(1) WWWR 24.1
Liquified natural gas, LNG, is to be shipped from the Alaskan Kenai Peninsula by an
ocean carrier to processing plant on Yaquina Bay, Oregon. The molecular composition
ofg commercial LNG is
Methane, CH4
Ethane, C2H6
Propane, C3H8
Carbon Dioxide, CO2
93.5 mol %
4.6 mol %
1.2 mol %
0.7 mol %
Determine
a.
b.
c.
d.
e.
the weight fraction of ethane;
the average molecular weight of the LNG mixture;
the density of the gas mixture when heated to 270 K and at 1.4 x 10 5 Pa;
the partial pressure of methane when the total pressure is 1.4 x 105 Pa;
the mass fraction of carbon dioxide in parts per million by weight.
(2) WWWR 24.12
Highly purified tetrachlorosilane (SiCl4) gas is reacted with hydrogen gas (H2) to produce
electronic-grade polycrystalline silicon at 800
and 1.5 x 105 Pa according to the
equation
SiCl4 (g) + 2 H2 (g)  Si (s) + 4 HCl (g)
There are concerns that the reaction experiences diffusional limitations at the growing Si
solid surface. Estimate the molecular diffusion coefficient for (a) SiCl4 in H2 and (b)
SiCl4 in a gas phase mixture containing 40 mol % SiCl4, 40 mol % H2, and 20 mol %
HCl. The Lennard-Jones parameters for SiCl4 (species A) are
.
(3) WWWR 24.13
An absorption tower has been proposed to remove selectively two pollutants, hydrogen
sulphide (H2S) and sulphur dioxide (SO2), from an exhaust gas stream containing
H2S
SO2
N2
3 vol %
5 vol %
92 vol %
Estimate the diffusitivity of hydrogen sulphide in the gas mixture at 350 K and 1.013 x
105 Pa. The critical temperature (TC) of H2S is 373.2 K and the critical volume (Vc) of
H2S is 98.5 cm3/mol.
(4) WWWR 24.15 (d)
Estimate the liquid diffusivity of the following solutes that are transferred through dilute
solutions:
a.
b.
c.
d.
oxygen in ethanol at 293 K;
methanol in water at 283 K;
water in methanol at 288 K;
n-butanol in water at 288 K.
Compare this value with experimental value reported in Appendix J.2. (n-butanol in
water at 288 K = 0.77 x 10-5 cm2/s)
(5) WWWR 24.22
Researchers are proposing the development of a “nano-channel reactor” for steam
reforming of methane (CH4) to fuel-cell hydrogen gas to power microscale devices.
As each channel diameter is so small, the gas flow is likely to be very small within a
given channel. Hence, gas diffusion processes may play a role in the operation of this
device, particularly during the mixing and heating steps. We are specifically interested in
evaluating the effective diffusion coefficient of methane gas (species A, MA = 16
g/g.mol) in water vapor (species B, MB = 18g/g.mol) at 300 and 0.5 atm total system
pressure. The diameter of the channel is 200 nm (1 x 109 nm = 1.0 m). A feed gas
containing 20 mol % CH4 in water vapor is fed to the nanochannel with a flux ratio
NA/NB = 0.25. What is the effective diffusion coefficient of CH4 in the nanochannel at the
feed gas conditions? Is the Knudsen diffusion important?
(1) Problem 24.1 (WWWR)
Assume 1 mole,
CH 4
C2 H 6
C3H8
CO2
mole fraction
0.935
0.046
0.012
0.007
mole
0.935
0.046
0.012
0.007
MW
16
30
44
44
Wt(g)
14.96
1.38
0.528
0.308
total = 17.176
1.38 g
 0.0803
17.176 g
total wt
(b) average molecular wt of mixture =
 17.176 g / mole
1 mole
PCH4  YCH4 P=(0.935)(2.0  105 Pa)
(c) partial pressure of methane
=1.87  105 Pa
(a) Wt fraction ethane C2 H6 
0.308 g
x ppm

17.176 g 1 000 000 ppm
x  17932ppm
(d) composition of CO2
(2) Problem 24.12 (WWWR)
From Appendix K.2
(a)
Si Cl4
MW
169.89
/K
358
H2
2.02
33.3


5.08 A

2.968 A
T 3/ 2  (800  273)3/ 2  35148
P  1.0 atm
1/ 2
 1
1 
1 
 1



 
  0.7085
M
M
169.89
2.02


 A
B

   B 5.08  2.968
 AB  A

 4.024 A
2
2
 AB
K
TK
 AB

1/ 2
A B
 358(33.3)  109.19
K K
1073

 9.83
109.19
 from Appendix K1,  D  0.74
Using Hirschfelder’s Eq.
0.001858(35148)(0.7085)
DSiCl4 -H2 
 3.837cm2s-1
2
1(4.624) (0.744)
(b) First, calculate DSiCl4 HCl
T 3/ 2  35148
1/ 2
 1
1 
1 
 1



 

 169.89 36.46 
 MA MB 

5.08  3.305
 AB 
 4.193 A
2
 AB
K
TK
 AB
1/ 2
 0.182
 358(360)  359
 2.99   D  0.9586
 DSiCl4 HCl =
Gas mixture
0.001858(35148)(0.182)
 0.705cm2s-1
1(4.193)2 (0.9586)
YSiCl4 = 0.40
Y'H2
YH2
= 0.40
Y'HCl
YHCl
= 0.20
DSiCl4  mixture
=
0.40
 0.667
0.40  0.20
0.20
=
 0.333
0.40  0.20
=
1
Y'H2
DSiCl4 -H2


Y'HCl
DSiCl4 -HCl
1
0.667 0.333

3.837 0.705
 1.547cm2s-1
(3) Problem 24.13 (WWWR)
H 2 S in gas mixture of SO2 and N2 at 350K, 1.0atm
Assume non-polar
(i) H 2 S -N2
 H2S  0.841(Vc )1/ 3  3.88
H
 0.77Tc  287.4
2S
K
T
3
2
 (350)
3
H2S
 6548
2
1
 1
1  2  1
1 

    

 34 28 
 MA MB 
P=1.0
 B
 AB  A
 3.78
2
 AB
K
 A B

TK
 AB
1
 DH 2 S  N2 
(ii) H 2 S - SO2

 /K
H2S
MW
34.0
3.88
287.4
SO2
64.0
4.29
2520
1
2
 0.212
 AB  4.085
 AB
3.88
3.681
287.4
91.5
Appendix K .1)
1
1 12

)
MA MB
 0.208cm2 s 1
2
P AB
0
3
K
TK
 /K
 0.255
 2.158  0  1.048( from
 1
1 



 MA MB 

 162.2
K K
0.001858T 2 (
 AB
2
N2
Mw
34
28
 269.1
 1.30  0  1.273

DH 2 S  SO 2 =0.121 cm 2 s 1
(iii) Mixture:
H 2 S 3% , SO2 5% , N 2 92% by volume
0.05
y ' SO2 
 0.0515
1  0.03
0.92
y ' N2 
 0.9485
1  0.03
1
DH 2 S mix 
 0.200cm 2 s 1
'
'
y SO2
y N2

DH 2 S  SO2 DH 2 S  N 2
(4) Problem 24.15(d) (WWWR)
Wilke-Chang:
7.48 108 ( B M B )1/ 2 T
DABL 
VA0.6
B
 B  1.14cp
T  288K
1 cp =0.001Pa-s
B  2.26, M B  18
(B M B )1/ 2  6.38
VC4 H9OH = 4VC +10VH  1V0  4(14.8)  10(3.7)  7.4
 VA0.6
 103.6
 16.19
 DABL 
7.4  108 (6.38)  288 
6
2 -1

  7.37  10 cm s
16.19
 1.14 
Hayduk-Laudie:
DABL  13.26 105  B 1.14VA 0.589
 B 1.14  (1.14)1.14  0.861
VA 0.589  (103.6)0.589  0.665
 DABL  13.26 105 (0.861)(0.065)  7.42 106 cm 2s -1
Compared to Table J.2 of 7.7  106 cm2 s-1
1 poise = 1 g/cm-s =100cp
1 cp = 0.01 poise
Pa = N/m2 = m/s2 x kg/m2
=kg/m-s2
Pa.s = kg/m-s
(5) Problem 24.22 (WWWR)
A = CH4 (20 mole %); B = H20 (80 mole %)
Use Hirschfelder Equation (24-33)
1/ 2
 1
1 


  0.3436
M
M
 A
B 
2
 AB  10.468
 AB
 220.4

1 N.m = 107 erg
T
 2.60   D  0.9878
 AB
 DAB  1.694cm 2s -1
erg/K
T
1.38 1016 107 573


 2.405 107 m
2
10 2
5
2 A P
2 (3.822 10 ) (0.5*1.01325 10 )
Kn 

d pore

2.405 107 m
 1.2  1
200 109 m
Since K n  1, Knudsen diffusion could be the dominant factor.
DK,A  4850(200 10-7 cm) 573 /16  0.58cm 2s -1
1   yA
N
1
1


;   1  B  5; y A  0.2.
DA , E
DA , B
DK , A
NA
 DA, E  0.58cm 2s-1
Knudson diffusion is important.