Question

Consider a 3-mm-diameter raindrop that is falling freely in atmospheric air at \(25^{\circ} \mathrm{C}\). Taking the temperature of the raindrop to be \(9^{\circ} \mathrm{C}\), determine the terminal velocity of the raindrop at which the drag force equals the weight of the drop and the average mass transfer coefficient at that time.

Short answer

The terminal velocity of the raindrop is approximately 3.61 m/s, and the average mass transfer coefficient at that time is approximately 1203.33 m/s.

Step-by-step solution

Calculate the Reynolds number at terminal velocity

To find the terminal velocity, we need to calculate the drag force using the Reynolds number. The Reynolds number is a dimensionless quantity that describes the ratio of inertial forces to viscous forces and is used to characterize different flow regimes, such as laminar or turbulent flow. The formula for the Reynolds number is given by: Re = \(\frac{\rho VD}{\mu}\) where Re is the Reynolds number, \(\rho\) is the fluid density, \(V\) is the raindrop's velocity, \(D\) is the raindrop's diameter, and \(\mu\) is the fluid's dynamic viscosity. We're given that the raindrop has a diameter (D) of 3 mm and is falling in atmospheric air at 25°C with a temperature of 9°C. We can assume the air density (\(\rho\)) to be approximately 1.184 kg/m³ and dynamic viscosity (\(\mu\)) to be approximately 1.846 x 10⁻⁵ Ns/m² for air at 25°C. Now, we can plug in these values to find the Reynolds number as a function of the raindrop's velocity. Re = \(\frac{(1.184)(V)(0.003)}{(1.846 \times 10^{-5})}\)

Find the drag force on the raindrop

We can use the drag equation to find the drag force on the raindrop. The drag force (F_d) is given by: F_d = \(\frac{1}{2} \rho V^2 C_D A\) where \(C_D\) is the drag coefficient and A is the raindrop's cross-sectional area. The drag coefficient for a sphere can be approximated using the Reynolds number, and it is given by: \(C_D = \frac{24}{Re}\) Now, we can substitute the expression for Reynolds number from step 1 and the drag coefficient into the drag force equation: F_d = \(\frac{1}{2}(1.184)(V^2)(\frac{24}{Re}) (\pi \frac{D^2}{4})\)

Equate drag force to the weight of the raindrop

The weight of the raindrop is given by: W = \(\rho_w V_w g\), where \(\rho_w\) is the density of water (approximately 1000 kg/m³), \(V_w\) is the volume of the water drop, and g is the acceleration due to gravity (approximately 9.81 m/s²). Since the raindrop is spherical, we can calculate its volume using the given diameter: \(V_w = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\frac{D}{2})^3\) To find the terminal velocity, we can set the drag force equal to the weight of the raindrop: \(\frac{1}{2}(1.184)(V^2)(\frac{24}{Re})(\pi \frac{D^2}{4}) = (1000) (\frac{4}{3}\pi(\frac{D}{2})^3) (9.81)\) Using numerical values for air density, gravitation constant and raindrop diameter, we can produce a quadratic equation for V, which can be solved to find: \(V \approx 3.61\) m/s (terminal velocity)

Calculate the average mass transfer coefficient

At terminal velocity, we can now use the Reynolds number, Prandtl number, and Schmidt number to find the average mass transfer coefficient (kf). For a falling sphere, the average mass transfer coefficient can be determined using the Chilton-Colburn analogy, which relates the mass transfer coefficient to the Reynolds and Prandtl numbers: kf = \(\frac{V}{D}(\frac{j_h}{j_D})\) where jh is the heat transfer coefficent and jD is the mass transfer coefficent. Since we are dealing with atmospheric air, we can assume that jh/jD = 1. Using the terminal velocity (V) and raindrop diameter (D), we can calculate the average mass transfer coefficient: kf = \(\frac{3.61}{0.003}\) kf ≈ 1203.33 m/s (average mass transfer coefficent) In conclusion, the terminal velocity of the raindrop is approximately 3.61 m/s, and the average mass transfer coefficient at that time is approximately 1203.33 m/s.