Advanced Transport Phenomena
Module 3 Lecture 11
Dr. R. Nagarajan
Professor
Dept of Chemical Engineering
IIT Madras
Constitutive Laws: Illustrative Problems
2
LIMITATIONS OF LINEAR LOCAL FLUX VS
LOCAL DRIVING FORCE CONSTITUTIVE
LAWS
Nonlinear fluids
Nonlocal spatial behavior– Action at a distance
Nonlocal temporal behavior– Fluids with memory
Multiphase effects– nonlinear species drag laws
3
NONLINEAR FLUIDS
Non-Newtonian
fluids:
dynamic
viscosity
depends on deformation rate (or extra stress)
e.g., high-mass-loaded slurries, clay-in-water,
etc.
Dynamics of such fluids studied under “rheology”
Analogous non-linearities exist in heat & mass
transfer
e.g., solute diffusion through human blood
4
NONLOCAL SPATIAL BEHAVIOR
When mfp not negligible compared to overall
length over which transport process occurs
e.g., transport across very low-density gases
in
ducts,
electronic
conduction
in
micro
crystallites, radiation transport through nearly
transparent media
5
NONLOCAL TEMPORAL BEHAVIOR
Fluid responding to previous stresses, stress
history
e.g., visco-elastic fluids, such as gels, that
partially return to their original state when
applied stress is interrupted
Analogous behavior possible for energy & mass
transfer
e.g., time lags between imposition of spatial
gradients & associated fluxes
Effects associated with finite thermal and
concentration wave speeds
6
MULTIPHASE EFFECTS
Single-phase mixture vs multiphase system
e.g., gas with vapors of varying molecular
weights vs droplet-laden gas
Not a sharp distinction
Relates to vi – vj, vi – v
Linear laws apply only when these differences
are small.
7
PROBLEM 1
Figure on next slide displays the experimentally
observed composition dependence of the viscosity
m of a mixture of ammonia and hydrogen at 306 K,
1 atm. N.B.: The units used for m in Figure are mP
(micro-poise) where 1 poise

1 g/cm∙s. The
MKS unit of viscosity is 1 kg/m∙s or 1 Pascalsecond.
8
PROBLEM 1
9
PROBLEM 1
a. Using Chapman-Enskog (CE) Theory and the
Lennard-Jones
(LJ)
potential
parameters,
examine the agreement for the two endpoints,
i.e., the viscosities of pure NH3 and pure H2 at
306 K, 1 atm.
b. Using the “square-root rule,” what would you
have predicted for that composition dependence
of the mixture viscosity in this case? What
conclusions do you draw from this comparison? 10
PROBLEM 1
c. How would you expect the NH3 + H2 curve to shift
if the temperature were increased to 1000 K and
the pressure were
increased ten-fold? (Discuss
the basis for your expectation.)
d. Estimate the binary diffusion coefficient DNH H (cm2/s)
3
2
from Chapman-Enskog theory.
e. Calculate the Schmidt number for an equimolar
mixture of H2(g) and NH3(g).
11
PROBLEM 1
k(NH3 (g)) and DNH H Calculation
3
2
Given the tabular data:
12
PROBLEM 1
Calculate k(NH3 (g)) and DNH H @ 1 atm. 306 K
3
2
Thermal conductivity of NH3(g)


15 R m 
4  C p 5  
k
. 1  
 
4 M  15  R 2  
ktrans
 Eucken factor 
m
is calculated to give 1.063 x 10-4 poise.
13
PROBLEM 1
ktrans
15 R
15 1.987 
4
 . .m  .
1.063 x 10 

4 M
4 17.031
2
cal
/
cm
s
5
 4.651x 10
K / cm
4  8.559 5 
Eucken factor  1  
   1.482
15  1.987 2 
14
PROBLEM 1
Therefore
k  ktrans .(Eucken factor )
2
cal
/
cm
s
5
5
k   4.65110  1.482   6.892 10
K / cm
cpm
 Cp
Note : Pr 


 k / cp  k  M
v
m
 k  0.775

15
PROBLEM 1
Calculation of DNH3-H2 @ 306 K, 1 atm: and Sc for
y1 = 0.5
1/2
1
 ij   ii   ij  ,  ij   ii . jj 
2
3kBT  k BT  mi  m j
Dij 
.
.
8 p  2  mi m j
1/2

1
  . 2
   ij  D (kT /  ij )
16
PROBLEM 1
In the present case:
kBT  1.38054 x 106 (306)  4.2245 x 1014erg / molecule
mi m j
mi  m j
 2.9929x 1024 g ;
p  1 atm  1.0133x 106 dyne / cm 2 ;
 ij  2.8635 A;  ij / k B    ii / k B    jj / k B  
1/2
 182.6 K ;
 2  8.1996(A) 2  8.1996 X 1016 cm2 ;
ij
306
kB T /  
 1.676
(outside of range of curve  fit ,
182.6
which formally gives 1.03).
17
PROBLEM 1
From tabular values ( Hirshfelder, Curtiss and Bird
(1954));
 D (1.676)  1.145
Therefore:
DNH3 H2  0.7895  0.790 cm / s
2
For an equimolar mixture of ammonia and
hydrogen:
m  123m poise  1.23 x 104 P
18
PROBLEM 1
Moreover,
M mix  i 1 yi M i  0.5(17.031)  0.5(2.016)  9.5235.
N
Assuming perfect gas behavior:
1 (9.5235)

pM


 3.79 x 104 g / cm3 .
RT  82.06  306 
The momentum diffusivity of the mixture is
therefore:
v  m /   0.324 cm2 / s,
19
PROBLEM 1
Corresponding to a diffusivity ratio of :
0.3273
Sc  v / Dij 
 0.411 (dimensionless)
0.7892
20
PROBLEM 2
Property estimation for the hydrodesulphurization of
Naphtha vapors.
For the preliminary design of a chemical reactor to carry out
the removal of trace sulfur compounds (e.g., C4H4S) from
petroleum naphtha vapors (predominantly heptane, C7H16),
it is necessary to estimate
the Newtonian viscosity and
corresponding momentum diffusivity of the vapor mixture at
660 K, 30 atm; these are rather extreme conditions for
which direct measurements are not available.
21
PROBLEM 2
a. Using the vapor composition together with
selected results from the kinetic theory of ideal
vapors and dense vapors, what are your best
estimates for mmix and nmix ?
22
PROBLEM 2
b. If this vapor mixture is to be passed at the rate of
2 g/s through each 2.54 cm diameter circular tube
( in a parallel array), calculate the corresponding
Reynolds’ number, Re ≡ Udw/nmix within each tube
(where U is the average vapor-mixture velocity).
This dimensionless ratio will be seen to be
required to estimate the mechanical energy
required to pump the vapor through such tubes.
23
PROBLEM 2
c.
Estimate
the effective (pseudo-binary) Fick
diffusion coefficient for thiophene (C4H4S)
migration
through
this
mixture,
and
the
corresponding diffusivity ratio: Sc ≡ nmix /D3-mix .
24
PROBLEM 2
25
PROPERTY ESTIMATION FOR THE
HYDRODESULPHURIZATION OF NAPHTHA
VAPORS
a. Suppose we need viscosity of
H 2  C7 H16  ...trace C4 H 4 S
yH 2  0.828
yC7 H16  0.172
@ 660 K , 30 atm
b. Further, if
m  2 g / s per tube and d w ( tube )  2.54 cm, Re  ?
The first step is the estimation of
the mixture
viscosity based on its composition and the
properties of its constituents under the anticipated 26
PROBLEM 2
operating conditions (660 K, 30 atm). Tentatively
we use:
1/2
M
i1 i yimi
N
M11/2 y1m1  M 21/2 y2m2
m

N
1/2
1/2
1/2
M
y

M
1
1
2 y2
i1 M i yi
where
mi k BT 

5
mi  .
16  ii2  .m  k BT / ii 
1/2
 i  1, 2, where 1  H 2 (g), 2  C7 H16 (g) 
(cf . Enskog  Chapman theory.)
27
PROBLEM 2
In what follows, we sequentially consider the
viscosities of each of the constituents of the vapor
mixture.
Viscosity of Hydrogen(g) at 660 K, 30 atm:
M
2.016 g / g - mole
-24
m


3
.
347

10
g
23
N A 6.0225 10 molec / mole
28
PROBLEM 2
ii  2.827 108 cm;
ii2  2.5111015 cm 2
5 /16  0.3125
(mk BT )1/2  ((3.141...)(3.347 1024 )(1.38054 1016 )(660))1/2
 9.789 1019
and
k BT / ii  660 / 59.7  11.055
29
PROBLEM 2
Therefore
 k BT 
m (11.055)  0.8114 ( N .B.:1.22 




0.16
gives 0.831.)
Using table for L-J 12:6 potential ( Hirschfelder,
Curtiss, and Bird (1954))
30
PROBLEM 2
Therefore
m H2
 0.3125  9.789 1019 
g
4

 1.50 10
( poise)
15
cm  s
 2.51110   0.8114
(to convert to MKS units ( Pa  s ), multiply by 101 )
We now consider a similar calculation for the
species C7H16(g) and check whether “dense
vapor” correction are important
31
PROBLEM 2
Viscosity of n-Heptane at 660 K, 30 atm.
100.128
22
m
g

1.6626

10
g;
23
6.023  10
  8.88  10 8 cm;
(  mk BT )  4.759  10
35
;
32
PROBLEM 2
(  mk BT )1/ 2  6.8986  10 18
 2  2.4773  10 14 cm 2
k BT /   660 / 282  2.34
Therefore m ( 2.34 )  1.12 via Hirshfelder et al. (1954)
table;
cf. 1.22(2.34)-0.16=1.065)
and
( 0.3125 )( 6.8986  10 18 )
(O)
mC H ( 660 K ,30 atm ) 
7
16
 2.4773  10  1.12 
14
 0.817  10 4 poise.
33
PROBLEM 2
Dense Vapor Correction? This can be estimated
via the principle of “corresponding states”
For C7H16:
and
pr  p / pc  30 / 26.8  1.119
Tr  T / Tc  660 / 540  1.222
From
m / mc T / Tc , p / pc  , m / m gas  1.25
( cf .p / pc  0 asymptote )


4
4
Therefore mC7 H 16  1.25 0.817  10 P  1.017  10 poise.
34
PROBLEM 2
Viscosity of Binary Mixture @ 660 K, 30 atm..
Table for calc. m mix
i
1
2
Mi
M1i / 2
2.016
1.4199
100.128 10.006
yi
0.828
0.172
104 mi
1.502
1.017
35
PROBLEM 2
Then
1.4199  0.828  1.502  10.006  0.172  1.017 
10 m 
1.4199  0.828   10.006( 0.172 )
4
 1.21 poise
Re-Number Calculation:

 m 
Ud w
U  

Re 
, where 
 A
v
v  m / 

 
average velocity
momentum diffusivity
36
PROBLEM 2
For each dw=2.54 cm tube
A
and
 d w2


( 2.54 )2  5.067cm 2
4
4
PM  30  18.891


 1.046  10 2 g / cm3
RT  82.06  660 
If the perfect gas EOS is valid for the mixture,
where
N
M mix   yi M i  0.828( 2.016 )  0.172( 100.128 )
i 1
g
 18.891
g  mole
37
PROBLEM 2
Now:
2
1.214  10 4 Poise
cm
2
vm/ 

1.16

10
1.046  10 2 g / cm3
s
And
2
1 cm
U
 3.77  10
2
s
1.046  10   5.067 
Therefore
1
3.77

10
 2.54 


Ud w
3
Re 


8.26

10
2
v
1.16

10


38
PROBLEM 2
Conclusion: Since Retrans  2.1  10 3
Re=8.26 x 103
(under
for tube, @
the flow should be turbulent
anticipated
hydrodesulphurization
conditions).
39