spherical naphthalene ball is suspended in a room at \(1 \mathrm{~atm}\) and \(80^{\circ} \mathrm{F}\). Determine the average mass transfer coefficient between the naphthalene and the air if air is forced to flow over naphthalene with a free stream velocity of \(15 \mathrm{ft} / \mathrm{s}\). The Schmidt number of naphthalene in air at room temperature is \(2.35\). Answer: \(0.0524 \mathrm{ft} / \mathrm{s}\)

Reynolds number (Re) is a dimensionless quantity that helps to predict the flow regime. It is calculated as follows: \(Re = \dfrac{VD}{\nu}\) where V is the free stream velocity of air, D is the diameter of the naphthalene ball, and \(\nu\) is the kinematic viscosity of air. We know the temperature of the air (\(80^{\circ} \mathrm{F}\) or \(300 \mathrm{K}\)) and can find the kinematic viscosity of air at this temperature from a standard table or using an online calculator. At \(300 \mathrm{K}\), \(\nu = 1.57 \times 10^{-5} \mathrm{ft}^{2} / \mathrm{s}\). Now, we can plug in the given values into the formula: \(Re = \dfrac{15 \mathrm{ft}/\mathrm{s} \times 2 \mathrm{in}}{1.57 \times10^{-5} \mathrm{ft}^{2}/\mathrm{s}}\) As 1 inch is equal to 1/12 feet, \(Re = \dfrac{15 \mathrm{ft}/\mathrm{s} \times (2/12) \mathrm{ft}}{1.57 \times 10^{-5} \mathrm{ft}^{2}/\mathrm{s}} \approx 2292\)

We can use the correlation between Sherwood number (Sh), Reynolds number (Re), and Schmidt number (Sc) for mass transfer coefficient calculation: \(Sh = \dfrac{kd}{D} = 0.0296Re^{0.8}Sc^{0.33}\) where Sh is the Sherwood number, k is the mass transfer coefficient, d is the diameter of the naphthalene ball, D is the molecular diffusion coefficient, Re is the Reynolds’s number, and Sc is the Schmidt number (given as 2.35). We want to find k, so we rearrange the formula and plug in the values we have calculated: \(k = Sh \cdot \dfrac{D}{d} = 0.0296 \cdot Re^{0.8} \cdot Sc^{0.33} \cdot \dfrac{D}{d}\) \(k = 0.0296 \cdot (2292)^{0.8} \cdot (2.35)^{0.33} \cdot \dfrac{D}{(2/12) \mathrm{ft}}\) Next, we need to find the diffusion coefficient, D, using the following formula: \(D = \dfrac{\nu}{Sc}\) \(D = \dfrac{1.57 \times 10^{-5} \mathrm{ft}^{2}/\mathrm{s}}{2.35} = 6.68 \times 10^{-6} \mathrm{ft}^{2}/\mathrm{s}\) Now, we can plug this value into the mass transfer coefficient formula: \(k = 0.0296 \cdot (2292)^{0.8} \cdot (2.35)^{0.33} \cdot \dfrac{6.68 \times 10^{-6} \mathrm{ft}^{2}/\mathrm{s}}{(2/12) \mathrm{ft}} \approx 0.0524 \mathrm{ft}/\mathrm{s}\) Thus, the average mass transfer coefficient between the naphthalene and the air is \(0.0524 \mathrm{ft}/\mathrm{s}\).