Miscellaneous notes on mass transfer coefficient models

P.C. Chau (UCSD, 1999)
Chapter 6 Supplementary Notes
We provide here the derivations needed to clarify Section 6.4 in Middleman. Example 6.4.1 makes use of
mass transfer coefficients in a channel flow. The concentration is “understood” to be the cup-mixing
average, but the story is longer than that. We illustrate here with flow through a tube. After reading these
<strong>notes</strong>, you should repeat as an exercise the derivations with a channel flow.
❏ Mass transfer coefficient model of flow through a tube
If we have a fluid flow through a tube where mass transfer occurs at the wall, it is not
uncommon that we find people write the steady-state mass balance equation of species A as
dC A
dz = 2k
U R (C s –C A ) (1)
where z is the axial distance down the tube, R is the radius, and U – is the average velocity. There is
no chemical reaction, and the mass transport is described by the average or overall mass transfer
coefficient – k . As written, the surface concentration of the species A at the wall, Cs, is a constant
and is presumed to be higher than that in the fluid, C A = C A (z). Thus Eq. (1) describes how the
concentration C A increases along the tube.
The curious (and confusing) thing is that the same general form holds whether the flow is
laminar or turbulent, and whether the tube is empty or filled with a packing material. The
derivation details are in the Appendix. We just point out here that Eq. (1) is not entirely correct,
although we can still use it. The meaning of each quantity is very different under different
circumstances, and we have to be careful with their interpretations. One more time: read the
Appendix.
With the two boundary conditions C A (0) = C Ao , and C A (L) = C AL , we can integrate Eq. (1) to
obtain
ln C s –C AL
=– 2kL C s –C AL
or
= exp – 2kL
(2)
C s –C Ao UR C s –C Ao UR
The handling of the negative sign from the integration is a bit odd; we simply follow how most
textbooks do the integration. What is even odder is how the RHS of Eq. (2) is rearranged to
C s –C AL
4Sh
= exp –
D
C s –C Ao Sc Re D (D/L)
(3)
There is no trick here. It is just a matter of multiplying a fraction top and bottom with the same
quantities until we re-mold them into dimensionless groups:
2kL
U R = k D D AB
D AB
ν
ν 2L
U D R =
4Sh D
Sc Re D (D/L)
where we have introduced the tube diameter D = 2R, binary diffusivity D AB , and kinematic
viscosity ν. Among the fractions in the middle, the first one is the Sherwood number (based on k –
and D), the second and the third are the reciprocal of the Schmidt and Reynolds numbers. The
fourth becomes the D/L ratio in the denominator. The factor of 4 comes from the substitution of
R = D/2.
Because of the mass confusion that we may create with three probable choices of length scales
in R, D, and L, we use the subscript D in the Sherwood and Reynolds numbers to denote that they
are defined on the basis of using the tube diameter D.
Back to Eq. (3). One reason why we choose to use this form in dimensionless groups is that
most mass transfer data are published in correlations using dimensionless groups. It certainly is
arguable if Eq. (3) is easier for us to grasp the physical picture, at least at the learning stage.
1

P.C. Chau (UCSD, 1999)
_____________
Appendix
Derivation 1. Turbulent flow in a tube
In a fully developed turbulent flow, the velocity profile is "flat," and at steady state, we have
U – = U – (z), and C A = C A (z). A steady state differential mass balance would lead to the step
0= (UC A ) z –(UC A ) z+dz πR 2 +k(C s –C A )2πRdz (A1)
where the first two terms are the convective transport in and out of the differential volume, and the
last term is the mass transport from the wall with some local mass transfer coefficient k. Recall
that the laminar boundary layer "grows" with distance and hence k = k(z).
We next do a Taylor expansion on the z+dz term and if we further assume that U – = constant,
the mass balance becomes
U dC A
dz = 2k
R (C s –C A )
(A2)
We now integrate this equation between the two boundaries C A (0) = C Ao , and C A (L) = C AL :
C L dC A
= 2
(C s –C A ) UR
C o
0
L
k dz
(A3)
The result is
ln C s –C AL
C s –C Ao
=– 2kL
UR
(A4)
where we have introduced the definition
k = 1 L
0
L
k dz
(A5)
as the average (or overall) mass transfer coefficient.
With this result, we could have written Eq. (A2) with the overall mass transfer coefficient as
in Eq. (1), and the integration would still lead to (A4) since k – is a constant. Because we could get
away with this sloppiness in the final result, it is one reason why we see people write Eq. (1) even
in research papers. In other words, Eq. (1) is really wrong; (A2) is the proper form. This comment
applies to derivations below and we will not repeat it.
Derivation 2. Laminar flow in a tube
The derivation for the case of laminar flow is not as straightforward. Now, the axial velocity
and concentration of species A are functions of radial position, u z = u z (z) and C A = C A (r, z). At
steady state, an exercise of differential mass balance leads to
0= u z (r)C A (r,z) z –u z (r)C A (r,z) z+dz πR 2 +n A,w (z) 2πRdz (A6)
In contrast to Eq. (A1), we have written the radial and axial dependence explicitly. We also have
used the notation n A,w (z) to denote the local wall flux; the introduction of the mass transfer model
will be delayed. As analogous to (A2), the next step after (A6) is
d
dz u z(r)C A (r,z) = 2 R n A,w(z)
(A7)
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P.C. Chau (UCSD, 1999)
However, we cannot integrate (A7) as we did with (A2). We need to do an area average by
integrating (A7) across the tube cross-section
d
dz
0
R
u z (r)C A (r,z) r dr
R
= 2 n
R A,w (z) r dr
0
=Rn A,w (z)
(A8)
We cannot go further with the LHS since C A (r,z) remains an unknown function. We work around
that by defining the cup-mixing average
C A =
0
R
u z (r)C A (r,z) r dr
0
R
u z (r) r dr
(A9)
With the average velocity evaluated as
U = 2π
πR 2
0
R
u z (r) r dr
(A10)
we can rewrite (A8), or rather (A7), in terms of the cup-mixing average as
or
d
dz U C R 2
A
2
U dC A
dz = 2 R n A,w(z)
=Rn A,w (z)
(A11)
We now finally can introduce the definition n A,w = k(C – s – C – A) in terms of the local mass transfer
coefficient k = k(z). Eq. (A11) becomes
dC A
dz = 2k
UR (C s –C A )
(A12)
and its integration, following (A3) and (a5), should once again arrive at the form
ln C s –C AL
C s –C Ao
=– 2kL
UR
(A13)
These last two steps are equivalent in form to Eqs. (1) and (2). Texts or research papers
generally do not write the equations with the overhead bar on the concentrations, but we should
know better that they represent the cup-mixing average.
Furthermore, reflect on the derivation steps, and we should observe that we can only arrive at
(A12) if we use the steady state balance and if we delay introducing the mass transfer coefficient
model. You'll find that the concept of cup-mixing average is also used in heat transfer.
3

P.C. Chau (UCSD, 1999)
❏ Height of a theoretical unit and the log-mean concentration difference
There are two more definitions that we can derive based on the result in Eq. (2). The first is to
write the argument of the exponential function as
2kL
UR = L
(4)
H TU
where H TU is the height of a theoretical transfer unit defined as
H TU = UR
2k
(5)
The actual definition depends on different problems. The general idea is to interpret the argument in
Eq. (4) as a ratio of two length scales: the tube length L and the "transfer length" H TU . Loosely
speaking, the mass transport process is more effective if the H TU is small.
The second definition comes from the need to calculate the overall rate of mass transfer in the
tube as in
R =U(C A,L –C A,o ) πR 2 (6)
We may ask how the overall mass transfer coefficient may be used or may be related to this
calculation. One easy rearrangement is to substitute for the average velocity U – in (6) using Eq. (2).
We also need to manipulate the simple concentration difference in (6) to become a messier looking
form involving the surface concentration, and the result is
R = πR 2 2kL
R
(C s –C AL )–(C s –C Ao )
ln C (7)
s –C AL
C s –C Ao
If we define the big ugly concentration mess as ∆C LM , the "logarithmic mean," or simply the logmean
concentration difference,
∆C LM = (C s –C AL )–(C s –C Ao )
ln C (8)
s –C AL
C s –C Ao
Eq. (7) becomes the deceptively simple looking
R =(2πRL)k∆C LM (9)
We can certainly argue if we need Eq. (9) if we can use (6). We find better use of log-mean
difference in heat transfer design calculations.
❏ Mass transfer correlations in laminar flow
One big confusion is the use of mass transfer coefficients in laminar flow problems. We can
solve a problem using only Chapter 5, or we can rearrange everything as in Section 6.4. What it
comes down to is style and personal preference. Either approach should lead us to the correct
answer. What follows are the derivations to the two important correlations.
4

P.C. Chau (UCSD, 1999)
• Equation (6.4.6)
This equation is derived from the short contact time approximation at the gas/liquid interface
in a laminar liquid film (Section 5.1). We begin with Eq. (5.1.25), the local flux, 1
N y (x) = – (C o –C a )
D AB v m
πx
(10)
We now state that Eq. (10) is to be written in terms of a local mass transfer coefficient k = k(x):
It is clear that we need to define
N y (x) = k (C a –C o ) (11)
k=
D AB v m
πx
(12)
The next step is to evaluate the average mass transfer flux over some length L. In other words, we
want to find the average mass transfer coefficient
k = 1 L
0
L
k dx
(13)
and the result is 2 k = D ABv m
π
1/2 1
L
0
L
x –1/2 dx
=2 D ABv m
πL
1/2
(14)
We are ready to define the Sherwood number based on this average mass transfer coefficient and the
liquid film thickness δ:
Sh δ = kδ
D AB
(15)
The final result involves substituting for the mass transfer coefficient with Eq. (14) and the typical
multiplying a fraction top and bottom with the same quantities until we re-mold them into
dimensionless groups:
Sh δ =2 D AB 3
πL 2 U 1/2 δ
= π 4 3 D AB
2 L
U δ2
2
D AB
= π 6 Uδ ν δ
ν D AB L
= 6 π Re δ Sc δ L
D AB
1/2
1/2
1/2
(16)
which is Eq. (6.4.6). Note that in the derivation, we have to use, for film flow, v m = 3/2 U – . The
product of the Reynolds number and the Schmidt number is the Peclet number:
1 Recall that C o is the concentration of species A “deep” in the liquid and C a is the liquid
concentration at the gas/liquid interface. The sign of the flux is dictated by the coordinate system.
As derived in Section 5.1, the y-direction points into the liquid film. So a negative flux means the
transport of the species A is out of the liquid film into the gas phase. And vice versa.
2 Middleman defines first the contact (or exposure) times t = x/v m and T = L/v m . The average
mass transfer coefficient is evaluated as a time average over the “age” T, and we can arrive at the
identical result in Eq. (14). This is actually a very common approach.
5

P.C. Chau (UCSD, 1999)
Pe = Re δ Sc = Uδ
D AB
(17)
• Equation (6.4.9)
This equation is derived from the short contact time approximation at the solid/liquid interface
in a laminar liquid film (Section 5.2). We begin by stating that the local flux in Eq. (5.2.13) is to
be put in terms of the local mass transfer coefficient:
N y (x) = D bC s
Γ(4/3)
α
9D b x
1/3 =kC s (18)
where now the coordinate y is based on the solid surface. The local mass transfer coefficient has to
be defined as
k= D b α 1/3 x
Γ(4/3) 9D –1/3 (19)
b
and the average mass transfer coefficient can be evaluated as
k = 1 L
0
L
k dx
= 3 2
D b
Γ(4/3)
α
9D b 1/3 L –1/3 (20)
We now need to get a handle on α. For film flow, the velocity profile is usually derived with
the coordinate based on the gas/liquid interface, z = δ – y,
u x (z) = 3 2 U 1– z δ
2
(21)
Thus the velocity gradient at the solid surface is
α = du x
=– du x
dy y=0 dz
z=δ
= 3U δ
(22)
We finally are ready to find the Sherwood number as defined in Eq. (15):
Sh δ =
δ 3
D AB 2
= 1.165 Uδ
ν
D b
Γ(4/3)
ν
D b
δ
L
1/3
3U 1
δ 9D b L
(23)
1/3
which is Eq. (6.4.9) with the substitution of the definitions of dimensionless groups.
• Equation (6.4.12)
This equation is derived from the long contact time approximation at the solid/liquid interface
in a laminar liquid film (Section 5.3). The basis is the result in Eq. (5.3.32). If we take the P*
terms to be large, the P* will cancel out and leave us with the factor 1.6.
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