Chemical Engineering Chapter 4 Cdpi Cdpi Cdpi Cdpi Cdpi Cdpj Know For Conversion Maximum Should Minimum

subject Type Homework Help
subject Pages 9
subject Words 1342
subject Authors H. Scott Fogler

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CDP4-I (a)
CDP4-I (b)
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CDP4-I (c)
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CDP4-I (e)
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CDP4-J (a)
b) We know, ε = 0, so
2/1
0)1( WPP
!
"=
=>
2/1
)1(201 W
!
"=
=> α = 9.98 x 10-4
c) For conversion to be maximum, α should be minimum
Also, we know that α is proportional to
26
21
G=
for turbulent flow
ppipepc DDDA
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Hence for minimum α, Dp and Dpipe should be increased. d) Yes, we can
Now,
1
2
1
6
1
2
2
6
2
2
1==
ppipe
ppipe
DD
DD
!
!
6
4
2
1=
pipe
pipe
D
D
and Dp1= 0.5 cm
=>Dp2 = 0.044 cm
f) We’re given,
2
1
12
1
~
p
p
pD
D
kk
D
k=!
Hence
4.11
5.0
2==
k
g) Rate law:
Stoichiometry:
yXCC AA )1(
0!=
Mole balance:
0A
A
F
r
dW
dX !
=
Combining:
2/1
2
2
02 )1)(1(
A
WXCk
dX
!
""
=
(for ε = 0)
CDP4-K
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b) From polymath: Maximum flow rate = 1750 mol/s
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CDP4-L
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CDP4-M No solution
CDP4-N
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CDP4-O No solution
CDP4-P
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CDP4-Q
CDP4-R
CDP4-S
CDP4-T
CDP4-U
CDP4-V

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