Transcript Convective Mass Transfer

HW/Tutorial Week #11
WWWR Chapter 28
• Tutorial #11
• WWWR # 28.3, 28.13
& 28.25
• To be discussed on
April 7, 2015.
• By either volunteer or
class list.
Convective Mass Transfer
• 2 types of mass transfer between moving
fluids:
– With a boundary surface
– Between 2 moving contacting phases
• Analogy with heat transfer
Boundary Surfaces
• Convective mass transfer coefficient
N A  kc cAs  cA 
• Hydrodynamic boundary layer
– Laminar flow – molecular transfer
– Turbulent flow – eddy diffusion
Example 1
Dimensional Analysis
• Defining dimensionless ratios
– Schmidt number
momentum


 Sc 

mass
DAB DAB
– Lewis number
thermal
k
 Le 
cP DAB
mass
– Sherwood number
from
N A   DAB
to
dcA
dy
y 0
kc L  d c A  c As  / dy y 0
Sh 

c A,s  c A, / L
DAB
ratio of molecular mass-transfer resistance to
convective mass-transfer resistance
Example 2
• Transfer to stream flowing under forced
convection
– Using Buckingham- theory, 3  groups:
a
b c
(i)
1  DAB  D kc
L
1  
 t
2
kc L
 1 
DAB
a
 M 
c L 
  3  L   
t
 L 
b
(ii)
2  D  D v
d
AB
e
f
Dv
2 
DAB
(iii)
3  D  D 
g
AB
 3 
h
i

DAB
– The correlation relation is in the form:
Sh = NuAB = f (Re, Sc)
• Transfer to natural convection phase
– 3  groups:
a
b c
1  DAB L  kc
(i)
kc L
 1 
D AB
(ii)
2  D L  
d
e
AB
DAB
2 

f
 3  D L  g A
L g A
 3 
DAB
(iii)
g
h
AB
3
i
defining an analogous GrAB
 DAB  L3 g A  L3 g A L3 g A
 

 2 3  

 Gr AB
2
2

v
   DAB 
– The correlation relation is in the form:
Sh = f (GrAB, Sc)
Mass, Heat and Momentum
Analogies
• Similarities between the transport
phenomenon
• 5 conditions:
–
–
–
–
–
No reaction to generate heat/mass
No radiation
No viscous dissipation
Low mass-transfer rate
Constant physical properties
• Reynolds analogy
– Between momentum and energy, if Pr = 1
– Between momentum and mass, if Sc = 1
– From the profiles,
  c A  c A, s 
y  c A, s  c A, 
we get
  vx 
 
y 0 
y  v 
 vx
kc 
v y
y 0
y 0
– Combined with coefficient of skin friction
2 vx / y y 0
0
Cf  2

v / 2
v2
to get

kc C f

v
2
which is analogous to
Cf
h

v c p
2

– For turbulent flow, we use Prandtl’s mixing
length hypothesis
from velocity fluctuation and shear stress
dvx
dvx
'
vx   L
 
 vx' v y'
dy
dy
we find
dvx
     εM 
dy
from concentration fluctuation and
instantaneous transfer
dc A
'
cA  L
N A, y  cA' v'y
dy
we get
N A, y
dc A
 DAB   D 
dy
with the analogous heat transfer equation
qy
dT
  c p    H 
A
dy
• Prandtl and von Karman analogies
– Effect of turbulent core and laminar sublayer
– In the sublayer, for momentum
 s


vx 
and mass
c
A, s

 cA  
N A, y , s
DAB

we get
vx 


DAB

c A, s  c A 
N A, y , s

– In the core, using Reynolds analogy,

  v
N A, y  kc cA   cA, 
s

 vx 
 c
A
 cA,
– Combining both turbulent and laminar
equations
c A, s  c A,  
 

 v  v x  
 1
N A, y
s 
 DAB 
and simplify to
Cf / 2
kc

v 1  v x  / v Sc 1



– At the laminar sublayer,
vx 
Cf
5
v
2
substitute to get the Prandtl analogy
Cf / 2
kc

v 1  5 C f / 2 Sc 1
– Multiply by vL/DAB and rearrange, we get
Sh 
C
f
/ 2 Re Sc
1  5 C f / 2 Sc 1
–
With a buffer layer between the laminar
sublayer and turbulent core, we use the von
Karman analogy
for heat transfer
C f / 2 Re Pr
Nu 
1  5 C f / 2 Pr  1  ln1  5 Pr  / 6

for mass transfer
Sh 
C

f
/ 2 Re Sc
1  5 C f / 2 Sc 1  ln1  5 Sc  / 6
kc
Sh


Re Sc v
Cf / 2

 5

1  5 C f / 2 Sc  1  ln 1  Sc  1 
 6


• Chilton-Colburn analogy
– Modification to Reynolds’ analogy, for all Pr
and Sc
– j factor for mass transfer
kc 2 / 3 C f
jD  Sc 
v
2
– For fluids within 0.6 < Sc < 2500, we know
Sh x  0.332Re Sc
0.5
x
1/ 3
– Divide by RexSc1/3,
Sh x
0.332

1/ 3
Re xSc
Re 0x.5
– Substitute in Blasius solution,
 kc x    DAB  2 / 3 kc Sc 2 / 3 C f
Sh x


 Sc 
 

1/ 3
Re xSc
v
2
 DAB  xv    
– So the analogy is
jH  jD 
Cf
2
Example 3
Example 4
Example 5