3.batch reactor
TRANSCRIPT
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BATCH REACTOR Interpretation of rate data
A. SARATH BABU
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• Simplest reactor – open / closed vessel
• Reactants are placed inside the reactor and allowed to react over time
• Products and unconverted reactants are removed and the process is repeated
• Closed system - unsteady state operation
• Fitted with a stirrer
• May have a jacket / cooling or heating coils inside the reactor
• Generally constant volume / some designed at constant pressure
• Materials of construction – different linings
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Batch Reactor Contd. . . 5
Batch Reactor Cont. . . 7
• Used in variety of applications
• Typically for liquid phase reactions that require long reaction times
• Used only when small amount of product is required
• Favored when a process is in developmental stage or to produce expensive products
• Used to make a variety of products at different times
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(1) Each batch is a closed system.(2) The total mass of each batch is fixed. (3) The reaction (residence) time t for all
elements of fluid is the same.(4) The operation of the reactor is inherently
unsteady-state; for example, batch composition changes with respect to
time.(5) It is assumed that, at any time, the batch
is uniform (e.g., in composition, temperature, etc.), because of efficient stirring.
Characteristics of a Batch Reactor
Batch Reactor Contd ... 9
Advantages:
• High conversions can be obtained
• Versatile, used to make many products
• Good for producing small amounts
• Easy to Clean
Dis-advantages:
• High cost of labor per unit of production
• Difficult to maintain large scale production
• Long idle time (Charging & Discharging times) –
leads to periods of no production
• No instrumentation – Poor product quality
GUIDELINES FOR SELECTING BATCH PROCESSES
• Production rates:
— Sometimes batch process, if the plants have production capacity less than 10x106 lb/yr (5x106 kg/hr).
— Usually batch process, if the plants have production capacity less than 1x106 lb/yr (0.5x106 kg/hr).
— Where multiproduct plants are produced using the same processing equipment.
• Market forces:
— Where products are seasonal (e.g., fertilizers).
— Short product lifetime (e.g., organic pigments).
• Operational problems:
— Long reaction times (when chemical reactions are slow).
— Handling slurries at low flowrates.
— Rapidly fouling materials (e.g., materials foul equipment so
rapidly that shutdown and frequent cleaning are required).
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General Mass Balance Equation:
Input = output + accumulation + rate of
disappearance
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General Mass Balance Equation:
Input = output + accumulation + rate of
disappearance
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Design Equation
General Mass Balance Equation:
Input = output + accumulation + rate of disappearance
0 = 0 + dNA/dt + (-rA) V
General Design Equation-(1/ V) dNA/dt = (-rA)
General Design equation in terms of conversion(NAo/ V) dxA/dt = -rA
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Design Eqn. for variable volume batch reactorCAo/(1+AxA) dxA/dt = -rA
Design Eqn. in terms of Total Pressure(1/RT) dPT /dt = (-rA)
Design Eqn. for CVBR-dCA/dt = -rA
CAo dxA/dt = -rA in terms of conversion
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Constant Volume Batch Reactor
A
A
C
C A
A
r
dCt
0
t t = CA0 X
area
Ax
A
A
A r
dx
C
t
00
Stoichiometric Table – Batch Systems
B NB0 -(b/a)NA0xA NB= NA0(MB-(b/a)xA)
R NR0 +(r/a)NA0xA NR= NA0(MR+(r/a)xA)
S NS0 +(s/a)NA0xA NS= NA0(MS+(s/a)xA)
I NI0 0 NI = NI0
Total NT0 NT = NT0 + NA0δxA
Where: MI = NI0/NA0
δ = (r/a + s/a – b/a – 1)
aA + bB rR + sS
For CVBR: CA = CA0(1-xA); CR = CA0[MR+(r/a)xA]
Species Initial Change Final moles A NA0 -NA0xA NA= NA0(1-xA)
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Constant Volume Batch Reactor -rA = -dCA/dt = CA0 dxA/dt
1. Zero Order Reaction:-rA = -dCA/dt = k
tC
C
A dtkdCA
A 00
CA0 - CA = kt CA0xA = kt
Strictly homogenous reactions do not follow zero order. Apparentlythe reaction order is made zero w.r.t. a reactant.
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2. First Order Reaction: A Products-rA = -dCA/dt = kCA
tC
C A
A dtkC
dCA
A 00
-ln (CA/CA0) = kt -ln(1-xA) = kt
-rA = k CA0.6 CB
0.4 ??Unimolecular – Collision theory ??
Example: N2O5 2NO2 + ½O2
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First Order Reaction kineticsInfluence of k
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3. Second Order Reaction: 2A ProductsA + B Products CA0 = CB0
-rA = -dCA/dt = kCA2
tC
C A
A dtkC
dCA
A 02
0
1/CA – 1/CA0 = kt xA/(1-xA) = kCA0t
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4. Second Order Reaction: A + B Products CA0 CB0
-rA = -dCA/dt = kCACB
t
K(CB0-CA0)
ktCCCC
CC
x
xMAB
AB
AB
A
A )(ln1
ln 000
0
AB
AB
CC
CC
0
0ln
Example: CH3COOC2H5 + NaOH CH3COONa+C2H5OH
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5. Third Order Reaction: 3A Products 2A + B Products CA0 = 2CB0
A + B + C Products CA0 = CB0 = CC0
-rA = -dCA/dt = kCACBCC = kCA3
Example: 2NO + H2 H2O +N2O 2NO + Cl2 2NOCl
tkCC AA
21120
2
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6. Third Order Reaction:
2A + B Products CA0 2CB0
-rA = -dCA/dt = kCA2CB
2lnln
2
00
0 ktM
C
M
C
C
C
M
C
C
AA
B
AA
B
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7. Third Order Reaction:
A + B + C Products CA0 CB0 CC0
-rA = -dCA/dt = kCACBCC
))((
)/ln(
))((
)1/1ln(
0000
0
0000
0
CBAB
ABBB
CABA
AA
CCCC
xMMC
CCCC
xC
ktCCCC
xMMC
ACBC
ACCC
))((
)/ln(
0000
0
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8. nth Order Reaction: nA Products
-rA = -dCA/dt = kCAn
ktnCC nA
nA
)1(1110
1
ktnCC nA
nA )1(1
01
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The reciprocal of rate approaches infinity as CA → 0
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Integrated forms – Constant density
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Note that for a II order reaction with a large ratio of feed components, the order degenerates to a first order (pseudo first order).
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Differential Method of analysis
CA
t
-rA
f(c)
k
)(),( CfkCkfdt
dCr AA
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Differential Method of analysis
ln(-rA)
ln(CA)
ln k
nA
AA Ck
dt
dCr
n )ln()ln()ln( AA Cnkr
If –rA = kCAaCB
b, how to use DM?
• Use stoichiometric ratio of reactants• Use method of excesses• Use method of least squares
)ln()ln()ln()ln( BAA CbCakr
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Integral Method of analysis
• Guess the reaction order
• Integrate and Derive the equation
• Check whether the assumed order is correct or not by plotting the necessary graph
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Differential Method Integral Method
• Easy to use and is recommended for testing specific mechanism• Require small amount of data• Involves trial and error• Cannot be used for fractional orders• Very accurate
• Useful in complicated cases
• Require large and more accurate data• No trial and error• Can be used for fractional orders• Less accurate
Generally Integral Method is attempted first and if not successful, the differential method is used.
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Method of Excesses Consider –rA = kCA
aCBb
• Perform the experiment with CB0 >> CA0 and measure CA as a function of t.
–rA = kCAaCB
b = kCB0b CA
a = kCAa
Use either differential method or integral method and evaluate k’ & a
• Perform the experiment with CA0 >> CB0 and measure CB as a function of t.
–rA = kCA0a CB
b = k’CBb
Use either differential method or integral method and evaluate k’’ & b
Require multiple experiments
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Method of Half lives
)1()1(1110
1 nktn
CC nA
nA
At t = t1/2, CA = CA0/2
kn
Ct
nnA
)1(
)12( 110
2/1
knCnt
n
A )1(
)12(ln)ln()1()ln(
1
02/1
ln(t1/2)
ln(CA0)
K’
(1-n)
Require multiple experiments
For I Order reactions: t1/2 = ln(2)/k t1/2 does not depend on CA0
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Check the value of dimensionless rate constant kCA0(n-
1)t for each order at t = t½
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Method of Half lives
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At t = t1/n, CA = (1- 1/n) CA0
Method of Fractional lives
The ratio of any two fractional lives is characteristic of the order.
kn
Ct
nnA
)1(
)12( 110
2/1
kn
Ct
nn
A
)1(
)123(
110
3/1
)123(
)12(1
1
3/1
2/1
n
n
t
t1
23ln
2ln
3/1
2/1 nfort
t
The half-life, or half-period, of a reaction is the time necessary for one half of the original reactant to disappear.
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Method of Initial Rates
The order of the reaction with respect to an individual component can be determined by making an initial rate measurement at two different initial concentrations of this species while holding all other concentrations constant between the two runs.
Advantage of the initial rate method is that complex rate functions that may be extremely difficult to integrate can be handled in a convenient manner. Moreover, when initial reaction rates are used, the reverse reactions can be neglected and attention can be focused solely on the reaction rate function for the forward reaction.
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Method of Initial Rates
-rA0= k (CA0)n ln(-rA0) = ln(k) + n ln(CA0)
(CA0)1
(CA0)2
(CA0)3
Time, t
(-rA0)1 = slope at (CA0)1, t = 0
(-rA0)2 = slope at (CA0)2, t = 0
(-rA0)3 = slope at (CA0)3, t = 0
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Method of Initial Rates
ln CA0
ln (-rA0) Slope gives ord
er
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CA/CA0
DA = kCA0n-1 t
Comparison of Different orderReactions in a Batch reactor
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Variable Volume Batch Reactor
dt
dVC
dt
dCV
Vdt
VCd
Vdt
dN
Vr i
iiii
1)(11
dt
dV
V
C
dt
dCr iii Volume Change with time ??
Fractional Change in Volume or Expansion factor(ЄA):
fedmolesofnoTotal
completedisreactionwhenmolesofnototalinChangeA .
.
0
01
A
AA
X
XxA V
VV
Expansion factor can be obtained if we know the initial volume and the volume at any X. Similarly X can be obtained given expansion factor.
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00 T
T
N
N
V
V
AAAAAT
A
T
T xxyxN
N
N
N 111 00
0
0
0AA y 1a
b
a
s
a
r
Example: A 3R, starting with pure ASince pure A, yAO = 1.Also δ = 3/1 -1 = 3-1 = 2.
ЄA = (3 - 1)/1 = 2With 50% inerts: yAO = 0.5 & δ = 3/1 -1
= 2. ЄA= (4-2)/2 = 1 = 0.5 (3-1) = 1
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Variable Volume Batch Reactor
dt
dx
x
C
dt
dxN
xVdt
dN
Vr A
AA
AAA
AA
AA
1)1(
11 00
0
AA
AA
AA
AAAA x
xC
xV
xN
V
NC
1
)1(
)1(
)1( 0
0
0
AA
ARAR x
xarMCC
1
])/([0
AA
A
A
A
x
x
C
C
1
1
0 0
0
/1
/1
AAA
AAA CC
CCx
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CVBR VVBR
Ax
AAA
A
A xr
dx
C
t
00 )1(
xA
t / CA0
A
A
C
C
AA rdCt0
/
1 /-rA
t
CA CA0
)1(
1
AAA xr
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Variable Volume Batch Reactor
1. Zero Order Reaction:
tx
AA
AA dtk
x
dxC
A
00
0 1 Ln
(1+
ЄAx
A)
t
kdt
dx
x
Cr A
AA
AA
10
tkxC AAAA )1ln(0
tkVVC AA )/ln( 00
kЄA/CA0
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Variable Volume Batch Reactor
2. First Order Reaction:
tx
A
A dtkx
dxA
00 1
-ln
(1-x
A)
t
AA
AAA
A
AA
AA x
xCkkC
dt
dx
x
Cr
1
)1(
100
ktxA )1ln(
* Performance equation is similar to that of CVBR
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Variable Volume Batch Reactor
3. Second Order Reaction:
t
A
x
A
AAA dtCkx
dxxA
0
0
02)1(
)1(
2
020
1
)1(
1
AA
AAA
A
AA
AA x
xCkkC
dt
dx
x
Cr
tkCxxx AAAAAA 0)1ln()1/()1(
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Variable Volume Batch Reactor
4. Higher Order Reactions:
tnA
x
nA
An
AA dtCkx
dxxA
0
10
0
1
)1(
)1(
n
AA
AAnA
A
AA
AA x
xCkkC
dt
dx
x
Cr
1
)1(
100
Analytical integration would be difficult.
Resort to either graphical / numerical integration.
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Constant Volume Batch Reactor (PT vs. t)
)(1
000
0
0
0TT
TA
AAAA PP
Pdt
d
V
N
dt
dx
V
Nr
AAT
T
T
T xN
N
P
P 100
1
1
0T
T
AA P
Px
dt
dP
RTdt
dP
PV
N
dt
dP
PV
N TT
T
TT
TA
A
000
0
00
0 1
Design equation for CVBR in terms of PT
A Fractional Volume / Pressure Change ??
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CVBR – Concentrations in terms of PT
)1()1()1(0
0
0
00 A
AA
AAAA x
RT
px
V
NxCC
1
1
0T
T
AA P
Px
)()(1
00
00
0000 TT
A
ATT
TATAAA PP
y
yPP
PPyxp
/)( 0TT PP
0
00
0
0 ]/)[()1(
RT
PPpx
RT
pC TTA
AA
A
0
00 ]/)[(
RT
PPabp
CTTB
B
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CVBR – Complex reactions: 1. First Order Reversible Reaction:A R
)()1( 0201210 AAAARAA
AA xMCkxCkCkCkdt
dxCr
2
1
0
0Re
)1(
)(
k
k
xC
xMC
C
CK
AeA
AeA
AeC
)()(
)1()1( 1
1 AAe
AeA
A xMxM
xkxk
dt
dx
tkxM
M
xx
x
AeAAe
Ae1)(
)1(ln
Integrating:
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CVBR – Complex reactions:
2. Irreversible Reactions in parallel: A BA C
AAAA
A CkkCkCkdt
dCr )( 2121
tkkC
C
A
A )(ln 210
Integrating:
AB
B Ckdt
dCr 1 A
CC Ck
dt
dCr 2
21 / kkdC
dC
C
B 2
1
0
0
k
k
CC
CC
CC
BB K1/k2
CC
CB t
CC
CB
CA
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CVBR – Complex reactions:
3. Homogenous Catalytic Reactions: A RA + C R + C
CAAA
A CCkCkdt
dCr 21
tktCkkC
CobsC
A
A )(ln 210
Integrating:t
kobs
k2
k1
CC
kobs
0
lnA
A
C
C
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CVBR – Complex reactions:
4. Auto Catalytic Reactions: A + R R + R
)( 0 AARAA
A CCkCCkCdt
dCr
)ln(ln 00
AA
A CCC
C
Integrating:t
kC0
CA = CR =0.5 C0
-rA
tkCCCC
CCC
AA
AA0
00
00
)(
)(ln
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CVBR – Complex reactions:
5. Irreversible Reactions in series: A B C CB0 = CC0 = 0
CB, max and topt - ??
CA/CA0CC/CA0
CB/CA0
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AA
A Ckdt
dCr 1
tkAA
A
A eCCorktC
C1
00
)(ln Integrating:
tktkAB ee
kk
CkC 21
12
01
BAB
B CkCkdt
dCr 21
tkAB
B eCkCkdt
dC1
012
Solving:
BAAC CCCC 0
12
12 )/ln(0
kk
kkt
dt
dCopt
B
)(
2
1
0
max,12
2kk
k
A
B
k
k
C
C
If k1 = k2, find topt and CR, max
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Consecutive I-order reactionsConc. vs. time for various ratios of k2/k1
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Consider the irreversible reactions in series: A R S
Concept of Rate Determining Step (RDS)
AA
A Ckdt
dCr 1
RA
RR CkCk
dt
dCr 21 R
SS Ck
dt
dCr 2
I. When k1 >> k2
dt
dC
ekk
ke
kkk
e
dt
dCCk
dt
dCr R
tktk
tkR
RS
S ~112
1
12
1
12
22
Overall rate of product formation is dominated by reaction - 2
II. When k2 >> k1
dt
dCe
kk
k
dt
dC
dt
dCr AtkkASS ~)1( )(
12
2 21
Overall rate of product formation is dominated by reaction - 1
Overall rate of a reaction is always governed by the slowest step, which is known as the rate determining step (RDS).
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6. Reactions with shifting order: A R
A
AAA Ck
Ck
dt
dCr
2
1
1
tkCCkCC AAAA 1020 )()/ln(
The order shifts from low to high (zero to one)as the reactant concentration drops.
t/(CA0-CA)
CA
-rA k1
k2
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7. Reactions with shifting order: A R
AA
A Ckkdt
dCr 21
tkCkk
Ckk
A
A2
21
021ln
The order shifts from high to low (one to zero)as the reactant concentration drops.
CA
-rA
k1
k2
t
CA
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Guggenheim's Method for First-Order Reactions
• A special method to obtain the rate constant for a first-order reaction when an accurate value of the initial reactant concentration is not available.
• Requires a series of readings of the parameter used to follow the progress of the reaction at times t1, t2, t3, etc. and at times t1 + ∆, t2 + ∆, t3 + ∆ etc.
• The time increment ∆ should be two or three times the half life of the reaction.
For a I order reaction: ln(1-xA) = -kt xA = 1 - e-kt
At t1 and t1 + ∆, (xA)t1 – (xA)t1+∆ = e-kt1 (1-ek∆)
Similar equations are valid at times t2, t3, etc. In all cases, the right side will be a constant, since the time increment is a constant.
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• applicable to systems that give apparent first-order rate constants.
• also applicable to irreversible first order reactions in parallel and reversible reactions that are first-order in both the forward and reverse directions.
• the technique provides an example of the advantages that can be obtained by careful planning of kinetics experiments instead of allowing the experimental design to be dictated entirely by laboratory convention and experimental convenience.
• Guggenheim's technique can also be extended to other order reactions, but the final expressions are somewhat cumbersome.
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Example:
Note that k can be determined without a knowledge of the dilatometer readings at times zero and infinity.
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Batch reactors are charged with reactants, closed, and heated to the reaction temperature, maintained isothermally for the duration of the reaction. After the reaction is completed, the mixture cooled, and the reactor opened, the product is discharged and the reactor is cleaned for the next batch. In industrial operations, the cycle time is constant from one batch to the next.
The time required for filling, discharging, heating, cooling, and cleaning the reactor is referred to as the turnaround time (tt).
The total batch cycle time tb is the reaction tr time plus the turnaround time tt.
tb = tr + tt
The total batch cycle time tb is used in batch reactor design to determine the productivity of the reactor.
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dt
dx
V
N
dt
dN
Vr AAAA
01
Design of Batch Reactor
How can you call the above equation as Design equation of a Batch Reactor ??
Can we Design the Batch Reactor using the Above equation ??
What do you mean by Design ??
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Design Problem
The reaction 2A R takes place in a batch reactor. Pure A is to be taken initially in the reactor. It is required to produce 3 tons of R per day. The molecular weight of R is 120. The density of A is 0.8 kg/lit. The expected conversion of A is 75%. A time of 30 min must be allowed for filling the reactor and 45 min for discharging and cleaning the reactor. Kinetic calculations show that a reaction time of 4hr 45 min is needed for the required conversion.
What size reactor must be used??
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SolutionTotal batch time = ½ + ¾ + 4¾ = 6 hrs.Number of batches / day = 4Required production /batch = ¾ tons = 750kg750 kg of A is required, if xA = 100%For 75% conversion: Amount of A to be fed /batch = 750/0.75 = 1000kgVolume of 1000 kg of A = 1250 lit. Size of the vessel = 1250 lit.
What is the use of the Design equation ??
Is the above Design always valid ??
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ANY CLARIFICATIONS ?
Gauss, KarlI have had my results for a long time;
but I do not yet know how I am to arrive at them.