BATCH REACTOR Interpretation of rate data
A. SARATH BABU
1
• Simplest reactor – open / closed vessel • Reac Reacta tant ntss are are place placed d ins insid idee the the react reactor or and allowed to react over time • Prod Product uctss and and uncon unconve vert rted ed react reactan ants ts are are removed and the process is repeated • Cl Clos osed ed syst system em - uns unstea teady dy stat statee oper operat ation ion • Fitted with a stirrer rer • May May have have a jac jacke kett / coo cooli ling ng or heat heatin ingg coils inside the reactor • Gene Genera rall llyy con const stan antt vol volum umee / some some designed at constant pressure • Mater Materia ials ls of cons constr truc ucti tion on – diff differe erent nt linings 2
3
4
B t hR t C td
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B t hR t A i ti
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• 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 ma make a va variety iety of pr produ oduct ctss at at different times BatchR hReactorCont.
7
Characteristics of a Batch Reactor (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; 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. 8
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 B t hR t C td
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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 pl ants 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).
General Mass Balance Equation: Input = output + accumulation + rate of disappearance
11
General Mass Balance Equation: Input = output + accumulation + rate of disappearance 12
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 13
Design Eqn. for CVBR -dCA/dt = -rA CAo dxA/dt = -rA
in terms of conversion
Design Eqn. for variable volume batch reactor CAo /(1+ε AxA) dxA/dt = -rA Design Eqn. in terms of Total Pressure (1/δ RT) dPT /dt = (-rA) 14
Constant Volume Batch Reactor C A
− d C A t = ∫ − r A C A 0
t C A0
x A
=
d x A
∫ − r 0
A
A 0
t
t = C a X are
15
Stoichiometric Table – Batch Systems aA + bB → rR + sS Species
AB R S I
Total
Initial
N NAB00 NR0 NS0 NI0
NT0
Change
Final moles
-NA0 xA A0 xA -(b/a)N +(r/a)NA0 xA +(s/a)NA0 xA 0
NNAB== N ) NAA00(1-x (MBA-(b/a)x A) NR= NA0 (MR+(r/a)xA) NS= NA0 (MS+(s/a)xA) NI = NI0
NT = NT0 + NA0 δxA
Where: MI = NI0 /NA0 δ = (r/a b/a –);1)C = C [M +(r/a)x ] For CVBR: CA+=s/a CA0–(1-x A R A0 R A
Constant Volume Batch Reactor -rA = -dCA/dt = CA0 dxA/dt 1. Zero Order Reaction: -rA = -dCA/dt = k C A
t
∫ − dC = ∫ k dt A
C A 0
0
CA0 - CA = kt CA0 xA = kt Strictly homogenous reactions do not follow zero order. Apparently the reaction order is made zero w.r.t. a reactant. 17
2. First Order Reaction: A → Products -rA = -dCA/dt = kCA C A
∫
C A 0
− dC A C A
t
= ∫ k dt 0
-ln (CA/CA0 ) = kt -ln(1-xA) = kt Example: N2O5
2NO2 + ½O2
-rA = k CA0.6 CB0.4 ?? Unimolecular – Collision theory ?? 18
First Order Reaction kinetics Influence of k
19
3. Second Order Reaction: 2A → Products A + B → Products CA0 = CB0 -rA = -dCA/dt = kCA2 C A
∫
C A 0
− dC A 2 A
C
t
= ∫ k dt 0
1/CA – 1/CA0 = kt xA/(1-xA) = kCA0 t
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4. Second Order Reaction: A + B → Products CA0 ≠ CB0 -rA = -dCA/dt = kCACB
C BC A0 C B 0C A
ln
K(CB0 -CA0 )
t
M − x A C BC A0 ln = ln = (C B 0 − C A0 )kt 1 − x A C B 0C A Example: CH3COOC2H5 + NaOH
CH3COONa+C2H5OH 21
5. Third Order Reaction: 3A → Products 2A + B → Products CA0 = 2CB0 A + B + C → Products CA0 = CB0 = CC0 -rA = -dCA/dt = kCACBCC = k CA3 ′
1 2 C A
−
1 2 C A0
= 2k ′t
Example: 2NO + H2 2NO + Cl2
H2O +N2O 2NOCl
22
6. Third Order Reaction: 2A + B → Products CA0
≠
2CB0
-rA = -dCA/dt = kCA2CB ln
C B C A
−
M C A
=
ln
C B 0 C A0
−
M C A0
2
−
M kt 2
23
7. Third Order Reaction: A + B + C → Products CA0
≠
CB0 ≠
CC0
-rA = -dCA/dt = kCACBCC ln (1 / 1 − x A ) C A0 ln( (C A0 − C B 0 )(C A0 − C C 0 )
+
+
ln (M B / M B − x A ) C B 0 ln( (C B 0 − C A0 )(C B 0 − C C 0 )
C C 0 ln(M C / M C − x A ) (C C 0 − C B 0 )(C C 0 − C A0 )
= kt
24
8. nth Order Reaction: nA → Products -rA = -dCA/dt = kCAn 1 n −1 C A
−
1− n A
C
1 n −1 C A0
= ( n − 1) kt
1− n A 0
− C = (n − 1)kt
25
The reciprocal of rate approaches infinity as C A → 0 26
Integrated forms – Constant density
27
Note that for a II order reaction with a large ratio of feed components, the order degenerates to a first order (pseudo first order).
28
Differential Method of analysis
CA
-rA
t
− r A =
k
f(c)
− dC A dt
= f (k , C ) = k f (C )
29
Differential Method of analysis
− r A =
− dC A
= k C
n A
dt ln(− r A ) = ln(k ) + n ln( C A )
n ln(-rA)
If –rA = kCAaCBb, how to use DM?
ln k
• Use stoichiometric ratio of reactants • Use method of excesses • Use method of least squares
ln(CA)
ln(−r A ) = ln(k ) + a ln( C A ) + b ln(C B ) 30
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32
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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34
Integral Method
Differential Method
• Easy to use and is recommended for testing
• Useful in complicated cases
specific mechanism • Require small amount of data • Involves trial and error • Cannot be used for fractional orders • Very accurate
• 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. 35
Method of Excesses Consider –rA = kCAaCBb • Perform the experiment with CB0 >> CA0 and measure CA as a function of t. –rA = kCAaCBb = kCB0 b CAa = k′ CAa 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 = kCA0 a CBb = k’′ CBb Use either differential method or integral method and evaluate k’’ & b Require multiple experiments
36
Method of Half lives 1 n −1 A
C
−
1 n −1 A 0
C
= (n − 1)kt
(1-n)
At t = t1/2 , CA = CA0 /2 t 1/ 2
=
1− n A 0
C (2
(n ≠ 1)
n −1
− 1)
(n − 1)k
ln(t1/2 ) K’ ln(CA0 )
(2n −1 − 1) ln(t 1/ 2 ) = (1 − n) ln(C A0 ) + ln n − k ( 1 ) For I Order reactions reactions:: t1/2 = ln(2)/k t1/2 does not depend on CA0 Require multiple experiments 37
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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Method of Fractional lives The half-life, or half-period, of a reaction is the time necessary for one half of the original reactant to disappear.
At t = t1/n , CA = (1- 1/n) CA0 t 1/ 2 t 1/ 2 t 1/ 3
=
C A1−0n (2 n −1 − 1) (n − 1)k n −1
− 1) = n −1 − (3 1) 2 (2
t 1/ 3 t 1/ 2 t 1/ 3
= =
1− n A 0
C ( 3
n −1
− 1)
2 (n − 1)k
ln 2 for n = 1 ln 3 2
The ratio of any two fractional lives is characteristic of the order.
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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 43 reaction.
Method of Initial Rates
(CA0 )1
(CA0 )2 (-rA0 )1 = slope at (CA0 )1, t = 0 (CA0 )3
(-rA0 )2 = slope at (CA0 )2, t = 0 (-rA0 )3 = slope at (CA0 )3, t = 0 Time, t
-rA0 = k (CA0 )n
ln(-rA0 ) = ln(k) + n ln(CA0 )
44
Method of Initial Rates
ln (-r A0 )
e p o o S l
r e r d o s e v g i
ln CA0 45
46
CA /CA0
DA= kCA0 n-1 t
Comparison of Different order Reactions in a Batch reactor
47
Variable Volume Batch Reactor r i = r i =
1 dN i
V dt dC i dt
+
=
1 d (C iV )
V
C i dV V dt
dt
dC i dV = V + C i V dt dt 1
Volume Change with time ??
Fractional Change in Volume or Expansion factor(ЄA): ε A
=
Change in total no. of moles when reaction is completed Total no. of moles fed ε A
=
V x A =1 − V X A =0 V X A =0
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. 48
V V 0
=
N T N T 0 ε A
N T N T 0
=1 +
N A 0 N T 0
δ x A
=δ y A0
=1 +δ y A 0 x A =1 +ε A x A δ =
r a
+
s a
−
b a
−1
Example: A → 3R, starting with pure A Since pure A, y = 1. Also δ = 3/1 -1 = 3-1 = 2. ЄA = (3 - 1)/1 = 2 With 50% inerts: y = 0.5 & δ = 3/1 -1 = 2. ЄA= (4-2)/2 = 1 = 0.5 (3-1) = 1 AO
AO
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Variable Volume Batch Reactor − 1 dN A
− r A = C A
V dt
=
C R C A C A0
N A
= =
V
=
− N A0 dx A C A0 −1 = = 1 + ε A x A V 0 (1 + ε A x A ) dt
N A0 (1 − x A ) V 0 (1 + ε A x A )
C A0 [ M R
=
dx A dt
C A0 (1 − x A ) 1 + ε A x A
+ (r / a) x A ]
1 + ε A x A 1 − x A 1 + ε A x A
x A
=
1 − C A / C A0 1 + ε AC A / C A0
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CVBR
t
C A
t =
d C r − / − A A ∫
C A0
C A 0
x A
=
VVBR
d x A
∫ − r (1 + 0
A
ε A x A
)
t / CA0 t
1
1 /-rA
−
CA
CA0
r A (1 + ε A x A )
xA 51
Variable Volume Batch Reactor 1. Zero Order Reaction: − r A = x A
C A0
C A0
dx A
1 + ε A x A dt
dx A
= k
t
∫ 1 + ε x = ∫ k dt 0
A A
)
)
A A
x x Є
A A
Є + + 1 1 ( ( n n L L
kЄA/CA0
0
C A0 ln(1 + ε A x A ) = k ε At
t
C A0 ln(V / V 0 ) = k ε At 52
Variable Volume Batch Reactor 2. First Order Reaction: − r A =
C A0
dx A
1 + ε A x A dt
x A
dx A
= kC A = k
A
1 + ε A x A
t
∫ 1 − x = ∫ k dt 0
C A0 (1 − x A )
0
− ln(1 − x A ) = kt
)
)
A A
x x - 1 1 ( ( l n l n - -
t
* Performance equation is similar to that of CVBR 53
Variable Volume Batch Reactor 3. Second Order Reaction: − r A = x A
dx A
1 + ε A x A dt
(1 + ε A x A )dx A
∫ 0
C A0
(1 − x A )
2
= kC
2 A
C A0 (1 − x A ) = k 1 ε + x A A
2
t
= ∫ k C A0 dt 0
x A (1 + ε A ) /(1 − x A ) + ε A ln(1 − x A ) = kC A0t
54
Variable Volume Batch Reactor 4. Higher Order Reactions: − r A = x A
dx A
1 + ε A x A dt
(1 + ε A x A )
∫ 0
C A0
n −1
(1 − x A )
n
= kC
dx A
n A
C A0 (1 − x A ) = k 1 ε + x A A
n
t
= ∫ k C An−01dt 0
Analytical integration would be difficult. Resort to either graphical / numerical integration. 55
Constant Volume Batch Reactor (PT vs. t) P T
=
P T 0 ε A
N T N T 0
=1 +ε A x A
Fractional Volume / Pressure Change ??
x A =
− r A = =
1 P T ε A
−1 P T 0
N A0 dx A V 0
dt
N A0
=
N A0 d
V 0 dt ε A P T 0
dP T
V 0ε A P T 0 dt
1
=
( P T − P T 0 )
N T 0 dP T V 0δ P T 0 dt
=
1
dP T
δ RT 0
dt
Design equation for CVBR in terms of P T
56
CVBR – Concentrations in terms of P T C A = C A 0 (1 − x A ) =
N A 0 V 0
(1 − x A ) =
p A 0
RT 0
(1 − x A )
1 P T
x A = −1 ε A P T 0 y A0 1 p A0 x A = y A0 P T 0 ( P T − P T 0 ) = ( P T − P T 0 ε A P δ y A0 T 0 = ( P T − P T 0 ) / δ p A 0 p A 0 −[( P T − P T 0 ) / δ ] (1 − x A ) = C A = RT 0 RT 0 C B =
p B 0 −b
a
[( P T − P T 0 ) / δ ] RT 0 57
CVBR – Complex reactions: 1. First Order Reversible Reaction:A − r A = C A0 K C = dx A dt
dx A
C Re C Ae
dt
=
= k 1C A − k 2C R = k 1C A0 (1 − x A ) − k 2C A0 ( M + x A )
C A0 ( M + x Ae ) C A0 (1 − x Ae )
= k 1 (1 − x A ) −
ln 58
x Ae − x A
=
=
k 1 (1 − x Ae ) ( M + x Ae )
Integrating: x Ae
⇔R
( M + 1) ( M + x Ae )
k 1t
k 1 k 2
( M + x A )
CVBR – Complex reactions: 2. Irreversible Reactions in parallel: A →B C A →C
CC
A
− r A =
− dC A dt
= k 1C A + k 2C A = (k 1 + k 2 )C A
Integrating: − ln r B
=
dC B dC C
dC B dt
= k 1C A
= k 1 / k 2
C A C A0
CB
= (k 1 + k 2 )t
r C =
dC C dt
= k 2C A
CB
t
K1/k2
− C B 0 k 1 = C C − C C 0 k 2 C B
CC 59
60
CVBR – Complex reactions: 3. Homogenous Catalytic Reactions: A →R A + C →R + C C − ln dC A − C − r = = k C + k C C A
A
dt
1
A
2
A
A 0
C
kobs
Integrating: C A − ln C A 0
t
= ( k 1 + k 2C C )t = k obs t kobs
k2 k1 61
CVBR – Complex reactions: 4. Auto Catalytic Reactions: A + R →R + R dC A − − r A = = kC AC R = kC A (C 0 − C A ) dt
Integrating: C A − ln C A 0
kC0
t
+ ln( ln( C 0 − C A )
C A 0 (C 0 − C A ) = ln C A (C 0 − C A0 )
-rA
= kC 0t CA = CR =0.5 C0 62
CVBR – Complex reactions: 5. Irreversible Reactions in series: A → B → C CB0 = CC0 = 0 CC/CA0
CA/CA0
CB/CA0
C
and t
- ??
63
− r A =
− dC A dt
= k 1C A C A − ln C A 0
Integrating: r B
=
dC B dt
dC B dt
= kt
(or ) C A
= C A0 e
−k 1t
= k 1C A − k 2C B
+ k 2C B = k 1C A0e −k t 1
Solving: C B = C C dC B dt
k 1C A 0 −k 1t (e − e −k 2t ) k 2 − k 1
= C A0 − C A − C B
= 0 ⇒ t opt
ln( ln( k 2 / k 1 ) = k 2 − k 1
If k = k , find t
and C
C B , max C A 0
k 2
k 1 = k 2
( k 2 −k 1 )
64
Consecutive I-order reactions Conc. vs. time for various ratios of k 2/k1
65
Concept of Rate Determining Step (RDS) Consider the irreversible reactions in series: A → R → S − r A =
− dC A dt
= k 1C A
r R
= dC R = k 1C A − k 2C R
r S
dt
=
dC S dt
= k 2C R
I. When k1 >> k2
− k t dC S dC e dC R ~ = k 2C R = − r S = − R 1 − k k dt dt dt − k t − k t 2 1 k − k e − k − k e 2 1 2 1 1
2
1
Overall rate of product formation is dominated by reaction - 2 II. When k2 >> k1 r S
dC k dC = dC S = − A 2 (1 − e ( k −k )t ) ~ − A dt dt k 2 − k 1 dt 1
2
Overall rate of product formation f ormation 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). 66
6. Reactions with shifting order: A → R − r A = − dC A = k C A 1
dt
1 + k 2C A
The order shifts from low to high (zero to one) as the reactant concentration drops. ln(C A0 / C A ) + k 2 (C A0 − C A ) = k 1t
k1
-rA k2
t/(CA0 -CA)
CA 67
7. Reactions with shifting order: A → R − r A = − dC A = k + k C A 1
dt
2
The order shifts from high to low (one to zero) as the reactant concentration drops. ln
-rA
k 1 + k 2C A0 k 1 + k 2C A
k2
= k 2t
CA
k1 CA
t 68
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 t 1 + ∆, 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 t 2 , t 3 , etc. In all cases, the right side will be a constant, since the time increment is a constant. 69
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. •
70
Example:
Note that k can be determined without a knowledge of the dilatometer readings at times zero and infinity.
71
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 72
Design of Batch Reactor What do you mean by Design ?? − r A =
− 1 dN A V dt
=
N A0 dx A V
dt
Can we Design the Batch Reactor using the Above equation ?? How can you call the above equation equ ation as Design equation of a Batch Reactor ?? 73
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. conversion. What size reactor must be used?? 74
Solution Total batch time = ½ + ¾ + 4¾ = 6 hrs. Number of batches / day = 4 Required production /batch = ¾ tons = 750kg 750 kg of A is required, if xA = 100% For 75% conversion: Amount of A to be fed /batch = 750/0.75 = 1000kg Volume of 1000 kg of A = 1250 lit. ∴ Size of the vessel = 1250 lit. Is the above Design always valid ?? What is the use of the Design equation e quation ??
75
ANY CLARIFICATIONS ? Gauss, Karl
I have had my results for a long time; but I do not yet know how I am to arrive at them. 76
