kinetics & reactor design i
TRANSCRIPT
1
UNIVERSITY OF GEZIRA
COLLEGE OF ENGINEERING &
TECHNOLOGY
Department Applied Chemistry &
Chemical Technology
Student's Lecture Notes
IN
CHEMICAL REACTOR DESIGN
Prepared by
Dr. Babiker Karama Abdalla
Associate Professor of Chemical Engineering
E-Mail Address: [email protected]
2010-2011
2
Out Lines
• Chemical Kinetic
• Chemical Reaction Engineering
(CRE)
• Mole Balances
• Rate of Reaction.
• Mole balance Equation
• Types of Reactors
• Conversion and Reactor Sizing
• Rate laws & Stoichiometry
• Stoichiometric Tables
• Isothermal Reactor Design
• Membrane Reactors
• Semi-Batch & Reactive Distillation
Reactors
• Collection & Analysis of Rate Data
• Multiple Reactions
• Steady-State Nonisothermal Reactor
Design
3
1.1Chemical Kinetic:
Chemical kinetic is concerned with the rates of chemical
reaction. Chemists use kinetics to understand fundamentals
and speeds of the reaction pathways to design new or better
ways to achieve desired product:
Chemical Engineers use kinetics to design a reactor for
specific reaction or reactions.
The objective of chemical kinetics is to enable us to
produce the rate of the reaction.
1.2 Chemical Reaction Engineering (CRE):
- CRE is concerned with rational design or analysis of
performance of chemical reactors.
- Chemical reactors are devises in which change in
composition of matter occurs by chemical reaction.
- CRE is to enable us to predict by rational design the
performance of a reactor.
1.3 Kinetics of CRE:
Three levels of system size can be considered for
comparing the kinetic of the CRE.
1. Microscopic or Molecular: Where
molecules react and can be characterized by (C,
T, P or ρ)
2. Local macroscopic: Where large amounts
reactants with gradient in Ci or T with P.
4
3. Global macroscopic: Like a bed of solid
reacting with a flow fluid where gradient of Ci
and T on the bed beside the local gradient.
1.4- Mole Balances:
1.4.1 Rate of Reaction.
1.4.2 Mole balance Equation.
1.4.3 Batch Reactors.
1.4.4 Continuous flow reactors.
1- CSTR
2- FR
3- PBR
Level 2
Level 1
Level 3
5
- General Mole Balance:
In - out + generation = accumulation
FAW – FA (W+ΔW) + r\AΔW = 0
A
A
F
FA
AA
A
rWWr
dw
dF
molemass
mass
moleWr
0
\
\
A\ dF &
time
A catalyst of
catalyst of time
A )(
Example 1:
Gas phase reaction
(1) At constant value
A
AAA rdt
dC
dt
VNd
dt
dN
V
/1
(2) At constant pressure i.e. NA=CAV
dt
V ln
11
dC
dt
dCr
rdt
dV
V
C
dt
dC
dt
VCd
Vdt
dN
V
AAA
AAAAA
Example 2:
1st order reaction AB in PFR v is constant. Derive the
equation relating V, CA, k and v .find the value of reactor to
reduce exiting concentration CA to 10% of entering CA0 where
min
10 3dmv
And k = 0.23 min-1
m.b. of PFR A
A
rdV
dF
1st order reaction -rA = kCA
6
v is constant there v0 = v
A
A
V
V
C
CA
A
A
AA
A
AAA
C
CLn
k
vV
dvC
dC
k
dVC
dC
k
vkC
dN
dCv
rAdN
dCv
dv
vCd
dV
dF
o
A
A
0
0
0
o
0
0
0
0
v-
m 0.1 , L 100 i.e 10010ln23.0
10
1.0ln
min 23.0
10 3
3
0
0min*3 dmdm
C
Cdm
VA
A
2. Conversion and Reactor Sizing:
- Conversion (X):
XAX
fedA of moles
reactedA of molesX
Da
dC
a
c B
a
b A
dDcCbBaA
A
2.1 Design Equations:
][
reactedA of 0at treaactor tofed
initiallyA of
at treactor in the
A of
.(consumed) reactedA of
fedA of moles
reactorA of *
fed
A of
consumed
A of
00
0
XNNNN
molesmolesmoles
XNmoles
molesmolesmoles
AAA
A
7
Moles of A in the reactor after X conversion
AB
A
AAAA
kCrA
VrAdt
Da
dC
a
cB
a
bA
Vrdt
XNXNNNo
A
A
dN-
thencomponent key theisA ,
dN
systemboth in the balance mole
100
In both systems we are looking for (t) then
AAAA
A
A
AA
AAA
dt
dC
dt
VNd
dt
dN
V
Vrdt
dX
Vrdt
dXcombing
dt
dXN
dt
dN
XNNN
/(1
][N
reactor batch ofequation design
N-
0
0
00
A
A
The design equation can be expressed as
X(t)
0
)(
0
A
0
dX int
Ndtor
A
A
tX
A
AA
AA
r
dXC
Vregratig
Vr
dX
r
dXNVdt
8
-Flow systems:
0
3
03A
A
0
0
00
00
0000
*C
](F ],[][][
system theleaving
A of flowmolar
consumedA
of rate feed fedA rate feedolar
time
reactedA of moles.
fedA of moles
reactedA of .
time
fedA of moles.
vCF
s
dmv
cm
moles
s
molF
XFFFXFF
molarm
XF
molesXF
AA
A
AAAAA
A
A
for gas systems CA0 can be calculated form ideal gas equation
0
00
0
0
0 RT
py
RT
pC
A
A
A
A
-CSTR:-
assuming
exit
XF
then
VXF
then
XFFF
VFF
Da
dC
a
cB
a
bA
A
A
AA
AAA
AAA
0
0
00
0
CSTRV
is CSTRfor equation design
9
-PFR:-
Using the mole balance equation of the PFR and multiply by (-
1) then
X
A
APFR
AAA
AAAA
AAAA
AA
r
dXFV
rdV
dXdXFdF
dXFXdFdFdF
XFFFF
rdV
dF
0
A
0
00
000
000
F then
0
-PBR:
using the same analyses as for PFR now
x
A
APFR
AA
r
dXFV
rdW
dXF
A
0 \
\
0
Use of the design equation:
the rate of reactant A, -rA is function of the concentration, CA
which is function of the conversion CA (1-x)
xkCr
xkCr
AA
AA
1
111
)1(
0
0
10
mol
dm
A
31
Reactor sizing:
- Consider the gas phase decomposion reaction
A→ B + C
The reaction was 422.2 k press at 10 atm initially A was
charged with equal molar ratio with an inert
1/-rA→ ∞ as X → 1
if A B + C
1/ - rA → ∞ as X → Xe
sizing a CSTR
A
A
r
XFV
0
X 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85
sdm
molrA
.3
0.0053 0.0052 0.005 0.0045 0.004 0.0033 0.0025 0.0018 0.00125 0.001
mol
dm
rA
31
189 192 200 222 250 303 400 556 800 1000
1000
300
600
400
200
0.2 0.4 0.6 0.8 1
conversion (x)
11
Sizing a PFR
X
A
APFRr
dxFV
00
Using five –point equation formula Δx=0.2
8.06.04.02.00
14241
30
AAAAxA
Arrrrr
xFV
Also using Simpson’s rules
800
600
400
200
0.2 0.4 0.6 0.8
800
600
400
200
0.2 0.4 0.6 0.8
12
comparing CSTR and PFR
600
400
200
0.2 0.4 0.6
PFR
1
VPFR dm3 250
X
13
Reactor in Series
FA1 = FA0 –FA0X1
FA2 = FA0 –FA0X2
FA3 = FA0 –FA0X3
Now Reactor value are given as
3
0
2
0
2
21
1
0
3
12
2
01
V
A
A
A
A
A
AA
X
A
r
dxF
r
xxF
r
FFV
rA
dxFV
-2CSTRs in series
V2 V3
V1
FA2
X2
F3
X3
FA1
X1
X=0
FA0
V2
V1
X2 = 0.8
X1=0.4
FA0
14
2
0
1
0
122
11 V ,
A
A
A
Ar
XXF
r
XFV
-2PFR’s:
V2
V1
X2 = 0.8
X1=0.4
800
600
400
200
0.2 0.4 0.6 0.8
0.2 0.4 0.6 0.8
800
600
400
200
15
Order of Sequencing reactors:
(A)
(B)
800
600
400
200
0.2 0.4 0.6 0.8
(A)
X1
V2
X2 = 0.8
FA0
X1
X2 = 0.8
FA0
16
-Relative Rate of reaction:
A+b/aB→ c/aC+d/aD
The rate of formation of C=c/a (rate of disappearance of A)
rC=c/a(-rA)= -c/a rA
From the Stoichiometry of the reaction
-rA/a = -rB/b = rC/c = rD/d
Space time: ( τ)
X
A
Ar
dxC
v
V
00
0
800
600
400
200
0.2 0.4 0.6 0.8
(B)
17
space velocity:
0
0
0
0
0
0
00
0
0
00
0
A
A
0
0
0
0A0
0
0
C x X
C 0X when
:
F,
:
1
)(
A
A
A
A
C
CA
A
C
CA
A
A
A
A
A
A
AA
A
AAAA
A
AA
A
x
A
r
dC
r
dCvV
C
dCdX
c
C
C
CC
vC
vCvC
F
FFX
vvwhen
F
FFX
Cvr
dxV
vvwhen
SV
V
vSV
τ
CA CA0
18
3. Rate laws & Stoichiometry:
- Homogeneous reaction
- Heterogeneous reaction
- Irreversible reaction
A+B → C +D
- Reversible reaction
A+B ↔ C+D
3.1 Reaction rate constant ( K)
-rA = [ KA(T)] [fn(CA , CB, …..)]
the Arrhenius equation
kA(T) = Ae-E/RT
A : pre exponential factor or frequency factor
E : activation energy J/mol , cal/mol
R : gas constant = 8.314 J/mol.k = 1.987 cal/mol.k
T : absolute temperature k
ln kA = lnA – E/R(1/T)
Straight line equation with slope (E/R)
Also can be determined by
12
1
2
121
2
2
2
1
1
11
ln
11ln
)1
(lnln
)1
(lnln
TT
K
KR
E
TTR
E
K
K
TR
EAk
TR
EAk
19
3.2 Reaction order & Rate Law
-rA= kACαA C ββ
n = α + β
Example : gas – phase reaction
2NO + O2 2NO2
-rNO = kNO C2NO CO2
CO + Cl2 COCl2
-rCO = k Cco CCl23/2
2
22
\2
222
1
22
O
ONON
Pt
CK
CkOrN
ONON
Rate constants correspond to :
Zero–order (-rA)= kA
{k} = mol /dm3.s
1st order
-rA= kACA
{k} = S-1
2nd order
-rA= kAC2A
{k} = dm3/mol .S
3rd order
-rA= kAC3A
{k}= (dm3/mol)2 1/s
20
nth order
-rA= kACnA
{k}= (dm3/mol)n-1 1/s
-3.3 Elementary rate laws & Molecularity
elementary rate law if the reaction order is identical with the
stoichiometric coefficient
H2+ I2 2HI
-rH2= kCH2CI2
Reversible Reactions :
aA + bB ↔ cC + dD
at equilibrium
CC
CCb
Bc
a
Ae
d
De
C
Ce
CK
units of Kc is (mol/dm3)d+c-b-a
2C6H6 C12H10+H2
2B D+ H2
6621012
2
B
2101266
2
k forward
2
HCHHC
r
HHCHC
B
B
k
BB
k
C
rB reverse = k-B CD CH2
rB≡ rB Net = rB forward + rB reverse
rB = - kB C 2B + k-BCDCH2
21
multiplying by (– 1)
C
C
CC
e
e
B
HD
C
C
B
B
C
HD
BBB
HD
B
B
BBHDBBBB
CCK
Kk
k
K
CCkr
CCk
kkCCkkr
2
2
22
2
2
22
-3.4 Non elementary rate laws & reactions
H2 + Br2 2HBr
Reaction rate law :
2
2/1
2
/
1
BrHBr
BrH
HBrCCk
CCkr
Also
CH3CHO CH4+CO
C CHOCHCHOCH kr2/3
33
Catalytic gas –phase decomposing of cumene
C6H5CH (CH3)2↔ C6H6 + C3H6 + C3H6
C B + P
Langmuir – Hinshellwood kinetics:
BBCC
PPBC
c PKPK
KPPPkr
1
/\
KP press equilibrium constant (atm,bar,kPa).
22
KC,KB adsorption constant (atm-1,bar-1,kP-1a).
K rate constant (mole cumene /kg cat.s.atm).
At equilibrium
C
CC
CC
CCC
C
PB
P
BCC
PPBC
P
PPK
KPK
KpPPr
1
/0
-3.5 Stiochiometric Tables:
aA + bB cC + dD
taking A as basis and assuming irreversible reaction
A + b/a B c/a C + d/a D
3.5.1 Batch Systems
t= t
NA
NB
NC
ND
NI
t=0
NA0
NB0
NC0
ND0
NI0 Batch Systems
NA =NA0 – NA0X
= NA0(1-X)
23
Species Initial mol Change mol Remaining mol
A NA0 -(NA0X) NA=NA0-NA0X
B NB0 -b/a (NA0X) NB =NB0-b/a(NA0X)
C NC0 c/a (NA0X) NC = NC0 +c/a(NA0X)
D ND0 d/a (NA0X) ND = ND0 + d/a(NA0X)
I NI0 --------- NI = NI0
Total NT0 NT = NT0+(d/a+c/a-b/a-1)NA0X
Moles of B reacted = (moles B reacted/moles A reacted).moles
A reacted = b/a(NA0X) = NB = NB0 – b/a NAoX
δ = d/a + c/a – b/a -1 increase or decrease of on total No of mole
NT0 = NT0 + δNA0X
Known CA = NA/V if V is a constant
Xa
dCC
Xa
cCC
V
Xa
bN
C
XCV
XNC
V
Xa
dN
C
V
Xa
cN
V
Xa
bN
C
N
N
N
N
NN
N
V
XNadN
V
N
V
XNacN
V
NC
V
XNabN
V
N
V
XN
V
NC
DAD
CAC
BA
B
A
A
A
DA
D
CABA
B
A
I
A
D
C
AA
B
B
ADDACCC
ABBAAA
0
0
0
0
0
0
00
0
0
0
0
0
0
0
0
0000
000
0
0
0
C
ID
C
D
B
11
VV valumeconsatant for
C ,
, , N
,
/C,
/
/C ,
1
24
3.5.2 Flow system
For flow system CA can be expressed as
CA = FA/v = [moles/time]/[dm3/time] = moles/dm3
v
Fa
dF
v
F
v
XFa
cF
v
FC
v
Fa
bF
v
FX
v
F
v
FC
XADD
ACC
C
ABBAA
A
0000
000
D
B
C ,
C , 1
IDC
0
0
, ,
0
0
0
0
0
0
0
0
similarly
y
y
C
C
vC
vC
F
F
A
B
A
B
A
B
A
B
B
A+b/a B→c/a C+d/aD
FA
FB
FC
FD
FI
FA0
FB0
FC0
FD0
FI0
entering
25
Species Initial molar
feed
Change
mol/t
Remaining mol
A FA0 -FA0X NA=FA0(1-X)
B FB0=θBFA0 -b/a FA0X NB =(θB-b/a*X)
C FC0= θCFA0 c/a FA0X FC= FA0(co + c/a*x)
D FD0= θDFA0 d/a FA0X FD = FA0 (θD+d/a*X)
I FI0= θIFA0 --------- FI = θIFA0
Total FT0 FT = FT0+(d/a+c/a-b/a-1)FA0X
FT=FT0+δFA0
For liquid without phase charge v = v0
CA=FA0/V0(1-X)=CA0(1-X)
CB=CA0(θB-b/c*X)
Volume charge with reaction
N2 + 3N2 ↔ 2NH3
For gas system using equation of state
PV = Z0NTRT
at t = 0
P0V0 = Z0NT0RT0
Dividing
V = V0(P0/P)T/T0(Z/Z0) NT/NT0
Volume V as function of X
NT = NT0+δNA0X
Dividing by NT0
NT/NT0= 1+ (NA0/NT0)δX = 1+ δyA0X
δ = yA0δ=(d/a+ c/a – b/a -1)NA0/NT0
V= V0 ( P0/P) T/T0(Z/Z0)(1+δ X)
26
Assuming no change on Z I.e. Z0 =Z
V = V0 (P0/P) (1+δ X) T/T0
Also for variable volume
CT = FT/V = P/ZT
CT0 = FT0/V0 = P0 /Z0RT0
Combining assuming no change on Z
V = V0 (FT/ FT0) P0/P (T/T0)
FT = FT0+ FA0δX substitute
V = V0(FT0+FA0δX)/FT0(P0/P)(T/T0)
T
T
P
P
X
XCCj
CyF
F
T
T
P
P
XF
F
X
F
FCC
T
T
P
P
XFF
XFCC
T
T
P
P
F
FCCj
T
T
P
P
F
F
v
F
T
T
P
P
F
Fv
F
v
T
T
P
PXVV
T
T
P
PXyV
T
T
P
PX
F
FVV
jj
A
TA
T
A
T
A
jj
T
A
Tj
AT
jjA
Tj
T
j
T
T
jT
T
T
j
A
T
A
0
0
AA
0
0
T
0
0
0
0
0
0000
0
j
0
0
0
0
0
0
0
0
0
1
C & y Recalling
1
`
levelesboth Fby dividing
F Cj
j speciesany for
1
11
0
000
0
0
0
0
00
0
0
0
00
0
0
0
0
0
0
0
0
27
Assuming δj is stoichiometric number, then
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
)1(1
)1(1
)1(1
)1(1
)1(
1
1
11
000
0
00
0
00
0
00
0
00
P
P
T
T
X
C
P
P
T
T
Xv
F
v
F
v
FC
P
P
T
T
X
Xa
d
CP
P
T
T
Xv
Xa
dF
v
Xa
dF
v
FC
P
P
T
T
X
Xa
c
CP
P
T
T
Xv
Xa
cF
v
Xa
cF
v
FC
P
P
T
T
X
Xa
b
CP
P
T
T
Xv
Xa
bF
v
Xa
bF
v
FC
P
P
T
T
X
XC
P
P
T
T
Xv
XF
v
XF
v
FC
IAIAAI
I
D
A
DADA
D
D
C
A
CACA
C
C
B
A
BABA
B
B
A
AAA
A
4.Isothermal Reactor Design
4.1 Procedure:
1- Mole Balance (M.B)
dt
dNdvrFF A
VAAA 0
2- Design equations:
Batch
X
A
AAV
dXVr
dt
dXN
0A00
N t &
CSTR A
A
r
XFV
0
28
PFR
X
A
AAr
dXr
dV
dXF
0A00
FV &
(3)
No
given
yes )(
xfrA
1- Rate law: -rA = k(CACB- CCC0/Kc)
2- Stoichiometry:
liquid phase or
constant value
Gas phase Isothermal
isobaric
CA= CA(1-X)
CB = CA0(θB – b/a
*X)
ΘB = yB0/yA0
0
0
1
10 P
P
T
T
X
XCC
AA
0
0 1
)1(
A
AA
y
X
XCC
6 Combine gas law and Stoichiometry to find –r then
proceed to (7 ).
7 Evaluate the above equations to find tbatch, VCSTR or VPFR
29
4.2 French Menu Analogy
0A
A
F
r
dv
dx
CA=FA/V
CA=NA/V
-rA =kCA
V= FA0X
-rA 0A
A
N
Vr
dt
dx
e
eBA
k
CCCkr AA
A
Ck
kCr
1
2-Rate law
3-Stoichiometry
flow
FA=FA0(1-X)
NA=NA(1-k)
Bate
CSTR Batch 1-M.B PFR
30
Liquid
constant
flow rate
Ideal gas variable
flow rate
Ideal gas variable
volume
Liquid or gas
constant
volume
v= v0
0
0
01
T
T
P
PXvv
0
0
01
T
T
P
PXvv
v =v0
XCCAA
10
T
T
P
P
X
XCC
A
A
0
01
10
T
T
P
PXCC
AA
0
0
10
XCC
AA 1
0
4-Combining 1st order G.P. in PFR
From M.B From R.L from Stoichiometry
T
T
p
P
X
XC
FF
r
dv
dXA
AA
A 0
0A
A
1
1
F
k
kC
0
000
T
T
P
P
X
X
v
k
dV
dX 0
00 1
1
Integrating for const Temp & press.
X
Xk
vV
1
1ln10
Batch system:
- No in or out flow
- Well mixed.
- For liquids density change is negligible.
- For gases volume remains constant.
31
M.B AA r
dt
dN
V
1
In forms of concentration:
AA
AA
A
AA
rdt
dC
rdt
dC
dt
V
Nd
dt
dN
vdt
dN
v
0
0
11
consider A B
2nd order in A
R.L -rA= kC2A
Combining M.B with R.L
-dCA/dt = kC2A
-dCA/kC2A= dt
CA= CA0 at t = 0 and reaction was isothermal
tCCk
dtdC
k
AA
tC
C
A
AA
A C
0
0
111
102
Reaction times for batch systems:
M.B V
Ndt
dx
A
A
0
R.L 1st order
-rA = kCA
2nd order
-rA=kC2A
Stoichiometry
2nd
order isothermal
liquid – phase batch
reaction
32
XCV
NC
A
A
A 1
0
0
Combining Xk
dt
dx 1 2
10
XkCdt
dxA
Integrate
Xkt
1
1ln
1
XkC
Xt
A
1
0
- CSTR:
exitA
A
CSTRr
XFV
0
where v =v0
A
AA
A
st
A
AA
A
AA
kC
CC
kCr
r
CC
v
V
r
CCvV
A
0
0
- order 1
0
0
Solving for CA then k
CC
A
A
1
0
No volume change XCCAA
10
Combining k
kX
1
33
τk is Damkoher Number (Da)
Da is the ratio of the rate reaction of A to the rate of
convective transport of A at the entrance of the reactor.
1st order k
Cv
VkC
F
VDa
A
A
A
A
0
0
0
0
0
2nd order 0
0
0
0
2
A
A
AkC
Cv
VCkDa
90% X 10 Da
10% X 1.0
Da
Assuming 2 reactors in series
Ck
CCv
r
FF
BM
k
CC
A
AA
A
AA
A
A
22
0
2
11
21
2
21
0
1
V
2rector for .
1
Solving for CA2
221122
111
01
2 kk
C
k
CC
AA
A
Suppose we have n reactor in series
n
n
A
n
A
A
for
Da
C
k
CC
n
Da1
1-1X
conversion the
)1(1
00
34
Rate consumption of A in the nth reactors:
n
A
AAk
CkkCr
nn )1(
0
CSTR in parallel:
Reactors are of the same size then
ii A
iAi
r
XFV
0
X1 = X2 = …. = Xn=X
Rate will be the same
AAAA rrrrn
....21
volume of reactor in compared to total number of reactors
Vi = V/n
Total moles flow will be related as
n
FF
A
Ai
0
0
Now
ii A
i
A
iA
r
X
r
X
n
F
n
V
00 AF
V ,
2nd order in CSTR
Combining M.B with R.L then
cA
iA
r
XFV
2
0
For constant density
35
Da
DaDaX
kC
kCkCX
kC
kCkCkCX
kCXkCX
XkC
Xthen
k
CC
v
V
A
AA
A
AAA
AA
A
A
AA
c
2
4121
2
4121
2
22121
021kC
)1(
0
00
0
000
000
0
0
22
2
A
2
2
0
Tubular reactors:
- Gas phase.
- Turbulent
- No Radial dispersion.
- No Radial Gradient of temp, velocity & conc.
1- D.E
FA0dx/dv = -rA
if no press drop & heat exchange with surrounding
x
A
Ar
dxFV
00
assuming 2nd order reaction & combing
x
C A
Ak
dxFV
0 20
for constant temp press gas phase
36
dxX
X
k
FV
dx
Xk
XFV
X
XC
Xv
XF
Xv
F
v
FC
C
C
A
A
X
A
APFR
A
AAA
A
2
2
2
0 22
2
00
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
integrating we obtain
x
XX
kC
vV
A1
11ln12
2
20
0
since no change on the tube radius the X-section area AC is
constant and
x
XX
AkC
vL
CA
C1
11ln12
2
20
0
Pressure Drop In Reactors
For ideal gas
0
i
0
0
0
1C
operation isothermalfor Design
1
0
0
P
P
X
XC
T
T
P
PXv
XF
v
FC
ii
A
iiAi
i
Now we must determine the ratio P/P0 as function of volume.
Using the differential form of mole balance.
Consider the 2nd order isomerization
A B
Carried in a packed bed
37
M.B FA0 dx/dw = -r\A
R.L -r\A = kC2
A
Stoichiometry for gas phase
T
T
P
P
X
XCC
A
A
0
01
10
R.L can be written as
2
0
0
\
1
10
T
T
P
P
X
XCkr
A
A
combing and using isothermal design case
PXFP
P
X
X
v
kC
dw
dx
P
P
X
XCk
dw
dxF
A
A
A
11
2
0
2
0
2
0
2
1
1
1
1
0
0
0
- Flow through a packed bed using the Ergun equation
G
DDPg
G
dz
dp
PPC
75.111501
3
where
G: PU: superficial mass velocity [ g/cm2 .s]
g: conversion factor for gravitational acceleration
Dp: partial diameter
p: pressure
u: superficial velocity (cm/s)
z: length along the packed bed
ф: porosity : void volume /Total bed volume
μ: viscosity of gases passing through the bed g/cm.s
p: gas density g/cm3
38
Using the Ergun equation the only variable is the gas density P
at S.S mass flow rate m (kg/s) is the same all through the bed
0
0
0
0
0
0
3
0
0
0
0
3
0
00
0
0
0
0
00
0
00
0
dz
dP
75.111501
75.111501
T
T
PPC
T
T
PPC
T
T
T
T
F
F
T
T
P
Pthen
GDDgP
Glet
F
F
T
T
P
PG
DDgP
G
dz
dp
Combining
F
F
T
T
P
PP
v
vPP
F
F
T
T
P
Pvv
PvvP
mm
The catalyst weight W can be related to z by
W =(1-ф)Ac z*Pc
Pb =(1-ф)Pc bulk density
dW =Pb Ac dz
0
0
00
0
0
1
2
/20
PPA
F
F
T
T
PP
P
dw
dp
CC
T
T
For single reaction Ergun equation can be expressed in terms
of X
39
XPP
P
T
T
dW
dP
now
yF
XF
F
XF
FFXFF
F
F
A
TT
T
T
A
TAT
T
T
1/2
F WHREE1
1
0
0
0
A
0
0
0
0
0
0
000
0
Note when ε =0 ΔP will be independent of conversion
ε= +ve pressure will damp
ε= -ve pressure will increase.
Now dP/dW=F2(X,P)
Solution isothermal system with ε= 0
dW
PPd
dW
PPd
P
P
ingrearr
PP
P
dW
dP
2
0
0
0
0
0
/
/2
arg
/2
Integrating with P = P0 at W=0
(P/P0)2 = 1-αW
(P/P0) = (1-αW)1/2
Also
2/1
00
21
P
z
P
P
Membrane Reactor
There are two main types of catalytic membrane reactors
40
Inert membrane reactor with catalyst pellets on the feed side
[MRCF]
Catalytic Membrane Reactor [CMR
e .g C6H12 C6H6 + 3H2
symbolically A B + C
M.B
For A FB│v - FA│v+Δv-RBΔv+γBΔv = 0
As before dFB/dv = rB - RB
-R.L
feed
H2 H2
H2
Catalytic
membrane
product
CMR
H2
H2
Feed
H2 H2
H2
catalyst C6H6
Inert membrane
IMRCF
41
AC
AB
C
CBAA
rr
rr
k
CCCkr
- Transport Theory membrane:
RB = kCCB
- Stoichiometry:
CB
CBA
T
CT
T
BT
T
ATA
rrr
FFF
F
FC
F
FC
F
FCC
T
CB
F
C , C , 000
- Combining & Summarizing:
CFFFF
F
F
F
F
K
C
F
FCKr
rdW
dF
F
FCKr
dV
dF
rdV
dF
BAT
T
C
T
B
C
T
T
ATCA
AC
T
BTCA
B
AA
0
0
0
42
Unsteady State Operation
A,B Heat
Reactive distillation
CA0 CA
A
B
Heat
Semi Batch
43
Steady State Operation of CSTR
Startup of a CSTR :
Assuming semi-batch condition with reaction
A + B C +D
B is fed gradually to A
- M.B FA0 – FA + rAV = dNA/dt
to liquid phase v =v0 & V = V0 using τ = V0/v0
then CA0 – CA + rAτ = dCA/dt
1st order (-rA = kCA)
tK
K
CC
CC
K
dt
dC
A
A
A
A
A
11exp11
1
0
0
Suppose t s is time for 99% conversion of CAs
CAs = CA0 /1+τk
Using CA = 0.99 CAs
ts = 4.6 τ/1+τk
slow reactions ts = 4.6 τ
for fast reactions ts = 4.6/k
Semi batch Reactor:
Elementary liquid phase reaction
A+B C
B
A
44
M.B for A
0 +0 + rA V(t) = dNA/dt
inters of concentration
dt
dVC
dt
dCV
dt
VCdVr A
AAA
over all mass balance
P0v0- 0+0 = d(Pv)/dt
Constant density P0=P & dV/dt = 0
V=V0 at t = 0 integrating
V = V0 + v0t
Substitute in the M. B
dt
dCVVrCv A
AA 0
M.B for A can be rewritten as
AAA C
V
vr
dt
dC 0
M.B for B with feed rate of FB0
V
CCvr
dt
ngsubstituti
FVrdt
dCVC
dt
dV
dt
VCd
FVrdt
dN
BB
B
BBB
BB
BBB
0
0
0
0BdC
gintegratin & Vfor
Interims of conversion:
A+ B C+ D
For A
45
NA= NA0- NA0X
For B
t
ABBBXNdtFNN
i 0 00
Constant molar feed rate
NB= NBi+FB0t –NA0X
M.B on A rAV =dNAdt
Assuming 2nd order
)(1
/-
DC BAfor
1
Nby dividingk
C
1
000
00
0
00
0
0
0
00
0
A
00
2
A
00
00
00
B
00
eABeA
eAeA
C
BA
DC
BA
DC
BA
DC
C
CDCBA
e
C
A
ABB
A
D
A
C
ABBA
AAA
DCBAA
XNtFXN
XNXNK
NN
NN
V
N
V
NV
N
V
N
CC
CCK
kCCCCkr
X
tvV
k
XNXNtFNXk
dt
dX
combining
tvV
XNC
tvV
XNC
tvV
XNtFN
V
N
tvV
XN
V
NC
dt
CCCCkVr
ee
ee
ee
ee
ee
ee
i
i
46
Rearranging
12
141)1(
1
0
0
0
0
0
0
0
0
2
2
C
A
B
C
A
B
C
A
B
C
e
e
e
eC
BC
A
K
N
tFK
N
tFK
N
tFK
X
or
X
XXK
FK
Nt
Reactive Distillation
Vrdty
dX
XNXa
bNN
XNN
dt
dNVr
volume
DCBA
OHCOOHCHOHCHCOOHCH
A
ABAB
AA
AA
k
k
0
00
0
2
1
A
23333
N
B&A feadmolar equal 1
1
00
A on
Balance on D with FD rate of evaporation
0+FD+rDV = dND
47
Integrating
V
dtFN
XK
XXkN
Vr
V
dtFXN
V
NC
V
XN
V
NC
V
XN
V
NC
V
XNC
K
CCCCkr
dtFXNN
t
D
AC
A
A
t
DAD
D
ACC
ABB
A
A
C
DCBA
t
DAD
0
22
0
A
0
0
0
0
0
0
0
0
11
1
1
- R.L
Find volume as f(X or t)
D
WDD
WDD
FP
MF
dt
dV
dt
PVdMF
00
specify the rate of evaporation of D
Case 1:
Immediate evaporation
dt
dxNVrVrF AADD 0
48
Integration
V= V0-αNA0X
V = V(1+εLX)
Where εL = -αCA0 = MwDCA0/ρ
Combining with CD = 0
xx
x
x
xkC
dt
dx
L
L
L
A
1
1ln
1
)1(
kC
1t
gintegratinby for t solving
1
)1(
0
0
A
2
Case 2: Inert gas is bubbled through reactor
Assuming Raoult’s law
D
DCBAD
D
DCBA
I
D
D
D
D
I
T
TDIDIF
TDD
V
DCBD
DV
DD
FV
NNNNk
dt
dN
FV
NNNNk
dt
Coupling
Fy
yF
y
FF
FyFFFF
FyF
P
DP
NNNN
N
P
DPlxgy
)(
)(dN
M.R with
1
1
A
00
49
5. Collection & Analysis of Rate Data
Objective of this chapter is to obtain and analysis reaction
rate data to obtain the rate law of specific reaction.
Two type of reactor are discussed:
1- batch reactor – Homogeneous
2- Differential reactor – Solid fluid.
Two techniques of data acquisition are presented:
1- Concentration –time –batch.
2- concentration only – differential reactor
Six method of data analysis are used
1- Differential method.
2- Integral method.
3- Half-live method.
4- Initial rates method.
5- Linear regression method.
6- Non –linear (least-sequences analysis)
5.1 Batch Reactor Data:
5.1.1 Differential Method:
Assuming the decomposition reaction.
A product
-rA = kCαA
or for A + B product
-rA = kCαACβ
B
the reaction can be run in excess of B and CB will remain
constant then
-rA = k1CαA
where K1 = kCβB ≈ kCβ
B0
50
then α will be determined and the reaction will be run again
in excess of A and
-rA = k\\CβA
where K\\ = kCβA ≈ kCβ
A0
the β will be determined now α k β are determined. Then kA
can obtained
Smol
dmrk
CC BA
AA /
13
Also using the differential method
AA
A
AA
A
Ckdt
dC
kdt
dCC
lnlnln
α
Ln(-dCA/dt
lnCA
51
PA
P
A
AC
dt
dC
k
-dCA/dt can be obtained by
1- Graphical differentiation
2- Numerical differentiation
3- Polynomial differentiation
1. Graphical Method
- plotting ∆CA/∆t as function of t
Equal Area Differentiation
1-Tabulate xi&yi
2- Calculate ∆xn= xn- xn-1 & ∆y= yn-yn-1
3- Calculate Δyn/∆xn as estimate of average slope in interval xn-
1 to xn
4- plot these values versus xi e.g
-dCA/dt
CAP
PA
PA
AC
dtdCk
52
23
23
32
y is x&
xx
yx
5-Draw smooth curve
Dy/dx ∆y/∆x ∆y ∆x yi xi
(dy/dx)1 - - - y1 x1
(dy/dx)2 (∆y/∆x)2 y2-y1 x2-x1 y2 x2
(dy/dx)3 (∆y/∆x)3 y3-y2 x3-x2 y3 x3
(dy/dx)4 (∆y/∆x)4 y4-y3 x4-x3 y4 x4
(∆y/∆x)5 y5-y4 x5-x4 y5 x5
2- Numerical Method
if data is equally spaced i.e.
t1-t0 = t2 - t1= ∆t
t0 t1 t2 t3 t4 t5 time (min)
CA0 CA1 CA2 CA3 CA4 CA5 Conc. mol/dm3
Using 3-point formula
Initial point
341dC
point last
1dC e.g
1dC pointsinter
43
543
5
24
3
210
0
A
A
11
A
AAA
t
AA
t
iAiA
t
AAA
t
A
CCCtZdt
CCtZdt
CCtZdt
tZ
CCC
dt
dC
i
53
3- Polynomial Fit:
Fit the concentration – time data to an nth order polynomial:
CA = a0 +a1t +a2t2 + ….+antn
Then dCA/ dt = a1 + 2a2t + 3a3t2+….+nantn-1
Now knowing dCA/dt at different t we can do the following
t3 t2 t1 t0 time
CA3 CA2 CA1 CA0 Conc.
3dt
dCA
2dt
dCA
1dt
dCA
0dt
dCA
Derivative
Now plotting ln(-dCA/dt) versus ln CA then the reaction order
is obtained and the intercept will give lnkA since
AA
A Ckdt
dClnlnln
5.2 Integral Method
Is used when reaction order known and rate constant or
activation energy is to be obtained for the reaction
A product
dCA/dt=rA
for zero- order
dCA/dt = -k
integrating with CA=CA0 at t=0
then
CA= CA0-kt
54
1st order
-dCA/dt = kCA
Integrating with CA = CA0 at t=0
lnCA0/CA = kt
if reaction is 2nd order then
C A
A kdt
dC 2
integrating with same initial conditions
ktCC
AA
0
11
lnCA0/CA
t
CA0
CA
Zero order
t
55
5.3- Initial rates:
When it is difficult to run experiments of concentration at
different times, then the initial rate method is used. In which,
different runs at different initial concentrations are done. The
initial rate (-rA0) are calculated. Since rate law is [-rA0=kCαA0]
Plotting (ln(-rA0)) versus (lnCA0) will give α
5.4- Method of half–Live:
Half lives (t1/2) of a reaction is the time required for a reactant
to drop to half of its initial concentration. If two reactants are
used.
Assume
A products
CAAA kr
dt
dC
integrating with CA = CA0 at t = 0
AC
1
0
1
AC
t
56
11
1
11
1
1
1
1
11
0
0
0
A
A
A
AA
C
C
kt
kt
C
CC
solving for CA = 1/2CA0 at t = t1/2
then
C Ak
t1
1
2/1
0
1
1
12
the concentration can fit to any 1/n value of its initial
concentrations. Then
0
0
ln11
1lnln
1
1
1
21
2/1
1
1
/1
A
A
n
Ck
t
kt
Cn
plotting lnCA0 versus lnt1/2 with give slope equal to (1-α) & then
α =1- slope
57
5.5 Differential Reactor:
There are similar to using batch reactor with small initial
concentration. The reactor is concentrated to be gradient less
with low conversion
S.S M-B. on A
1-α
lnt1/2
lnCA0
Catalyst
inert
FA0
inert
FAe
ΔL ΔL
FA0
CA0
FAe
FP
CP
58
W
FF
WF
catalystrate
FF
e
e
e
AAA
AA
AA
0
0
0
\
\
A0F
0mass masscatalyst
reaction of `
In terms of concentrations
W
vCCvr eAA
A
00\
Interims of conversion product
W
Cv
W
CCvr
W
F
W
XFr
PAAA
PA
A
e 00\
00
0
- vconstant for
-r\A can be determined by the concentration of the partial CP
Assuming CAb in the concentration in the bed then
–r\A = - r\
A (CAb)
let CAb=(CA0- CAe)/2
since very small reaction is taking place then
CAb ≈ CA0
-r\A= -r\
A(CA0)
5- least squares Analysis
- linearization of the rate law
if rate law depend one concentration of more than one
species also if it is to determine other parameter like ( α,A& E)
mole balance on a constant volume batch
59
22110
210
2A1
0
A
0
A
a & a , ln
ln x,lnCx , dC-
lny
lnlnlndC-
ln
rates initial of method using
00
00
xaxaay
ka
Cdt
let
CCkdt
krdt
dC
B
BA
BAAA
CC
if N experiments are carried out then
yj=a0+a1x1j+a2x2j
The values of the parameters a0, a1 and a2 are found by
N
j
jj
N
j
j
N
j
jj
N
j
j
N
j
jj
N
j
j
N
j
jj
N
j
j
N
j
j
N
j
j
N
j
j
XaXXaXayX
XXaXaXayX
XaXaNay
1
22
22
1
1
1
120
1
2
1
222
2
1
12
1
110
1
1
1
2
1
2110
1
6- Nonlinear least – squares
For the nonlinear least square method we want to minimize
the sum of the estimate differ
i.e. if we have rm: measured rate and
rC: calculated rates then we want to minimize
N
i
ii
KN
crmr
kN
s
1
222
Where
60
s2= ∑(r1M-riC )2
N= Numbers of runs
K= number of parameters to be determined
riM= measured rate for run i
riC = calculated rate for run i
6. Multiple Reactions
There are 3types of basic multiple reactions
1- Series
2- Parallel
3- Independent
1\ Parallel (competing)
e.g. oxidation of ethylene to ethylene oxide avoiding complete
combustion to CO2 & H2O
2- series (consecutive ) reactions
K1
B
C
A K2
O
CH2=CH2+O
2
CH2-CH2
2CO2+2H2O
61
CBA kk 21
e.g. reaction of ethylene oxide with NH3 to form mono-, di-
and tri-ethanolamine
it is desired to produce di-ethanolamine and tri
* Multiple reaction combining both is
A+B C+D
A+C E
e.g. formation of butadiene from ethanol
C2H5OH C2H4+H2O
C2H5OH CH3CHO+H2
C2H4+CH3CHO C4H6+H2O
3- independent reactions
A B
C D+E
- in the parallel reaction
product undesierd
product desired
uA
DA
u
D
k
k
CH2- CH2+NH3→HOCH2CH2NH2→(HOCH2CH2)2NH→(HOCH2CH2)3N
O
62
Maximizing Desired product in parallel Reaction:
C
C
Auu
ADD
k
k
kr
kr
uA
DA
u
D
2
1
undesired
desired
Rate of consumption of A is
C
CC
Au
D
u
D
AuADuDA
k
k
r
r
kkrrr
21
21
DuS
is (S)parameter seletivity rate
Case 1:- α1>α2 i.e. α1-α2=a
Reactor
Separator D
u
Total
cost
cost Separator cost
Reactor cost
63
Ca
Au
D
u
DDu
k
k
r
rS
To maximize D:-
1-use high CA e.g. for gas –phase use pure A at high pressure
Possible – for liquid – phase use minimum diluents possible.
Use batch or PFR reactors . CSTR is not used because of CA
exit.
Case 2:- α2>α1 i.e. α2-α1=a
CCCC
a
Au
D
Au
D
Au
AD
u
DDu
k
k
k
k
k
k
r
rS
122
1
to maximize D use low CA by diluting or use inert. CSTR is the
best if the activation energy is known then
e
RTEE
u
D
u
D uD
A
A
k
k /
Case 3: ED>ED
To maximize D run the reactor at high possible temp.
Case 4: Eu>ED
To maximize D run at low possible temp.
Reaction with reactants
CC
CC
CC
BAu
D
u
DDu
BAu
k
BAD
k
k
k
r
rS
kruBA
krDBA
u
D
2121
22
u
D
64
Case1: α1>α2 , β1>β2
α1-α2=a , β1-β2=b
CCb
B
a
Au
D
u
DDu
k
k
r
rS
to maximize D use high CA & CB possible
* PFR or batch and high pressure for gas – phase
Case 2: α1>α2 , β1<β2 a=α1-α2 & b= β2-β1
CC
b
Bu
a
AD
u
DDu
k
k
r
rS
to maximize SDu use high possible CA and low CB
* use semi batch reactor containing large A &B is fed slowly
* PFR with side stream of B
* series of small CSTR with A fed to 1st reactor and small
amount of B fed to each
A
B
B
Pure A
65
Case 3: α1<α2 & β1<β2 a=α2-α1 & b=β2-β1
CCb
B
a
Au
D
u
DDu
k
k
r
rS
to maximize SDu lower possible conc. Of A&B
* use CSTR
* PFR with large recycle
* feed diluted with inert
* low possible pressure if gas – phase
Case 4: α1<α2 , β1>β2 , a= α2-α1 & b=β1-β2
CC
a
Au
b
BD
u
DDu
k
k
r
rS
to maximize SDu use high possible CB and low possible CA
* semi batch will B and A feed slowly
* PFR with side streams of A
A
A
A
B
B
B
66
* semis of small CSTR with B and A fed at small amounts to
each reactor
Series reactions:
In case of consecutive reactive reactions it is important to
control the time of reaction or the space time
CBA
COCHOCHOHCHCH
kk
kk
21
21
23232
desired product is acetaldehyde (B) reaction is 1st order one
(A) &(B)
ek
A
A
bb
AA
A
AA
A
C
Cegrating
PvvPvw
Ckdw
dCCombining
vCtrystoichiome
CkrRate
rdw
\1
0
0
A
A
00
\
10
0A
1
\
\A
C
0at w C using int
///
v
F
- law
dF balance mole 1
ek
ABB
BAB
B
BA
BBB
B
CkCd
dC
gu
CCdw
dCcombining
Cvtrystoichiome
CC
rrrRate
rdw
\1
012\
A
210
0B
21
2\
1\\
\B
k
above from C sin
kkv
F
kk
- law
dF balance mole 2
67
using the integration factor
12
1B
B
0
\
1\
\2
\1
0
\12
0
\2
C
interance at the 0C and 0 w sin
kkCk
v
wgu
Ckd
Cd
ee
ee
kk
A
kk
A
k
B
differentiating equation CB to find maximum concentration
2
1
21
0
2
1
21
\
11
12
1
\
ln
ln1
0\
2\
10
k
k
kk
vW
k
k
kk
kkkk
Ck
d
dC
opt
opt
kkAB
ee
Optimum conversion of A
τ\1 τ
\2 τ
\3 τ\4 τ
\
Ci
A B C
68
21
1
21
1
\1
0
0
2
1
2
1
1
lnexp
1
kk
k
opt
kk
k
opt
k
A
AA
opt
k
kX
k
kX
C
CCX e
Multiple Reactions:
Consider the gas phase reactions
NO+2/3NH3 5/6N2+H2O ……..(1)
2NO N2+O2 …….(2)
O2+1/2N2 --> NO2 ……..(3)
For gas phase concentration with no pres drop and isothermal
Cj=cT0Fi/FT
FT = FNO+FNH3+FN2+FH2O+FO2+FNO2
Mole balances for each component
69
T
O
T
N
TT
NO
TT
NO
T
NHTNH
N
OCNNONONH
N
T
NO
T
NHT
oH
NONH
oH
T
NO
T
NHT
NH
NONH
NH
T
NOT
T
NO
T
NHT
NO
NOONNHNO
F
F
F
Fk
F
Fk
F
F
F
FCCk
dv
dF
CkkCCkdv
dFN
F
F
F
FCk
dv
dF
CCkOrHdv
dFO
F
F
F
FCk
dv
dF
CCkrrNHdv
dF
F
FCk
F
F
F
FCk
dv
dF
CkCkCrNOdv
dF
CC
C
22
00
3
03
2
22
23
2
3
0
2
3
2
3
0
3
3
3
0
3
0
23
3
3
2
2
2
5.1
5.1
1
3
2
2
5.1
122
5.1
5.2
1
5.1
122
5.1
5.2
1
5.1
1133
2
2
2
5.1
5.2
1
2
2
5.1
2
1
6
5
2
1
6
5rN 4
H 3
3
2
3/23
2 NH 2
2
2 NO 1
2
3
3
ON
3
ON
2
2
3
3
2
NO2
T2
O
32
O
2
22
0
2
2
2
22
00
2
2
2
dF
dF 6
FCk
dF
dF O 5
T
O
T
NT
NO
T
O
T
NT
T
O
F
F
F
FCk
dv
rrdv
NO
F
F
F
FCk
Fdv
rrrdv
70
7. Steady-State Non-isothermal Reactor Design
The Energy Balance:
Consider a flow system at steady – state with the following
energy balance equation
n
i
n
iiiiiS
HFHFWQ1 1
000
Suppose the following reaction is taking place
A+(b/a)B (c/a)C+(d/a)D
∑HioFio = HA0FA0 +HB0FB0 +HC0FC0+HD0FD0+HI0FI0
∑HiFi = HAFA+HBFB+HCFC+HDFD+HIFI
also
FA=FA0(1-X) , FB =FA0(θB-bx/a)
FC =FA0(θC-cx/a) , FD =FA0(θD-dx/a) , FI = QIFA0
Substitute these values in the two equation & subtract
# X
nowreaction ofHeat
000
00
000
0
111
11
ARX
n
i
IIIA
n
i
ii
n
i
iOiO
ABCDRX
ABCD
IIIDDD
CCCBBBAA
A
n
i
ii
n
i
iOiO
FTHHHFFHFH
THTHa
bTH
a
cTH
a
dTH
HHa
bH
a
cH
a
d
HHHH
HHHHHHFFHFH
More stable form
0000XFTHHHFWQ
ARXiIiAS
The enthalpies Hi of species i are expressed
71
T
T pRiidtCTHH
i1
0
CPi = αi+βiT+υiT2 [J/mol of i.k]
The change on enthalpy (Hi-Hi0) with change on phase
T
Pi
T
T PiRi
T
T PliRiii
iT
i
RR
dTC
dTCTdTCTHH HH
0
0
0
00
Substituting in E.B
n
i
T
T ARXpiASi
iXFTHdTCFWQ
10
000
The heat of reaction at T is given by
T
T PRii
ABCDRX
RidtCTH
THTHa
bTH
a
CTH
a
dTH
H0
Substituting we will have
ABCD pppp
ARBRCRDRX
CCa
bC
a
CC
a
d
TTa
bT
a
cT
a
dTH HHHH
0000
Combing these equations
T
T PRRXRXR
dTCTTH H0
this will gave the heat of reaction of any temp. in terms of heat
of reaction & reference temp TR(298k)
72
Mean Heat Capacity
∆HRX(T) = ∆H0RX(TR) + ∆C\ P(T-TR)
where
0
0
00
|
|
ii
|
i
T
Ti P
P
iiP
T
Ti P
R
T
T P
P
TT
dTCC
where
TTCdTC
Similary
TT
dTCC
i
i
R
Substitution in S.S. E.B
0\0\
000 RPRRXAiP
iASTTCTXFTTCFWQ Hi
Variable Heat Capacities:
dtTTTTH
TTC
H RXRX
iiiPi
20
2
Integrating gives
ABCD
ABCD
ABCD
RRRRXRX
a
b
a
c
a
d
a
b
a
c
a
d
a
b
a
c
a
d
where
TTTTTTTHTH
32
33220
Similarly the heat capacity of the heat input can be evaluated
73
32
033
022
01
2
1
0
00
TTTTTTdtC
dtTTdtC
iiii
ii
n
i
T
T Pi
T
T iiiiii
n
i
T
T Pi
i
i
Substituting these integration in the S.S.EB equation we get
Application to continuous Reactors (CSTR) A B 1- CSTR design eqn. V= FA0X/-rA 2- Rate law -rA = kCA With k= Aexp[-E/RT] 3- Stoichiometry CA = CA0(1-x ) 4- Combining V=(v0/Aexp[-E/RT])(x/1-x) Case 1 x , v0 , CA0 , Fi0 are known and V will be calculated
5.1.1 solve for T using the E.B equation 5.1.2 calculate k 5.1.3 then calculate V
Case 2 v0 , CA0 , V and Fi0 are known X and T will be calculated
5.2.1 solve the E.B equation for XEB as function of T 5.2.2 solve the combined equation for XMB as for of
032
32
033
022
0
0
033
022
0
0
0
TTTTTTTHXF
TTTTTTFWQ
RRXA
iiii
iiAS
74
5.2.3 find values of X and T that satisfy the two
equations
PFR & PBR A↔B 1- Mole balance dx/dv = -rA/FA0 2- Rate law -rA = k(CA- CR/kC)
12
0
2
1
1
11exp
11exp
TTR
HTkk
TTR
Ekk
RX
C
3- Stoichiometry
T
TXCC
T
TXCC
AB
AA
0
0
0
01
4- Combining
X
X
XEB
XMB
T T
75
T
T
k
XXkCr
C
AA01
0
5- solve the E.B equation to find X & T as before then solve the M.B of PFR or PBR REFERNCES:
1. H. Scott Fogler, “Elements of Chemical Reaction Engineering” 3rd Edition, Prentice Hall International Inc., 1999.
2. Octave Levenspiel, “Chemical Reaction
Engineering”, 3rd Edition, John Wiley and Sons Inc., New York, 1998.
3. Ronald W. Missen, Charles A. Mims and Bradley
A. Saville, “Introduction to Chemical Reaction Engineering and Kinetics”, John Wiley and Sons Inc., New York, 1999.
4. Froment, G. F. and Bischoff, K. B., "Chemical
Reactor Analysis and Design", 2nd Edition, Wiley Series in Chem. Eng. 1990.