1. 반응공학 1 Chapter 4 Isothermal Reactor Design
Chemical Reaction Engineering 1
반응공학 1

2. Objectives
• Describe the algorithm that allows the reader to solve chemical reaction engineering problems through logic rather than memorization.
• Sizing batch reactors, semi-batch reactors, CSTRs, PFRs, and PBRs for isothermal operation given the rate law and feed conditions.
• Studying a liquid-phase batch reactor to determine the specific reaction rate constant needed for the design of a CSTR.
• Design of a tubular reactor for a gas-phase reaction.
• Account for the effects of pressure drop on conversion in packed bed tubular reactors and in packed bed spherical reactors.
• The principles of unsteady operation and semi-batch reactor.

3. Fig. 4-1
Isothermal
Reaction
Design
Algorithm
for Conversion

4. 4.1 Design structure for isothermal reactors
Algorithm for isothermal reactor design
1. Mole balance and design equation
2. Rate law
3. Stoichiometry
4. Combine
5. Evaluate
We can solve the equations in the combine step either
A. Graphically (Chapter 2)
B. Numerical (Appendix A4)
C. Analytical (Appendix A1)
D. Software packages (polymath)

5. Algorithm for Isothermal Reactors

6. French Menu Analogy

7. Scale-up of Liquid-Phase Batch Reactor
to the Design of a CSTR
Pilot plant
Operation
Full-scale
Production
Laboratory
Experiment
PAST
Full-scale
Production
Microplant
(Lab-bench-scale unit)
High cost of a pilot-plant leads to jump pilot plant operation
FUTURE
To make this jump successfully requires
a through understanding of the chemical kinetics and transport limitations.

8.   ( ) Batch Operation CA=CAo(1-X) dX CAo =kCAo2(1-X)2 dt 1 dX = dt
Algorithm for isothermal reactor design
1. Mole balance and design equation
2. Rate law
3. Stoichiometry
4. Combination
5. Analytical Evaluation
CA=CAo(1-X) dX
CAo
=kCAo2(1-X)2 dt 1 X t  dX 
= dt
kCAo
(1-X)2
( ) 1 X t=
kCAo
1-X

9. Batch Reaction Times Mole balance Rate law Stoichiometry (V=V0)
Combine
Integration

10. Batch Reaction Times

11. Batch Reaction Times (Table 4-2)
The order of magnitude of time
to achieve 90% conversion
For first- and second-order irreversible batch reactions
1st-order
k (s-1)
2nd-order
kCA0 (s-1)
Reaction time tR
10-4
10-3
Hours
10-2
10-1
Minutes 1 10
Seconds
1,000
10,000
Milliseconds

12. Reaction Time in Batch Operation (Table 4-3)
2nd order
Isothermal
Liquid-phase
Batch reaction
This time is the time t needed to reduce the reactant concentration in a batch reactor from an initial value CA0 to some specified value CA.
Typical cycle times for a batch polymerization process
tt = tf + te + tR + tc
Activity
Time (h)
Heat to reaction temperature, te
Charge feed to the reactor and agitate, tf
Carry out reaction, tR
Empty and clean reactor, tc
Varies (5-60)
Total cycle time excluding reaction
Decreasing the reaction time with a 60-h reaction is a critical problem. As the reaction time is reduced, it becomes important to use large lines and pumps to achieve rapid transfer and to utilize efficient sequencing to minimize the cycle time

13. Example 4-1: Design a Reactor Producing Ethylene Glycol
Design a CSTR to Produce 200 million pounds of ethylene glycol per year by hydrolyzing ethylene oxide. However, before the design can be carried out , it is necessary to perform and analyze a batch reactor experiment to determine the specific reaction rate constant (kA). Since the reaction will be carried out isothermally, kA will need to be determined only at the reaction temperature of the CSTR. At high temperature there is a significant by-product formation, while at temperature below 40oC the reaction does not proceed at a significant rate; consequently, a temperature of 55oC has been chosen. Since the water is usually present in excess, its concentration may be considered constant during the course of the reaction. The reaction is first-order in ethylene oxide.
O CH2-OH
CH2-CH2 + H2O CH2-OH
H2SO4
A B C
Catalyst

14. Example 4-1: Determining k from Batch Data
In the lab experiment, 500mL of a 2 M solution (2 kmol/m3) of EO in water was mixed with 500mL of water containing 0.9 wt % sulfuric acid catalyst. At T=55oC, the CEG was recorded with time.
Time Concentration of EG
(min) (kmol/m3)

15. Problem Solving Algorithm
Example 4-1 Determining k from Batch Data
Problem statement. Determine the kA
Sketch
Identify
C1. Relevant theories
Rate law:
Mole balance:
C2. Variables
Dependent: concentrations
Independent: time
C3. Knowns and unknowns
Knowns: CEG = f(t)
Unknowns:
1. CEO = f(t)
2. kA
3. Reactor volume
C4. Inputs and outputs: reactant fed
all at once a batch reactor
C5. Missing information: None
Assumptions and approximations:
Assumptions
1. Well mixed
2. All reactants enter at the same time
3. No side reaction
4. Negligible filling and emptying time
5. Isothermal operation
Approximations
1. Water in excess (CH2O~constant)
E. Specification. The problem is neither overspecified nor underspecified.
F. Related material. This problem uses the mole balances developed in Chap. 1 for a batch reactor and the stoichiometry and rate laws developed in Chap. 3.
G. Use an Algorithm.

16.   A + B  C NC=NAoX=NAo-NA
dCA
= -kCA dt CA
dCA t  
CAo - = k dt ln
= kt
CA = CAoe-kt
CA0 CA CA
A B  C
NC=NAoX=NAo-NA
CC=NC/V=NC/Vo= CAo-CA=CAo- CAoe-kt =CAo(1-e-kt)

17. Rearranging and taking the logarithm of both side yields
We see that a plot ln[(CA0-CC)/CA0] as a function of t will be a straight line with a slope –k.

18. Design Equation for a CSTR
Design of CSTR
Design Equation for a CSTR
Mole balance
the space time
For a 1st-order irreversible reaction, the rate law is
Rate law
Combine
Solving for the effluent concentration of A, we obtain
Relationship between space time and conversion for a 1st-order liquid-phase rxn

19. Reaction Damköhler number
The Damkohler is a dimensionless number that can give us a quick estimate of the degree of conversion that can be achieved in continuous-flow reactor.
For 1st-order irreversible reaction
For 2nd-order irreversible reaction
Da  0.1 will usually give less than 10% conversion.
Da  10.0 will usually give greater than 90% conversion.
For first order reaction, X = = k Da
1 + k
1 + Da

20. CSTRs in Series
CA1, X1
CA2, X2
CA0 v0
-rA1, V1
-rA2, V2
For first-order irreversible reaction with no volume change (v=v0) is carried out in two CSTRs placed in series. The effluent concentration of A from reactor 1 is
From a mole balance on reactor 2,

21. CSTRs in Series X= 1-CA/CAo 1 = 1 - (1+Da)n
Solving for CA2, the concentration exiting the second reactor, we get
If instead of two CSTRs in series we had n equal-sized CSTRs connected in series (t1 = t2 = … = tn = ti = (Vi/v0)) operating at the same temperature (k1 = k2 = … = kn = k), the concentration leaving the last reactor would be
The conversion and the rate of disappearance of A for these n tank reactors in series would be
X= 1-CA/CAo 1
= 1 -
(1+Da)n

22. Conversion as a Function of Reactors in Series
for different Damkohler numbers for a first-order recation
Da=k=1
Da=k=0.5
Da=k=0.1
Da  1, 90% conversion is achieved in two or three reactors;
thus the cost of adding subsequent reactors might not be justified
Da ~0.1, the conversion continues to increase significantly with each reactor added

23. CSTRs in Parallel A balance on any reactor i, gives the individual
reactor volume
-rA1, V1, X1
-rAi, Vi
-rAn, Vn
FA0
FA0 i 1 i n
The volume of each individual reactor, Vi, is related to the total volume, V, of all the reactors, and similar relationship exists for the total molar flow rate

24. CSTRs in Parallel Substituting these values into Eq (4-12) yields
FA01
-rA1, V1
-rAi, Vi
-rAn, Vn
FA0
FA0i 1 i n
-rA1, V1, X1
-rAi, Vi, Xi
FA0n
The conversion achieved in any one of the reactors in parallel is identical to what would be achieved if the reactant were fed in one stream to one large reactor of volume V
-rAn, Vn, Xn

25. A Second-Order Reaction in a CSTR
For a 2nd-order liquid-phase reaction being carried out in a CSTR, the combination of the rate law and the design equation yields
We solve the above eq. for X:
(4-14)
For const density v=v0, FA0X=v0(CA0-CA)
(4-16)
The minus sign must be chosen in the quadratic equation because X cannot be greater than 1.
Using our definition of conversion, we have
(4-15)

26. A Second-Order Reaction in a CSTR60 6
0.67
0.88

27. Example 4-2: Producing 200 Million Pound/Year in a CSTR
A 1 lb mol/ft3 solution of ethylene oxide (EO) in
water is fed to the reactor together with an equal
volumetric solution of water containing 0.9 wt%
of the catalyst H2SO4. The specific reaction rate
constant is min-1.
(a) If 80% conversion is to be achieved, determine
the necessary CSTR volume.
(b) If two 800-gal reactors were arranged in parallel,
what is the corresponding conversion?
(c) If two 800-gal reactors were arranged in series,

34.  ( ) ( ) Tubular Reactors For Liquid-Phase Reaction = FAo V/vo=  =
Rate law
- 2nd-order gas-phase reaction
- Turbulent
- No dispersion
- No radial gradients in T, u, or C
PLUG-FLOW REACTOR
PFR mole balance
For Liquid-Phase Reaction X dX 
= FAo
kCAo2(1-X)2 vo
( ) X =
kCAo
1-X 1
( ) X
must be used when there is a DP or heat exchange between PFR & the surrounds. In the absence of DP or heat exchange, the integral form of the PFR design equation is used.
V/vo=  =
kCAo
1-X
kCAo
Da2 X= =
1+kCAo
1 + Da2
(Da2=Damköhler number for a 2nd-order RXN)
For n-th order RXN, Dan=kCAon-1

35. For Gas-Phase Reaction

36. Conversion as a Function of Distance Down the Reactor
v=vo(1+X)
the reactant spends more time
the reactant spends less time ö
V(m3)
The volumetric flow rate decreases with increasing conversion, and the reactant spends more time in the reactor than reactants that produce no net change in the total number of moles.

37. Change in Gas-Phase Volumetric Flow Rate Down the Reactor
v=vo(1+X)

38. Example 4-3: Determination of a PFR Volume
Determine the PFR volume necessary to produce 300 million pounds of ethylene a year
from cracking a feed stream of pure ethane. The reaction is irreversible and follows an
elementary rate law. We want to achieve 80% conversion of ethane, operating the
reactor isothermally at 1100K at a pressure of 6 atm.
C2H6 (A)  C2H4 (B) + H2 (C)
FB=300x106 lb/year=0.340 lb-mol/sec
FB=FAoX
FAo=FB/X=0.340/0.8=0.425 lb-mol/sec

44. Pressure Drop in Reactors
 In liquid-phase reaction
- the concentration of reactants is insignificantly affected by even
relatively large change in the total pressure
- ignore the effect of pressure drop on the rate of reaction
when sizing liquid-phase chemical reactors
- that is, pressure drop is ignored for liquid-phase kinetics calculations
 In gas-phase reaction
- the concentration of the reacting species is proportional to
the total pressure
- the effects of pressure drop on the reaction system are a key factor
in the success or failure of the reactor operation
- that is, pressure drop may be very important for gas-phase reactions

45. Pressure Drop and Rate Law
• For an ideal gas,
- determine the ratio P/P0 as a function of V or W
- combine the concentration, rate law, and design equation
- the differential form of the mole balance (design equation) must be used
기상반응에서는 반응 성분
의 농도가 반응압력에 비례
하므로 압력강하에 대한
고려가 필수적이다. To
(4-18) T

46. Pressure Drop and Rate Law
• For example,
- the second order isomerization reaction in a packed-bed reactor
2A  B + C
-the differential form of the mole balance
- rate law
-from stoichiometry for gas-phase reactions

47. Pressure drop and the rate law
• Then, the rate law
- the larger the pressure drop from frictional losses, the smaller the reaction rate
• Combining with the mole balance and assuming isothermal operation (T=To)
• Dividing by FA0
(4-20)

48. Pressure Drop and Rate Law
• For isothermal operation (T =T0)
a function of only conversion and pressure
-Another equation is needed to determine the conversion as a function of catalyst weight
- that is, we need to relate the pressure drop to the catalyst weight
(4-21)

49. Flow Through a Packed Bed
• The majority of gas-phase reactions are catalyzed by passing the reactant through a packed of catalyst particles
• Ergun equation: to calculate pressure drop in a packed porous bed
(4-22)
laminar
turbulent
G=ru=superficial mass velocity [g/cm2s]; u=superficial velocity [cm/s]; Dp=diameter of particle in the bed [cm]; f=porosity=volume of void/total bed volume; 1- f =volume of solid/total bed volume
• The gas density is the only parameter that varies with pressure on the right-hand side. So, calculate the pressure drop through the bed laminar turbulent

50. Flow through a Packed Bed
• Equation of continuity
mo = m  
- steady state  the mass flow rate at any point is equal to the entering
mass flow rate
• Gas-phase volumetric flow rate
(3-41)
• Then,
(4-23)

51. Pressure Drop in a Packed Bed Reactor
• then, Ergun equation
• Simplifying
(4-24)
(4-25)
We need
• The catalyst weight,
(4-26)
Volume of
solid
Density of
solid catalyst

52. Pressure Drop in a Packed Bed Reactor
• then, Ergun equation
• Simplifying
(4-27) dy FT
Let y=P/Po
(4-28) 2y
FTo
(4-29)

53. Pressure drop in a packed bed reactordy
(1+X)
(4-30) 2y
ε < 0, the pressure drop (DP) will be less than ε = 0
ε > 0, the pressure drop (DP) will be greater than ε = 0
• For isothermal operation
and
(4-31)
• The two expressions are coupled ordinary differential equations. We can solve them simultaneously using an ODE solver such as Polymath.
• For the special case of isothermal operation and ε = 0, we can obtain an analytical solution.
• Polymath will combine the mole balance, rate law and stoichiometry

54. Pressure Drop in a Packed Bed Reactordy
(1+X)
(4-30) 2y
Analytical Solution
If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes dy
Isothermal (T=To) with ε = 0 2y dy
dy2 2y
= - 
= -  dW dW
At W=0, y=1 (P/Po=1)
y2= 1- w

55. Pressure Drop in a Packed Bed Reactor
If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes
Pressure ratio only for ε = 0 y=
(4-33)
(4-26)
Pressure as a function of reactor length, z y=
=f (z)
(4-34)

56. Example 4-4: 1½” schedule 40 x1000-ft L (ap=0.018), DP<10%
Pressure Drop in Pipes
Pressure drop along the length of the pipe
Integrating with P=P0 at L=0, and assuming that f = constant
Rearranging, we get
Example 4-4: 1½” schedule 40 x1000-ft L (ap=0.018), DP<10%

57. Analytical Solution for Reaction with Pressure Drop
Conversion as a function of catalyst weight
Combining
2nd-order isothermal reaction
A B
Mole balance:
Separating variable and Integrating
Rate law:
Stoichiometry: Gas-phase isothermal with e=0
(4-38) y=

58. Reaction with Pressure Drop
Conversion as a function of catalyst weight
(4-38)
Catalyst weight for 2nd-order isothermal reaction in PFR with DP
(4-39)

59. Reaction with Pressure Drop
Conversion as a function of catalyst weight
For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop.

60. Example 4-5 and Example 4-6 지금까지 배운 지식을 활용하여 적들이 방심하고 있는 사이에
이 예제들은 집에서 풀어 봐야지...

61. Spherical Packed-Bed Reactors
Spherical Ultraformer Reactor (Amoco) for dehydrogenation reaction such as
Paraffin Aromatic H2
 Spherical reactor
- minimize pressure drop
- inexpensive
- the most economical shape for high pressure

62. Coordinate system and variables
used with a spherical reactor

63. Synthesizing a Chemical Plant
 Always challenge the assumptions, constraints, and boundaries of the problem
 The profit from a chemical plant will be the difference between income from sales and the cost to produce the chemical
Profit = (value of products) – (cost of reactants)
– (operating costs) – (separation costs)
 The operating cost: energy, labor, overhead, and depreciation of equipment

64. Production of Ethylene Glycol2
H2, C2H4
C2H6
separator
C2H6  C2H4 + H2 1
V=81 ft3, X=0.8
402 million
lbC2H6 /yr 3
C2H4 4
Air
W=45,440 lb, X=0.6
C2H4+ ½ O2  C2H4O Ag 5
O2, C2H4, N2, C2H4O 6
separator
H2O
C2H4O
absorber
C2H4O(aq) 7 8
H2O, 0.9wt% H2SO4
200 million
lb EG/yr 9
V=197 ft3, X=0.8
CH2OH
C2H4O + H2O 
Cat.

65. Synthesizing a Chemical Plant
 Ethylene glycol = $0.38/lb (2x108 lb/yr)
Ethane = $0.04/lb (4x106 lb/yr)
Sulfuric acid = $0.043/lb (2.26x108 lb/yr)
Operating cost = $8x106/yr
 Profit = ($0.38/lb x 2x108 lb/yr) – ($0.04/lb x 4x108 lb/yr)
-($0.043/lb x 2.26x106 lb/yr) – ($8x106/yr)
= $52 million
 How the profit will be affected by conversion, separation, recycle stream, and operating costs?

66. Using CA (liquid) and FA (gas) in the mole balance and rate laws
 More convenient to work in terms of the number of moles or molar flow rate rather than conversion.
Membrane reactor, multiple reaction, and unsteady state
 Must write a mole balance on each species when molar flow rates (Fi) and concentrations (Ci) are used as variables

67. Mole balance for liquid-phase reactions
For liquid-phase reaction with no volume change  Concentration is the preferred variable
Mole balance for liquid-phase reactions
Batch
CSTR
PFR
PBR

68. Gas Phase  For gas phase reactions  Total molar flow rate
 need to be expressed in terms of the molar flow rates
(1-11) y
(3-42)
 Total molar flow rate dy - FT =
(4-28, T=To) dW 2y
FTo
= FA + FB + FC + FD + FI

69. Algorithm for Gas Phase Reaction
 Mole balances:
Batch
CSTR
PFR

70. Algorithm for Gas Phase Reaction
 Rate law:
 Stoichiometry:
Relative rate of reaction: y
Concentration:
Total molar flow rate:

71. Algorithm for Gas Phase Reaction
 Combine:
- Specify the parameter values: kA, CT0, a, b, T0, a, b, c, d
- Specify the entering number: FA0, FB0, FC0, FD0, and final value: Vfinal
 Use an ODE solver

72. Example 4-7: Gas-Phase Reaction in a Microreactor
Microreactors
 Microreactors are used for the production of special chemicals, combinatorial chemical screening, lab-on-a-chip, and chemical sensors.
Example 4-7: Gas-Phase Reaction in a Microreactor
The gas phase reaction, 2NOCl 2NO + Cl2, is carried out at 425oC and 1641 kPa
(16.2 atm). Pure NoCl is to be fed, and the reaction follows an elementary rate law.
It is desired to produce 20 tons of NO per year in a micro reactor system using a
bank of ten microreactors in parallel.
Each microreactor has 100 channels
with each channel 0.2 mm square and
250 mm in length. Plot the molar flow
rates as a function of volume down the
length of the reactor. The volume of
each channel is 10-5 dm3.

73. Rate constant and activation energy (given): k=0
Rate constant and activation energy (given): k=0.29 dm3/mol-sec at 500K with E=24 kcal/mol
To produce 20 tons/year of NO at 85% conversion would require a feed rate of mol/s.
Solution
For one channel,
FAo=22.6 mol/s
FB=19.2 mol/s, X=0.85
2NOCl  2NO + Cl2
2A  2B + C
A  B +1/2C
1. Mole balance:
2. Rate law:
dFA
-rA=kCA2, k=0.29 dm3/mol-sec
= rA dV
dFB
= rB dV
dFC
= rC dV

74. rB rC rA 1/2 -1 1 rB = -rA, rC= -0.5rA dFA dV dFB dV dFC dV
3. Stoichiometry: Gas phase with T=To and P=Po, then v=vo(FT/FTo)
3-1. Relative rate rA rB rC = = -1 1
1/2
rB = -rA, rC= -0.5rA
3-2. Concentration
By applying Equation (3-42)
CA=CTo(FA/FT), CB=CTo(FB/FT), CC=CTo(FC/FT) with FT=FA+FB+FC
4. Combine
5. Evaluate
-rA=kCA2=kCTo2(FA/FT)2
CTo=Po/RT=(1641)/(8.314)(698K)
=0.286 mol/dm3=0.286 mmol/cm3
dFA
= -kCTo2(FA/FT)2 dV
dFB
= kCTo2(FA/FT)2
Use Polymath to solve the ODE dV
dFC
= 0.5kCTo2(FA/FT)2 dV

75. FA FB FC

76. Membrane Reactors
 Membrane reactors can be used to achieve conversions greater than the original equilibrium value. These higher conversions are the result of Le Chatelier's Principle; you can remove one of the reaction products and drive the reaction to the right. To accomplish this, a membrane that is permeable to that reaction product, but is impermeable to all other species, is placed around the reacting mixture
 By having one of the products pass throughout the membrane, we drive the reaction toward completion

77. What kinds of membrane reactors are available?
 Membrane reactors are most commonly used when a reaction involves some form of catalyst, and there are two main types of these membrane reactors: the inert membrance reactor and the catalytic membrane reactor.
 The inert membrane reactor allows catalyst pellets to flow with the reactants on the feed side (usually the inside of the membrane). It is known as an IMRCF, which stands for Inert Membrane Reactor with Catalyst on the Feed side. In this kind of membrane reactor, the membrane does not participate in the reaction directly; it simply acts as a barrier to the reactants and some products.
 A catalytic membrane reactor (CMR) has a membrane that has either been coated with or is made of a material that contains catalyst, which means that the membrane itself participates in the reaction. Some of the reaction products (those that are small enough) pass through the membrane and exit the reactor on the permeate side.

78. Membrane Reactors C6H12  C6H6 + 3H2
 Inert membrane reactor with catalyst pellets on the feed side (IMRCF)
C6H12  C6H6 + 3H2
H2 molecule is small enough to diffuse through the small pore of the membrane while C6H12 and C6H6 cannot..

79. Membrane Reactors C6H12  C6H6 + 3H2
 Catalytic membrane reactor (CMR)
C6H12  C6H6 + 3H2
Membrane reactors are commonly used in dehydrogenation reactions (e.g., dehydrogenation of ethane), where only one of the products (molecular hydrogen) is small enough to pass through the membrane. This raises the conversion for the reaction, making the process more economical.

80. Dehydrogenation Reaction
 According to The DOE, an energy saving of 10 trillion Btu per year could result from the use of catalytic membrane reactors as replacements for conventional reactors for dehydrogenation reactions such as the dehydrogenation of ethylbenzene to styrene.
A  B + C
CH2CH3
CH=CH
+ H2 (B)
C4H10  C4H8 + H2 (B)
C3H8  C3H6 + H2 (B)

81. Basic Algorithm for Membrane Reactor (Example 4-8)
RB=kcCB
A  B + C FA FB DV FC
membrane
V+DV RB
KC=0.05
P=8.2 atm
T=227oC
FA0=10mol/min
There are two “OUT” terms
1. Mole balance:
for a differential mole balance on A
in the catalytic bed at steady state
for a differential mole balance on B
in the catalytic bed at steady state
for a differential mole balance on C
in the catalytic bed at steady state

82. Basic Algorithm for Solving Reaction in the Membrane Reactor
2. Rate law:
3. Transport out the sides of the reactor:
kc is a transport coefficient. kc=f(membrane & fluid properties, tube diameter…)  constant
4. Stoichiometry:

83. Basic Algorithm for Solving Reaction in the Membrane Reactor
5. Combining and Summarizing:
6. Parameter evaluation:
CT0=0.2 mol/L, k=0.7 min-1, KC=0.05 mol/L, kc=0.2 min-1
FA0=10 mol/min, FB0=FC0=0
7. Numerical solution:
Solve with POLYMATH or MATLAB

84. Effects of Side Stream, RB=kcCB
in a membrane reactor
Reactor volume, V [L] FA FB FC 10 5
100
200
300
400
500
kc=0.20 min-1
Conversion
X=(10-4)/10=0.6
Reactor volume, V [L] FA FB FC 10 5
100
200
300
400
500
kc=20 min-1
Large side stream
Reactor volume, V [L] FA FB FC 10 5
100
200
300
400
500
kc= min-1
Little side stream

85. Unsteady-State Operation of Stirred Reactors
 Determine the time to reach steady-state operation
 Predict the concentration and conversion as a function of time B C
A, B A
 Semibatch reactor
(b) ammonolysis, chlorination, hydrolysis
(c) acetylation reaction, esterification reaction

86. Utilizing the definitions of FA and NA, we have
Startup of a CSTR
Time to Reach Steady State for a First-Order Reaction in a CSTR
To determine the time to reach steady-state operation of a CSTR, we begin with the general mole balance equation applied to a well-mixed CSTR.
(4-45)
Utilizing the definitions of FA and NA, we have
Conversion does not have any meaning in startup because one cannot separate moles reacted from moles accumulated. Consequently, we must use concentration as the variable in our balance equation. For liquid-phase reactions V =V0 and for a constant overflow, v =v0. After dividing by v0 and replacing V/v0 by the space time t, we find that
(4-46)

87. Startup of a CSTR For a first-order reaction: Solution (4-47)
Letting ts be the time necessary to reach 99% of the steady state concentration, CAS:
Rearranging (4-46) for CA=0.99CAS yields
(4-48)

88. Startup of a CSTR Time to reach steady state in an isothermal CSTR
For most first-order system, steady state is achieved in three to four space times
Time to reach steady state in an isothermal CSTR
(4-48)
For slow reactions :
For rapid reactions:
(4-49)
(4-50)

89. Semibatch Reactors A A B B Mole balance on specials A:
volume as a
function of time
(4-55) B B A
(4-56) A
Mole balance on specials A:
[ in ] [ out ] [ gen. ] = [ acc. ]
Mole balance on specials B:
[ in ] [ out ] [ gen. ] = [ acc. ]
(4-51)
(4-52)
(4-57)
Mass balance of all specials:
[ in ] [ out ] [ gen. ] = [ acc. ]
(4-53)
(4-54)
(4-58)

90. Mole balance of A, B, C, and D
Example 4-9: Isothermal Semibatch Reactor with 2nd–order Reaction
CNBr + CH3NH2  CH3Br + NCNH2
A B  C D
Isothermal elementary reaction
in a semibatch reactor
t=0, CA=0.05 gmol/, CB=0.025 gmol/ℓ,
v0=0.05ℓ/s, k=2.2ℓ/s·mol, V0=5ℓ
Mole balance of A, B, C, and D C A D B
Conversion, X

91. Concentration-time Trajectories
in Semibatch Reactor CA CC CB
0.05
0.04
0.03
0.02
0.01
0.00
100
200
300
400
500
CNBr + CH3NH2  CH3Br + NCNH2
(A) (B) (C) (D)
t=0, CA=0.05 gmol/, CB=0.025 gmol/ℓ,
v0=0.05ℓ/s, k=2.2ℓ/s·mol, V0=5ℓ
Concentration
Time

92. Reaction Rate-time Trajectories Reaction rate [mole/s•L)
in Semibatch Reactor
0.0025
CNBr + CH3NH2  CH3Br + NCNH2
(A) (B) (C) (D)
t=0, CA=0.05 gmol/, CB=0.025 gmol/ℓ,
v0=0.05ℓ/s, k=2.2ℓ/s·mol, V0=5ℓ
0.0020
0.0015
Reaction rate [mole/s•L)
0.00 50
100
150
200
250

93. Reactor Equations in terms of Conversion
in semibatch reactor B A
The number of moles of A remaining at any time , t
(4-59)
where X is the mole of A reacted per mole of A initially in the vat.
The number of moles of B remaining at any time , t
(4-60)
For a constant molar feed rate
(4-61)

94. Reactor Equations in terms of Conversion
in semibatch reactor B A
The concentration of A and B
A mole balance on specials A:
(4-62)
The number of moles of C and D cab be taken directly from the stoichiometric table
(4-63)
Combine
(4-66)
The rate law (reversible 2nd-order reaction)
(4-65)
(4-66) can be solved numerically

95. Equilibrium conversion
in semibatch reactor B A
For reversible reactions carried out in a semibatch reactor, the maximum attainable conversion (i.e., the equilibrium conversion) will change as the reaction proceeds because more reactant is continually to the right.
The rate law (reversible 2nd-order reaction)
(4-68)
(4-69)
Equilibrium conversion in a semibatch reactor
(4-70)

96. Reactive Distillation
 The distillation of chemically reacting mixtures has become increasingly common in chemical industries.
 Carrying out these two operations, reaction and distillation, simultaneously in a single unit results in significantly lower capital cost and operating costs.
 Reactive distillation is particularly attractive when one of the reaction products has a lower boiling point, resulting in its volatilization from the reacting liquid mixture.

97. Reactive Distillation
 An example of reactive distillation is the production of methyl acetate:
 By continually removing the volatile reaction product, methyl acetate, from the reacting liquid-phase reaction, the reverse reaction is negligible and the reaction continues to proceed towards completion in the forward direction.
 Reactive distillation is used with reversible, liquid phase reactions. Suppose a reversible reaction had the following chemical equation :

98. Reactive Distillation
 For many reversible reactions the equilibrium point lies far to the left and little product is formed :
 However, if one or more of the products are removed more of the product will be formed because of Le Chatlier's Principle :
 Removing one or more of the products is one of the principles behind reactive distillation. The reaction mixture is heated and the product(s) are boiled off. However, caution must be taken that the reactants won't boil off before the products.

99. Homework
P4-11B
P4-12B
P4-13B
P4-14C
Due Date: Next Week