Transcript Lec13_non

Lecture 13
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Lecture 13 – Tuesday 2/22/2011
A +2B  C
A + 3C  D
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Gas Phase
Multiple Reactions
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1. Number Every Reaction
2. Mole Balance on every species
3. Rates:
(a) Net Rates
of Reaction for every species
N
rA   riA
i 1
(b) Rate Laws for every reaction
r1 A   k1 A C A C B2
r2C   k 2 C C A2 C C3
(c) Relative Rates of Reaction for every reaction
For a given reaction i: (i) aiA+biB ciC+diD:
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riA
riB riC riD



 ai  bi ci
di
Reactor Type
Gas Phase
dN A
 rAV
dt
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Liquid Phase
dC A
 rA
dt
dN A
 rAV
dt
 0C A
dC A
 rA 
dt
V
dN B
 rBV  FB 0
dt
0 C B 0  C B 
dC B
 rB 
dt
V
Reactor Type
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Gas Phase
Liquid Phase
C A0  C A 
FA0  FA
V
 rA
V  0
dF A
 rA
dV
dC A
0
 rA
dV
dFA
 rA
dW
dC A
0
 rA
dW
 rA
Note: The reaction rates in the above mole balances are net rates.
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NB
CB 
V
NT P0 T0
V  V0
NT 0 P T
FB
CB 

FT P0 T0
  0
FT 0 P T
N B NT 0 P T0
CB 
NT V0 P0 T
FB FT 0 P T0
CB 
FT 0 P0 T
N B P T0
CB  CT 0
NT P0 T
FB P T0
CB  CT 0
FT P0 T
:
 FA   T0 
C A  CT 0   y 
 FT   T 
FT  FA  FB  FC  FD
Note: We could use the gas phase mole balances for liquids and
then just express the concentration as:
FA
CA 
0
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NA
CA 
V0
The complex liquid phase reactions follow elementary rate
laws
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
r

k
C
C
A  2B  C (1)
1A
1A A B
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
 r2C  k2CCC3 CA2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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(1) A  2B  C
(2) A  3C  D
dFA

(1)
 rA
dV
dFC
(3)
 rC
dV
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dFB
( 2)
 rB
dV
dFD
( 4)
 rD
dV
(5) rA  r1A  r2 A
(6) rC  r1C  r2C
(7) rB  r1B  r2 B
(8) rD  0  r2 D
(9) r1 A   k1 AC AC B2
(10) r2C   k 2C C A2 CC3
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Reaction 1 r
r1B r1C
1A


1  2 1
(11) r1B  2r1 A
(12) r1C  r1 A
Reaction 2
r2 A r2C r2 D


2 3 1
2
(13) r2 A  r2C
3
r2C
(14) r2 D  
3
2
rA  k1 AC AC  k 2C C A2CC3
3
rB  2k1 AC ACB2
2
B
rC  k1 AC ACB  k 2C C A2CC3
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k 2C 2 3
rD 
C ACC
3
(15) C A  FA 0
(16) C B  FB 0
(17) CC  FC 0
(18) C D  FB 0
 FC 
~
(19) SC D  if V  0.00001 then  else 0
 FD 
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FT  Liquid – Not Needed
(19)   Liquid – Not Needed
(20) CT 0  Liquid – Not Needed
( 21) k1 A  10
( 22) k 2C  20
( 23)   Liquid
( 24) CT 0  Liquid
( 25) V f  2500
( 26) FA0  200
( 28) FB 0  200
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( 26)  0  100
Same reactions, rate laws, and rate constants as example A
A  2B  C (1)
 r1A  k1ACACB2
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
 r2C  k2CCC3 CA2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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The complex liquid phase reactions take place in a 2,500 dm3
CSTR. The feed is equal molar in A and B with FA0=200
mol/min, the volumetric flow rate is 100 dm3/min and the
reation volume is 50 dm3.
Find the concentrations of A, B, C and D existing in the reactor
along with the existing selectivity.
Plot FA, FB, FC, FD and SC/D as a function of V
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(1) A + 2B →C
(2) 2A + 3C → D
r1 A   k1 AC AC B2
r2 C   k 2 C C A2 CC3
(1) A
(2) B
(3) C
(4) D
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0C A0  0C A  rAV  0
0CB 0  0CB  rBV  0
0  0CC  rCV  0
0  0CD  rDV  0
(5)-(14)
(15)-(18)
(19) SC / D
FC
0CC


FD  0.0001 0CD  0.0001
k1A , k2C , CA0 , CB0 , V , 0
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(1) A + 2B →C
(2) 2A + 3C → D
r1 A   k1 AC AC B2
r2 C   k 2 C C A2 CC3
(1) f FA   FA0  FA  rAV (=0)
(2) f FB   FB0  FB  rBV (=0)
(3) f FC   0  FC  rCV (=0)
(4) f FD   0  FD  rDV (=0)
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(15) C A  FA  0
(16) C B  FB  0
(17) CC  FC  0
(18) C D  FD  0
(19)
SC D 
FC
FD  0.00001
(1) A + 2B →C
(2) 2A + 3C → D
r1 A   k1 AC AC B2
r2 C   k 2 C C A2 CC3
(1) f CA   0CA0 0CA  rAV
(=0)
(2) f CB   0CB0 0CB  rBV
(=0)
(3) f CC   0 0CC  rCV
(=0)
(4) f CD   0 0CD  rDV
(=0)
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(15) SC D 
FC
FD  0.00001
Same reactions, rate laws, and rate constants as example A
A  2B  C (1)
 r1A  k1ACACB2
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
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 r2C  k2CCC3 CA2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
(1)
(2)
dFA
 rA
dV
dFB
 rB
dV
dFC
(3)
 rC  RC
dV
dFD
( 4)
 rD
dV
Same as CSTR (5)-(14)
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Gas : Isothermal T = T0
FA
(15) C A  CT 0
y (16) C B  CT 0
FT
FC
(17) CC  CT 0
y (18) C D  CT 0
FT
(19) FT  FA  FB  FC  FD
dy
  FT  T 
 FT




 




dW
2 y  FT 0  T0 
2 y FT 0
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FB
y
FT
FD
y
FT
C A  CT 0
FA
y 15
FT
CC  CT 0
FC
y 17
FT
CB  CT 0
FB
y 16
FT
CD  CT 0
FD
y 18
FT
FT  FA  FB  FC  FD 19
 FC 
FC
S
 if V  0.00001 then  else 0 20
FD
 FD 
y  1 21
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Same reactions, rate laws, and rate constants as example A
A  2B  C (1)
 r1A  k1ACACB2
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
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 r2C  k2CCC3 CA2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
Because the smallest molecule, and the one with the lowest
molecular weight, is the one diffusing out, we will neglect the
changes in the mass flow rate down the reactor and will take as
0  m

first approximation: m
dFC
dFA
A
 rA 1 C
 rC  RC 3
dV
dV
dFB
dFD
B
 rB 2  D
 rD 4 
dV
dV
We also need to account for the molar rate of desired product
C leaving in the sweep gas FCsg
dFCsg
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dV
 RC
We need to reconsider our pressure drop equation. When mass
diffuses out of a membrane reactor there will be a decrease in
the superficial mass flow rate, G. To account for this decrease
when calculating our pressure drop parameter, we will take the
ratio of the superficial mass velocity at any point in the reactor
to the superficial mass velocity at the entrance to the reactor.
 F  MW 
G
i
i


  0  0
G0

 Fi0  MWi 

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The superficial mass flow rates can be obtained by multiplying
the species molar flow rates, Fi, by their respective molecular
weights, Mwi, and then summing over all species:
G m AC1  Fi  MWi  AC1  Fi MWi 



G0 m0 AC1  Fi0  MWi  AC1  Fi0 MWi 

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Same (5)-(14)
dy
 FT

dW
2 y FT 0
Same (15)-(20)
dy
 FT
;

dV
2 y FT 0
21
RC  kC CC  CCSweep 
FCsg
V
dFCsg
dV
 FCsg
V  V
 RC V  0
 RC
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Large sweep gas velocity
CC  CCsg ,
then
CCsg  0
RC  kCC CC
Moderate to small sweep gas velocity

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
RC  kCC CC  CCsg
CCsg 

FCsg
 sg
 F0 sg  FCsg 

 sg   sg 0 

F
0
sg


Vary υsg to see changes in profiles
Same reactions, rate laws, and rate constants as example A
A  2B  C (1)
 r1A  k1ACAC
2
B
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
 r2C  k2CCC3 CA2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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The complex liquid phase reactions take place in a semibatch
reactor where A is fed to B with FA0=3 mol/min. The
volumetric flow rate is 10 dm3/min and the initial reactor
volume is 1,000 dm3.
The maximum volume is 2,000 dm3 and CA0=0.3 mol/dm3
and CB0=0.2 mol/dm3. Plot CA, CB, CC, CD and SS/D as a
function of time.
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(1) A + 2B →C
dN A
dt
dN B
dt
dN C
dt
dN D
dt
(2) 2A + 3C → D
FA0
B
 rAV  FA0
N A0  0
 rBV
N B0  CB0V0  2.000
 rCV
NC 0  0
 rDV
N D0  0
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Same (5)-(14)
Net Rates, Rate Laws and relative rates – are the same as Liquid
and Gas Phase PFR and Liquid Phase CSTR
V  V0  v0t 15
NA
CA 
V
NC
CC 
V
SC / D
16
18
NB
17
CB 
V
ND
19
CD 
V
 NC 
 else (0) 20
 if (t  0.0001) then
 ND 
0  10dm3 min V0  100dm3
FA0  3 mol min
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