CME 301 – Mass Transfer
Convective Mass Transfer
Dr. Chandra Mouli MR
Chemical Engineering Department
College of Engineering
Abu Dhabi University
Convective Mass Transfer
Convective mass transfer: Transfer of a
component between two phases that is
promoted by the motion of the fluid phase(s)
NA = kc ∆cA
Convective Mass Transfer
Coefficient
What is the unit of kc?
It is analogous to convective heat transfer:
q/A = h ∆T 2
kc and h are related to:
Properties of the fluid
Dynamic characteristics of the flowing fluid
Geometry of the system of interest
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Parameters for Analysis of Convective
Mass Transfer
Analogy Between Momentum, Heat and
Mass Transfer:
1. Momentum Diffusivity = ν = µ/ρ [=] L2/t
2. Thermal Diffusivity = α = kth/ρ Cp [=] L2/t
3. Mass Diffusivity = DAB [=] L2/t
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Concentration Boundary Layer
In the case of fluid flowing past a surface, there will
be a layer, sometimes extremely thin, close to the
surface wherein the flow is laminar Molecular
mass transfer will always be present in any
convective process through such boundary layer.
If the fluid flow is laminar, then all of the transport
will be by molecular means.
If the fluid flow is turbulent, eddies will move the
material physically higher mass transfer rates
are associated with turbulent conditions
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Velocity and concentration profiles:
v∞
[cA∞ – cAs]
y
v=v(y) [cA – cAs] = f(y)
At the height of the velocity/concentration boundary
layer, the bulk conditions will apply
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Two General Cases of Convective Mass
Transfer
1. Two immiscible fluid phases in contact
(transferred “A” is soluble in both phases)
2. Fluid contacting a solid surface (solid acts
as a source or a sink)
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1. Two immiscible fluid phases in contact (“A” is
soluble in both phases)
This process is known
as interphase
convective mass
transfer:
1. Gas-Liquid Contact
(e.g., absorption,
stripping, evaporation)
2. Liquid-Liquid Contact
(e.g., liquid-liquid
extraction)
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2. Fluid contacting a solid surface (solid
acts as a source or a sink)
z=δ z=0
Porous Solid
cA0 (Molecular Transfer)
L = characteristic
length of solid
cAs
NAz
NAz = -DAB cA
NAz = kc (cA0 – cAs)
+ cA V
Fluid Boundary
Layer
At steady state, cA0 and cAs are constant, and
At the thickness of the boundary layer (z = δ):
NAz(convection) = NAz (molecular diffusion) 9
dc A
kc ( c A0 − c As ) =
− DAB + c AV
dz
Multiply both sides by L, rearrange:
molecular mass
d [ c A − c As ]
−
kc L dz z =0 transfer resistance
=
Sh = = c A 0 − c As
DAB convective mass
L
transfer resistance
Sherwood Number
(dimensionless)
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Methods for Evaluating Convective Mass
Transfer
1. Dimensional analysis coupled with
Experiments
2. Exact boundary layer analysis
3. Analogy between momentum, energy and
mass transfer
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1. Dimensional Analysis of Convective
Mass Transfer
Target: finding dimensionless groups which
affect the mass transfer behavior.
Solution:
Sh = f(other dimensionless groups),
Experiments needed to determine
functionality
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Dimensionless Groups for Convective
Mass Transfer
Schmidt Number :
ν momentum diffusivity Analogous to Prandtl Number
=
Sc =
DAB mass diffusivity (Pr) in Heat Transfer
Lewis Number : Reynolds Number :
α thermal diffusivity =
v.L inertial flow
=
=
Le = Re
DAB mass diffusivity ν viscous flow
Analogous to Nusselt Sherwood Number :
Number (Nu) in Heat kc L convective mass transfer rate
=
Sh =
Transfer DAB diffusive mass transfer rate
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Peclet Number for Mass Transfer :
v.L
= =
PeAB Re.Sc
DAB
Grashof Number for Mass Transfer :
L3 g ∆ρ A
GrAB =
ρ
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Example 1. Forced Convection
Step 1: Define the effective variables
No. Variable Symbol Dimensions
1 Diameter D L
2 Fluid density ρ M.L-3
3 Fluid viscosity µ M.L-1. t -1
4 Fluid velocity v L . t -1
5 Diffusivity DAB L2. t -1
6 MT Coefficient kc L . t -1
6 variables 3 Dimensions
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Example 1. Forced Convection
For forced convection, the three
dimensionless groups [Sh, Sc, Re] are
sufficient to describe the mass transfer
behavior How can we determine
Sh = f(Sc, Re) this functionality?
Experiments can be done by varying only Re
and Sc and observing the variations in Sh
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Example 2: Natural Convection
Step 1: Define the effective variables
No. Variable Symbol Dimensions
1 Characteristic length L L
2 Fluid density ρ M.L-3
3 Fluid viscosity µ M.L-1. t -1
4 Buoyancy g∆ρΑ M.L-2. t -2
5 Diffusivity DAB L2. t -1
6 MT Coefficient kc L . t -1
6 variables 3 Dimensions
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Sh = f(Sc, GrA,B)
L.kc ρ DAB 1
π1 = = Sh π2 = =
DAB µ Sc
L ρ g ∆ρ A
3
π3 = = GrA, B Grashof No. for
µ 2
natural convection
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2. Exact Analysis of Concentration
Boundary Layer
Assuming:
(1) constant cT and DAB, (2) RA = 0, (3) st.st.
v.cA = DAB 2cA
∂c A ∂c A ∂ 2cA ∂ 2cA
vx + v y = DAB 2 + 2
∂x ∂y ∂x ∂y
Boundary Conditions:
(1) cA = cAs @ y = 0 (2) cA = cA∞ @ y = ∞
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Solution:
dc A Re x
= ( c A∞ − c As ) 0.332
dy y =0 x
⇒ Flux :
dc A Re x
N Ay = − DAB − DAB ( c A∞ − c As ) 0.332
=
dyy =0 x
= kc ( c As − c A∞ )
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DAB
⇒ kc = 0.332 Re x
x
kc x
or =Sh = 0.332 Re0.5 x (for Sc = 1)
DAB
If Sc ≠ 1
kc x
=
Sh = 0.332 Re x Sc
12 13
DAB
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Mean mass transfer coefficient (kc,av) over a
plate of width W and length L is obtained by
integration over area:
Moles transferred (wA) = f A kc (cAs - cA∞) dA
= WL kc,av (cAs - cA∞)
kc ,av x Local (average) Sh
=
ShL = 0.664 Re L Sc
12 13
No. at distance x
DAB
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3. Mass, Energy and Momentum
Transfer Analogies
If we have heat transfer coefficient (h) we can
calculate the mass transfer coefficient (kc), and
vice versa
All analogies require:
1. Constant physical and chemical properties
2. No generation of energy or mass (no homo. reaction)
3. Velocity profile is not affected by mass transfer
4. No viscous dissipation (no energy loss due to
momentum)
5. No homogeneous reaction occurs
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A. Reynolds Analogy
f = skin friction factor
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B. Chilton-Colburn Analogy
{Sc ≠1, Pr ≠1}
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Flat Plate
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Flow Through Pipes
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Specific Correlations for Fixed
Configurations
Chapter 28 gives detailed correlations for calculating
the mass transfer coefficients (Laminar and
Turbulent) for:
1. Flat Plates
2. Single Spheres
3. Spherical bubble swarms
4. Single Cylinders
5. Flow through pipes
6. Wetted wall column
7. Packed and fluidized beds
8. Stirred tanks
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Example
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Example
30
Example
31
