1. Isothermal Reactor Design
授課教師：林佳璋

2. Design Structure for Isothermal Reactors
Level 4 :the rate law must be determined by either finding it in books or journals or determining it
experimentally in the laboratory.
Without such a structure, one is faced with the possibility of choosing or perhaps memorizing the correct equation from multitude of equations that can arise for a variety of different combinations of
reactions, reactors, and sets of conditions.

3. ~We see how the algorithm is used to
formulate the equation to calculate
the PFR reactor volume for a first-
order gas-phase reaction.
~The pathway to arrive this equation
is shown by the ovals connected the
dark lines through the algorithm.
~The dashed lines and the boxes
represent other pathways for
solutions to other situations.
The algorithm for the pathway shown is
1.Mole balance, choose species A reacting in a PFR
2.Rate law, choose the irreversible first-order
reaction
3.Stoichiometry, choose the gas-phase concentration
4.Combine Steps 1, 2, and 3 to arrive Equation A
5.Evaluate. The combine step can be evaluated either
a. Analytically
b. Graphically
c. Numerically, or
d. Using software

4. Scale-Up of Liquid-Phase Batch Reactor Data to the Design of a CSTR
~One of the jobs in which chemical engineers are involved is the scale-up of laboratory
experiments to pilot-plant operation or to full-scale production.
~In the past, a pilot plant would be designed based on laboratory data.
~However, owing to the high cost of a pilot-plant study, this step is beginning to be
surpassed in many instances by designing a full-scale plant from the operation of a
laboratory-bench-scale unit called a microplant.
~To make this jump successfully requires a thorough understanding of the chemical
kinetics and transport limitations.
~In this section we show how to analyze a laboratory-scale batch reactor in which a
liquid-phase reaction of known order is being carried out.
~After determining the specific reaction rate, k, from a batch experiment, we use it in
the design of a full-scale flow reactor.

5. ~for most liquid-phase reactions, the density change with reaction is
usually small and can be neglected.
~for gas-phase reactions, the batch reactor volume remains constant.
For a constant-volume batch reactor, V=V0
used for analyzing rate data in a batch reactor
When analyzing laboratory experiments, it is best to process the data in terms of the measured variable. Because concentration is the measured variable for most liquid-phase reactions, the general mole balance equation applied to
reactions in which there is no volume change becomes

6. Consider the irreversible second-order reaction
mole balance
rate law
combining
evaluating
stoichiometry
t=0, X=0
This time is the reaction time (tR) needed to achieve a conversion X for a second-
order reaction in a batch reactor.

7. for first-order reactions
X=0.9
k=10-4 s-1
for second-order reactions
X=0.9
kCA0=10-3 s-1

8. Table 4-2 gives the order of magnitude of time to achieve 90% conversion for first- and second-order irreversible
batch reactions.
Flow reactor would be used for reaction with characteristic reaction times, tR, of minutes or less.
The total cycle time in any batch operation is
Table 4-3 shows typical cycle times for a batch polymerization process.
Batch polymerization reaction times may vary between 5 and 60 hours.

9. Example 4-1
It is desired to 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, k. Because the reaction will be carried out isothermally, the specific reaction rate will need to be determined only at the reaction temperature of the CSTR. At high temperature there is a significant by by-product formation, while at temperature below 40C the reaction does not proceed at a significant rate; consequently, a temperature of 55C has been chosen. Because 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.
In the laboratory experiment, 500 mL of a 2 M solution of ethylene oxide in water was mixed with 500 mL of water containing 0.9 wt% sulfuric acid, which is a catalyst. The temperature was maintained at 55C. The concentration of
ethylene glycol was recorded as a function of time (Table E4-1.1).
Using the data in Tale E4-1.1, determine the specific reaction
rate at 55C.

10. Solution
mole balance
rate law
Because water is present in such excess, the concentration of water at any time t is virtually the same as the initial concentration and the rate law is independent of the
concentration of H2O. (CB≈CB0)
stoichiometry
liquid phase, no volume change, V=V0

11. combining
evaluating
for isothermal operation, k is constant.
The concentration of ethylene glycol at any time can be obtained from the reaction stoichiometry:
slope-=-k= min-1
k=0.311 min-1

12. Design of Continuous Stirred Tank Reactors (CSTRs)
A single CSTR
Consider a first-order irreversible reaction for which the rate law is
For liquid-phase reactions, there is no volume change during the course of the reaction

13. for a first-order irreversible reaction
Damkohler number is the ratio of the rate of reaction of A to the rate of convective transport of A at the entrance to the reactor.
for a first-order irreversible reaction
if Da<0.1, then X<0.1
if Da>10, then X>0.9
for a second-order irreversible reaction
Da is a dimensionless number that can give us a quick estimate of the degree of conversion that can be achieved in
continuous flow reactor.

14. CSTRs in Series
Consider first-order no change in the volumetric flow rate (v=v0)
Consider n equal-sized CSTRs connected in series operation at the same temperature, the concentration leaving the last reactor would be

15. When the product of the space time and the specific reaction rate is relatively large, say, Da1, approximately 90% conversion is achieved in two or three reactors; thus the cost of adding subsequent reactors might
not be justified.
When the product k is small, Da~0.1, the conversion continuous to increase
significantly with each reactor added.
The rate of disappearance of A in the nth reactor is

16. CSTRs in Parallel
Consider the case in which equal-sized reactors are placed in parallel rather than series, and the feed is distributed
equally among each of the reactors
individual reactor volume
equal size
same temperature
identical feed rates
This result shows that 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.

17. A Second-Order Reaction in a CSTR
Observe from Figure 4-6 that at high conversions (say 67%) a 10-fold increase in the reactor volume (or increase in the specific reaction rate by raising the temperature) will increase the conversion only to 88%.
This observation is a consequence of the fact that the CSTR operates under the condition of the lowest value of the reactant concentration (i.e., the exit concentration), and consequently
the smallest value of the rate of reaction.

18. Example 4-2
Close to 12.2 billion metric tons of ethylene glycol (EG) were produced in 2000, which ranked it the twenty-sixth most produced chemical in the nation that year on a total pound basis. About one-half of the ethylene glycol is used for antifreeze while the other half is used in the manufacture of polyesters. In the polyester category, 88% was used for fibers and 12% for the manufacture of
bottles and films. The 2004 selling price for ethylene glycol was $0.28 per pound.
It is desired to produce 200 million pounds per year of EG. The reactor is to be operated isothermally. A 1 lb mol/ft3 solution of ethylene oxide (EO) in water is fed to the reactor (shown in Figure E4-2.1) together with an equal volumetric solution of water containing 0.9 wt% of the catalyst H2SO4. The specific reaction rate constant is min-1, as determined in Example 4-1.
(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 reactor were arranged in series, what
is the corresponding conversion?

19. Solution
The specified ethylene glycol (EG) production rate in lb mol/min is
(a)
mole balance
rate law
combining
stoichiometry

20. evaluating

21. (b)
The conversion exiting each of the CSTRs in parallel is 81%.

22. (c)
The two equal-sized CSTRs in series will give a higher conversion than two CSTRs in parallel of the same size
when the reaction order is greater than zero.

23. Plug-Flow Reactors no dispersion
no radial gradients in either temperature, velocity, or concentration
used when there is a pressure drop in the reactor or heat exchange between the PFR and the surroundings
used in the absence of pressure drop or heat exchange
Consider the reaction
AProducts

24. Liquid phase v=v0
Da2 is the Damkohler number for a second-order reaction
Gas phase T=T0 P=P0
k constant
CA0 constant

25. The fluid moves through the reactor at a constant volumetric flow rate (v=v0) as the conversion
increases.
The volumetric gas flow rate decreases as the
conversion increases.
The gas molecules will spend longer in the reactor than they would if the flow rate were constant, v=v0. As a result, this longer residence time would result in a higher conversion than if the flow were
constant at v0.
The volumetric flow rate will increase as the
conversion increases.
The molecules will spend less time in the reactor than they would if the flow rate were constant. As a result of this smaller residence time in the reactor the conversion will be less than what would result if
the volumetric flow rate were constant at v0.

26. Note that, at the end of the reactor, virtually complete conversion has
been achieved.

27. Example 4-3
Ethylene ranks fourth in the Unite States in total pounds of chemicals produced each year, and it is the number one organic chemical produced each year. Over 50 billion pounds were produced in 2000, and it sold for $0.27 per pound. Sixty-five percent of the ethylene produced is used in the manufacture of fabricated plastics, 20% for ethylene oxide, 16% for
ethylene dichloride and ethylene glycol, 5% for fibers, and 5% for solvents.
Determine the plug-flow reactor 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 1100 K at a pressure of 6 atm.
Solution
The molar flow rate of ethylene exiting the reactor is

28. mole balance
rate law
combining
stoichiometry
evaluating

29. It was decided to use a bank of 2-in
It was decided to use a bank of 2-in. schedule 80 pipes in parallel that are 40 ft in length. For pipe schedule 80, the cross-section are, Ac, is ft2. The number of
pipes necessary is

30. Pressure Drop in Reactors
~In liquid-phase reactions, the concentration of reactants is insignificantly
affected by even relatively large change in the total pressure.
Consequently, we can totally ignore the effect of pressure drop on the
rate of reaction when sizing liquid-phase chemical reactors.
~In gas-phase reactions, the concentration of the reacting species is
proportional to the total pressure; consequently, proper accounting for
the effects of pressure drop on the reaction system can, in many
instances, be a key factor in the success or failure of the reactor operation.
~This fact is especially true in “microreactors” packed with solid catalyst.
Here the channels are so small that pressure drop can limit the
throughput and conversion for gas-phase reactions.

31. Pressure Drop and Rate Law
For an ideal gas, the concentration of reacting species i is
Consider a second-order isomerization reaction
is being carried out in a packed-bed reactor, the differential form of the mole balance equation in terms of catalyst weight is
The rate law is
Note that the larger the pressure drop (i.e., the smaller P) from frictional losses, the smaller the reaction rate.

32. Flow Through a Packed Bed
T=T0
dominant for laminar flow
Flow Through a Packed Bed
Ergun equation
dominant for turbulent flow

33. Because the reactor is operated at steady state, the mass flow rate at any point down the reactor, m (kg/s), is equal to the entering mass flow rate, m0 (i.e., equation of continuity)

34. distance of z down the reactor is
For tubular packed-bed reactors, we are more interested in catalyst weight rather than the distance z down the reactor. The catalyst weight up to a
distance of z down the reactor is
Ac is the cross-sectional area
c is the density of solid catalyst
b is the bulk density of solid catalyst per volume of reactor bed
Note that when  is negative, the pressure drop P will be less (i.e., higher pressure) than that for =0. when  is positive, the pressure drop
P will be greater than when =0.

35. For isothermal operation
We have two coupled first-order differential equations that must be solved simultaneously.
Analytical Solution
This equation can used to substitute for the pressure in the rate law, in which case the mole balance can be written solely as a function of conversion and catalyst weight. The resulting equation can readily be solved
either analytically or numerically.

36. Pressure Drop in Pipes
For flow in pipes, the pressure drop along the length of the pipe is given by
D = pipe diameter, cm
u = average velocity of gas, cm/s
f = Fanning friction factor
G = u, g/cm2s
For the flow conditions given in Example 4-4 in a 1000-ft length of 11/2-in schedule 40 pipe (p = ), the pressure drop is less than 10%. However, for high volumetric flow rates through microreactor,
the pressure drop may be significant.

37. Example 4-4
Plot the pressure drop in a 60 ft length of 1 1/2-in. schedule 40 pipe packed with catalyst pellets 1/4-in. in diameter. There is lb/h of gas passing through the bed. The temperature is constant along the length of pipe at 260C. The void fraction is 45% and the properties of the
gas are similar to those of air at this temperature. The entering pressure is 10 atm.
Solution
For 11/2-in. schedule 40 pipe, Ac = ft2
For air at 260C and 10 atm,
From the problem statement,

40. Analytical Solution for Reaction with Pressure Drop
These graphs compare the concentrations, reaction rates, and conversion profiles for the cases of pressure drop and no
pressure drop.
We see that when there is pressure drop in the reactor, the reactant concentrations and thus reaction rate for reaction (for reaction orders greater than 0 order) will always be smaller than the case with no pressure drop. As a result of this smaller reaction rate,
the conversion will be less with pressure drop than without pressure drop.

41. Consider a second-order isothermal reaction
mole balance
rate law
stoichiometry

43. Example 4-5
Reconsider the packed bed in Example 4-4 for the case where a second-order reaction
2AB+C
is taking place in 20 meters of a 11/2 schedule 40 pipe packed with catalyst. The flow and packed-bed conditions in the example remain the same except that the are converted to SI
units; that is, P0 = 10 atm = 1013 kPa, and
Entering volumetric flow rate: v0 = 7.15 m3/h
Catalyst pellet size: Dp = m
Solid catalyst density: c = 1923 kg/m3
Cross-sectional are of 11/2-in. schedule 40 pipe: Ac = m2
Pressure drop parameter: 0 = 25.8 kPa/m
Reactor length: L = 20 m
We will change the particle size to learn its effect on the conversion profile. However, we
will assume that the specific reaction rate, k, is unaffected by particle size.
(a)First, calculate the conversion in the absence of pressure drop.
(b)Next, calculate the conversion accounting for pressure drop.
(c)Finally, determine how your answer to (b) would change if the catalyst particle diameter
were doubled.
The entering concentration of A is 0.1 kmol/m3 and the specific reaction rate is

44. Solution
The weight of catalyst in the 20 m of 11/2-in. schedule 40 pipe is
(a)
(b)

45. We see the predicted conversion dropped from 82. 2% to 69
We see the predicted conversion dropped from 82.2% to 69.3% because of pressure drop. It would be not only embarrassing but also an economic disaster if we had neglected pressure drop and the actual conversion had turned out to be significantly smaller.
(c)
By increasing the particle diameter we decrease the pressure drop parameter and thus increase the reaction rate
and the conversion.

46. When interparticle diffusion effects are important in the catalyst pellet this increase in conversion with increasing particle size will not always be the case.
For larger particles, it takes a longer time for a given number of reactant and product molecules to diffuse in and out of the catalyst particle where they undergo reaction. Consequently, the specific reaction rate decreases with increasing particle size k~1/Dp,
which in turn decreases the conversion.
At small particle diameter, the rate constant, k, is large, and at its maximum value,
but the pressure drop is also large, resulting in a low rate of reaction.
At large particle diameter, the pressure drop is small, but so is the rate constant, k and
the rate of reaction, resulting in low conversion.

47. Example 4-6
Approximately 7 billion pounds of ethylene oxide produced in the United States in The 1997 selling price was $0.58 a pound, amounting to a commercial value of $4.0 billion. Over 60% of the ethylene oxide are antifreeze (30%), polyester (30%), surfactant (10%), and solvent (5%). We want to calculate the catalyst weight necessary to achieve 60% conversion when ethylene oxide is to be made by the vapor-phase catalyst oxidation of ethylene with air.
Ethylene and oxygen are fed in stoichiometric proportions to a packed-bed reactor operated isothermally at 260C. Ethylene is fed at a rate of 0.30 lb mol/s at a pressure of 10 atm. It is proposed to use 10 banks of 11/2-in.-diameter schedule 40 tubes packed with catalyst with 100 tubes per bank. Consequently, the molar flow rate to each tube is to be 310-4 lb mol/s. The properties of the reacting fluid are to be considered identical to those of air at this temperature and pressure. The density of the 1/4-in.-catalyst particle is 120 lb/ft3 and the bed
void fraction is The rate law is

48. Solution
mole balance
rate law
stoichiometry
Gas-phase, isothermal

50. software
~Both the conversion and the volumetric flow increase
along the length of the reactor while the pressure decrease.
~For gas-phase reactions with orders greater than zero, this
decrease in pressure will cause the reaction rate to be less
than in the case of no pressure drop.
~60% conversion is achieved with 44.5-lb catalyst in each
tube.
~Note that the catalyst weight necessary to raise the
conversion the last 1% from 65% to 66% (3.5 lb) is 8.5
times more than that (0.41 lb) required to raise the
conversion 1% at the reactor’s entrance.
~Also, during the last 5% increase in conversion, the
pressure decreases from 3.8 atm to 2.3 atm.

51. The catalyst weight of 44.5 lb/tube corresponds to a pressure drop of approximately 5 atm. If we had erroneously neglected pressure drop, the catalyst weight would have
been found by integrating equation with y=1 to give
If we had used this catalyst weight in our reactor we would have had insufficient catalyst to achieve the desired conversion. For this catalyst weight, the conversion is only 53%.