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Applications on Molecular
Transport Equation
Lani B. Pestao
Instructor
Mass Transport
Calculate the mass flux of benzene through alayer of air 10 mm
thick at 25oC and 200 kN/m2
total pressure. The partial pressure of benzeneis 6 kN/m2 at the
left side of the layer and 1kN/m2 at the right side. The mass
diffusivity atthis temperature and pressure is 4.4x10-6 m2/s.Also
express the flux in English units.
9.2 Foust
Air
C6H6
T=25oC
PT=200kPa
PA1=6kPaPA2=1kPa
1 2
Solution:
A=C6H6
D= 4.4x10-6 m2/s
NA
A= -D
dCA
dxMass Transport Equation:
For gases: Express CA = f(PA)
Assume ideal gas: RTnVP AA
RT
P
V
n AA
RT
P
C
A
A
RT
dPdC
AA In Mass Transport eqn
10mm
Solution: dx
dCD
A
N AA
By variable separable:
dx
dP
RT
D
A
N AA
2A
1A
2
1
P
PA
x
x
AdP
RT
Ddx
A
N
AA P
RT
Ddx
A
N
1A2A12A PP
RT
Dxx
A
N
12
21
xx
PP
RTD
AN AAA
Solution:
Substituting given values:
12
21
xx
PP
RT
D
A
N AAA
0010
16
2983148
10446
..
x.
A
NA
m
kPa
KKkmole
mkPa
s
m
A
NA3
2
sm
kmolx.
A
NA
2
710888
sm
kmol10x888
A
N
2
7A
.
kmol1
gmol1000x
hrft
lbmol10x546
A
N
2
4A
.
In English units:
hr1
s3600x
2
ft1
m30480x
.gmol454
lbmol1x
Solution:
Get D at 200kPa:
DD
200kPa
gg
,., C2523
kPa325101C2523
oo T
P
T
P
x200kPakPa200C25g
6oD32510110x88
,..
Case II: IfD is not given
Get D from App. D-11(a) Foust
D (at 25oC and 1atm or 101.325kPa) = 0.088 cm2/s
= 8.8x10-6 m2/s
D (at 25oC and 200kPa) = 4.45 x 10 -6
Then substitute to the Mass Transport Equation
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Solution:
Get D at 200kPa:
DD
200kPa
gg
,KkPa.,KT
P
T
P
298
23
325101273
23
x200KPa
23
200298
23
6
298273
3251011077/
kPa,Kg
/
D.x.
Case II: IfD is not given
Get D from table 2-324 Perrys
D (at 0oC and 1atm or 101.325kPa) = 0.077 cm2/s
= 7.7x10-6 m2/s
D (at 25oC and 200kPa) = 4.45 x 10 -6
Then substitute to the Mass Transport Equation
Air
C6H6
T=25oC
PT=200kPa
PA1=6kPaPA2=1kPa
1 2
Solution:A=C6H6
D= 4.4x10-6 m2/s
Case III: Find PA at 5mm
3x=5mmPA3=?
13
31
xx
PP
RT
D
A
N AAA
3A
1A
3
1
P
PA
x
x
A dPRT
Ddx
A
N
0050
6
2983148
104410888 3
67
.
P
.
x.x.
A
PA3 = 3.5kPa
Another Solution:
tconsx
PA tan
0050
6P
010
613
A
..
PA3 = 3.5kPa
Mass Transport
Calculate the mass flux of methanol in waterat 77 (D=4.95x10-5
ft2/hr). The concentrationsof methanol at 2 points 0.1 in. apart
are 0.05and 0.10 lb-mole/ft3.
9.3 Foust
Methanol in water
T=77oF
CA1=0.1 lbmol/ ft3
CA2=0.05 lbmol/ ft3
1
2Solution:
A=Methanol
(D= 4.95x10-5 ft2/hr)
dx
dCD
A
N AA Mass Transport Equation:
0.1 in
12
21
xx
CCD
A
N AAA
fthr
mollb0002970
A
N2
A
.
1210
05010010954
5
.
..x.
A
NA
Solution:
Get D at 25oC or 77oF
Case II: IfD is not given
Get D from App. D-11(b) Foust
D (at 20oC or 68oF) = 1.28x10-5 cm2/s
= 1.28x10-9
m2
/s
C20
L
C25
L
oo T
D
T
D
(at 25oC) = 0.97 cP
(at 20oC) = 1.08 cP
Get viscosities of the water at 25 oC and 20oC
use 1. Table 2-318, Perrys
2. nomograph (Fig.2-32) and
For water, X=10.2 Y=13
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10.2
13
1.08 cP
0.97 cP
To continue...
D of methanol in water at 20oC
D (at 20oC )= 1.44 x 10-9
D = 1.44 x 10-5 cm2/s
Or: use table 2-325 instead: D(at 25oC or 77oF) = 1.6 x
10-5cm2/s
Methanol in water
T=77oF
5% methanol
2.5% methanol
1
2
Case III: If data is in wt.%
A=Methanol
(D= 4.95x10-5 ft2/hr)
lbmol10x5625132
050 3 ..
Basis: 1 lb solution
10mm
Pt. 1: 0.05 lb methanol
Pt. 2: 0.025 lb methanol
At pt. 1: = lbmole methanol =
3
ln
0.05
lbmol32 0.097341
62.3
So
nV ft
CA1 =
Assume density of water=density of solution=62.3lb/ft3
ln +
methanol water So
methanol water
wt wt
V Density Density
Determine from Table 2-30:
nsol
nsolOHnsol
mVV
2
'
'
'
lbmol8125732
0250 4 ..
At pt. 2: = lbmole methanol =
ft
lbmol048670
3621
32
0250
V
n
30H2
.
.
.
CA2 =
for water = 997.042 kg/m3 or 62.3 lb/ft3
Then use(for very dilute solutions):
12
21
xx
CCD
A
N AAA
Substitute values in:
8th ed: Table 1-7(Conversion): for water = 997.042 kg/m3 or
62.24lb/ft3
