Fundamentals of Transport Phenomena ChE 715

1. Fundamentals of Transport Phenomena
ChE 715
Lecture 20
Ch 9
• Convective Heat/Mass Transfer
C fi d Fl• Confined Flows
Spring 2011

2. Convective Heat/Mass Transfer
To, L
Conc., or temp change at wall
y Vx
Scenario—Flow thru flat plate:
Steady-state conservation equations: 2
v source term(s)A A AC D C⋅∇ = ∇ +
o y
x
x
2
v source term(s), where =
p
k
T T
C
α α
ρ
⋅∇ = ∇ +
Generalized non-dimensional form:
2
Pe source term(s)θ θ⋅∇ = ∇ +v
θ = Dimensionless concentration or temperature
Generalized non dimensional form
2
0
Re P
∇⋅ =
⋅∇ = −∇ + ∇
v
v v v
θ Dimensionless concentration or temperature
Will need velocity
profile and energy eqn
to solve prob
Pe = Peclet#; Re= Reynolds#
to solve prob

3. Convective Heat/Mass Transfer
Pe˜v⋅ ˜∇θ = ˜∇2
θ +source term(s)
Steady-state conservation equations:
˜∇⋅ ˜v = 0;
Re ˜v⋅ ˜∇˜v = − ˜∇ ˜P + ˜∇2
˜v
UL UL
Pe Pr Re
ν
α α ν
= = =
UL UL
Pe ScRe
i iD D
ν
ν
= = =
Sc: Schmidt #
Pr: Prandt’l #
Sc: Schmidt #
U,L characteristic velocity and length
UL UL U U
Pe
ii
or or
DD αα
= = =
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
Note,
Charac. vel. of convection
Charac vel ofii
L L
α ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Charac. vel. of
conduction/diffusion

4. Example Problem: Heating on a Plate
Energy Equation:
2 2
T T T
U α
⎛ ⎞∂ ∂ ∂
+⎜ ⎟
Insulated Insulated
Heated
qo
2 2
U
x x y
α= +⎜ ⎟
∂ ∂ ∂⎝ ⎠UTo L y
x
II
I III
K
BC’s:
⎧
II
Heated Plate moving at const. speed
0( , )T y T−∞ = ( , ) 0
dT
y
dx
∞ =; ( ,0) 0
dT
x
dy
= ;
0
0 0
( , )
/ 0
x or x KdT
x L
q k x Kdy
< >⎧
= ⎨
≤ ≤⎩
;
f l h E EDefine cross-sectional avg.
temperature:
1
L
Averaging each term in Energy Eq.:
2 y L
dT d T T
U
α
=
∂
+
0
1
( ) ( , )T x T x y dy
L
= ∫ 2
0y
U
dx dx L y
α
=
= +
∂

5. Example Problem: Heating on a Plate
Using BC’s at top and
2
2
0, 0
d T U dT
x or x K
dx dxα
− = < >
Regions
I & III
Using BC s at top and
bottom surfaces:
dx dxα
2
0
2
, 0
qd T U dT
x K
dx dx kLα
− = ≤ ≤
Region
II
Plug into eqn. for Region II:
dx dx kLα
Let:
2
x
L
ζ =
Pe
0T T
T
−
Θ =
Δ
2
0
2
q LUL d d
d d k Tα ζ ζ
Θ Θ
= +
Δ Let:
0q L
T
k
Δ =
K
L
λ =
k
For confined fluids in
general, a cross-sectional
dimension is usually they
appropriate length scale

6. Example Problem: Heating on a Plate
The dimensionless avg. temp. is governed by:
2
2
0, 0
d d
Pe or
d d
ζ ζ λ
ζ ζ
Θ Θ
− = < >
2
2
1, 0
d d
Pe
d d
ζ λ
ζ ζ
Θ Θ
− = − ≤ ≤
BC‘s: ( ) 0Θ −∞ = ( ) 0
d
dζ
Θ
∞ =
0ζ <
The resulting solution is:
( )1 2
( ) 1
Pe
Pee
e
Pe
ζ
λ
ζ −
Θ = −
( )( )
2 2
1
( ) 1 Pe
e
Pe Pe
ζ λ ζ
ζ −
Θ = − + 0 ζ λ< <
3 ( )
Pe
λ
ζΘ = ζ λ>

7. Convective Heat/Mass Transfer
Dimensional analysis:
θ = θ(˜x,Pe,Re,geometry)g y
Heat transfer coefficient (mass transfer homolog is Sh):
Nu
f
hL
k
=
Dimensionless temp. gradient
at tube wall
Dimensionless temp. difference
Nusselt #
( )S
b S
n∂θ ∂
θ θ
−
= =
−
i
kL
Sh
D
= =
Convective mass transport
Diffusive mass transport
Nusselt #
Sherwood #
For confined flows (flow in the z-direction)
z
A
v dAθ
θ ≡
∫
iD Diffusive mass transportSherwood #
For confined flows (flow in the z-direction), b
z
A
v dA
θ ≡
∫

8. Example Problem: Hollow-Fiber Dializer
Flux from fluid to wall:
[ ]( , ) ( ) ( ) ( , )i i ib iN R z k z C z C R z= −C r R
Cid Dialysate
M b [ ]( , ) ( ) ( ) ( , )ir ci ib iN R z k z C z C R zCi0
r
z
R
Ci(r,Z)
vz
Membrane
L
Flux of solute i through membrane
[ ]( ) ( )N R z k C R z C
Transport PDE for fluid inside
fiber:
L [ ]( , ) ( , )ir mi i idN R z k C R z C= −
where: kci(z) = mass transfer coeff.
kmi = permeability of
membrane to i
⎛ ⎞
f ber membrane to i
v i i i
z
C D C
r
z r r r
∂ ∂∂
∂ ∂ ∂
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
Integrating over r :
v ( , ) ( , )
R
i i
z i ir
C C
rdr RD R z RN R z
∂ ∂
∂ ∂
= = −∫0
( , ) ( , )z i ir
z r∂ ∂∫

9. Example Problem: Hollow-Fiber Dializer
C r R
Cid Dialysate
M b
( , ) ( , )
R
i i
z i ir
C C
v rdr RD R z RN R z
z r
∂ ∂
∂ ∂
= = −∫
Recalling that:
Ci0
r
z
R
Ci(r,Z)
vz
Membrane
L
( )
i z
A
ib
Cv dA
C z ≡
∫
∫
0
z r∂ ∂
L
2R R
bC dCd R U∂
∫ ∫
( )ib
z
A
v dA∫
The LHS becomes:
0 0
2
i ib
z i z
C dCd R U
v rdr C v rdr
z dz dz
∂
∂
= =∫ ∫
Th RH i l d i
U= avg. vel
[ ]
2
( ) ( , )ib ci
ib i
dC k
C z C R z
dz RU
= −
The RHS is evaluated using
the equation flux from fluid
to wall eqn:

10. Example Problem: Hollow-Fiber Dializer
Ci0
r R
Cid Dialysate
Membrane
2
0 0
2
R R
i ib
z i z
C dCd R U
v rdr C v rdr
z dz dz
∂
∂
= =∫ ∫
z Ci(r,Z)
vz
L
How?
0
( )
R
i zi z
A
ib R
Cv rdrCv dA
C z
v dA
≡ =
∫∫
∫ ∫
0
2
2
R
z R
zR
v rdr
U v rdr
R
≡ =
∫
∫
∫
0
2
2
R
zv rdr
UR
=∫
0
z
A z
v dA
v rdr∫ ∫ 0
0
R
rdr∫
0
2R
UR
Cv rdr C∫0
2
i z iCv rdr C=∫
2R R
i ibC dCd R U
d C d
∂
∫ ∫0 0
2
i ib
z i z
d U
v rdr C v rdr
z dz dz∂
= =∫ ∫

11. Example Problem: Hollow-Fiber Dializer
To eliminate C(r,z) the two equations
for flux are combined to give:C r R
Cid Dialysate
M b
f f m gCi0
r
z
R
Ci(r,Z)
vz
Membrane
L
[ ]( ) ( , ) ( )ci mi
ib i ib id
ci mi
k k
C z C R z C z C
k k
⎛ ⎞
− = −⎜ ⎟
+⎝ ⎠L
The final ODE for the bulk
t ti i
ci mi⎝ ⎠
[ ] 0
2
, (0)ib ci mi
ib id ib i
dC k k
C C C C
⎛ ⎞
= − − =⎜ ⎟
If kci is independent of z then:
concentration is: [ ] 0, (0)ib id ib i
ci mi
C C C C
dz RU k k
⎜ ⎟
+⎝ ⎠
0
( ) 2
expib id ci mi
i id ci mi
C z C k kz
C C RU k k
⎡ ⎤⎛ ⎞−
= −⎢ ⎥⎜ ⎟
− +⎝ ⎠⎣ ⎦
Overall mass transfer coeff.

12. Example Problem: Hollow-Fiber Dializer
How do we get the following equation?Ci0
r
z
R
Cid
C
Dialysate
Membrane
z Ci(r,Z)
vz
L
[ ]( ) ( , ) ( )ci mi
ib i ib id
ci mi
k k
C z C R z C z C
k k
⎛ ⎞
− = −⎜ ⎟
+⎝ ⎠
Flux from fluid to wall:
[ ]( ) ( ) ( ) ( )N R z k z C z C R z= −
Plug in Ci(R,z)
[ ]( , ) ( ) ( ) ( , )ir ci ib iN R z k z C z C R z=
Flux of solute i through membrane
[ ]( ) ( )N R k C R C
( )
( , ) ci ib mi id
i
ci mi
k C z k C
C R z
k k
+
=
+
rearrange
[ ]( , ) ( , )ir mi i idN R z k C R z C= −
[ ] [ ]( ) ( ) ( , ) ( , )ci ib i mi i idk z C z C R z k C R z C− = −[ ] [ ]( ) ( ) ( , ) ( , )ci ib i mi i idk C C k C C

13. Convective Heat/Mass Transfer
Materials from this slide onwards is covered in lecture 21
Heat/mass transfer coefficients in confined flows:
vz profile is known, assumed well-developed.
Entrance regime:
uorSh
ll d l d i
Entrance regime:
Nu ~ L/δ(z)
Nu
Well-developed regime:
Nu = constant ~ 3
z

14. Nusselt # in Tube with Specified Temperatures
T R
Tw At entrance
( )Tk ∂
∂η−
(step change)
Nu
hL
;T0 R
δ
E t
z=0
Fully Developed
( )w
b w
h
T T
∂η
=
−
Nu
k
=
( ) b wT TT∂
∂η δ
−
=
;
Entry
Region
Fully Developed
Region (FD)
Entry FD
( )w
∂η δ
~ at entrance
L
Nu
δ
∴
r
Nu ~ 1
Nu ~ L/δ(z)
logNu
Entrance Region
• Fluid enters at T = T0
• Step change at wall T = Tw
l
log z
Step change at wall T Tw
• Flow is laminar & fully developed for z>0
• Large Pe# (ignore axial conduction)
• Near to entrance (z=0)

15. Nusselt # in Tube with Specified Temperatures
Dimensional problem is: 2
T T⋅∇ = ∇v
⎡ ⎤2
2 1
r T T
U r
R z r r r
α⎡ ⎤ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞
− =⎢ ⎥⎜ ⎟ ⎜ ⎟
∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
0( ,0)T r T=BC’s: ; (0, ) 0
dT
z
dr
= ; ( , ) wT R z T=
Now nondimensionalizeNow, nondimensionalize
r
R
η = 0
0w
T T
T T
θ
−
=
−
z
z
ζ =
Δ
2
2
2 1
1
U
z R
θ α θ
η η
ζ η η η
⎛ ⎞∂ ∂ ∂
⎡ ⎤− = ⎜ ⎟⎣ ⎦Δ ∂ ∂ ∂⎝ ⎠0w
To be determined
z Rζ η η ηΔ ∂ ∂ ∂⎝ ⎠
2
22 1
1
UR
z
θ θ
η η
α ζ η η η
⎛ ⎞∂ ∂ ∂
⎡ ⎤− = ⎜ ⎟⎣ ⎦Δ ∂ ∂ ∂⎝ ⎠zα ζ η η η⎣ ⎦Δ ∂ ∂ ∂⎝ ⎠

16. Nusselt # in Tube with Specified Temperatures
Let 2
2 2UR UR
z R R Pe
α α
Δ = = ⋅ = ⋅
2
22 1
1
UR
z
θ θ
η η
α ζ η η η
⎛ ⎞∂ ∂ ∂
⎡ ⎤− = ⎜ ⎟⎣ ⎦Δ ∂ ∂ ∂⎝ ⎠
We get
( )2 1θ θ⎛ ⎞∂ ∂ ∂
2z UR
where Pe
R Pe
ζ
α
∴ = =
⋅
( 0) 0θBC’s: ; (0 ) 0
dθ
ζ ; (1 ) 1θ ζ
( )2 1
1
θ θ
η η
ζ η η η
⎛ ⎞∂ ∂ ∂
− = ⎜ ⎟
∂ ∂ ∂⎝ ⎠
( ,0) 0θ η =BC s: ; (0, ) 0
d
ζ
η
= ; (1, ) 1θ ζ =
Entrance Region
Use similarity solution
For ζ~0, temp.
changes occur near
Use new radial variable
based at the wall:
1χ η= −changes occur near
the tube wall
χ η

17. Nusselt # in Tube with Specified Temperatures:
Entrance Region
Then
1χ η= −
1
( ) ( )2 1
1 1 2 1
1
θ θ
χ χ χ
ζ χ χ χ
⎛ ⎞∂ ∂ ∂
⎡ ⎤− − + = − − −⎜ ⎟⎣ ⎦ ∂ − ∂ ∂⎝ ⎠
χ η∂ = −∂
1η χ= −
( ) ( ) ( )
2
2
2
1 1
2 1 1
1 1
θ θ θ
χ χ χ
ζ χ χ χ χ
∂ ∂ ∂
− = − + −
∂ − ∂ − ∂
( )
2
2
2
1
2
1
θ θ θ
χ χ
ζ χ χ χ
∂ ∂ ∂
− = −
∂ ∂ − ∂
Let
1
δ
δ
Order of magnitude analysis
1
R
δ ≡ <<
1 1θ∂ ⎛ ⎞
( ) ( )2 1
2 ~
θ
χ χ δ
ζ ζ
⎛ ⎞∂
− ⎜ ⎟
∂ ⎝ ⎠
2
2 2
1
~
θ
χ δ
∂ ⎛ ⎞
⎜ ⎟
∂ ⎝ ⎠
( )
1 1
~ 1
1
θ
χ χ δ
∂ ⎛ ⎞
⎜ ⎟
− ∂ ⎝ ⎠
( )2
χ χ<<
Comparing terms, term is negligible
θ
χ
∂
∂

18. Nusselt # in Tube with Specified Temperatures:
Entrance Region
Then
2
1 1
~δ
ζ δ
⎛ ⎞
⎜ ⎟
⎝ ⎠
2
2
2
θ θ
χ
ζ χ
∂ ∂
=
∂ ∂
3
~δ ζ 1/3
~δ ζ
Nusselt number (from 9.3-23):
ζ δ⎝ ⎠ζ χ∂ ∂
1/3
1/3z
Nu C Pe
R
−
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
C = constant
with order of
magnitude 1
Describes heat
transfer in
entrance region
To evaluate C
we solve the
energy eq.
g
2
2
2
θ θ
χ
ζ χ
∂ ∂
=
∂ ∂
( ,0) 0θ χ =BC’s: ; (0, ) 1θ ζ = ( , ) 0θ ζ∞ =;
ζ χ∂ ∂
Assume that θ = θ(s)
only, where: χ
Converting to
the similarity 2
2 2
2 ( ') 0
θ θ∂ ∂
+
y
( )
s
g
χ
ζ
≡ variable:
2 2
2
2 ( ') 0s g g
s s
+ =
∂ ∂

19. Nusselt # in Tube with Specified Temperatures:
Entrance Region
2
2 2
2
2 ( ') 0
d d
s g g
ds ds
θ θ
+ =
Need to have g2g’ = const. for s to be
only independent variableds ds y p
Let
( )2 31 3
' '
3 2
g g g= = ( )3 9
'
2
g =
Using
g(0)=0
1/3
9
2
g ζ
⎛ ⎞
=⎜ ⎟
⎝ ⎠
( )3 2
( ) 2 g(0) 0
1/3
1/3
2
9
s
χ
ζ
⎛ ⎞
∴ =⎜ ⎟
⎝ ⎠
2⎝ ⎠
9 ζ⎝ ⎠
The eqns. for
θ(s) are now: 2
2
3 0
d d
s
θ θ
+ =
( 0) 1sθ = =
2
3 0s
ds ds
+
( ) 0sθ = ∞ =
Let
d
p
θ
=
2
3
dp
s p= −
3
sd
p Ce
θ −
= =p
ds
= 3s p
ds
= 1p Ce
ds
= = −

20. Nusselt # in Tube with Specified Temperatures:
Entrance Region
3
0
1
s
d C e dsθ
∞
−
=∫ ∫
3
1
sd
p Ce
ds
θ −
= = − BC 1:
sθds
BC 2: 3
0
1
1 0
s
d C e dsθ
∞
−
=∫ ∫ 3
1
1
s
C
e ds
∞
−
= −
∫1 0
0
e ds∫
3 (1/3)
3
s
e ds
∞
− Γ
=∫ 33 s
dθ
∞
−
∫
The Nusselt
b is th :
0
3
(1/3)
s
s
e dsθ∴ =
Γ ∫
θ⎛ ⎞∂number is then:
0
0
2
2 6
2
(1/3) ( )w b
hR
Nu
k g
χ
χ
θ
χ θ
θ θ χ ζ
=
=
⎛ ⎞∂
−⎜ ⎟∂⎝ ⎠ ∂
= = = − =
− ∂ Γχ
θb = 0 since nonisothermal region is thin and almost
all fluid remains at inlet temp.

21. Nusselt # in Tube with Specified Temperatures:
Entrance Region
(1/3)=2.6789Γ
1/3
1/3
Nu 1.357
R
Pe
z
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
(Tw const.)
For circular tube with
Same form as in previous equation with C=1.357
specified flux (qw) at
the wall: 1/3
1/3
Nu 1.640
R
Pe
z
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
(qw const.)
both have same form with different constant value
z⎝ ⎠