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1 1 CH701 - Chemical Reaction Engineering - II L T P C 3 - 2 4 Ref. Books: 1. Chemical Reaction Engineering – Leven Spiel, 3 rd Edition, Wiley. 2. Introduction to Chemical Reaction Engineering – H. Scott Fogler, 4 th Edition, PHI Pub. 3. G.F. Froment and K.B. Bischoff, Chemical Reactor Analysis and Design, Jhon Wiley & Sons ü Feedback ü Audibility in class ü Visibility of slides and board work ü Theory and fundamental discussion ü Discussion part for applications and others ü Practice of examples ü Your suggestions to learn better 2 Your suggestions to improve class room teaching – learning process Think and Reply for following Question: Question: Write various most attractive and inspiring reactions from natural phenomena. • To go further to learn in the field of Reaction Engineering, I would like to draw your consciousness for the reactions occurring naturally and they are the most attractive and inspiring all of us to think in different direction. •Can you take this challenge and give your reply to this question with smallest possible answer with suitable example…We will discuss answer in class today… •Are you inspired to think? 3 ü Why we need to learn CRE? ü How CRE is useful during your professional career? ü What is the significance of reactor in overall plant or process? ü How reactor performance effect overall performance of the plant or process? ü How can we improve the reactor performance? ü How lab scale reactor differ from industrial reactor? ü Is rate of reaction dependent on type of reactor configuration? ü How different contacting pattern effects the performance of the reactor? ü When you can select PFR/MFR? 4 ü Before we move forward, can you tell me, at least one new observation from Nature where reaction engineering is doing wonderful job, before starting of every lecture of CRE-II? ü Oxygen generation from plant…Some thing different? Can you take a lead to grow a plant in this semester to neutralize our own CO 2 generation..Can any body calculate our carbon footprint in a day or month or year? How many such plants or trees need to be grow? Homework.. 5 ü Industrial Examples of types of reactors § Ammonia production § Hydrogen production through steam reforming § Sulfur dioxide to sulfur trioxide § Removal of CO 2 from gas mixture (like in ammonia synthesis) § ??? § ?? § ? 6

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CH701 - Chemical Reaction Engineering - II

L T P C3 - 2 4

Ref. Books:1. Chemical Reaction Engineering – Leven Spiel, 3rd Edition, Wiley.2. Introduction to Chemical Reaction Engineering – H. Scott Fogler, 4th

Edition, PHI Pub.3. G.F. Froment and K.B. Bischoff, Chemical Reactor Analysis and Design,

Jhon Wiley & Sons

ü Feedbackü Audibility in classü Visibility of slides and board workü Theory and fundamental discussionü Discussion part for applications and othersü Practice of examplesü Your suggestions to learn better

2

Your suggestions to improve classroom teaching – learning process

Think and Reply for following Question:

Question: Write various most attractive and inspiring reactions from natural phenomena.

• To go further to learn in the field of Reaction Engineering, I would like to draw your consciousness for the reactions occurring naturally and they are the most attractive and inspiring all of us to think in different direction.•Can you take this challenge and give your reply to this question with smallest possible answer with suitable example…We will discuss answer in class today…

•Are you inspired to think?

3

ü Why we need to learn CRE?

ü How CRE is useful during your professional career?

ü What is the significance of reactor in overall plant or process?

ü How reactor performance effect overall performance of the plant orprocess?

ü How can we improve the reactor performance?

ü How lab scale reactor differ from industrial reactor?

ü Is rate of reaction dependent on type of reactor configuration?

ü How different contacting pattern effects the performance of the reactor?

ü When you can select PFR/MFR?

4

ü Before we move forward, can you tell me, at least onenew observation from Nature where reaction engineeringis doing wonderful job, before starting of every lecture ofCRE-II?

ü Oxygen generation from plant…Some thing different?Can you take a lead to grow a plant in this semester toneutralize our own CO2 generation..Can any bodycalculate our carbon footprint in a day or month or year?How many such plants or trees need to be grow?Homework..

5

ü Industrial Examples of types of reactors§ Ammonia production§ Hydrogen production through steam reforming§ Sulfur dioxide to sulfur trioxide§ Removal of CO2 from gas mixture (like in

ammonia synthesis)§ ???§ ??§ ?

6

2

üHow can we design a reactor?

üWhich information is required to startreactor design?

7

üWhat you have learn in CRE-I?

üWhy we need to learn CRE-II?

üWhat we are going to learn in CRE-II?

- Non-ideality of flow patterns

- Heterogeneous systems

- Without catalyst

- Fluid-Fluid reactions

- Solid-Fluid reaction

- With catalyst

- Heterogeneous solid catalytic reactions

üWhich are different types of reactors you have seen/learned/observed

during Industrial Training?

Introduction to CRE-II

8

Question and Answer based Discussion – A true Dialogue rather than Monologue

A Vedic Perspective – Bhagvad Gita and Srimad Bhagvatam were taught in Question and Answer fashion…A better of learning..

I believe, if you have question, probably I may or may not be able to answer, but you can find answer from anywhere and keep your mind running for it. But if you don’t have question, your learning may not be complete..

Are you ready to learn?9

Non Ideal Flow

Characteristics of ideal Plug flow reactor (PFR)

qNo axial dispersionqNo over takingqNo back mixingqConcentration changes gradually

along the lengthqFlat velocity profile throughout one

particular cross sectionqAll fluid elements spent same

amount of time within the reactorqNo bypassingqSteady state reactor

Characteristics of ideal mixed flow reactor (MFR)

q Complete mixing. Uniform concentration within reactor.

q Concentration within the reactor is same as of leaving the reactorq No bypassingqSteady state reactor

3

Factors affecting contacting patternI. Residence Time Distribution (RTD) : Different fluid

element spent different amount of time within the vessel.II. Degree of aggregation : Tendency of fluid to clump and

for a group of molecules to move about together.III.Earliness or lateness of mixing

1. Residence Time Distribution (RTD)Deviation from the two ideal flow patterns can be caused by channeling of fluid, by recycling of fluid, or by creation of stagnant regions in the vessel.

E, EXIT AGE DISTRIBUTION, RTD

It is evident that elements of fluid takingdifferent routes through the reactor maytake different lengths of time to passthrough the vessel. The distribution ofthese times for the stream of fluid leavingthe vessel is called the exit agedistribution E, or the residence timedistribution RTD of fluid. E has the unitsof time-l.

It is convenient to represent the RTD in such a way that the area under the curve is unity.

E, EXIT AGE DISTRIBUTION, RTD (cont….)

ØThe fraction of exit stream of age between t and t + dt is, E dt. ØRestriction of E curve : fluid only enters and only leaves

the vessel one time. This means that there should be no flow ordiffusion or eddies at the entrance or at the vessel exit. We callthis the closed vessel boundary condition.

ØWhere elements of fluid can cross the vessel boundary morethan one time we call this the open vessel boundary condition.

Experimental Methods for Finding EXIT AGE DISTRIBUTION (E)

q The simplest and most direct way of finding the E curve uses a physical or nonreactive tracer.

q For finding ‘E’ curve some sorts of experiments can be used, which is shown in Figure below.

q Because the pulse and the step experiments are easier to interpret, the periodic and random harder, here we only consider the pulse and the step experiment.

‘E’ curve from pulse experiment

Let us find the ‘E’ curve for avessel of volume V m3

through which flows v m3/sof fluid. For thisinstantaneously introduce Munits of tracer (kg or moles)into the fluid entering thevessel, and record theconcentration-time of tracerleaving the vessel. This isthe ‘Cpulse’ or ‘C’ curve.

4

From ‘C’ curve to ‘E’ curveTo find the E curve from the ‘C’ curve simply changethe concentration scale such that the area under thecurve is unity. Thus, simply divide the concentrationreadings by M/v.

‘F’ curve (Cumulative age distribution curve)

‘F’ curve directly give ideaabout the what fraction offluid has spent time lessthan t and greater than t.For ex. Consider the ‘F’curve shown in figure. Asper ‘F’ curve fraction of fluidspent time less than 8 minis 0.8, means 80% of fluidhas spent time less than 8minute and 20% of fluid hasspent time greater than 8minute.

‘F’ curve (Cumulative age distribution curve) from step

experimentLet us find the ‘F’ curve fora vessel of volume V m3

through which flows v m3/sof fluid. Now at time t = 0switch from ordinary fluid tofluid with tracer ofconcentration Cmax, andmeasure the outlet tracerconcentration Cstep versust. This is called ‘Cstep’ curve.

‘F’ curve from ‘cstep’ curveSimply divide Cstep by Cmax

Convolution integral theorem

It shows relation between Cin, Cout and E curve

Convolution means Intricacy, Complexity, Involvedness, Sophistication

5

OR

We say that Cout is the convolution of E with Cinand we can write,

Relation between ‘E’ AND ‘F’

Application of convolution integral theorem

If the input signal Cin, ismeasured and the exit agedistribution functions Ea , E,b andEc are known, then C1 is theconvolution of Ea with Cin and soon, thus

2. Degree of aggregationFlowing material is in some particular state of aggregation, depending on its nature. In the extremes these states can be called microfluids and macrofluids

6

Two-Phase Systems

These lie somewhere between the extremes of macro and microfluids.

SINGLE-PHASE SYSTEMS

A stream of solids always behaves as a macrofluid, but forgas reacting with liquid, either phase can be a macro- ormicrofluid depending on the contacting scheme being used.

3. Earliness or lateness of mixing The fluid elements of a single flowing stream can mix with each other either early or late in their flow through the vessel.

Conversion of non-ideal reactor

To evaluate reactor behavior in general we have to know four factors:1. the kinetics of the reaction2. the RTD of fluid in the reactor3. the earliness or lateness of fluid mixing in the reactor4. whether the fluid is a micro or macro fluid

Conversion in non-ideal reactor for macrofluid

For macrofluids, imagine little clumps of fluid staying for different lengths of time in the reactor (given by the E function). Each clump reacts away as a little batch reactor, thus fluid elements will have different compositions. So the mean composition in the exit stream will have to account for these two factors, the kinetics and the RTD.

Examples for RTD A pulse input to a vessel gives the results shown in Fig. l.(a) Check the material balance with the tracer curve to see whetherthe results are consistent.

(b) If the result is consistent, determine τ, V and sketch the Ecurve.

7

Example : Find conversion in real reactor and compare with ideal

reactorExample :

Models for non-ideal reactors

Models for non-ideal reactors

Compartment model

Dispersion model

Tank in series model

COMPARTMENT MODEL

• In the compartment models, we consider the vessel and the flow through it as follows:

• By comparing the E curve for the real vessel with the theoretical curves for various combinations of compartments and through flow, we can find which model best fits the real vessel.

DIRAC DELTA FUNCTION• In mathematics, the Dirac delta function,

or ƍ function, is a generalized function, ordistribution, on the real number line that iszero everywhere except zero, with an integralof one over the entire real line.

• The delta function is sometimes thought of asan infinitely high, infinitely thin spike at theorigin and total area under the spike (curve)one.

• It was introduced by theoretical physicist PaulDirac.

• It is often referred to as the unit impulsesymbol.

8

Dirac delta function for RTD Studies

The Dirac Delta Function, ƍ(t – t0), which says that the pulse occurs at t = to, as seen in figure.

The two properties of this function which we need to know are:

Ideal Plug flow reactor• Response of ideal impulse function

(Dirac Delta function) input would be identical but it would be observed at time equal to mean residence time.

• Area under the curve would be 1.

Plug flow reactor with non-ideality of dead zone

• Response of ideal impulse function (Dirac Delta function) input would be identical to input but it would be observed at time equal to (Vp/ν).

• Where, Vp and ν are volume of plug flow reactor and actual volumetric flow rate respectively.

• V = Vp + Vd, Where Vd is the volume of dead zone.

Plug flow reactor with non-ideality of by-passing

• Response of ideal impulse function (Dirac Delta function) input would be two peaks. First peak would be observed at time zero and second peak would be at V/ νa.

• Where, νa & νb are active volumetric flowrate and bypassing volumetric flowrate respectively.

• ν = νa + νb

IDEAL MIXED FLOW REACTORMIXED FLOW REACTOR WITH NON-IDEALITY OF

DEAD ZONE

9

MIXED FLOW REACTOR WITH NON-IDEALITY OF BY-PASSING

TWO plug flow reactors in parallel

TWO MIXED FLOW REACTORS IN PARALLEL

Plug flow reactor & mixed flow reactor in series

1.First PFR THEN MFR

2.FIRST MFR THEN PFR

MFR-PFR IN SERIES WITH DEAD ZONE

MFR-PFR in series with bypassing and dead zone

10

Diagnosing reactor ills(PFR) Diagnosing reactor ills(PFR) Cont.

Diagnosing reactor ills (MRF) Diagnosing reactor ills (MFR) Cont.

Dispersion Model

Some Dimensionless terms

• Dimensionless time,

• Dimensionless exit age distribution,

• Dimensionless variance,

11

Fitting dispersion model for small extent of dispersion (D/uL<0.01)

• If we impose an idealized pulse onto flowing fluid then dispersion modifies this pulse as shown in figure.

• For small extent of dispersion (D/uL<0.01) the spreading of curve does not significantly changes its shape as it passes through measuring point. Under this conditions relation between tracer concentration versus dimensionless time is simple and gives symmetric curves.

• This represents a family of gaussian curves, also called error curve or normal curves.

12

Fitting dispersion model for large extent of dispersion(D/uL>0.01)

Dispersion number from Step Input

13

Large extent of dispersion

Step response curves for large deviations from plug flow in closed vessels.

D/uL from pulse curve

• For the following data of tracerconcentration versus time. Calculatedispersion number of the vessel. Assumeclosed boundary condition.

• Since C curve is unsymmetrical, let usassume large dispersion(D/uL>0.01). Let ‘sstart with variance matching procedure.

For closed vessel and large dispersion

D/uL for fixed-bed reactor • Find the vessel dispersion number in a fixed-bed

reactor packed with 0.625 cm catalyst pellets. The

catalyst is laid down in a haphazard manner above a

screen to a height of 120 cm, and fluid flows downward

through this packing. A sloppy pulse of radioactive

tracer is injected directly above the bed, and output

signals are recorded at two levels in the bed 90 cm

apart. The following data apply to a specific

experimental run. Bed voidage = 0.4, superficial velocity

of fluid (based on an empty tube) = 1.2 cm/sec, and

variances of output signals are found to be σ12 = 39

sec2 and σ22 = 64 sec2. Find D/uL.

14

Chemical Reaction and Dispersion

• Reaction, A⟶ Product• -rA = kCA

n

• input = output + disappearance by reaction + accumulation• For Component A,(out-in) bulk flow + (out-in)axial dispersion + disappearance by reaction +

accumulation = 0

Comparison of real and ideal PFR for first order reaction

Comparison of real and ideal PFR for second order reaction

For first order reaction conversion

15

Tank In Series Model

Comparison bet’n Tank in Series and Dispersion Model

• This model can be used whenever the dispersion model is used; and for not too large a deviation from plug flow both models give identical results, for all practical purposes.

• In this model, we have to find out number of tank in series, which will give same performance as of real reactor.

• For N = 1, It is ideal CSTR• For N = infinite, It is ideal PFR

Pulse experiments three tank in series

• RTD will be analyzed from the exit of third tank and the pulse will be injected to first tank.

• Fraction of fluid leaving the third reactor between time t and t+∆t,

16

E(θ) versus θ curve for different values of N

Relation bet’n N and dimensionless variance

Conversion from Tank In Series (First order reaction, microfluid

and macrofluid)

17

Conversion from Tank in Series (any order, Microfluid)

Conversion from Tank in Series (any order, Macrofluid)

Effect of degree of segregation on conversion

For plug flow and batch reactor degree of segregation does not affect conversion.

Putting this value in equation

Which is same as of microfluid. Thereforedegree of segregation for first orderreaction does not affect conversion in MFR

18

Fluid-Fluid Reactions: Kinetics and Design

103

Chap 23 Fluid-Fluid Reactions: Kinetics

104

105

Questions to cover content of this chapter1. Why we should considered fluid-fluid reactions?

Or

1. What are the advantages of such reactions?

2. Give examples of fluid-fluid reactions?

3. Which type of reactions provide more resistance under same reaction conditions? Homogeneous or Heterogeneous? Why?

4. Addition of reaction along with mass transfer will enhance or decrease performance of operation…How???

5. What is two film theory? 106

Continue…

6.What do you mean by rate controlling step? 7.List various resistance present in fluid-fluid reactions?8.Represent various rate expression for individual rate controlling steps.9.Represent overall rate expression to consider all the resistance present in two phase reactions.

• What do you mean by fluid-fluid reactions?

• Can we consider liquid-liquid reaction as heterogeneous reactions?

107

• What do you mean by fluid-fluid reactions?– Gas –Liquid reactions

– Liquid-Liquid reactions

• Can we consider gas-gas reactions as heterogeneous reactions?

108

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• Applications of fluid-fluid reactions in Chemical Engineering…do you have any idea?

• Have you seen oxidation,hydrogenation, chlorination,carbonation reactions during yourindustrial training? Can you giveindustrial examples of suchreactions?

109

When we can suggest fluid-fluid reactions?ü Product of reaction may be a desired material. For example of liquid-liquid reaction is the nitration of organics with a mixture of nitric and sulfuricacids to form materials such as nitroglycerin. The chlorination of liquidbenzene and other hydrocarbons with gaseous chlorine is an example ofgas-liquid reactions.

üIn the inorganic field we have the manufacture of sodium amide, a solid,from gaseous ammonia and liquid sodium:

ü To facilitate the removal of an unwanted component from a fluid. Thus,the absorption of a solute gas by water may be accelerated by adding asuitable material to the water which will react with the solute beingabsorbed.

üTo obtain a vastly improved product distribution for homogeneous multiplereactions than is possible by using the single phase alone. 110

111

• What do you mean by fluid-fluid reaction kinetics?

• Why we need to learn this chapter? How it will beuseful?

• At the end of this chapter, what should be your learningoutcomes?ü Should be able to choose type of reaction

(homogeneous or heterogeneous) and reactionconditions

ü Should be able to find rate controlling step or stephaving maximum resistance for reaction to occur

ü Should be able to develop rate expression for fluid-fluid reaction

112

113

Let’s Recap (Questions)

Factors to be considered for fluid-fluid reactions:

v The Overall Rate Expression. Since materials in thetwo separate phases must contact each other beforereaction can occur, both the mass transfer and thechemical rates will enter the overall rate expression.

v Equilibrium Solubility. The solubility of the reactingcomponents will limit their movement from phase tophase. This factor will certainly influence the form of therate equation since it will determine whether the reactiontakes place in one or both phases.

vThe Contacting Scheme. In gas-liquid systemssemibatch and countercurrent contacting schemespredominate. In liquid-liquid systems mixed flow (mixer-settlers) and batch contacting are used in addition tocounter and concurrent contacting. 114

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THE RATE EQUATIONØ For convenience in notation consider G/L reactions (similarlycorrelations can be developed for L/L reactions).

Ø Assume that gaseous A is soluble in the liquid but that Bdoes not enter the gas (or B is non volatile).

ØThus A must enter and move into the liquid phase before itcan react, and reaction occurs in this phase alone.

ØNow the overall rate expression for the reaction will have toaccount for the mass transfer resistance (to bring reactantstogether) and the resistance of the chemical reactions step.

ØRelative magnitude of these resistances can vary greatlymeans a whole spectrum of possibilities to consider (one sidepure mass transfer – gas film or liquid film control to purekinetic of reaction control)….???What it menas? 115

Following second-order reaction considered for further analysis:

For notation consider a unit volume of contactor Vr with its gas, liquid, and solid volume fractions

116

117

Can any one explain this? What it means logically?

The Rate Equation for Straight Mass Transfer (Absorption) of A

118

What is two film theory?

with Henry's law pAi = HACA

Eliminate the unknown interface conditions pAi and CAi

For gas phase

For liquid phase

119

The Rate Equation for Mass Transfer of A with Chemical Reaction in Liquid Phase

120

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121

Let’s Recap (Questions)

1.What is two film theory? 2.What do you mean by rate controlling

step?3.List various resistance present in fluid-

fluid reactions?

The Rate Equation for Mass Transfer and Reaction

122

Case A: Instantaneous reaction with low CBCase B: Instantaneous reaction with high CBCase C: Fast reaction in liquid film, with low CBCase D: Fast reaction in liquid film, with high CBCase E and F: Intermediate rate with reaction in thefilm and in the main body of the liquidCase G: Slow reaction in main body but with film resistanceCase W: Slow reaction, no mass transfer resistance

123

The absorption of A from gas is larger when reaction occurs within the liquid film than for straight mass transfer.

Thus for the same concentrations at the two boundaries of the liquid film we have

Liquid Film Enhancement Factor

124

How to find value of E????E >/= (greater or equal to) 1.

How to find numerical values. Figure 23.4 shows that E is dependent on two quantities:

MH stands for the Hatta modulus,

125 126

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127

Give answer, thought for the day or life?

1.Can you make an analogy of rate controlling step while you are in traffic jam or at crossings which is about to start..

2. What is Alchemy? Search and think? Basic question is can we turn Iron into Gold? Is it possible logically?

At steady state the flow rate of B toward the reaction zone will be b timesthe flow rate of A toward the reaction zone. Thus,

Case A: Instantaneous Reaction with Respect to Mass Transfer.

where kAg and kAl , kBl are the mass transfer coefficients in gas and liquid phases.

Reaction will occur at a planebetween A-containing and B-containing liquid.

Change in pA or CB will move the plane one way or the other (see Fig.).

128

But mass transfer coefficient can be given by,

Eliminating the unmeasured intermediates x, x0, pAi, CAi

For the special case of negligible gas-phase resistance, for example, if you used pure reactant A in the gas phase, then pA = pAi or kg = ∞, in which case Eq.

129

Case B: Instantaneous Reaction; High CB

130

Case C: Fast Reaction; Low CB

Case D: Fast Reaction; High CB, Hence Pseudo First-Order Rate with Respect to A.

131

Cases E and F: Intermediate Rate with Respect to Mass Transfer.

132

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133

Case G: Slow Reaction with Respect to Mass Transfer.

Case H: Infinitely Slow Reaction.

134

Review of the Role of the Hatta Number, MH

To tell whether reaction is fast or slow, a film conversionparameter is define

Assume that gas-phase resistance is negligible,

1. If MH > 2, reaction occurs in the film and we have Cases A, B, C, D.2. If 0.02 < MH < 2, we then have the intermediate Cases E, F, G.3. If MH < 0.02, we have the infinitely slow reaction of Case H.

135

Let’s Recap (Questions)

1.What is roll of Enhancement factor? 2.What is maximum and minimum value of

E?3. If solubility of gas is very high in a solvent

which type of resistance may be rate controlling?

4.How rate of consumption effects accumulation and mass transfer in fluid-fluid reactions?

136

When MH is large, select a contacting device whichdevelops or creates large interfacial areas; energy foragitation is usually an important consideration in thesecontacting schemes. …. spray or plate columns shouldbe efficient devices for systems with fast reaction (orlarge MH)

On the other hand, if MH is very small, a large volume ofliquid is required. Agitation to create large interfacialareas is of no benefit here…. bubble contactors shouldbe more efficient for slow reactions (or small MH).

Similarly selection of contacting device can be basedon solubility of gas with given liquid….????? How???Just correlate MH and HA

Selection of Contacting Device

137

Solubility Data

138

Clues to the Kinetic Regime from Solubility Data

For reactions which occur in the film, the phase distribution coefficient H can suggest whether the gas-phase resistance is likely to be important or not.

For slightly soluble gases, HA is large; liquid film resistance termis large – Liquid film controls …Select such contacting devicewhich provides more liquid film..like bubble contactor

For highly soluble gases, HA is small; gas film resistance term islarge – Gas film controls…Select such contacting device whichprovides more gas film..like spray or plate contactor

24

139

Example 23.1 FINDING THE RATE OF A G/L REACTION

Air with gaseous A bubbles through a tank containing aqueous B. Reactionoccurs as follows:

For this systemkAga = 0.01 mol/hr.m3.Pa, fl = 0.98kAla = 20 hr-1 , HA = 105 Pa.m3/mol, very low solubilityDAl = DBl = 10-6 m2/hr, a = 20 m2/m3

For a point in the absorber-reactor wherepA = 5 X 103 Pa and CB = 100 mol/m3

(a) locate the resistance to reaction (what % is in the gas film, in the liquid film, in the main body of liquid)(b) locate the reaction zone(c) determine the behavior in the liquid film (whether pseudo first-order reaction, instantaneous, physical transport, etc.)(d) calculate the rate of reaction (mol/m3 hr)

140

This chapter has only analyzed second-order reactions,however, this problem deals with a third-order reaction.Since no analysis is available for other than second-orderreactions, let us replace our third-order reaction with asecond order approximation. Thus,

To find the rate from the general expression, first evaluate Eiand MH.

SOLUTION

141

first evaluate Ei and MH.

Since (Ei)first guess > 5 MH, For any other value of Pai, Ei value will be higher than this. So for any Ei value, have Ei > 5 MH.

Therefore, from Fig. 23.4 we have pseudo first-order reaction in the film

142

143

Thus, E = MH = 100

Thus,(a) 2/3 of the resistance is in the gas film, 1/3 is in the liquid film (b) the reaction zone is in the liquid filmI (c) reaction proceeds by a pseudo first-order reaction of A, at the interface(d) the rate is –rA‴ = 33 mol/hr.m3

144

Assignment – 1Solve only one problem. Use last digit of your roll number and solvecorresponding problem. Problem numbers are given from 0 to 9.

Gaseous A absorbs and reacts with B in liquid according to Ag + Bl → Rl -rA= kcAcBIn a packed bed under conditions wherekAga = 0.1 mol/hr.m2 of reactor.Pa fl= 0.01 m3 liquid/m3 reactorkAla = 100 m3 liquid /m3 reactor.hr DAl = DBl = 10-6 m2/hra = 100 m2/m3 reactorAt a point in the reactor where pA = 100 Pa and CB = 100 mol/m3 liquid•calculate the rate of reaction in mol/hr.m3 of reactor•describe the following characteristics of the kinetics :

(a)location of the major resistance (gas film, liquid film, main body of liquid)(b)behavior in the liquid film (pseudo first-order reaction, instantaneous, second-order reaction, physical transport)

25

145

for the following values of reaction rate and Henry’s law constant

k, m3 liquid/mol.hr HA, Pa.m3 liquid/mol0 10-1 101

1 10 10-5

2 106 104

3 10 103

4 10-4 15 10-2 16 108 1

7 Redo example 1 with just one change. Let us suppose that CB is very low, or CB = 1.

146

8 At high pressure CO2 is absorbed into a solution of NaOH in a packed column. The reaction is as follows :CO2 + 2NaOH → Na2CO3 + H2O with -rAl = kCACB---- -------(A) (B)Find the rate of absorption, the controlling resistance, and what is happening in the liquid film, at a point in the column where pA = 105 Pa and CB = 500 mol/m3.Data : kAga = 10-4 mol/m2.s.pa HA = 25000 pa.m3/mol

KAl = 1 x 10-4 m/s DA = 1.8 x 10-9 m2/sa = 100 m-1 DB = 3.06 x 10-9 m2/sk = 10 m3/mol.s f1 = 0.1

This problem was adapted from Danckwerts (1970).

147

9 Hydrogen sulfide is a absorbed by a solution of methanolamine (MEA) in a packed column. At the top of the column, gas is at 20 atm and it contains 0.1% of H2S, while the absorbent contains 250 mol/m3 of free MEA. The diffusivity of MEA in solution is 0.64 times that of H2S. The reaction is normally regarded as irreversible and instantaneous.H2S + RNH2 → HS- + RNH+

3------ -------(A) (B)For the flow rates and packing usedkAIa = 0.03 s-1

kAga = 60 mol/m3.s.atmHA = 1 x 10-4 m3atm/mol, Henry’s law constant for H2S in water.•Find the rate of absorption of H2S in MEA solution.•To find out whether it is worthwhile using MEA absorbent, determine how much faster is absorption with MEA compared to absorption in pure water.This problem was adapted from Danckwerts (1970).

Chap 24 Fluid-Fluid Reactors: Design

• What is the use of fluid-fluid reactionkinetics?

• How we can select and design a reactorusing information of fluid-fluid reactionkinetics?

148

149

Characteristics a contactors, ü have widely different G/L volume ratios, ü interfacial areas, ü kg and kl and concentration driving forces.

Characteristics of system (fluid),ü the solubility of gaseous reactant,ü the concentration of reactants, etc.

The location of the main resistance in the rate equation-will suggest that you use one class of contactor and notthe other.

First line of this chapter by Levenspile,

First choose the right kind of contactor, then find the size needed.

150

26

151

Table 24.1 shows some of the characteristics of these contactors.

152

Factors to Consider in Selecting a Contactor

(a) Contacting pattern. We idealize these as shown in Fig. 24.2.Towers approximate plug G/plug L.Bubble tanks approximate plug G/mixed L.Agitated tanks approximate mixed G/mixed L.

Towers have the largest mass transfer driving force and in this respect have an advantage over tanks. Agitated tanks have the smallest driving force.

153

(b) kg and kl : For liquid droplets in gas kg is high, kl islow. For gas bubbles rising in liquid kg is low, kl is high.

(c) Flow rates: Packed beds work best with relative flowrates of about Fl/Fg= 10 at 1 bar. Other contactors aremore flexible in that they work well in a wider range ofFl/Fg values

(d) If the resistance is in the gas and/or liquid films, alarge interfacial area "a,“ is required thus most agitatedcontactors and most columns.

If the L film dominates, stay away from spray contactors. Ifthe G film dominates stay away from bubble contactors.

(e) If the resistance is in the main body of the L, largefl = Vl/Vr is required Stay away from towers. Use tankcontactors. 154

(f) Solubility. For very soluble gases, those with asmall value of Henry's law constant H (ammonia, forexample), gas film controls, avoid bubble contactors.For gases of low solubility in the liquid, thus high Hvalue (O2, N2, as examples) liquid film controls, so avoidspray towers.

(g) Reaction lowers the resistance of the liquid film, soFor absorption of highly soluble gases, chemicalreaction is not helpful. For absorption of slightly solublegases, chemical reaction is helpful and does speed upthe rate.

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Nomenclature. Following symbols are used in various correlations

ACS = cross-sectional area of column.a = interfacial contact area per unit volume of reactor (m2/m3).fl = volume fraction of liquid i = any participant, reactant or product, in the reaction.A, B, R, S = participants in the reaction.U = Carrier or inert component in a phase, hence neitherreactant nor product.T = Total moles in the reacting (or liquid) phase.YA = pA/pU - moles A/mole inert in the gas XA = CA/Cu --- moles A/mole inert in the liquidFg

′, Fl′ = molar flow rate of all the gas and the liquid (mol/s).

Fg = Fg′ pUl π, upward molar flow rate of inert in the gas (mol/s).

Fl = Fl′ CUl CT; downward molar flow rate of inert in the liquid

phase(mol/s).

156

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157

STRAIGHT MASS TRANSFERSince the approach for reacting systems is astraightforward extension of straight mass transfer, let usfirst develop equations for absorption alone of A by liquid

then go to reacting systems.

158

Plug Flow G/Plug Flow L-Countercurrent Flow in a TowerTo develop the performance equation, we combine the rate equation with thematerial balance. Thus for steady-state countercurrent operations we have for a differential element of volume

159

Integrating for the whole tower gives

160

Design procedure for straight mass transfer

161 162

For dilute systems

CA << CT and pA << π, so Fg′≅ Fg, and Fl′≅ Fl. In this situation the differential material balance becomes

and for any two points in the absorber

The rate expression reduces to

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163

General integrated rate expression for dilute solution

164

MASS TRANSFER PLUS NOT VERY SLOW REACTION

Here we only treat the reaction

Assume that the rate is fast enough so that no unreacted A enters the main body of the liquid.

This assumes that the Hatta modulus is not very much smaller than unity.

165

Plug Flow G/Plug Flow L-Mass Transfer + Reaction in a Countercurrent Tower

For a differential slice of absorber-reactor, material balance

For Dilute Systems, pU = π and CU = CT in which case the above expressions simplify to

166

Rearranging and integrating

167

Objective 1: Find Volume of reactor (Vr)

To Solve for Vrpick a few pA values, usually pA1, pA2, and one intermediate value areenough, and for each p, find the corresponding CB. Evaluate the rate foreach point from

integrate the performance equation graphically

168

Objective 2: Find outlet concentration of gas with given Volume of reactor (Vr)

Solve such cases by trial and error, by assuming outlet concentration and calculate volume of reactor as per earlier calculations. If calculated volume of reactor is equal to given volume of reactor, assumptions is correct. Otherwise change assumed value and repeat the calculations till

Vr, cal = Vr,given

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169

Plug Flow G/Plug Flow L-Mass Transfer + Reaction in a Cocurrent Tower

Here simply change Fl to -Fl (for upflow of both streams)or Fg to -Fg (for downflow of both streams) in theequations for countercurrent flow. Be sure to find theproper C, value for each p,.

The rest of the procedure remains the same.

170

Mixed Flow G/Mixed Flow L-Mass Transfer + Reaction in an Agitated Tank Contactor

Since the composition is the same everywhere in the vessel make anaccounting about the vessel as a whole. Thus,

In symbols these equalities become

171

To find Vr the solution is direct; evaluate –rA‴ from known stream compositions and solve Eq.

To find CA and pA given Vr guess, pA,out evaluate CB,out then -rA then Vr .

Compare the calculated Vr value with the true value. If different, guess another PAOUt .

If Vr is to be found and the exit conditions are known, thenthe procedure is direct. Pick a number of pA, values andintegrate graphically.

If pAout and CBout , are to be found in a reactor of knownvolume Vr then require a trial and error solution. Simplyguess Cbout and then see if Vr,calculate = Vr,given

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Plug Flow G/Mixed Flow L-Mass Transfer + Reaction in BubbleTank Contactors

A differential balance for the loss of A from the gas because G is in plugflow, and an overall balance for B because L is in mixed flow.

For rising gas,

Balance about the whole reactor gives

173

Integrating gas phase Eq. along the path of the bubble and also using overall mass balance for reactor gives

174

If Vr is to be found and the exit conditions areknown, then the procedure is direct. Pick anumber of pAi values and integrate graphically.

If pAout and CBout, are to be found in a reactor ofknown volume Vr then it require a trial and errorsolution. Simply guess CBout and then see ifVr,calculate = Vr,given

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175

Mixed Flow G/Batch Uniform L-Absorption + Reaction in a Batch Agitated Tank Contactor

Since this is not a steady-state operation, composition andrates all change with time, as shown in Fig. 24.4.

176

At any instant the material balance equates thethree quantities shown below and thus in general

177

To Find the Time Needed for a Given Operation

Choose a number of CB values, say CB0, CBf and an intermediate CB value.

For each CB, value guess pA,0utNext calculate MH, Ei and then E and -rA‴′.

This may require trial and error, but not often.

See if terms I and III are equal to each other

and keep adjusting pAOut, until they do.178

As a shortcut: if pA << π and if E = MH (pseudo first order case from E Vs MH curve) then E is independent of pA in which case I and III combine to give

Next combine terms II and III to find the processing time

This time can be compared with the time required for100% conversion of component A. This situation isrepresented by pAout = 0 at all times.

179

Actual min. time spend by component A is given by

Combining t and tmin gives the efficiency of utilization of A. Thus

% mole of unreacted A leaves reactor = (trect - tmin )/tmin

180

Example 24.1:TOWERS FOR STRAIGHT ABSORPTIONThe concentration of undesirable impurity in air (at 1 bar = lo5 Pa) is to be reducedfrom 0.1% (or 100 Pa) to 0.02% (or 20 Pa) by absorption in pure water. Find theheight of tower required for countercurrent operations.DataFor consistency let us use SI units throughout.For the packing, kAg.a = 0.32 mol/(hr.m3.Pa) , kAl.a = 0.1 /hrSolubility is given by Henry's law constant, HA = pAi/Cai = 12.5 Pa.m3/molFlow rate per unit meter squared cross section of tower, Fg/Acs = 1×105 mol/hr.m2 and Fl/Acs = 7×105 mol/hr.m2

Molar density of liquid remains constant through out the column, CT = 56000 mol/m3

Given figure shows information known at starting of design

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181

Solution:It is a case of dilute solutions, use the simplified form of the material balance. So for any point in the tower pA and CA are related by

182

183

Example 24.2:TOWERS FOR ABSORPTION WITH CHEMICAL REACTIONThe concentration of undesirable impurity in air (at 1 bar = lo5 Pa) is to be reducedfrom 0.1% (or 100 Pa) to 0.02% (or 20 Pa) by absorption with chemical reactionwith component B. Add a high concentration of reactant B, CB1 = 800 mol/m3 orapproximately 0.8 N. Material B reacts with A extremely rapidly.

Find the height of tower required for countercurrent operations.DataFor consistency let us use SI units throughout.Assume that the diffusivities of A and B in water are the same, kAl = kBl = klFor the packing, kAg.a = 0.32 mol/(hr.m3.Pa) , kAl.a = 0.1 /hrSolubility is given by Henry's law constant, HA = pAi/Cai = 12.5 Pa.m3/molFlow rate per unit meter squared cross section of tower, Fg/Acs = 1×105 mol/hr.m2 and Fl/Acs = 7×105 mol/hr.m2

Molar density of liquid remains constant through out the column, CT = 56000 mol/m3

Given figure shows information known at starting of design

184

Solution:

Step 1. Express the material balance and find CB2 in the exit stream.Step 2. Find which of the many forms of rate equation should be used.Step 3. Determine the tower height.

Step 1. Material balance. For dilute solutions with rapid reaction Eq. 6 gives for any point in the tower, pA3, CB3

185

At both ends of the tower kAgpA< klCB; therefore, gas-phase resistance controlsand use a pseudo first-order reaction.

Why only mass transfer control?? Check with overall rate?

186

Example 24.3:TOWERS FOR LOW CONCENTRATION OF LIQUID REACTANT; CASE A

Repeat Example 24.2 using a feed with CB1 = 32 mol/m3, instead of 800 mol/m3, see Fig

32

187

Solution:

188

At both ends of the tower kAgpA> klCB, therefore, the reaction takes placewithin the liquid film and use following rate Eq.,

189

Example 24.4:TOWERS FOR INTERMEDIATE CONCENTRATIONS OF LIQUID REACTANTRepeat Example 24.2 using a feed in which CB1 = 128 mol/m3.Following figure shows conditions known at starting of design.

190

Solution:

191

At the top kAgpA< klCB; hence, Eq. 16 in Chapter 23 must be used. At thebottom kAgpA> klCB; hence, Eq. 13 in Chapter 23 must be used.

Now find the condition at which the reaction zone just reaches theinterface and where the form of rate equation changes. This occurs where

192

Example 24.5:TOWERS FOR ABSORPTION WITH CHEMICAL REACTION using overall rateThe concentration of undesirable impurity in air (at 1 bar = lo5 Pa) is to be reducedfrom 0.1% (or 100 Pa) to 0.02% (or 20 Pa) by absorption with chemical reactionwith component B. Add a high concentration of reactant B, CB1 = 800 mol/m3 orapproximately 0.8 N. Material B reacts with A extremely rapidly.

Find the height of tower required for countercurrent operations using general overaall rate expression .DataFor consistency let us use SI units throughout.Assume that the diffusivities of A and B in water are the same, kAl = kBl = klFor the packing, kAg.a = 0.32 mol/(hr.m3.Pa) , kAl.a = 0.1 /hrSolubility is given by Henry's law constant, HA = pAi/Cai = 12.5 Pa.m3/molFlow rate per unit meter squared cross section of tower, Fg/Acs = 1×105 mol/hr.m2 and Fl/Acs = 7×105 mol/hr.m2

Molar density of liquid remains constant through out the column, CT = 56000 mol/m3

Given figure shows information known at starting of design

33

193

Solution:

Evaluate E at various points in the tower. For this we need to first evaluate MH and Ei. At the Top of the Tower. From Fig. 23.4

194

Guess the value of pAi. It can be anywhere between 0 Pa (???) (gas film controls) up to 20 Pa (???) (liquid film controls).

Let us guess no gas-phase resistance.Then pAi = pA in which case

E = Ei = 500

195

Our guess was wrong, so let's try again. Now guess the other extreme, pAi = 0, meaning that the total resistance is in the gas film.

Then from Eq. (ii) we seethat Ei = ∞, E = ∞ and the rate equation becomes

Thus our guess was correct

196

Example 24.6:REACTION OF A BATCH OF LIQUIDReduce the concentration of B in the liquid (Vl = 1.62 m3, CT = 55555.6 mol/m3)of an agitated tank reactor by bubbling gas (Fg = 9000 mol/hr, π = l05 Pa)containing A (pAin = 1000 Pa) through it. A and B react as follows:

(a) How long must we bubble gas through the vessel to lower the concentration from CB0 = 555.6 to CBf = 55.6 mol/m3?(b) What percent of entering A passes through the vessel unreacted?

Additional DatakAg.a = 0.72 mol/(hr . m3. Pa) , fl = 0.9 m3 liquid/m3 totalkAl.a = 144 hr-1 , DA = DB = 3.6 X m2/hr, a = 100 m2/m3

HA = l03 Pa. m3/mol , k = 2.6 x l05 m3/mol. hr

197

Solution:

198

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199 200

Assignment – IV

Solve problem in group manner. Each group is having four problems. LikeGroup I:Prob. 1-4, Group II:Prob. 5-8, Group III:Prob. 9-12, Group IV:Prob. 13-16.

We plan to remove about 90% of the A present in a gasstream by absorption in water which contains reactant B.Chemicals A and B react in the liquid as follows:A (g→ l) + B (l) → R(l) , -rA = kCACBB has a negligible vapor pressure, hence does not go intothe gas phase. We plan to do this absorption in either apacked bed column, or an agitated tank contactor.(a) What volume of contactor is needed ?(b) Where does the resistance of absorption reaction lie ?

201

Data :For the gas stream :Fg = 90000 mol/hr at π = 105 paPAin = 1000 paPAout = 100 paPhysical DataD = 3.6 x 10-6 m2/hrCU = 55 556 mol H2O/m3 liquid, at all CBFor the packed bedFl = 900 000 mol/hr kAla = 72 hr-1

CBin = 55.56 mol/m3 a = 100 m2/m3

kAga = 0.36 mol/hr.m3.pa fi= Vl/V = 0.08For the agitated tankFl = 9000 mol/hr kAla = 144 hr-1

CBin= 5556 mol/m3 (about 10% B) a = 200 m2/m3

kAga = 0.72 mol/hr.m3.pa fl = Vl/V = 0.9202

Henry’s law For Type of Contactor

Constant:HA Reaction: k, T= Tower,Pa.m3/mol m3/mol.hr Countercurrent

A= AgitatedTank1 0.0 0 A2 18 0 T3 1.8 0 T4 105 ∞ T5 105 2.6 x 107 A6 105 2.6 x 105 A7 103 2.6 x 103 T8 105 2.6 x 107 T9 103 2.6 x 105 T

203

10 Danckwerts and Gillham, in Trans. I. Chem. E.,m 44, 42, March 1966, studied the rate of CO2absorption into an alkaline buffered solution of K2CO3. The resulting reaction can be represented as:CO2 (g→ l) + OH-(l) → HCO3

- with –rA = kCACB(A) (B)

In the experiment pure CO2 at 1 atm was bubbled into a packed column irrigated by rapidlyrecirculating solution kept at 200C and close to constant CB. Find the fraction of entering CO2absorbed.DataColumn : Vr = 0.6041 m3 fl = 0.08 a = 120 m2/m3

Gas : π = 101 325 Pa HA = 3500 Pa.m3/mol Vo =0.0363 m3/sLiquid : CB = 300 mol/m3 DAl = dbi = 1.4 x 10-9 m2/sRates : k = 0.433 m3/mol.s kAl a = 0.025 s-1

This problem is by Barry Kelly.11 A column packed with 5 cm polypropylene berl saddles ( a= 55 m2/m3) is being designed for theremoval of chlorine from a gas stream (G = 100 mol/s.m2, 2.36 mol% Cl2) by countercurrent contactwith an NaOH solution (L = 250 mol/s. m2, 10% NaOH, CB = 2736 mol/m3) at about 40-450C and 1atm. How high should the tower be for 99% removal of chlorine ? Double the calculated height totake care of deviations from plug flow.Data :The reaction Cl2 + 2NaOH → product is very very fast and irreversible for these very high flow rates(close to the limits allowed) an extrapolation of the correlations in Perry 6th ed. Section 14, giveskga = 133 mol/hr.m3.atm HA = 125 x 106 atm.m3/molkla = 45 hr-1 Diffusivity=D = 1.5 x 10-9 m2/s.

204

Repeat Example 5 with the following two changesHenry’s Law ConstantSecond Order Reaction Rate, HA, Pa.m3/mol Constant k, m3/mol.hr

12 105 2.6 x 105

13 105 2.6 x 109

14 105 2.6 x 103

15 103 2.6 x 1011

16 104 2.6 x 108

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205

Fluid-Particle Reactions (Solid as reactant):

Kinetics and Reactor Design

206

Chapter 25

Fluid-Particle (Solid as reactant) Reactions: Kinetics

This chapter treats the class of heterogeneous reactions in which a gas orliquid contacts a solid, reacts with it, and transforms it into product. Suchreactions may be represented by

207 208

Those in which the solid does not appreciably change in size duringreaction are as follows.1. The roasting (or oxidation) of sulfide ores to yield the metal oxides.For example, in the preparation of zinc oxide the sulfide ore is mined,crushed, separated from the gangue by flotation, and then roasted in areactor to form hard white zinc oxide particles according to the reaction

Similarly, iron pyrites react as follows:

2. The preparation of metals from their oxides by reaction in reducingatmospheres. For example, iron is prepared from crushed and sizedmagnetite ore in continuous-countercurrent, three-stage, fluidized-bed reactorsaccording to the reaction

3. The nitrogenation of calcium carbide to produce cyanamide

4. The protective surface treatment of solids such as the plating of metals.

209

The most common examples of fluid-solid reactions in which the size ofsolid changes are the reactions of carbonaceous materials such as coalbriquettes, wood, etc. with low ash content to produce heat or heatingfuels. For example, with an insufficient amount of air, producer gas isformed by the reactions

With steam, water gas is obtained by the reactions

210

Still other examples are the dissolution reactions (NaOH flaks and aq. HCl), the attack of metal chips by acids, and the rusting of iron.

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211

Two factors in addition to those normally encountered in homogeneous reactions: 1. modification of the kinetic expressions resulting from the mass transfer

between phases and 2. contacting patterns of the reacting phases.

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25.1 SELECTION OF A MODEL

Every conceptual picture or model or mechanism for the progress of reactioncomes with its mathematical representation, its rate equation.

Selection of a model (or mechanism) means acceptance of rate equation

If a model corresponds closely to what really takes place, then its rateexpression will closely predict and describe the actual kinetics (good model); if amodel differs widely from reality, then its kinetic expressions will be useless (badmodel).

Remember that the most elegant and high-powered mathematical analysisbased on a model which does not match reality is worthless (like currentresearch on Large Hydron Collidor to search for God Particle???) for theengineer who must make design predictions . What we say here about a modelholds not only in deriving kinetic expressions but in all areas of engineering.

The requirement for a good engineering model is that it be the closestrepresentation of reality which can be treated without too many mathematicalcomplexities (like simply considering all natural activities we canunderstand existence of God…or compare electricity and bulb).

213

It is of little use to select a model which very closely mirrors reality but whichis so complicated that we cannot do anything with it.

Unfortunately, in today's age of computers, this all too often happens.

For the non-catalytic reaction of particles with surrounding fluid, consider twosimple idealized models,1. progressive-conversion model and2. Shrinking unreacted-core model

214

Progressive-Conversion Model (PCM):

Reactant gas enters and reacts throughout the particle at all times, mostlikely at different rates at different locations within the particle. Thus, solidreactant is converted continuously and progressively throughout the particleas shown in Fig. 25.2

215

Shrinking-Core Model (SCM):

Reaction occurs first at the outer skin of the particle. The zone of reactionthen moves into the solid, leaving behind completely converted material andinert solid. We refer to these as "ash." Thus, at any time there exists anunreacted core of material which shrinks in size during reaction, as shown inFig. 25.3.

216

Burning Incense stick, it is better represented by which model?PCM and SCM????

37

217

Comparison of Models with Real Situations:

In slicing and examining the cross section of partly reacted solid particles, weusually find unreacted solid material surrounded by a layer of ash. Theboundary of this unreacted core may not always be as sharply defined as themodel pictures it; nevertheless, evidence from a wide variety of situationsindicates that in most cases the shrinking-core model (SCM) approximatesreal particles more closely than does the progressive conversion model(PCM). Observations with burning coal, wood, briquettes, and tightly wrappednewspapers also favor the shrinking-core model.

Since the SCM seems to reasonably represent reality in a wide variety ofsituations, we develop its kinetic equations in the following section. In doingthis we consider the surrounding fluid to be a gas. However, this is done onlyfor convenience since the analysis applies equally well to liquids.

218

SHRINKING-CORE MODEL FOR SPHERICAL PARTICLES OFUNCHANGING SIZEThis model was first developed by Yagi and Kunii (1955, 1961), whovisualized five steps occurring in succession during reaction (see Fig. 25.4).

Step 1. Diffusion of gaseous reactant A through the film surrounding theparticle to the surface of the solid.

Step 2. Penetration and diffusion of A through the blanket of ash to thesurface of the unreacted core.

Step 3. Reaction of gaseous A with solid at this reaction surface.

Step 4. Diffusion of gaseous products through the ash back to the exteriorsurface of the solid.

Step 5. Diffusion of gaseous products through the gas film back into the mainbody of fluid.

In some situations some of these steps do not exist. For example, if nogaseous products are formed, steps 4 and 5 do not contribute directly to theresistance to reaction.

219 220

Also, the resistances of the different steps usually vary greatly one from theother. In such cases we may consider that step with the highest resistance tobe rate-controlling.

Diffusion Through Gas Film Controls

Whenever the resistance of the gas film controls, the concentration profile forgaseous reactant A will be shown as in Fig. 25.5. From this figure we see that nogaseous reactant is present at the particle surface; hence, the concentrationdriving force, CAg - CAs, becomes CAg and is constant at all times duringreaction of the particle. Now since it is convenient to derive the kinetic equationsbased on available surface, we focus attention on the unchanging exterior surfaceof a particle S,.

dNB = bdNA,

221 222

Where, ρB = Molar density of B in the solidV = Volume of a particle,

No. of moles of B present in a particle is

The decrease in volume or radius of unreacted core accompanying thedisappearance of dNB moles of solid reactant is then given by

Rate of reaction in terms of the shrinking radius of unreacted core

where kg is the mass transfercoefficient between fluid and particle

Unreacted core shrinks with time.

38

223

Time for complete conversion of a particle be τ. Then by taking rC = 0

The radius of unreacted core in terms of fractionaltime for complete conversion is obtained bycombining above Eqs. gives

In terms of fractional conversion

224

Diffusion through Ash Layer Controls

225

Above Figure illustrates the situation in which the resistance to diffusionthrough the ash controls the rate of reaction.

Consider a partially reacted particle as shown in above Fig. Both reactant A andthe boundary of the unreacted core move inward toward the center of theparticle. But for G/S systems the shrinkage of the unreacted core is slower thanthe flow rate of A toward the unreacted core by a factor of about 1000, which isroughly the ratio of densities of solid to gas.

Because of this it is reasonable for us to assume, in considering theconcentration gradient of A in the ash layer at any time, that the unreactedcore is stationary.

226

For GIS systems the use of the steady-state assumption allows greatsimplification in the mathematics which follows. Thus the rate of reaction of Aat any instant is given by its rate of diffusion to the reaction surface,

For convenience, let the flux of A within the ash layer be expressed by Fick's lawfor equimolar counterdiffusion, though other forms of this diffusion equation willgive the same result. Then, noting that both QA and dCA/dr are positive

where De is the effective diffusion coefficient of gaseous reactant in the ashlayer.

Often it is difficult to assign a value beforehand to this quantity because theproperty of the ash (its sintering qualities, for example) can be very sensitive tosmall amounts of impurities in the solid and to small variations in the particle'senvironment.

227

In the second part of the analysis we let the size of unreacted core changewith time. For a given size of unreacted core, dNA/dt is constant; however, asthe core shrinks the ash layer becomes thicker, lowering the rate of diffusion of A.

228

Chemical Reaction Control

39

229

where k″ is the first-order rate constant for the surface reaction.

With rc =0

230

RATE OF REACTION FOR SHRINKING SPHERICAL PARTICLESWhen no ash forms, as in the burning of pure carbon in air, the reactingparticle shrinks during reaction, finally disappearing. This process isillustrated in Fig. below.

231

For a reaction of this kind we visualize the following three steps occurring insuccession.Step 1. Diffusion of reactant A from the main body of gas through the gas filmto the surface of the solid.

Step 2. Reaction on the surface between reactant A and solid.

Step 3. Diffusion of reaction products from the surface of the solid through the gas film back into the main body of gas. Note that the ash layer is absent and does not contribute any resistance.

As with particles of constant size, let us see what rate expressions result whenone or the other of the resistances controls

232

Chemical Reaction Controls

When chemical reaction controls, the behavior is identical to that of particlesof unchanging size. Use same correlations derived for unchanging size.

Gas Film Diffusion ControlsFilm resistance at the surface of a particle is dependent on numerous factors,such as the relative velocity between particle and fluid, size of particle, and fluidproperties. These have been correlated for various ways of contacting fluid withsolid, such as packed beds, fluidized beds, and solids in free fall. As anexample, for mass transfer of a component of mole fraction y in a fluid to free-falling solids Froessling (1938) gives

233 234

During reaction a particle changes in size; hence kg also varies. In generalkg rises for an increase in gas velocity and for smaller particles.

Stokes Regime (Small Particles): At the time when a particle, originally of size R0, has shrunk to size R,

40

235

Since in the Stokes regime,

Combining and integrating, gives

The time for complete disappearance of a particle is thus

It well represents small burning solid particles and small burning liquid droplets.

236

EXTENSIONS

Particles of Different Shape: Conversion-time equations similar to thosedeveloped above can be obtained for various-shaped particles, and Table 25.1summarizes these expressions.

Combination of Resistances: The above conversion-time expressionsassume that a single resistance controls throughout reaction of the particle.However, the relative importance of the gas film, ash layer, and reaction stepswill vary as particle conversion progresses.

For example, for a constant size particle the gas film resistance remainsunchanged, the resistance to reaction increases as the surface of unreactedcore decreases, while the ash layer resistance is nonexistent at the startbecause no ash is present, but becomes progressively more and moreimportant as the ash layer builds up. In general, then, it may not be reasonableto consider that just one step controls throughout reaction.

237

To account for the simultaneous action of these resistances is straightforwardsince they act in series and are all linear in concentration.

Thus on combining all three resistance correlations with their individualdriving forces and eliminating intermediate concentrations we can show thatthe time to reach any stage of conversion is the sum of the times needed ifeach resistance acted alone, or

238

239

In an alternative approach, the individual resistances can be combineddirectly to give, at any particular stage of conversion,

240

the relative importance of the three individual resistances varies asconversion progresses, or as rC, decreases.

On considering the whole progression from fresh to completely convertedconstant size particle, we find on the average that the relative roles ofthese three resistances is given by

For ash-free particles which shrink with reaction, only two resistances, gasfilm and surface reaction, need to be considered. Because these are bothbased on the changing exterior surface of particles, we may combine themto give at any instant

41

241

Limitations of the Shrinking Core Model:

The assumptions of this model may not match reality precisely. For example,reaction may occur along a diffuse front rather than along a sharp interfacebetween ash and fresh solid, thus giving behavior intermediate between theshrinking core and the continuous reaction models. This problem isconsidered by Wen (1968), and Ishida and Wen (1971).

Also, for fast reaction the rate of heat release may be high enough to causesignificant temperature gradients within the particles or between particle andthe bulk fluid. This problem is treated in detail by Wen and Wang (1970).

242

Exceptions to SCM: The first comes with the slow reaction of a gas with avery porous solid. Here reaction can occur throughout the solid, in whichsituation the continuous reaction model may be expected to better fitreality. An example of this is the slow poisoning of a catalyst pellet, asituation treated in Chapter 21.

The second exception occurs when solid is converted by the action ofheat, and without needing contact with gas. Baking bread, boilingmissionaries, and roasting puppies are mouthwatering examples of suchreactions. Here again the continuous reaction model is a betterrepresentation of reality. Wen (1968) and Kunii and Levenspiel (1991) treatthese kinetics.

243

DETERMINATION OF THE RATE-CONTROLLING STEPThe kinetics and rate-controlling steps of a fluid-solid reaction are deducedby noting how the progressive conversion of particles is influenced by particlesize and operating temperature.

The following observations are a guide to experimentation and to theinterpretation of experimental data.

Temperature: The chemical step is usually much more temperature-sensitive than thephysical steps; hence, experiments at different temperatures should easilydistinguish between ash or film diffusion on the one hand and chemicalreaction on the other hand as the controlling step.

Time: Following Figures show the progressive conversion of spherical solidswhen chemical reaction, film diffusion, and ash diffusion in turn control.

Results of kinetic runs compared with these predicted curves should indicatethe rate controlling step. Unfortunately, the difference between ash diffusionand chemical reaction as controlling steps is not great and may be maskedby the scatter in experimental data.

244

245

Gas Film Controlling

Ash Film Controlling

Reaction Rate Controlling

246

42

247 248

Particle Size: Time needed to achieve the same fractional conversion forparticles of different but unchanging sizes is given by

t α R1.5 to 2.0 for film diffusion controlling (the exponent drops as Re. no. rises)t α R2 for ash diffusion controllingt α R for chemical reaction controlling

Thus kinetic runs with different sizes of particles can distinguish betweenreactions in which the chemical and physical steps control.Ash Versus Film Resistance: When a hard solid ash forms during reaction, the resistance of gas-phase

reactant through this ash is usually much greater than through the gas filmsurrounding the particle. Hence in the presence of a nonflaking ash layer, filmresistance can safely be ignored. In addition, ash resistance is unaffected bychanges in gas velocity.

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Predictability of Film Resistance:

The magnitude of film resistance can be estimated from dimensionlesscorrelations such as Eq.

Thus an observed rate approximately equal to the calculated ratesuggests that film resistance controls.

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Overall Versus Individual Resistance: If a plot of individual rate coefficients ismade as a function of temperature, asshown in Fig., the overall coefficientgiven by Eq. cannot be higher than anyof the individual coefficients.

With these observations we can usually discover with a small, carefully plannedexperimental program which is the controlling mechanism.

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Consider gas-solid reaction of pure carbon particles with oxygen:

Since no ash is formed at any time during reaction, we have here a case ofkinetics of shrinking particles for which two resistances at most, surfacereaction and gas film, may play a role. In terms of these, the overall rateconstant at any instant from Eq

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Note that when film resistance controls, the reaction is rather temperatureinsensitive but is dependent on particle size and relative velocity between solidand gas. This is shown by the family of lines, close to parallel and practicallyhorizontal.

In extrapolating to new untried operating conditions, we must know when tobe prepared for a change in controlling step and when we may reasonablyexpect the rate-controlling step not to change.

For example, for particles with nonflaking ash a rise in temperature and to alesser extent an increase in particle size may cause the rate to switch fromreaction to ash diffusion controlling.

For reactions in which ash is not present, a rise in temperature will cause ashift from reaction to film resistance controlling.

On the other hand, if ash diffusion already controls, then a rise in temperatureshould not cause it to shift to reaction control or film diffusion control.

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Assignment VSolve problem in group manner. Each group is having four problems. Like Group IV: Prob. 1-4, Group III:Prob. 5-8, Group II:Prob. 9-12, Group I:Prob. 13-16.

1 Calculate the time needed to burn to complete spherical particles of graphite (Ro = 5 mm, ρB = 2.2 gm/cm3, k’’ = 20 cm/sec) in an 8% oxygen stream. For the high gas velocity used assume that film diffusion does not offer any resistance to transfer and reaction. Reaction temp. = 9000C.2 Spherical particles of zinc blende of size R = 1 mm are roasted in an 8% oxygen stream at 9000C and 1 atm. The stoichiometry of the reaction is----------------2ZnS + 3O2 → 2ZnO + 2SO2Assuming that reaction proceeds by the shrinking core model calculate the time needed for complete conversion of a particle and the relative resistance of ash layer diffusion during this operation.Data:Density of solid, ρB = 4.13 gm/cm3 = 0.0425 mol/cm3

Reaction rate constant , k’’ = 2 cm/secFor gases in the ZnO layer , DA = 0.08 cm2/secNote that film resistance can safely be neglected as long as a growing ash layer is present.-----------------On doubling the particle size from R to 2R the time for complete conversion triple. What is the contribution of ash diffusion to the overall resistance for particles of size.3 R ? 4 2R?

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Spherical solid particles containing B are roasted isothermally in an oven with gasof constant composition. Solids are converted to a firm nonflaking productaccording to the SCM as follows :A(g) + B (s) → R(g) + S(s) , CA = 0.01 kmol/m3, ρB = 20 kmol/m3

From the following conversion data (by chemical analysis) or core size data (byslicing and measuring) determine the rate controlling mechanism for thetransformation of solid.5 dp , mm XB t,min

1 1 41.5 1 6

6 dp , mm XB t,sec1 0.3 21 0.75 5

7 dp , mm XB t,sec1 1 2001.5 1 450

8 dp , mm XB t,sec2 0.875 11 1 1

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9 Uniform-sized spherical particles UO3 are reduced to UO2 in a uniform environment with the following results:t, hr 0.180 0.347 0.453 0.567 0.733XB 0.45 0.68 0.80 0.95 0.98If reaction follows the SCM, find the controlling mechanism and a rate equation to represent this reduction.

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Chapter 26

Fluid-Particle Reactors: Design

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Three factors control the design of a fluid-solid reactor;

1. reaction kinetics for single particles,2. size distribution of solids being treated, and3. flow patterns of solids and fluid in the reactor.

Where the kinetics are complex and not well known, where the products ofreaction form a blanketing fluid phase, where temperature within the systemvaries greatly from position to position, analysis of the situation becomesdifficult and present design is based largely on the experiences gained bymany years of operations, innovation, and small changes made on existingreactors.

The blast furnace for producing iron is probably the most important industrialexample of such a system.

Here only greatly simplified idealized systems considered in which thereaction kinetics, flow characteristics, and size distribution of solids areknown.

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Types of contacting in gas-solid operations.

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Solids and Gas Both in Plug Flow: Gas and solid flow in plug flow manner.Such operations are nonisothermal.

Examples of such case are,Countercurrent flow as in blast furnaces and cement kilns [Fig. 26.1(a)],crossflow as in moving belt feeders for furnaces [Fig. 26.1(b)],cocurrent flow as in polymer driers [Fig. 26.1(c)].

Solids in Mixed Flow: The fluidized bed [Fig. 26.1(d)] is the best example of a reactor with mixedflow of solids.

The gas flow in such reactor is difficult to characterize and however treatedas mixed flow.

Such operation is considered isothermal, because solids have high heatcapacity

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Semibatch Operations: The ion exchange column of Fig. 26.1(e) is an example of the batch treatmentof solids in which the flow of fluid closely approximates the ideal of plug flow.

Batch Operations: The reaction and dissolution of a batch of solid in a batch of fluid, such as the acid attack of a solid, is a common example of batch operations.

Analysis and design of fluid-solid systems are greatly simplified if thecomposition of the fluid can be considered to be uniform throughout thereactor. Since this is a reasonable approximation where fractional conversionof fluid-phase reactants is not too great, where fluid backmixing isconsiderable, or where solids wander about the reactor, sampling all the fluidas in fluidized beds,

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Particles of a Single Size, Plug Flow of Solids, Uniform GasComposition

The contact time or reaction time needed for any specific conversion of solidis found directly from the equations of Table 25.1.

Mixture of Particles of Different but Unchanging Sizes, Plug Flow ofSolids, Uniform Gas CompositionConsider a solid feed consisting of a mixture of different-size particles. The sizedistribution of this feed can be represented either as a continuous distributionor as a discrete distribution.

Way of measuring size distributions, gives discrete measurements.

Let F be the quantity of solid being treated in unit time. Since the density ofsolid may change during reaction, F is defined as the volumetric feed rate ofsolid in the general case. Where density change of the solid is negligible, F canrepresent the mass feed rate of solid as well. In addition, let F(Ri) be thequantity of material of size about Ri fed to the reactor. If Rm is the largestparticle size in the feed, we have for particles of unchanging size

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Following Figure shows the general characteristics of a discrete size distribution

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When in plug flow all solids stay in the reactor for the same length of time tP.From this and the kinetics for whatever resistance controls, the conversionXB(Ri) for any size of particle Ri can be found.

Then the mean conversion XB (bar) of the solids leaving the reactor can beobtained by properly summing to find the overall contribution to conversionof all sizes of particles. Thus,

where R(tP = τ) is the radius of the largest particle completely converted in the reactor.

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Note: Smaller particle requires a shorter time for complete conversion.

Hence some of feed particles, those smaller than R (tP = τ), will becompletely reacted. But if we automatically apply our conversion-timeequations to these particles we can come up with X, values greater thanunity, which makes no sense physically.

Thus the lower limit of the summation indicates that particles smaller thanR(tp = τ) are completely converted and do not contribute to the fractionunconverted, (1 – XB (bar))

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Example 26.1: CONVERSION OF A SIZE MIXTURE IN PLUG FLOW

A feed consisting30% of 50-pm-radius particles40% of 100-pm-radius particles30% of 200-pm-radius particlesis to be fed continuously in a thin layer onto a moving grate crosscurrent to aflow of reactant gas. For the planned operating conditions the time required forcomplete conversion is 5, 10, and 20 min for the three sizes of particles. Find the conversion of solids on the grate for a residence time of 8 min in the reactor(see Fig.).

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Solution: From the statement of the problem we may consider the solids to be in plugflow with t, = 8 min and the gas to be uniform in composition. Hence for amixed feed Eq. 2 is applicable, or

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Because for the three sizes of particles

τ α R, so chemical reaction controlling

Hence the fraction of solid converted equals 93.2%. Note that the smallest size of particles is completely converted anddoes not contribute to the summation of Eq.

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Mixed Flow of Particles of a Single Unchanging Size, Uniform GasComposition

Consider fluidized bed reactor as shown inFig. with constant flow rates of both solidsand gas into and out of the reactor.

Assume uniform gas concentration andmixed flow of solids,

No elutriation of fine particles.

The conversion of reactant in a singleparticle depends on its length of stay in thebed.

However, the length of stay is not the samefor all the particles in the reactor; hence wemust calculate a mean conversion XB(bar)of material.

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Thus, for the solids leaving the reactor

where E is the exit age distribution of the solids in the reactor

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Thus, for mixed flow of the single size of solid which is completely converted in time τ,

For gas film resistance controlling,

which on integration by parts gives

or in equivalent expanded form, useful for large t/τ, thus for very highconversion

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For chemical reaction controlling:

Integrating by parts using the recursion formula, found in any table of integrals,we obtain

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For ash resistance controlling: Integration leads to a cumbersome expression which on expansion yields [seeYagi and Kunii (1961)

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Comparison of holding times needed to effect a given conversion formixed flow and plug flow of a single size of solid.

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Example 26.2: CONVERSION OF A SINGLE-SIZED FEED IN A MIXED FLOW REACTOR:

Yagi et al. (1951) roasted pyrrhotite (iron sulfide) particles dispersed inasbestos fibers and found that the time for complete conversion was related toparticle size as follows:

Particles remained as hard solids of unchanging size during reaction. Afluidized-bed reactor is planned to convert pyrrhotite ore to the correspondingoxide. The feed is to be uniform in size, I = 20 min, with mean residence time7 = 60 min in the reactor. What fraction of original sulfide ore remainsunconverted?

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Since a hard product material is formed during reaction, film diffusion can be ruledout as the controlling resistance. For chemical reaction controlling Eq. shows that

whereas for ash layer diffusion controlling

As the experimentally found diameter dependency lies between these twovalues, it is reasonable to expect that both these mechanisms offer resistanceto conversion.

Using in turn ash diffusion and chemical reaction as the controlling resistanceshould then give the upper and lower bound to the conversion expected.

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For ash layer diffusion controlling Eq. gives

The solids in a fluidized bed approximate mixed flow; hence, for chemical reaction controlling, Eq. , with t/τ = 20 min/60 min = 1/3 gives

Hence the fraction of sulfide remaining is between 6.2% and 7.8%, or on averaging

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Mixed Flow of Mixture of Particles of Unchanging Size, Uniform Gas Composition

Most of the industrial applications involved mixture of different size of particle and asingle-exit stream (no elutriation of the fines).

Consider the reactor shown in Fig. 26.6. Since the exit stream is representative ofthe bed conditions, the size distributions of the bed as well as the feed and exitstreams are all alike, or

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where W is the quantity of material in the reactor and where W(Ri) is thequantity of material of size Ri in the reactor. In addition, for this flow the meanresidence time t(Ri) (bar) of material of any size Ri is equal to the meanresidence time of solid in the bed, or

Where

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For film diffusion controlling:

For chemical reaction controlling,

For ash diffusion controlling,

where τ(Ri) is the time for complete reaction of particles of size Ri.

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Example 26.3: CONVERSION OF A FEED MIXTURE IN A MIXED FLOWREACTOR

A feed consisting30% of 50-pm-radius particles40% of 100-pm-radius particles30% of 200-pm-radius particlesis to be reacted in a fluidized-bed steady-state flow reactor constructed from avertical 2-m long 20-cm ID pipe. The fluidizing gas is the gas-phase reactant, andat the planned operating conditions the time required for complete conversion is5,10, and 20 min for the three sizes of feed. Find the conversion of solids in thereactor for a feed rate of 1 kg solids/min if the bed contains 10 kg solids.

Additional Information:The solids are hard and unchanged in size and weight during reaction. A cyclone separator is used to separate and return to the bed any solids that maybe entrained by the gas.The change in gas-phase composition in the bed is small.

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Solution:

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Example 26.4: FINDING THE SIZE OF A FLUIDIZED BEDIn a gas-phase environment, particles of B are converted to solid product asfollows:

Reaction proceeds according to the shrinking core model with reaction control andwith time for complete conversion of particles of 1 hr.

A fluidized bed is to be designed to treat 1 ton/hr of solids to 90% conversion usinga stoichiometric feed rate of A, fed at CA0. Find the weight of solids in the reactor ifgas is assumed to be in mixed flow. Note that the gas in the reactor is not at CA0.Figure sketches this problem.

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Solution:In a CA0 environment with reaction controlling

Now for equal stoichiometry feed XA = XB (bar).

Thus, the leaving gas is at 0.1 CA0.

Since the gas is in mixed flow, the solids see this exit gas, τ = 10 hr.

By trial and error, find ratio of τ/t(bar) such that above equation can be satisfied. LHS=RHS

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Instantaneous Reaction:When reaction between gas and solid is fast enough so that any volumeelement of reactor contains only one or other of the two reactants, but notboth, then it may consider reaction to be instantaneous. This extreme isapproached in the high temperature combustion of finely divided solids.

In this situation prediction of the performance of the reactor is straightforwardand is dependent only on the stoichiometry of the reaction. The kinetics do not enter the picture.

Ideal Contacting Patterns for Instantaneous Reaction:Batch Solids:Following Figure shows two situations, one which represents a packed bed,the other a fluidized bed with no bypassing of gas in the form of large gasbubbles. In both cases the leaving gas is completely converted and remainsthat way so long as solid reactant is still present in the bed. As soon as thesolids are all consumed, and this occurs the instant the stoichiometricquantity of gas has been added, then the conversion of gas drops to zero.

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Countercurrent Plug Flow of Gas and Solids:Since only one or other reactant can be present at any level in the bed, there will

be a sharp reaction plane where the reactants meet. This will occur either at one end or the other of the reactor depending on which feed stream is in excess of stoichiometric.

Assuming that each 100 moles of solid combine with 100 moles of gas, Figs.below show what happens when we feed a little less gas than stoichiometric anda little more than stoichiometric.

Reaction to occur in the center of the bed so that both ends can be used as heatexchange regions to heat up reactants. This can be done by matching the gasand solids flow rates; however, this is inherently an unstable system and requiresproper control. A second alternative, shown in Fig. c, introduces a slight excessof gas at the bottom of the bed, and then removes a bit more than this excess atthe point where reaction is to occur.

Moving bed reactors for oil recovery from shale is one example of this kind ofoperation. Another somewhat analogous operation is the multistage counter flowreactor, and the four- or five-stage fluidized calciner is a good example of this. Inall these operations the efficiency of heat utilization is the main concern.

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Cocurrent and Crosscurrent Plug Flow of Gas and Solids: In cocurrent flow, shown in Fig. a, all reaction occurs at the feed end, and thisrepresents a poor method of contacting with regard to efficiency of heatutilization and preheating of entering materials.

For crosscurrent flow, shown in Fig. b, there will be a definite reaction plane inthe solids whose angle depends solely on the stoichiometry and the relativefeed rate of reactants. In practice, heat transfer characteristics may somewhatmodify the angle of this plane.

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Mixed Flow of Solids and Gas: Again in the ideal situation either gas or solid will be completelyconverted in the reactor depending on which stream is in excess.

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Extensions:Modifications and extensions of the methods presented here, for example,- to more complicated particle kinetics- to growing and shrinking particles in single reactors and in solid circulationsystems- to changing gas composition in a single reactor and from stage- to stage in multistage operations to deviations from ideal plug and mixed flow- to elutriation of fines from a reactor are treated elsewhere:

see Kunii and Levenspiel (l991), and Levenspiel (1996).

Arbitrary Flow of Solids: Since particles flow as macrofluids, with any RTD and with known gascomposition

where 1 - XB is given by proper equation for SCM reaction control,and for SCM ash diffusion control.

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Assignment V (Conti.)

A stream of particles of one size are 80% converted (SCM/ash diffusion control, uniform gas environment) on passing through a reactor. If the reactor is made twice the size but with the same gas environment , same feed rate, and same flow pattern of solids, what would be the conversion of solids? The solids are in1. Plug flow 2. mixed flow.

A solid feed consisting of20 wt% of 1 mm particles and smaller30 wt% of 2 mm particles50 wt% of 4 mm particlesPasses through a rotating tubular reactor somewhat like a cement kiln where it reacts with gas to give a hard nonflakable solid product (SCM/reaction control, t = 4 h for 4 mm particles).3 Find the residence time needed for 100% conversion of solids.4 Find the mean conversion of the solids for a residence time 15 min

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5. Particles of uniform size are 60 % converted on the average (shrinking coremodel with reaction controlling) when flowing through a single fluidized bed. If thereactor is made twice as large but contains the same amount of solids and withthe same gas environment what would be the conversion of solids?

6 Solids of unchanging size, R = 0.3 mm are reacted with gas in a steady flowbench scale fluidized reactor with the following result.Fo = 10 gm/sec , W = 1000 gm, XB = 0.75

Also, the conversion is strongly temp. Sensitive suggesting that the reaction stepis rate controlling . Design a commercial sized fluidized bed reactor (find W) totreat 4 metric tons/hr of solids feed of size R = 0.3 mm to 98% conversion.

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26.7. Solve Example 26.3 with the following modification: the kinetics of thereaction is ash diffusion controlled with T(R = 100 µm) 10 min.26.8. Repeat Example 26.4 if twice the stoichiometric ratio of gas to solid, still atCA0 is fed to the reactor.26.9. Repeat Example 26.4 if the gas is assumed to pass in plug flow through thereactor.26.10. Consider the following process for converting waste shredded fibers into auseful product. Fibers and fluid are fed continuously into a mixed flow reactorwhere they react according to the shrinking core model with the reaction step asrate controlling. Develop the performance expression for this operation as afunction of the pertinent parameters and ignore elutriation.

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Hydrogen sulfide is removed from coal gas by passing the gas through amoving bed or iron oxide particles. In the coal gas environment (consideruniform) the solids are converted from Fe203 to FeS by the SCM reactioncontrol, 7 = 1 hr. Find the fractional conversion of oxide to iron sulfideif the RTD of solids in the reactor is approximated by the E curves ofFigs. P26.11-P26.14.

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Thank You