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Lecture 10

PROCTECH 2EC3 Lecture Notes - Lecture 10: Bulk Density, Partial Pressure, Nitrogen Dioxide

6 pages99 views2018

Department
Process Automation Technology
Course Code
PROCTECH 2EC3
Professor
Kostas Apostolou
Lecture
10

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Week 6: Feb. 6th, 2018
6.90 Benzene and Hexane are being considered as solvents to extract acetic acid from aqueous mixtures. At
!"#
, distribution coefficients for the two solvents are
$%& "'"()
mass fraction acetic acid in
*
Benzene/mass fraction acetic acid in water and
$+& "'",-
mass fraction acetic acid in Hexane/mass
fraction acetic acid in water.
a. Based on the distribution coefficients only, which of the two solvents would you use and why?
Demonstrate the logic of your decision by comparing the quantities of the two solvents required to
reduce the acetic acid content in 100 kg of an aqueous solution from
!"*./0
to
,"*./0
.
Total:
12314& 15316******************************************************************************7 ,"" & 15316814
Acetic Acid: 9
"'!
:9
,""
:
3
9
"
:
14&
9
"',
:
153;<616****************************7 !" &
9
"',
:
153;<616
Water: 9
"'-
:9
,""
:
3
9
"
:
14&
9
"'(
:
153
9
"
:
16***************************************7 -" &
9
"'(
:
15
Benzene: 9
"
:9
,""
:
3
9
,
:
14&
9
"
:
153
=
,8;<6
>
16*****************************7 14&
=
,8;<6
>
16
Distribution Coefficient:
?@AA*BC@DEFGH*GB*<DIEFD*<DFJ*FH*%IHKIHI
?@AA*BC@DEFGH*GB*<DIEFD*<DFJ*FH*L@EIC & "'"()**** 7 MNO
P'2 & "'"() QRST& U'UUVW
XY&ZZ'ZW*[\ 7!" &
9
"',
:9
--'-)
:
3;<616Q 16&!"8"',15
;<6&!"8
9
"',
:9
--'-)
:
"'""() ]^^_Z'`*[\
Total:
14&--'-)3aab-'c8,"" ]^^T`'^W*[\
If Hexane is used instead of Benzene:
- Mass balance would be the same, but the distribution coefficient is different.
Distribution Coefficient:
MNO
P'2 & "'",- QRST& U'UUdZ*
15&--'-)*ef 7!" &
9
"',
:9
--'-)
:
3;<616Q 16&!"8"',15
;<6&!"89"',:9--'-):
"'"",- ]dYUZ^*[\
Total:
14&--'-)8,!"-a8,"" ]dYU`U*[\
Therefore, use Benzene because its less equipment.
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b. What other factors may be important in choosing between Benzene and Hexane?
6.99 A 50.0 L tank contains an air – carbon tetrachloride gas mixture at an absolute pressure of
,*g/1
, a
temperature of
!h#
, and a relative saturation of 30%. Activated carbon is added to the tank to remove the
iij6
from the gas by adsorption and the tank is then sealed. The volume of added activated carbon may be
*
assumed negligible in comparison to the tank volume.
a. Calculate
kllmO
at the moment the tank is sealed, assuming ideal gas behaviour and neglecting
adsorption that occurs prior to sealing.
nllmO*9ngo/pgj*noqrrsoq:
at time “zero” (start)
nllmO& tln
b. Calculate the total pressure in the tank and the partial pressure of carbon tetrachloride at a point
when held of the
iij6
initially in the tank has been adsorbed.
Start:
n & ,*g/1
,
nllmO&c"'h*11uf
vEGE@m &wx
yz &
9
2*@E?
:9
{P*|
:
}
P'P~4P€•‚ƒ„
„…†‡•ˆ
9
5PŠ*
:
]d'VW*•Ž
vllm6&w••Ox
yz &
9
{P'6*??+•
:9
{P*|
:
}
•4'5•€•„„‘’
„…†‡•ˆ
9
5PŠ*
:
]U'dY*XŒ•Ž
Now: Half
iij6
gone,
nllmO&*“
vllmO&
,
a
9
"',!
:
]U'U_`*•Ž
v@FC &d'W`*•Ž
nEGE@m &vEGE@m–—
˜&
9
,'(,c*1™jq
:}
ba'!b*š11uf
1™jq$
9
!"-*$
:
*
c"]ZYY*XX›\
nllmO&vllmOyz
˜&
9
"'"bc*1™jq
:}
ba'!b*š11uf
1™jq $
9
!"-*$
:
c"]ah'(*11uf
Relative Saturation: !"0 7w••O
w••O
œ& "'!" QnllmO&9"'!:nllmO
œ
nllmO
œ&,"•'~Šž4•Ÿ2424'P42
56 44•'6Pž¡]d_W*XX›\
So, nllmO&9"'!:9,b):]`U'TXX›\
¢*v@FC & ,'()8"',! ]d'W`*•Ž
¢*vEGE@m & ,')c3"'"bc ]d'Vd`*XŒ•Ž
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