6. Mass Transfer
Figure 1 Diffusion of component A into component B.
Fick’s Law of Diffusion:
∂ CA
m& A
= −D
∂x
A
where,
D = proportionality constant–diffusion coefficient, m2/s
m& A = mass flux per unit time, kg/s
CA = mass concentration of component A per unit volume, kg/m3
Analogy with Heat and Momentum Diffusion:
⎛ Q& ⎞
∂T
⎜⎜ ⎟⎟ = − k
∂x
⎝ A ⎠x
∂u
τ =μ
∂y
Fourier law of heat conduction:
Newton’s viscosity law:
The Mass Transfer Coefficient:
m& A = K A (C A1 − C A2 ) = −
D A (C A2 − C A1 )
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:
Momentum Equation:
u
∂u
∂u
∂2 u
+v
=ν
∂x
∂y
∂ y2
Energy Equation:
u
∂T
∂T
∂2 T
+v
=α
∂x
∂y
∂ y2
Concentration Equation:
u
∂ CA
∂ CA
∂2 CA
+v
=D
∂x
∂y
∂ y2
Boundary Conditions:
(a) Unlike temperature, the concentrations of species on
the two sides of a liquid–gas (or solid–gas or solid–
liquid) interface are usually not the same.
(b) An impermeable surface in mass transfer is
analogous to an insulated surface in heat
transfer.
Figure 4 Two common types of boundary conditions.
2
The Schmidt Number:
Sc =
ν
D
=
μ
ρD
Note:
The Schmidt number plays a role similar to that of the Prandtl number in convection heat transfer
problems ( Pr =
υ
).
α
The Lewis Number:
Le =
α
D
=
Sc
Pr
The velocity, thermal and concentration laminar boundary thickness:
δV
= Pr n
δT
δV
= Sc n
δC
δT
= Le n
δC
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:
Sh =
K (hmass ) x
D
Note:
The Sherwood number plays a role similar to that of the Nusselt number in convection heat transfer
problems:
hx
Nu =
= f (Re x , Pr)
k
Sh =
K (hmass ) x
= f (Re x , Sc)
k
3
Chilton-Colburn Analogy for Mass Transfer:
Nu
Nu
h
=
=
Pe Re Pr ρ C p U ∞
Heat transfer Stanton Number:
St =
Mass transfer Stanton Number:
St mass =
K (hmass )
Sh
=
Re Sc
U∞
St Pr 2 3 =
Cf
= St mass Sc 2 3
2
Consequently,
St
⎛ Sc ⎞
=⎜ ⎟
St mass ⎝ Pr ⎠
h
⎛ Sc ⎞
= ρ Cp⎜ ⎟
K (hmass )
⎝ Pr ⎠
23
23
⎛α ⎞
= ρ Cp⎜ ⎟
⎝D⎠
23
= ρ C p Le 2 3
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:
D H 2O − Air = 1.87 × 10 −10
T 2.072
P
(m s 2 ) 280 K < T < 450 K
Table 2 DH 2O − Air at 1 (atm)
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