1
CH351 Mass Transfer
Sitaraman Krishnan
9/26/07
2
Fick’s First Law: Steady State
Molecular Diffusion
J A,z
J A,z
∂c A
= −DAB
∂z
= flux of component A along z - axis, mol/(cm 2 ⋅ s)
DAB = diffusion coefficient, cm 2 /s
∂c A
= concentrat ion gradient along z - axis
∂z
3
From Kinetic Theory or Low Density Gases :
DAA *
DAA *
1
2
⎛κ T ⎞
⎜
⎟
= 3
⎜ m ⎟
2
2
⎠
3π d P ⎝
= self diffusion coefficient
2
3
3
d = diameter of (spherical) molecules
m = mass of a molecule
Fuller, Schetler and Giddings equation :
DAB =
0.00143T 1.75
P (M AB )
1
2⎡
⎢⎣
(Σv )A
1
3
+ (Σv )B
1
3⎤
⎥⎦
2
4
Wilke - Chang Equation for
Diffusion in Liquids :
DAB =
7.4 × 10
−8
(ΦBMB )
ηBVA
12
T
0.6
Species A is diffusing in solvent B
5
Partial Pressures, Molar
Concetrations, Total Pressure
nA
pAV = n A RT ⇒ pA =
RT
V
pA = c A RT
pB = cB RT
(pA + pB ) = (c A + cB )RT
P = cRT
• pi = partial pressure of
component i
• ci = molar conc. of
component i
• P = total pressure
(atm, bar, N/m2, mmHg,
Torr)
• c = total molar
concentration of gas
(mol/cm3, mol/m3, …)
6
Concentrations, Partial
Pressures, and Mole Fractions
pA = c A RT
pB = cB RT
P = cRT
c A pA
Mole fraction of A = y A =
=
c
P
cB pB
Mole fraction of B = y B =
=
c
P
• At constant total pressure, P, and constant
temperature, T, total molar concentration, c,
is constant.
• For each species, mole fraction ∝
concentration ∝ partial pressure
7
Fick’s First Law in terms of Mole
Fractions
J A,z
∂c A
= −DAB
∂z
∂y A
= −cDAB
∂z
DAB P ∂y A
=−
RT ∂z
JB,z
∂cB
= −DBA
∂z
∂y B
= −cDBA
∂z
DBA P ∂y B
=−
RT ∂z
fluid
element
8
direction of bulk flow
direction of
diffusional
flow
cA
position
N A,z = c Av A,z
What is the net
molar flux of A
that a stationary
observer would
measure?
v A,z = velocity of A relative to
stationary observer
9
What is the net molar flux of A that
an observer who moves with the
fluid element would measure?
Moving observer will measure only
flux due to diffusion.
J A,z = c A (v A,z − v z * )
= −DAB
∂c A
∂z
10
Total Flux, NA,z = JA,z + cAvz*
J A,z = c A (v A,z − v z * )
v z * = y Av A,z + y Bv B,z
N A,z = c Av A,z
mol cm
mol
×
=
3
2
s
cm
cm ⋅ s
• vA,,z = velocity of A
measured by a
stationary observer
• vB,z = velocity of B
measured by a
stationary observer
• vz* = molar
average velocity
11
General Form of Fick’s First Law
N A,z = J A,z + c Av z *
= diffusive flux + flux due to bulk motion
N A,z = J A,z + y A (N A + NB )
N A,z = −cDAB
∂y A
+ y A (N A,z + NB,z )
∂z
12
Equimolar Counter Diffusion
(EMCD)
• A simplifying
assumption that
N A,z = −NB,z
0
∴ N A,z = J A,z + y A (N A,z + NB,z )
N A,z = J A,z = −DAB
∂c A
∂z
– allows easier
solution of the
general form of
Fick’s first law
– makes analytical
solution possible
13
Problem 1 (EMCD assumption)
Mixture of two gases, A and B, in a tube with conc. gradient
yA,1
pA,1
P = constant, T = constant
yA,2
pA,2
z-axis
z = z1 = 0
DAB P ∂y A
RT ∂z
DAB P ⎛ y A,1 − y A,2 ⎞
⎜⎜
⎟⎟
=
RT ⎝ z2 − z1 ⎠
N A,z = J A,z = −
N A,z
FLUX
z = z2
⎛ y A,1 − y A,2 ⎞
⎟⎟(z − z1 )
y A = y A,1 − ⎜⎜
⎝ z2 − z1 ⎠
CONCENTRATION PROFILE
14
Problem 2 (EMCD assumption)
Evaporation from surface of naphthalene ball
cA
b
r →∞
r
r+Δr
= cA ≈ 0
At steady state, N A
∂
∂r
R
⎛ 2 ∂c A
⎜r
∂r
⎝
NA
Air near surface is saturated
with naphthalene
∴ pA
r =R
= pA
vp
cA
r =R
vp
pA
s
=
= cA
RT
r =R
⎞
⎟=0
⎠
= DAB
(c
r
⇒
s
A
∂c A
= −DAB
∂r
− cA
R
cA − cA
s
b
cA − cA
b
)
b
=
r
R
r
⎛ mol ⎞
⎟
⎜
2
⎝ cm s ⎠
(c
s
2
A − cA
Evaporatio n rate = 4πR × DAB
R
⎛ mol ⎞
s
b
= 4π R DAB c A − c A
⎜
⎟
⎝ s ⎠
(
)
b
)
15
Unimolecular Diffusion
N A,z = J A,z + y A (N A,z + NB,z )
NB,z = 0
∴ N A,z = J A,z + y A N A,z
N A,z =
J A,z
1− y A
⎛ DAB P ⎞⎛ 1
= −⎜
⎟⎜⎜
⎝ RT ⎠⎝ 1 − y A
⎞ ∂y A
⎟⎟
⎠ ∂z
16
Solvent Evaporation (UMD)
pA2, yA2
z = z2
z
z = z1 = 0
pA1, yA1
Liquid, A
• At z = z1, air is saturated with liquid: pA1 = vap. pressure
• At z = z2, all A carried away by air: pA2 ~ 0
17
Solvent Evaporation
N A,z
⎛ DAB P ⎞⎛ 1 ⎞ ∂y A
= −⎜
⎟⎜⎜ ⎟⎟
⎝ RT ⎠⎝ y B ⎠ ∂z
⎛ DAB P ⎞⎛⎜ 1 ⎞⎟⎛ y A,1 − y A,2 ⎞
⎜⎜
⎟⎟
NA = ⎜
⎟⎜
⎟
⎝ RT ⎠⎝ y B,lm ⎠⎝ z2 − z1 ⎠
y B,2 − y B,1
y B,lm =
⎛ y B,2 ⎞
⎟
ln⎜⎜
⎟
y
B,1
⎝
⎠
FLUX
⎛ z − z1 ⎞
⎜⎜
⎟⎟
−
z
z
y A,2 ⎞ ⎝ 2 1 ⎠
⎛ 1− y A ⎞ ⎛ 1−
⎜
⎟=⎜
⎟
⎜ 1− y ⎟ ⎜ 1− y ⎟
A,1 ⎠
A,1 ⎠
⎝
⎝
CONC. PROFILE
18
Evaporation of Benzene
(bp 80 °C) at 25 °C and 70 °C
25 °C
70 °C
19
Flux Profiles (UMD)
20
Evaporation of Dichloromethane
(bp 40 °C) at 25 °C
21
Diffusion Into a Falling Liquid Film
2
⎡
ρgδ
⎛z⎞ ⎤
v y (z ) =
⎢1 − ⎜ ⎟ ⎥
2μ ⎢⎣ ⎝ δ ⎠ ⎥⎦
2
W
x
z
Liquid
vy(z)
cA(z)
y
Gas
cA,i
OBJECTIVE:
• To determine molar flux
(mol cm−1 s−1) at the
gas−liquid interface.
• The overall rate of
mass transfer (mol/s).
N A (y ) z =0 = local molar flux at interface ( z = 0)
at position y from the top entrance
22
N A,y ⋅ (WΔz ) − N A,y
y
y
+ N A,z z ⋅ (WΔy ) − N A,z
Liquid
Element
y+Δy
z+Δz
N A,y
z
N A,y
y + Δy
− N A,y
Δy
y
∂N A,y
∂y
N A,z
z + Δz
Δz
Δy
N A,z
W
N A,y
y
y + Δy
+
+
N A,y = −DAB
z
N A,z = −DAB
N A,z
y + Δy
z + Δz
⋅ (WΔz )
⋅ (WΔy ) = 0
z + Δz
− N A,z
Δz
∂N A,z
∂z
=0
∂c A
+ c Av y *
∂y
∂c A
+ c Av z *
∂z
∂c A
∂ 2c A
vy
= DAB
∂y
∂z 2
z
=0
23
∂ 2c A
∂c A
vy
= DAB
∂y
∂z 2
W
x
z
Liquid
vy(z)
cA(z)
Approximat e solution :
y
Gas
cA,i
⎛
⎜
⎜
z
c A (y,z ) = c A,i erfc ⎜
y
⎜ 4D
AB
⎜
v y,max
⎝
N A,z
z =0
∂c A
= −DAB
∂z
= c A,i
z =0
⎞
⎟
⎟
⎟
⎟
⎟
⎠
DABv y,max
πy
24
N A,z
z =0
∂c A
= −DAB
∂z
= c A,i
z =0
DABv y,max
πy
Moles of A transferred per second at position y
= (flux ) × (area )
(
= N A,z
z =0
)× (W ⋅ d y )
⎛
D
v
AB
y,max
⎜
= c A,i
⎜
πy
⎝
⎞
⎟ × (W ⋅ d y )
⎟
⎠
25
∴ total moles of A transferred per second
over length L of the tower
=
∫
L⎛
0
⎜c
⎜ A,i
⎝
= Wc A,i
DABv y,max ⎞⎟
× (W ⋅ d y )
⎟
πy
⎠
DABv y,max
= 2Wc A,i
π
∫
L
0
1
y
dy
DABv y,max L
= WL × 2c A,i
π
DABv y,max
πL
⎛ D ⎞⎛ 1 ⎞
= WL × 2c A,i ⎜ AB ⎟⎜ ⎟
⎝ π ⎠⎝ τ ⎠
where τ = time of contact between liquid element and gas
26
Mass Transfer Coefficient
• Fick’s law: flux ∝ conc. gradient
– const. of proportionality = diffusion coefficient
• Another approach to quantify flux:
– Flux ∝ conc. driving force, ΔcA
– const. of proportionality = mass transfer
coefficient, k
– Similar to Newton’s “law” of cooling
27
Fick' s law : N A = DAB
Δc A
Δz
Engineerin g approach : N A = k ⋅ Δc A
k = mass transfer coefficient
cm
k has units of
s
⎛ mol ⎞
⎛ cm ⎞
⎛ mol ⎞
NA ⎜
⎟ = k⎜
⎟ ⋅ Δc A ⎜
⎟
2
3
⎝ cm ⋅ s ⎠
⎝ s ⎠
⎝ cm ⎠
28
Evaporation from surface of
naphthalene ball
Conc. driving force for mass transfer
(
s
= Δc = c A − c A
R
b
)
Mass transfer coefficient, k =
NA
Δc
Solution of Fick' s law had given us :
vp
cA
cA
r =R
r →∞
pA
s
=
= cA
RT
= cA
b
NA
r =R
= DAB
(c
(c
s
A
− cA
R
s
A − cA
DAB
NA
R
∴k =
=
s
b
Δc A
cA − cA
(
)
b
b
)
)
=
DAB
R
29
Sherwood Number
• For evaporation from surface of a sphere, we obtained:
DAB
DAB
k=
=2
R
D
(D = sphere diameter)
k ⋅D
∴
=2
DAB
• kD/DAB is a dimensionless quantity that often appears in
mass transfer calculations.
• It is called the Sherwood number and is denoted by Sh.
• For diffusion from surface of a sphere into a stagnant
fluid, Sh = 2
30
Convective Mass Transfer
• Material is transported between
– a solid surface and a moving fluid (gas or
liquid)
– two relatively immiscible moving fluids (gas
and liquid, or liquid and liquid)
• Examples
– Mass-transfer of gas into a falling liquid film
– Mass-transfer of naphthalene into a flowing
gas stream
– Mixing in a stirred vessel
31
Turbulent Flow
• Laminar flow is characterized by streamlines
• Turbulent flow is characterized by chaotic flow of
packets of fluid called “eddies”
• Rate of mass transfer in laminar flow is
determined by molecular diffusivity, DAB
• Rate of mass transfer in turbulent flow is
determined by both molecular diffusivity, DAB
and turbulent diffusivity, Dt
32
Mass Transfer Correlations for
Convective Flows
• Dimensional Analysis
• Example
– The inner wall of a circular tube is coated with
species A (e.g., naphthalene)
– Mass transfer occurs to a fluid flowing through
the tube (e.g., air)
v (cm s−1)
cAs
cA
b
ρ (g cm−3)
μ cm−1 s−1)
kc (cm s−1)
DAB (cm2 s−1)
D (cm)
(
s
NA = kc ⋅ c A − c A
33
b
)
k c = convective mass transfer coefficient
k c = [constant ](D ) (ρ) (μ ) (v ) (DAB )
a
kc
D
a
0
1
0
gm
0
cm
1
s
−1
b+c =0
a − 3b − c + d + 2e = 1
− c − d − e = −1
5 variables, 3 equations
ρ
b
1
−3
0
b
c
μ
c
1
−1
−1
d
e
v
d
0
1
−1
DAB
e
0
2
−1
b = −c
− 3b + d + 2e = 1 − a + c
− d − e = −1 + c
34
0
⎡ 1
⎢
⎢− 3 1
⎢⎣ 0 − 1
⎡b ⎤ ⎡ 1
⎢ ⎥ = ⎢−
⎢d ⎥ ⎢ 3
⎢⎣ e ⎥⎦ ⎢⎣ 3
0 ⎤ ⎡b ⎤ ⎡ − c ⎤
⎥⎢ ⎥ ⎢
⎥
2 ⎥ ⎢d ⎥ = ⎢1 − a + c ⎥
− 1⎥⎦ ⎢⎣ e ⎥⎦ ⎢⎣ − 1 + c ⎥⎦
0
0 ⎤⎡ − c ⎤
⎥
⎢
⎥
− 1 − 2⎥ ⎢1 − a + c ⎥
1
1 ⎥⎦ ⎢⎣ − 1 + c ⎥⎦
⎡b ⎤ ⎡ − c ⎤
⎢ ⎥ ⎢
⎥
=
+
1
d
a
⎢ ⎥ ⎢
⎥
⎢⎣ e ⎥⎦ ⎢⎣− a − c ⎥⎦
Ax = B
A −1Ax = x = A −1B
35
k c = [constant ](D ) (ρ ) (μ ) (v ) (DAB )
a
b
= [constant ](D ) (ρ )
a
⎛ Dvρ ⎞
⎟⎟
= [constant ]⎜⎜
⎝ μ ⎠
a +1
−c
c
d
e
(μ )c (v )a+1 (DAB )−a−c
−a −c −1
(ρ)
(μ )
⎛ kc D ⎞
⎛ Dvρ ⎞
⎜⎜
⎟⎟ = [constant ]⎜⎜
⎟⎟
⎝ μ ⎠
⎝ DAB ⎠
a +1
a + c +1
(DAB )
⎛ μ ⎞
⎜⎜
⎟⎟
⎝ ρDAB ⎠
−a −c
1
D
a + c +1
Sh = [constant ]Re α Sc β
(
μ ρ ) momentum diffusivit y
Sc = Schmidt number =
=
DAB
mass diffusivit y
36
k c = [constant ](D ) (ρ )
a
−c
(μ )c (v )a+1 (DAB )−a−c
a
⎛ Dvρ ⎞
a+c
−a −c
−a −c
⎟⎟ (ρ )
(μ ) (DAB ) v
= [constant ]⎜⎜
⎝ μ ⎠
⎛ Dvρ ⎞
⎛ kc ⎞
⎟⎟
⎜ ⎟ = [constant ]⎜⎜
⎝v ⎠
⎝ μ ⎠
a
⎛ μ ⎞
⎜⎜
⎟⎟
⎝ ρDAB ⎠
a + c +1
kc
= [constant ]Re γ Sc δ
v
kc
2/3
γ
δ + (2 3 )
[
]
Sc
= constant Re Sc
v
kc
Sc 2 / 3 = JD = Chilton and Colburn J - factor
v
37
Turbulent Flow Inside Pipes
kc D
Sh =
DAB
Dvρ
Re =
μ
μ
Sc =
ρDAB
When Re > 2100 and 0.6 < Sc < 3000 :
Sh = 0.023 Re
0.83
Sc
0.33
In general, Sc for gases is in the range
0.5−3, and for liquids, Sc > 100.
38
Flow Past Single Spheres
kc D
Sh =
DAB
Dvρ
Re =
μ
μ
Sc =
ρDAB
For gases, when 1 < Re < 48000, and 0.6 < Sc < 2.7
Sh = 2 + 0.552Re 0.53Sc 1/ 3
For liquids, when 2 < Re < 2000 : Sh = 2 + 0.95 Re0.5 Sc 1/ 3
For liquids, 2000 < Re < 17000 : Sh = 0.347 Re0.62 Sc 1/ 3
39
Naphthalene Evaporation
• A sphere of naphthalene having a diameter of 1 cm is
suspended in air at 1 atm, and flowing at a velocity of 0.3
m/s. The diffusion coefficient of naphthalene in air at this
temperature is 6.92×10−6 m2/s. The molar density of
naphthalene solid is 8.86×10−3 mol/cm3. Calculate the
initial rate of evaporation of naphthalene from the
surface. Estimate the time required to reduce the
diameter from 1 to 0.5 cm.
40
Mass Transfer in Packed Beds
For gases in a packed bed of spheres, when 10 < Re < 10000,
JD =
kc
JD =
Sc 2 3
v'
0.4548
Re −0.4069
ε
Dpv ' ρ
μ
Re =
Sc =
ρDAB
μ
volume of void space
total volume of void space plus solid
The value of ε is usually between 0.3 and 0.5.
ε = void fraction in the bed =
Dp = diameter of the spheres
v ' = superficial velocity of gas
= average velocity in the empty tube without packing
=
volumetric flow rate
cross - sectional area of tube
41
Mass Transfer in Packed Beds
For liquids, when 0.0016 < Re < 55, and 165 < Sc < 70000
1.09
Re −2 3 (Wilson and Geankoplis )
JD =
ε
For liquids, when 55 < Re < 1500, and 165 < Sc < 10690
0.250
Re −0.31
JD =
ε
For non - spherical particles,
Dp = dia. of a sphere with the same surface area as the given solid particle
To calculate flux, use the surface area of the non - spherical particles.
42
Packed Bed of Benzoic Acid
Spheres
• Water at 26.1 °C flows at a rate of 5.514×10−7
m3/s through a packed bed of benzoic acid
spheres having a diameter of 6.375 mm. The
void fraction of the bed is 0.436. The tower
diameter is 0.0667 m and the tower height is 0.1
m. The solubility of benzoic acid in water is
2.948 ×10−2 kg mol/m3. Predict the mass transfer
coefficient kc and the outlet concentration of
benzoic acid in water.
43
cA,1 = 0
Water flow rate, Q = 5.514×10−7 m3/s
μ = 0.8718×10−3 Pa·s, ρ = 996.7 kg/m3
ε = 0.436
Dp = 6.375×10−3 m, Ap = 0.01198 m2
Tower dia. T = 0.0667 m; Height H = 1 m
DAB = 1.25×10−9 m2/s (Wilke-Chang)
cA,2 = ?
0.8718 × 10 −3
μ
Sc =
= 702.6
=
−9
ρDAB 996.7 1.245 × 10
(
)
Q
5.514 × 10 −7
−4 m
= 1.578 × 10
v' =
=
π
π 2
s
(0.0667 )2
T
4
4
Dpv ' ρ 6.375 × 10 −3 1.578 × 10 − 4 (996.7 )
Re =
=
= 1.150
−3
μ
0.8718 × 10
(
)(
)
44
Sc = 702.6 Re = 1.150 v ' = 1.578 × 10 −4 m/s
1.09
1.09
−2 / 3
(1.150 )−2 / 3 = 2.277
=
JD =
Re
0.0436
ε
(
)
k c = JD ⋅ v '⋅Sc −2 / 3 = (2.277 ) 1.578 × 10 − 4 (702.6 )
−2 / 3
= 4.447 × 10 −6 m/s
π 2
π
2
Total volume of bed = Vb = T H = (0.0667 ) (0.1) = 3.494 × 10 −3 m3
4
4
Void fraction = ε
∴ volume of spheres in the bed = Vb ⋅ (1 − ε )
Let number of spheres in the bed = Np
⎛π 3⎞
∴ Np ⋅ ⎜ Dp ⎟ = Vb ⋅ (1 − ε )
⎝6
⎠
(
Total surface area of spheres = A = Np ⋅ πDp
(
)
2
)
6Vb ⋅ (1 − ε ) 6 3.494 × 10 −3 (1 − 0.436 )
2
=
=
=
0
.
1855
m
Dp
6.375 × 10 −3
45
Mole balance :
(
s
kc ⋅ A ⋅ c A − c A
b
)
lm
= Q ⋅ (c A,2 − c A,1 )
m3 kg mol
⋅
s
m3
m 2 kg mol
⋅m ⋅
s
m3
(c
s
A
− cA
b
(
c
) =
lm
s
A
− cA
⎧⎪
ln⎨
⎪⎩
b
) − (c
1
s
cA
s
cA
(
(
s
A
− cA
− cA
b
− cA
) ⎫⎪⎬
) ⎪⎭
b
)
2
1
b
2
s
c A = 2.948 × 10 −2 kg mol/m 3
(c ) = conc. of benzoic acid at inlet = c = 0 kg mol/m
(c ) = conc. of benzoic acid at outlet = c
b
A
A,1
1
b
A
2
A,2
3
46
kc ⋅ A ⋅
(c A,2 − c A,1 )
⎛ c A − c A,1 ⎞
⎟
ln⎜ s
⎜c −c ⎟
A,2 ⎠
⎝ A
s
(
= Q ⋅ (c A,2 − c A,1 )
)
⎛ kc A ⎞
∴ c A,2 = c A − c A − c A,1 exp⎜ −
⎟
⎝ Q ⎠
−6
⎛
×
× 0.1855 ⎞
4
.
447
10
−2
−2
⎟
= 2.948 × 10 − 2.948 × 10 − 0 * exp⎜⎜ −
−7
⎟
5.514 × 10
⎝
⎠
s
(
s
)
= 2.287 × 10 −2 kg mol/m 3
47
Mass Transfer to Bubbles and
Particles
• Small particles in suspension
– Mass transfer from small gas bubbles to liquid
phase
– Mass transfer from liquid phase to the surface
of catalyst particles, microorganisms, liquid
drops, etc.
– Mass transfer coefficient depends on free fall
or rise of particles due to gravitational forces
48
•
Mass transfer coefficient, kc, in mass transfer to small particles is
affected by natural convection. Natural convection occurs when
there is significant density difference between the particles of the
dispersed phase (gas bubbles or liquid drops) and the fluid
(continuous phase). In such cases, kc is expected to depend on:
– diameter Dp of the particles of dispersed phase (e.g., O2 bubble)
– density ρc of the continuous phase (e.g., water)
– viscosity μc of the continuous phase (e.g., water)
– the buoyant force gΔρ where Δρ is the density difference between the
dispersed and continuous phases, and g is the acceleration due to
gravity (9.81 m/s2)
– the diffusion coefficient of the molecules of dispersed phase (oxygen) in
the continuous phase (water).
•
Using dimensional analysis, derive an expression for the mass
transfer coefficient in terms of the dimensionless numbers Sh and
Sc. The dimensionless group with the buoyant force variable, gΔρ, is
called the Grashof number, Gr. Derive an expression for Gr.
49
Mass Transfer to Small Particles
(< 0.6 mm)
1/ 3
⎛
⎞
2DAB
− 2 / 3 ⎜ Δρμ c g ⎟
+ 0.31Sc
kL =
⎜ ρ 2 ⎟
Dp
⎝ c ⎠
LOW DENSITY SOLIDS OR
SMALL GAS BUBBLES IN
AGITATED SYSTEMS
kL =' liquid - side' mass transfer coefficient
DAB = diffusivity of the solute A in solution
Dp = diameter of the gas bubble or solid particle
Sc = μ c (ρc DAB )
μ c = viscosity of the solution
g = 9.81 m/s 2
Δρ = magnitude of the density difference
between the dispersed and continuous phases (always positive)
50
Mass Transfer to Large Gas
Bubbles or Liquid Drops (> 2.5 mm)
1/ 3
AERATION OF PURE LIQUIDS IN
⎛ Δρμc g ⎞
⎟
MIXING VESSELS; SIEVE-PLATE
kL = 0.42Sc
2
⎜ ρ
⎟
COLUMNS
⎝ c ⎠
kL =' liquid - side' mass transfer coefficient
− 0 .5 ⎜
μc
Sc =
ρc DAB
μ c = viscosity of the solution
g = 9.81 m/s 2
Δρ = magnitude of the density difference
between the dispersed and continuous phases (always positive)
51
•
Mass Transfer to Particles in
Highly Turbulent Mixers
Turbulent forces become larger than gravitational forces. Mass
transfer coefficient is determined primarily by agitation power-input
and not by buoyant forces (natural convection).
1/ 3
⎛ (P / V )μ c ⎞
⎟
kL = 0.13Sc
2
⎜ ρ
⎟
c
⎝
⎠
kL =' liquid - side' mass transfer coefficient
−2 / 3 ⎜
μc
Sc =
ρc DAB
μ c = viscosity of the solution
ρc = denstity of the solution
52
Inter-phase Mass Transfer
• Mass transfer at gas−liquid or liquid−liquid
interfaces
• Consider mass transfer from naphthalene to air,
or from benzoic acid to water
– Mass transfer rate is determined by concentration
gradients in air or water
• In mass transfer at interface of two immiscible
liquids A and B (or gas A and liquid B)
• Mass transfer rate is determined by
concentration gradients on both sides of the
interface (in phase A and phase B)
53
Concentration Profile at Interface
∂ ⎛ 2 ∂c A ⎞
⎜r
⎟=0
∂r ⎝
∂r ⎠
• Consider evaporation
of napthalene from
surface of a sphere (1
cm dia.) at 45 °C and
1 atm total pressure.
∴
cA − cA
s
b
cA − cA
b
cA
cA
r =R
r →∞
= cA
s
= cA
s
R
=
r
b
c A = 2.798 × 10 −8 mol/cm 3
b
c A ≈ 0 mol/cm 3
R = 0.5 cm
54
55
‘Two-Film’ Theory & Overall
Mass Transfer Coefficient
Gas phase
pAb
Liquid phase
c Ai
pAi
c Ab
Mass transfer
from gas to liquid
gas−liquid
interface
(
56
)
⎛ mol ⎞
Flux on ' gas - side' = k g pA − pA
⎜
⎟
2
⎝ cm ⋅ s ⎠
mol
⎛
⎞
k g = gas - side mass transfer coefficient ⎜
⎟
2
⎝ atm ⋅ cm ⋅ s ⎠
b
(
i
)
⎛ mol ⎞
Flux on ' liquid - side' = k l c A − c A
⎜
⎟
2
⎝ cm ⋅ s ⎠
⎛ cm ⎞
k l = liquid - side mass transfer coefficient ⎜
⎟
⎝ s ⎠
i
b
The two fluxes across the interface are equal.
(
b
i
) (
i
⇒ N A = k g pA − pA = k l c A − c A
b
)
57
Assume that, at the interface, there is equilibrium
between the gas mixture and the liquid solution
i
∴ c A = H A ⋅ pA
i
H A is the equilibrium constant (e.g., Henry' s constant)
for species A
⎛ mol ⎞
⎜
⎟
3
⎝ atm ⋅ cm ⎠
58
⎛ b NA ⎞
⎛ b NA ⎞
i
⎟
⎟⎟ and pA = ⎜ pA −
Now, c A = ⎜⎜ c A +
⎟
⎜
k
k
l
g
⎝
⎠
⎠
⎝
⎛ b NA ⎞ ⎛ b NA ⎞
⎟ = ⎜ cA +
⎟⎟
∴ H A ⎜ pA −
⎜
⎟
⎜
k
kl ⎠
g ⎠
⎝
⎝
i
b
Solving for N A :
H A pA − c A
NA =
HA 1
+
k g kl
Define concentrat ion c A * = H A pA
b
b
c A * = saturation conc. of A in the liquid when the
b
partial pressure of A in the gas phase is pA .
59
b
∴ NA =
c A * −c A
HA 1
+
k g kl
' Over - all' mass transfer coefficient, K L =
NA
c A * −c A
b
=
1
⎛ HA 1 ⎞
⎜
+ ⎟
⎜ k g kl ⎟
⎝
⎠
1 HA 1
∴
=
+
KL
k g kl
1
=' mass transfer resistance'
mass transfer coefficient
Thus, mass transfer resistances across an
interface are additive.
60
Laminar Boundary Layer on a
Flat Plate (Rex < 5×105)
0.99v0
v0
y
vx
δx
0.99v0
Velocity
boundary
layer
x
x=0
xv 0 ρ
Rex =
μ
Flat plate
δx
4.96
=
x Rex 0.5
⎛ y
vx
= 1.5⎜⎜
v0
⎝ δx
⎛ y
⎞
⎟⎟ − 0.5⎜⎜
⎝ δx
⎠
⎞
⎟⎟
⎠
3
61
Laminar Boundary Layer on a
Flat Plate c b
A
v0
cA
y
δc
Conc.
boundary
layer
cA s
x
Flat plate
x=0
∂c A
∂y
(
b
= cA − cA
y =0
s
)
xv 0ρ
⎛ 0.332
1/ 2
1/ 3 ⎞
Rex Sc ⎟ where Rex =
⎜
μ
⎝ x
⎠
62
Local convection mass transfer coefficient is given by :
k c,x ≡
k c,x x
DAB
(c
N A,y
s
A
x
− cA
b
)
1/ 2
⎛ xv 0ρ ⎞
⎟⎟
= 0.332⎜⎜
⎝ μ ⎠
=
⎛ ∂c
− DAB ⎜⎜ A
⎝ ∂y
(c
s
A
− cA
⎞
⎟⎟
⎠ y =0
b
)
(definition)
1/ 3
⎛ μ ⎞
⎟⎟
⎜⎜
⎝ ρDAB ⎠
⇔ Shx = 0.332Rex
1/ 2
Sc 1/ 3
Mean mass - transfer coefficient for laminar flow over
L
1
a flat plate of width W and length L : k c =
k c,x d x
L0
∫
1/ 2
⎛ kc L ⎞
⎛ Lv ρ ⎞
⎜⎜
⎟⎟ = 0.664⎜⎜ 0 ⎟⎟
⎝ μ ⎠
⎝ DAB ⎠
1/ 3
⎛ μ ⎞
⎜⎜
⎟⎟
⎝ ρDAB ⎠
⇒ ShL = 0.664ReL
1/ 2
Sc 1/ 3
63
PROBLEM 18
k c = [constant ](D ) (ρ ) (μ ) (gΔρ ) (DAB )
a
kc
gm
cm
s
0
1
−1
D
a
0
1
0
ρ
b
1
−3
0
b
c
μ
c
1
−1
−1
d
e
gΔρ DAB
d
e
1
0
2
−2
−2 −1
b+c +d =0
a − 3b − c − 2d + 2e = 1
b = −c − d
a − 3b + 2e = 1 + c + 2d
− c − 2d − e = −1
5 variables, 3 equations
e = 1 − c − 2d
64
⎡0 1 0⎤ ⎡a ⎤ ⎡ − c − d ⎤
⎢
⎥⎢ ⎥ ⎢
⎥
−
=
+
+
1
3
2
1
2
b
c
d
⎢
⎥⎢ ⎥ ⎢
⎥
⎢⎣0 0 1⎥⎦ ⎢⎣e ⎥⎦ ⎢⎣1 − c − 2d ⎥⎦
⎡a ⎤ ⎡3 1 − 2⎤ ⎡ − c − d ⎤
⎢ ⎥=⎢
⎥
⎢
⎥
⎢b ⎥ ⎢ 1 0 0 ⎥ ⎢1 + c + 2d ⎥
⎢⎣e ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣1 − c − 2d ⎥⎦
⎡a ⎤ ⎡ 3d − 1 ⎤
⎢ ⎥ ⎢
⎥
=
−
−
b
c
d
⎢ ⎥ ⎢
⎥
⎢⎣e ⎥⎦ ⎢⎣1 − c − 2d ⎥⎦
Ax = B
A −1Ax = x = A −1B
65
k c = [constant ](D ) (ρ) (μ ) (gΔρ ) (DAB )
a
= [constant ](D )
3d−1
b
c
d
e
(ρ)−c −d (μ )c (gΔρ)d (DAB )1−c −2d
kc D
3d
−c − d
(μ )c (gΔρ)d (DAB )−c −2d
∴
= [constant ](D ) (ρ )
DAB
⎛ μ ⎞
⎟⎟
= [constant ]⎜⎜
⎝ ρDAB ⎠
c + 2d
(D ) (ρ)
3d
(gΔρ) (DAB )
d
−c − d+ c + 2d
−c − 2d+ c + 2d
(μ )
c −c − 2d
66
⎛ kc D ⎞
⎛ μ ⎞
⎜⎜
⎟⎟ = [constant ]⎜⎜
⎟⎟
⎝ DAB ⎠
⎝ ρDAB ⎠
c + 2d
(D )3d (ρ)d (μ )−2d (gΔρ)d (DAB )0
⎛ kc D ⎞
⎛ μ ⎞
⎜⎜
⎟⎟ = [constant ]⎜⎜
⎟⎟
⎝ DAB ⎠
⎝ ρDAB ⎠
c + 2d
⎛ D ρgΔρ ⎞
⎜
⎟
2
⎜ μ
⎟
⎝
⎠
3
Sh = [constant ]Sc αGr β
D 3 ρgΔρ
Gr = Grashof number =
2
μ
d
67
PROBLEM 17
(
s
Flux at surface = k c c A − c A
cA
s
⎛ 0.555 ⎞
=⎜
⎟ atm ×
⎝ 760 ⎠
b
) ≈ k (c
c
s
A
)
− 0 = kc c A
s
1
cm3 atm
× (273.15 + 45 ) K
82.06
K mol
−8 mol
= 2.797 × 10
cm3
68
Initially, D = 1 cm
cm
1 cm × 30
Dvρ Dv
s = 172.41
=
=
Re =
μ
ν
cm2
0.174
s
2
cm
0.174
μ
ν
s = 2.514
Sc =
=
=
2
ρDAB DAB
cm
0.0692
s
FLOW PAST A SINGLE SPHERE :
For gases, when 1 < Re < 48000, and 0.6 < Sc < 2.7
Sh = 2 + 0.552Re 0.53Sc 1/ 3 = 13.503
69
kc D
= 13.503
Sh =
DAB
cm2
13.503 × 0.0692
Sh ⋅ DAB
cm
s
∴ kc =
=
= 0.934
D
1 cm
s
⎛ mol ⎞
s
2
=
×
π
k
c
D
Initial rate of evaporatio n ⎜
⎟
c A
⎝ s ⎠
(
)
cm
2
2
−8 mol
(
)
= 0.934
× 2.797 × 10
×
π
×
1
.
0
cm
s
cm3
−8 mol
= 8.211× 10
s
70
⎛ cm3 ⎞
⎟
Initial rate of evaporatio n ⎜⎜
⎟
s
⎝
⎠
s
k c c A × πD 2
=
ρm
mol
8.211× 10
s
=
−3 mol
8.86 × 10
cm3
3
cm
= 9.267 × 10 −6
s
−8
71
4π 3
V=
R
3
dV
2 dR
= 4πR
dt
dt
s
2
s
k c c A × πD
k c c A × 4πR 2
dV
= −Q = −
=−
dt
ρm
ρm
s
2
4
k
c
R
×
π
d
R
∴ 4πR 2
=− c A
dt
ρm
kcc A
dR
⇒
=−
dt
ρm
s
72
kcc A
dR
=−
ρm
dt
s
0.33
0.53
⎧
Sh ⋅ DAB DAB
DAB ⎪
⎛ 2Rv ⎞ ⎛ ν ⎞ ⎫⎪
⎟⎟ ⎬
=
kc =
Sh =
⎟ ⎜⎜
⎨2 + 0.552⎜
ν ⎠ ⎝ DAB ⎠ ⎪
D
2R
2R ⎪
⎝
⎩
⎭
0.33
0.53
s
⎫⎪
⎧
⎛
⎞
⎛
⎞
D
c
ν
2
Rv
dR
1
⎛
⎞
⎛
⎞
⎪
⎟⎟ ⎬
∴
= −⎜ AB A ⎟ ⋅ ⎜
⎟ ⋅ ⎨2 + 0.552⎜
⎟ ⎜⎜
⎜
⎟
dt
⎝ ν ⎠ ⎝ DAB ⎠ ⎪⎭
⎝ ρm ⎠ ⎝ 2R ⎠ ⎪⎩
⎛ ρm ⎞
2R
⎜
⎟
∴dt = −
dR
0.33
⎜D c s ⎟ ⎧
0.53
⎝ AB A ⎠ ⎪
⎛ 2Rv ⎞ ⎛ ν ⎞ ⎫⎪
⎟⎟ ⎬
⎟ ⎜⎜
⎨2 + 0.552⎜
⎝ ν ⎠ ⎝ DAB ⎠ ⎪⎭
⎪⎩
73
⎛ ρm ⎞ R2
2R
⎜
⎟
⋅
dt = −
dR
0
.
33
⎜D c s ⎟ ⎧
0.53
⎝ AB A ⎠ R1 ⎪
0
⎛ 2Rv ⎞ ⎛ ν ⎞ ⎫⎪
⎟⎟ ⎬
⎟ ⎜⎜
⎨2 + 0.552⎜
⎝ ν ⎠ ⎝ DAB ⎠ ⎪⎭
⎪⎩
t
∫
∫
⎛ ρm ⎞ R1 2R
⎟⋅
t =⎜
dR
s
⎜ D c ⎟ Sh
⎝ AB A ⎠ R2
R1 = 1 cm and R2 = 0.5 cm
∫
74
xn
D
R
Re
Sc
Sh
2R/Sh
cm
1.00
0.98
0.96
0.94
0.92
0.90
0.88
0.86
0.84
0.82
0.80
0.78
0.76
0.74
0.72
0.70
0.68
0.66
0.64
0.62
0.60
0.58
0.56
0.54
0.52
0.50
cm
0.50
0.49
0.48
0.47
0.46
0.45
0.44
0.43
0.42
0.41
0.40
0.39
0.38
0.37
0.36
0.35
0.34
0.33
0.32
0.31
0.30
0.29
0.28
0.27
0.26
0.25
172.414
168.966
165.517
162.069
158.621
155.172
151.724
148.276
144.828
141.379
137.931
134.483
131.034
127.586
124.138
120.690
117.241
113.793
110.345
106.897
103.448
100.000
96.552
93.103
89.655
86.207
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
2.514
13.503
13.380
13.257
13.132
13.006
12.878
12.749
12.619
12.487
12.354
12.220
12.084
11.946
11.806
11.665
11.521
11.376
11.229
11.080
10.928
10.775
10.618
10.459
10.298
10.134
9.966
cm
0.0741
0.0732
0.0724
0.0716
0.0707
0.0699
0.0690
0.0682
0.0673
0.0664
0.0655
0.0646
0.0636
0.0627
0.0617
0.0608
0.0598
0.0588
0.0578
0.0567
0.0557
0.0546
0.0535
0.0524
0.0513
0.0502
⎧1
⎫
y d x = ⎨ (y 1 + y n ) + (y 2 + y 3 + L + y n −1 )⎬ ⋅ (Δx )
⎩2
⎭
x1
∫
Integral 0.0157
t
71875.03
1197.92
19.97
cm
s
min
h
75
Evaporation in Stagnant Air
4π 3
V=
R
3
dV
2 dR
= 4πR
dt
dt
s
4πDAB c A R
dV
= −Q = −
=
ρm
dt
s
4πDAB c A R
dR
∴ 4πR
=−
ρm
dt
2
DAB c A
dR
⇒R
=−
ρm
dt
s
76
R2
s
⎡R ⎤
DAB c A
(t − 0)
⎢ ⎥ =−
ρm
⎣ 2 ⎦ R1
2
∴t =
ρm
2DAB c A
−3 ⎛
(
2
R1
s
− R2
2
)
mol ⎞
8.86 × 10 ⎜
3 ⎟
cm ⎠
⎝
2
2
2
0.5 − 0.25 cm
=
2
⎛ cm ⎞
−8 ⎛ mol ⎞
⎜
⎟
2 × 0.0692⎜
× 2.797 × 10 ⎜
3 ⎟
⎟
⎝ cm ⎠
⎝ s ⎠
= 429147 .35 s
= 4.97 days
(
)(
)
77
9/28/07
• Dimensional analysis
– Application in developing correlations
• Correlations for convective mass
transfer coefficients
• Mass transfer calculations for packed
beds
78
10/3/07
• Inter-phase mass transfer
– Two-Film theory
– Gas-side and liquid-side mass transfer
coefficients
– Overall mass transfer coefficient
• Review
– Dimensional analysis (HW Problem 18)
– Mass transfer coefficient (HW Problem 17)
• Use of correlation
79
HW Problems
• Mass transfer in packed beds (benzoic
acid): Problem 19, HW # 5
• Kinematic viscosity (slide 35):
– ν = μ / ρ (cm2/s)
– Re = D v ρ/μ = D v / ν
– Sc = μ / (ρ DAB) = ν / DAB
• Problem 14: “Write 3 sentences explaining the
answers to part 3 of problems 11 and 13.”
80
10/5/07
• Inter-phase mass transfer
• Review of topics covered
• Simple quiz on fundamental concepts?
– Definitions, units, < 5-min problems, etc.
81
Review
• Slides 42 to 46
• Correlations
Sh = [constant ]Re Sc
α
β
Turbulent flow through pipes :
Sh = 0.023 Re
0.83
Sc
0.33
82
• Quiz
–16 questions, < 40 min
–Questions at the beginning require
less time than those at the end
• Course evaluation
–CH351: Mass Transfer and
Stagewise Operations
–Instructor: Sitaraman Krishnan
83
Benzoic acid spheres in a
packed bed
cA,1
Saturation solubility of benzoic
acid in water = cAs
Mass transfer driving force at
inlet = cAs – cA,1
= (2.948×10−5 − 0) mol/cm3
cA,2
Mass transfer driving force at
outlet = cAs – cA,2
= (2.948×10−5 − cA,2) mol/cm3