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Boundary Layer Theory: Mass and Heat/Momentum Transfer
Mass Transfer
Lecture 10, 22.11.2017, Dr. K. Wegner
Mass Transfer – Boundary Layer Theory 9-2
9.1 Fluid-Fluid Interfaces (lecture of 15.11.17)
9. Basic Theories for Mass Transfer Coefficients
9.2 Fluid-Solid Interfaces
Fluid-fluid interfaces are typically not fixed and are strongly affected by the flow leading to heterogeneous systems that make it difficult to development a general theory behind the MT correlations (Cussler Table 8.3-2).
Solids typically have fixed and well-defined surfaces, allowing to develop theoretical foundations for the empirical MT correlations (Cussler Table 8.3-3). In contrast to fluid-fluid interfaces the models are far more rigorous but, unfortunately, computationally INTENSIVE.
Mass Transfer – Boundary Layer Theory 9-3
In addition to this, fluid-solid interfaces have been investigated intensely with respect to heat transfer. We can make use of this due to the analogy between heat momentum and mass transfer.
9.2.1 MT from a plate (boundary layer theory)
MT correlation (Table 8.3-3):3121
DLU646.0
DkLSh
ν
ν== ∞
Example: A sharp-edged, flat plate that is sparingly dissolvable is immersed in a rapidly flowing solvent.
L: plate length; U∞: bulk fluid velocity
Mass Transfer – Boundary Layer Theory 9-4
Prandtl first introduced the concept of boundary layers.
Goal: Calculate the MTC for this fluid–solid interface
Literature:R.B. Bird, W.E. Steward, E.N. Lightfoot, “Transport Phenomena”, 2nd ed. J. Wiley&Sons 2011.H. Schlichting, K. Gersten, "Boundary Layer Theory", 8th ed., Springer 1999.
x
y
The transition from zero velocity at the plate to the velocity of the surrounding free stream takes place in the boundary layer.
Mass Transfer – Boundary Layer Theory 9-5
1. Calculate the velocity profile in the B.L.
2. Calculate the concentration profile in the B.L.
y 0cj D |y =
∂= −
∂
and set it equal to k∆c to obtain k.
3. Calculate the flux at the interface
x
y
Procedure:
Mass Transfer – Boundary Layer Theory 9-6
The relative magnitude of the fluid flow (momentum) B.L. and the concentration B.L. is given by
kinematic viscosityScD diffusivityν
= =
Sc > 1: Momentum B.L. > Concentration B.L. (typical)
Sc = 1: Momentum B.L. = Concentration B.L.
Sc < 1: Momentum B.L. < Concentration B.L. (rare case)
Mass Transfer – Boundary Layer Theory 9-7
Laminar B.L.Continuity equation:
(1)
Momentum equation in x (for steady flows):
0zw
yv
xu
=∂∂
+∂∂
+∂∂
= 0 (2 dim.)
∂∂
+∂∂
⋅ν+∂∂⋅
ρ−=
∂∂⋅+
∂∂⋅ 2
2
2
2
yu
xu
xp1
yuv
xuu
→0, const. shear stress on plate
(2)
=0 for fully developed, inviscid flow with U∞=const.
Mass Transfer – Boundary Layer Theory 9-8
(Negligible pressure changes in y (thin layer)
y = 0: u = 0 and v = 0 y = ∞: u = U∞
Momentum equation in y (for steady flows):
∂∂
+∂∂
⋅ν+∂∂⋅
ρ−=
∂∂⋅+
∂∂⋅ 2
2
2
2
yv
xv
yp1
yvv
xvu
As v typically is small in the BL (for lam. case here =0), this reduces to:
0yp=
∂∂
(3)
Boundary conditions:(fluid adheres to plate)
Mass Transfer – Boundary Layer Theory 9-9
Digression: How did Prandtl come up with that?
The B.L. thickness, δ, is related to Re by
Equations (1) and (2) are solved by introducing the variable
∞
⋅ν=η
Ux
y
xURe~
x21
⋅ν
=δ
∞
−
η=⋅ν
=δ
∞Ux
yRexy~y 21
Mass Transfer – Boundary Layer Theory 9-10
The corresponding stream function is: )(fxU η⋅ν=Ψ ∞
Eq. (2) can be simplified by formulating it in terms of a stream function Ψ. In order to satisfy the continuity equation (1), the velocity components must be:
yu
∂Ψ∂
=x
v∂Ψ∂
−=
2
2
yu
yuv
xuu
∂∂⋅ν=
∂∂⋅+
∂∂⋅ 3
3
2
22
yyxyxy ∂Ψ∂
⋅ν=∂Ψ∂
⋅∂Ψ∂
−∂∂Ψ∂
⋅∂Ψ∂
Thus,
where f(η) is the dimensionless stream function.
yyu
∂η∂
⋅η∂Ψ∂
=∂Ψ∂
= fUUx
1fxU ′⋅=⋅ν
⋅′⋅ν= ∞∞
∞
Mass Transfer – Boundary Layer Theory 9-11
1 2 1 21 12 2
U Uν f fx x∞ ∞ν ν ′= − + η⋅
fx
U21f
Ux
yU21
xu
3′′⋅η−=′′⋅
ν−=
∂∂ ∞
∞
∞
fx
UUyu 21
′′
ν
=∂∂ ∞
∞ fx
Uyu 2
2
2′′′⋅
ν
=∂∂ ∞
1 2
1 23
1 12 2
U yν f U x fx x x
U
∞∞
∞
ν∂Ψ ′= − = − + ν ⋅ ⋅ ∂ ν ⋅
and
( )ffxU
21 21
−′⋅η
ν= ∞
Mass Transfer – Boundary Layer Theory 9-12
Now the equation of motion (2) becomes:
0fff2 =⋅′′+′′′
at η = 0: f = 0 and f’ = 0η = ∞: f’ = 1 (see Table)
Blasius obtained the solution in the form of a series expansion at η=0 and η → ∞ (“Blasius series”) and the two forms are matched at a suitable η. See attached Table from H. Schlichting (“Boundary Layer Theory”), which gives the complete values of u and v for every x and y. The velocity profiles have been calculated.
Mass Transfer – Boundary Layer Theory 9-14
Blasius similarity solution of the velocity distribution in a laminar boundary layer on a flat plate.P.K. Kundu, I.M. Cohen, “Fluid Mechanics” 2nd ed. 2002, Academic Press
∞Uu
xUyν
=η ∞
Mass Transfer – Boundary Layer Theory 9-15
Mass Transfer in Boundary Layers
The diffusion equation in the B.L. is:
∂∂
+∂∂
=∂∂
+∂∂
21
2
21
211
yc
xcD
ycv
xcu
B.C.’s: y = 0: c1 = c10y = ∞: c1 = c1∞
Introduce dimensionless variables:∞∞∞
==−−
=Uvv,
Uuu,
ccccc **
101
101*1
∂∂
+∂∂
=∂∂
+∂∂
⇒∗∗
∞
∗∗
∗∗
21
2
negligibleusually
21
211
yc
xc
UD
ycv
xcu
B.C.s:y = 0: c1*=0y= ∞: c1*=1
(4)
(5)
Mass Transfer – Boundary Layer Theory 9-16
Compare to the momentum equation (2):
∂
∂+
∂
∂ν=
∂∂
+∂∂ ∗∗
∞
∗∗
∗∗
2
2
negligibleusually
2
2
yu
xu
Uyu
vxu
u B.C.’s: y = 0: u*=0 and v*=0y= ∞: u*=1
The two equations have the same form!
Sc = 1: In this special case where ∞∞
=ν
UD
U
the solution to the concentration profile is given by c1*=u*
or ν = D
The boundary layers for flow and concentration are the same.
Mass Transfer – Boundary Layer Theory 9-17
For this case we write Fick’s first law 0y1 |
ycDj =∂∂⋅−=
( )1010y
1 ccycDj −∂∂⋅−= ∞
=
∗
( ) ( )1010y
1010y
ccyu
Ucc
yu
j −∂∂ν
−=−∂∂⋅ν−= ∞
=∞∞
=
∗
( )1010
2
221
ccfx
UUU
j −
η∂∂
ν
ν−= ∞
=η
∞∞
∞
( )101Tablefromf
21
cc332.0xUj −⋅⋅
ν−= ∞
′′
∞
with c1*:
Mass Transfer – Boundary Layer Theory 9-18
As j = k (c10 - c1∞)21
xU332.0k
ν⋅= ∞
21
xU
Dx332.0
DxkSh
⋅ν⋅⋅
νν⋅=
⋅= ∞
2121
ReSc332.0xUD
332.0 ⋅⋅=
ν⋅
ν⋅= ∞
Here, Sc = 1: 21Re332.0Sh ⋅=
This is the Local Mass Transfer Coefficient to be distinguished from the average MTC over a plate of length L (Cussler Table 8.3-3):
3121 ScRe646.0Sh ⋅⋅=
,
Mass Transfer – Boundary Layer Theory 9-19
A more realistic case is with Sc >> 1 which is typical in liquids and particle / gas systems.
Sc >> 1: (S.K. Friedlander, “Smoke, Dust and Haze”, 1st ed., 1977)
Examples: Dissolution of solid compounds in flowing liquid streams.Flow (deposition) of pollutants over lakes or leaves
Mass Transfer – Boundary Layer Theory 9-20
More specifically, if the B.L. (displacement) thickness is:1/2
x* 1.72U∞
= ⋅
νδ
For a wind blowing at a speed of 10 km/h, δ ≈ 0.5 mm. E.g., at 2 cm from the leading edge of the flat surface:
Goal: To obtain the mass transfer coefficient
mm56.0m1033.072.13600sm10000
mkg1s Pa1015m02.0
72.1 33
6
* =⋅⋅=
⋅⋅⋅=δ −
−
(Blasius, Schlichting)
Mass Transfer – Boundary Layer Theory 9-21
Again at steady-state 2
*1
2*1*
*1*
yc
UD
ycv
xcu
∂∂
=∂∂
+∂∂
∞
Expand and
near the wall (η→0) into Taylor series:
( )η′==∞
fUuu* ( ))(f)(f
xU21
Uvv
21* η−η′⋅η
ν==
∞∞
η=η+≈+η⋅′′′+η⋅′′+′=
∞
= a332.00...2
)0(f)0(f)0(fUuu
2*
disregard higher order terms
221
221
3221
*
xU4a
2332.00
xU21
...)0(f31)0(f
21)0(f
xU21
Uvv
η
ν=
η+
ν≈
+η′′′+η′′+
ν==
∞∞
∞∞
(5)
Mass Transfer – Boundary Layer Theory 9-22
Substitution in the above differential equation (5) gives:
Assume that c1* is a fct only of η: )(gc*1 η=
Then:
gx2
1xc*
1 ′⋅η
−=∂∂
2
*1
2*12
21*1
yc
UD
yc
Ux4a
xca
∂∂
=∂∂
η
⋅ν
+∂∂
η∞∞
gx
Uy
gyc 21*
1 ′⋅
ν
=∂η∂
η∂∂
=∂∂ ∞
gx
Uyc2
*1
2
′′⋅
ν
=∂∂ ∞
ν∞ ⋅
1/2Uη = yx
with
Mass Transfer – Boundary Layer Theory 9-23
Introducing the above variables in the diffusion equation:
or∞=η
=η 01g0g
==
Solution: Set
00
gP=η
η∂∂
=where
gx
UUDg
xU
Ux4ag
x2a
212
212
′′ν
=′
ν
η
⋅ν
+
′η
− ∞
∞
∞
∞
0gD4
ag 2 =′⋅η⋅
ν+′′ with
0PSc4aP gP 2 =⋅⋅η+
η∂∂
⇒′=
Integrate: Sc12a
PPln 3
0
⋅η−=
Mass Transfer – Boundary Layer Theory 9-24
From the B.C’s.: at η = 0, g = 0:
From the B.C’s.: at η = ∞, g = 1, hence:
η−⋅==
η∂∂ Sc
12aexpPPg 3
0
dr rSc12aexpPg
0
30 ∫η
⋅−=
1
0m
30 dr rSc
12aexpP
−
∞
⋅−= ∫
Mass Transfer – Boundary Layer Theory 9-25
and
∞
⋅ ∂ ∂
∫
1 3
3η=00
1 3Γ(1 3)
a Scg 12=η exp -m dm
⋅
1 3a Sc12=
0.89
Remember (Gamma function):
[ ]∫∞
−−=−Γ−=Γ
0
1n dxxxexp)1n()1n()n(
Mass Transfer – Boundary Layer Theory 9-26
( )0 ∞
∂⋅ ⋅ ∂
*1
1 1y=0
c= D c - cy
( )ν∞
∞ ∂ ⋅ ⋅ ∂
1 2
10 1η=0
U g= D c - cx η
( )νν
∞∞
∞
1 2 1 3
10 1UD a= Sc c - c
U x 0.89 12
( )∞∞
⋅ ⋅
1 3-2 3 -1 2
10 1
k
U 0.332= Sc Re c - c0.89 12
0y
1
ycDj
=∂∂⋅−=
Mass Transfer – Boundary Layer Theory 9-27
2/13/2 ReScU34.0k −−∞⋅=So
This is the local mass transfer coefficient while the overall is found by:
L1
Avy=00
-2/3 1/210 1
c1j = - D dxL y
= 0.68 U Sc Re (c - c )−∞ ∞
∂ ∂
⋅ ⋅
∫
Mass Transfer – Boundary Layer Theory 9-28
Let’s calculate Sh to compare it with our Table of correlations
2/3 1/2
2/3 1/2
2/3 1/2 1/2
1/3 1/2
1/3 1/2
kL L0.68 Sc Re UD D
L D0.68 uD u L
uL0.68D
0.68 Sc Re
− −∞=
ν=
ν
ν = ν
= ⋅
So we obtained the mass transfer coefficient from theory!
Mass Transfer – Boundary Layer Theory 9-29
In general 2/13/2 ReScconstUk
cUj −−
∞∞⋅==
∆
Applicable for flow around spheres, cylinders etc.
Mass transfer correlation over a wide range of Re:
3/12/1 ScRe6.02Sh +=
Mass Transfer – Boundary Layer Theory 9-30
9.3 Theories for concentrated solutions
The MTC’s are based typically on dilute solutions but they are VERY successful even in concentrated ones as, typically, the volume average velocity normal to the fluid-fluid interface is rather small.
When the MTC’s fail, it is observed typically that “the k depends strongly on concentration” especially when mass transfer is fast.
The flux is written as ( ) 0i11i11 vccckN +−=
( ) ( )2211i11i1 NVNVccck ++−=
Mass Transfer – Boundary Layer Theory 9-31
However, this is difficult and we also have the previous unknowns again (film thickness, contact or residence times).Instead we are looking for corrections to the dilute mass transfer coefficient by the film theory:
0k MTC in concentrated solution=
MTC in dilute solutionk
This framework can be used to build new film, penetration and surface-renewal theories.
Mass Transfer – Boundary Layer Theory 9-32
The total flux for a concentrated solution across a film:
01
11 vc
dzdcDn +−= with B.C.’s: z = 0: c1 = c1i
z = L c1 = c1
For concentrated solutions, n1 and v0 are constant, so the above equation can be integrated from z=0 to L to give
[ ]D/Lvexpv/ncv/nc 0
01i1
011 ⋅=
−−
Rearranging:
Compare: ( ) 0i11i11 vccckN +−=
[ ] ( ) 0i11i10
0
10z1 vccc1D/Lvexp
vNn +−−⋅
===
Mass Transfer – Boundary Layer Theory 9-33
Then the mass transfer coefficient is
[ ] 1D/Lvexpvk 0
0
−⋅=
For dilute solutions:LDk0 =
Taking the ratio k/k0, we eliminate the dependency in L:
[ ] 1k/vexpk/v
kk
00
00
0 −=
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