6. Mass Transfer Figure 1 Diffusion of component A into component B. Fick’s Law of Diffusion: m ∂ C & A = −D A A ∂ x where, D = proportionality constant–diffusion coefficient, m2/s m& A = mass flux per unit time, kg/s 3 CA = mass concentration of component A per unit volume, kg/m Analogy with Heat and Momentum Diffusion: ⎛ Q& ⎞ ∂T Fourier law of heat conduction: ⎜ ⎟ = −k ⎜ A ⎟ ∂ x ⎝ ⎠x ∂u Newton’s viscosity law: τ = μ ∂ y The Mass Transfer Coefficient: D A(C − C ) A2 A1 m& A = K A(CA − CA ) = − 1 2 Δx D K = Δx Figure 2 Analogy between convection heat transfer and convection mass transfer. 1 Mass Convection: (a) external flow (b) internal flow Figure 3 The development of a concentration boundary layer. The Governing Equations: ∂u ∂u ∂2 u Momentum Equation: u + v =ν ∂ x ∂ y ∂ y2 ∂T ∂T ∂2 T Energy Equation: u + v = α ∂ x ∂ y ∂ y2 ∂ C ∂ C ∂ 2 C Concentration Equation: u A + v A = D A ∂ x ∂ y ∂ y 2 Boundary Conditions: (a) Unlike temperature, the concentrations of species on (b) An impermeable surface in mass transfer is the two sides of a liquid–gas (or solid–gas or solid– analogous to an insulated surface in heat liquid) interface are usually not the same. transfer. Figure 4 Two common types of boundary conditions. 2 The Schmidt Number: ν μ Sc = = D ρ D Note: The Schmidt number plays a role similar to that of the Prandtl number in convection heat transfer υ problems ( Pr= ). α The Lewis Number: α Sc Le = = D Pr The velocity, thermal and concentration laminar boundary thickness: δ δ δ V = Pr n V = Sc n T = Le n δ T δ C δ C Note: 1 n = for most applications in all three relations. 3 Figure 5 when the molecular diffusivities of momentum, heat, and mass are equal to each other, the velocity, temperature, and concentration boundary layers coincide. The Sherwood Number: K(h x Sh = mass) D Note: The Sherwood number plays a role similar to that of the Nusselt number in convection heat transfer problems: h x Nu = = f (Re ,Pr) k x K(h ) x Sh = mass = f (Re , Sc) k x 3 Chilton-Colburn Analogy for Mass Transfer: Nu Nu h Heat transfer Stanton Number: St = = = Pe Re Pr ρ C p U ∞ Sh K(hmass ) Mass transfer Stanton Number: Stmass = = Re Sc U ∞ C St Pr 2 3 = f = St Sc 2 3 2 mass Consequently, 2 3 St ⎛ Sc ⎞ = ⎜ ⎟ Stmass ⎝ Pr ⎠ 2 3 2 3 h ⎛ Sc ⎞ ⎛ α ⎞ 2 3 = ρ C p ⎜ ⎟ = ρ C p ⎜ ⎟ = ρ C p Le K(hmass) ⎝ Pr ⎠ ⎝ D ⎠ Mass Convection Correlations: Table 1 Sherwood number relations in mass convection for specified concentration at the surface corresponding to the Nusselt number relations in heat convection for specified surface temperature 4 Diffusion of Water Vapor in Air: Marrero and Mason empirical formula: T 2.072 D = 1.87 ×10−10 (m s 2 ) 280 K < T < 450 K H 2O− Air P Table 2 D at 1 (atm) H 2O− Air 5 Example 1: Dry air at 15°C and 92 kPa flows over a 2-m-long wet surface with a free stream velocity of 4 m/s. Determine the average mass transfer coefficient. Solution: 6 Example 2: 7 Example 3: A 2-in.-diameter spherical naphthalene ball is suspended in a room at 1 atm and 80°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 ft/s. The Schmidt number of naphthalene in air at room temperature is 2.35. Solution: 8.