Chapter 2-Mass Reactor Model (102 P)

7/27/2019 Chapter 2-Mass Reactor Model (102 P)

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1

Chapter 2. Mass Balance, Reactor and Flow Models

Materials balance

........+

+

=

vol.control

thewithin

generation

massofRate

volume.

controlthe

ofoutmassof

flowofRate

volume

controlthe

toinmassof

flowofRate

vol.control

thewithinmass

ofonaccumulati

ofRate

Accumulation = Inflow Outflow + Generation

PART I. REACTION KINETICS AND REACTOR MODELS


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Accumulation

- Accumulation =

-

- Accumulation =

- Typically V is constant: Accumulation =

- For the steady-state case, mass remains constant (dc/dt = ).

- Note the units of accumulation are mass per time (M T

-1

).

== Mc ;


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Loading

- Loading = W(t); rate of mass loading, M T-1.

- Loading =

Q: volumetric flow rate of all water sources entering the system (L3

T-1, m3/hr)

Cin(t): inflow concentration of these sources (M L-3, g/m3)

- =)(tcin


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Outflow

- Outflow =

Q: volumetric flow rate

Cout: outflow concentration (M L-3).

- We assume a well-mixed system, so that Cout = .

- Outflow = Q C


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Reaction

- Lets assume first order reaction

- Reaction =

where k: a first order reaction constant (T-1).

- Reaction =


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Settling

- Settling =

where v: apparent settling velocity (L T-1), As: surface area of the

sediment or equivalent.

- In our cases for reactor study, this term settling is not considered.


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Accumulation = Inflow Outflow + Generation

= tCA

For component A:


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Types of Reactors: (1) Batch, (2) Complete Mix, CFSTR, (3) Plug flow,

(4) Cascade of CFSTR, and (5) Packed Bed


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Characteristic of each reactor

- Each reaction type has its unique characteristic.

- The mass balance is composed of

- Lets discuss more detail.


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Batch Reactor

-What is batch reaction or reactor?


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Batch Reactor:

=Vdt

dCA =dt

dCA

Accumulation = In Out + Gen

- Batch reactors are never at steady state, unless the reactions

have reached equilibrium:


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Reactions in Batch Reactor Analytical solution by Integration.

- Zero Order A B, rA

=

- Accumulation = In Out + Gen

- V drops out, leaving dCA/dt = -K


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Plot CA vs. t to get K


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Vrdt

dCV A

A =

==t

t

A

A

ACor

C

C

0

Batch Reactor

- First order, r A =

Using mass balance on batch reactor.

=dt

dCA

=tC

C A

AdtK

C

dCtA

A 0

-

0

- Accumulation = In Out + Gen


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Plot

0

t


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=

t

0

C

C2A

A

dtK-C

dCtA

0A

Batch Reactor

2nd Order, rA = (Type 1)

VrVdt

dCA

A = =dt

dCA


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Plot


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=Ar

[ ]== Ar

dt

Ad

Batch Reactor

2nd Order Reaction Type 2 Batch Reactor

aA + bB Products

Remember,

Then from Mass Balance

We need to define [B] in terms of [A] and constants

Assume

orderondforrradt

dC

ar A

A sec,11

===


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b

BB

a

AA toto =

tab

0ab

0t AABB +=

)( 00 AABaKAdt

dAab

ab += =+

tA

A ab

ab

dtaKAABA

dAt

0000)(

( )+

+=

+C

bxa

xln

a

1

bxax

dx

aKt

Aa

bA

a

bB

A

Aa

bB

t

o

A

A

oooo

=

+ln

1

I am going to stop using [ ]


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aKt

Aa

bAa

bB

A

Aa

bAa

bB

A

Aa

bB ooo

o

too

t

oo

=

+

+

lnln1

aKtA

B

AabA

abB

A

AabB o

o

too

t

oo

=

+

ln

1

aKtAB

BA

Aa

bB ot

ot

oo

=

ln

1t

B


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aKtAB

BA

Aa

bB ot

ot

oo

=

ln

1

oo Aa

bBaKPlot -ln(AtB0/BtA0) versus t to get

We can also solve for At

If a = b,

tAa

bBaK

AB

BA

ot

ot )(ln 00 =

)(

0

)(

])[(

oo

oo

Aa

bBakt

o

Aa

bBakt

ooot

ebAaB

eAbBaAA

=

)(

0

)(])[(

oo

oo

ABakt

o

ABakt

ooot

eAB

eABAA

=


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=Ar

= tBB0

== Ardt

dA

=dt

dA ( )

+

+=

o

11

111 A

KK

KAKK

dt

dA

Batch

Reversible 1st Order: K1AB

K-1

Assume Bo = 0 then Bt = Ao -At


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( )

+=

+

t

o

A

A

t

011

011

1

dtKK

AKK

kA

dA

( )tKKAKKK

Aln 11

A

A011

1 t

0

+=

+

( )

+

+=

o

11

111 A

KK

KAKK

dt

dA

( )tKKKK

AKA

AKK

KAt

11

11

010

0

11

1

ln

+=

+

+


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( )tKK

KK

AKA

AKK

KAt

11

11

010

0

11

1

ln

+=

+

+

( ) 0111eq AKKKA0dt

dA ++==

11

01eq

KK

AKA

+=

( )( )tKK

AA

AAln 11

eq0

eqt+=

( )tKK

11

100

11

1t

11eKK

K1AA

KK

KA

+

+

+

+

=

At equilibrium, At = Aeq

Solving for At as a function of constants and time only.

tKK

eqoeqteAAAA

)( 11)( ++=


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=Ar

Saturation Reaction, e.g., 0 1order

Ks[A]


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]A[K

]A[Kr

SA +

=

[ ] [ ] == dtAd

KAat S

Saturation Reaction, e.g., 0 1order

If [A] >> KS then ?? order

If [A]


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]A[K

]A[Kr

dt

]A[d

SA +

==

1A

K

K

dt

]A[d

S

+

=

=

+

t

0

A

A

S dtKdA1A

Kt

o

Using Mass Balance

=+t

o

t

o

A

A

A

A

S KtdAA

dAK

KtAAA

AK ot

o

tS =+ln KtAA

A

AK to

t

oS =+ln

( )t

K

K

K

AA

A

A

SS

to

t

o =

+ln ( )SS

Sot

o

K

K

tK

AA

t

A

A

=

+

ln


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( )S

to

S

t

o

K

KAA

tKt

A

A

+=1

ln

How to plot?


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[ ]=dt

Ad

=

=

2

1

A

A

r

r

=dt

dA =tA

[ ]== Br

dt

Bd ( )=+

t

0

tKK01

B

B

dteAKdB 21t

0

Parallel 1st order:K1

A BK

2 C


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( )=

+t

0

tKK01

B

B

dteAKdB 21t

0

)1(111

0

at

t

atate

aae

adte

=

=

( )( )tKKt eKK

AKBB 211

21

010

++

=

( )( )tKKt eKK

AKBB 211

21

010

++

+=


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CBA21 KK

[ ][ ]AKr

dt

Ad1A == =tA

[ ]21 BB

rrdtBd +=

[ ] =dtBd tKo12 1eAKBK

dtdB =+

Series Reactions - 1st Order

Where, =1Br =2Brand


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CBA21 KK

Series Reactions - 1st Order

( ) ( )tQYtPdtdY

=+ = Qdt1

Y

BY =

tK

oeAKQ1

1

=

=Pdt

e

This is a first order linear differential equation of the form

P( t ) = K2, a constant

tK

o12

1

eAKBKdt

dB

=+


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=

t

dtK

e 02

tKe 2=

= dteAKeeYtKtK

tK12

201

1 ( )

= dteAKeY tKKtK 122 01

Boundary Conditions

Possible at t = 0, B = 0

Also possible at t = 0, B = B0 Lets use this

( )

tKKtK )(AK KKK


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11,,0 )(0122 ==== tKKtK eandeBBtAt

KKK

AKB

12

010 +

= ( )1201

0KK

AKBK

=

( )

( ) ( )

+

=

12

01

012

01

12

2

KK

AKB

KK

eAKeB

tKKtK

t

( )

( ) ( )

+=

12

01

12

010

12

22

KK

AK

KK

eAKeeBB

tKKtKtK

t

( )

= dteAKeY tKKtK 122 01

( )[ ]tKtKKtKtKt eeeKK

AKeBB 21222

12

010

+=

( )

( )tKtKtKt eeKK

AKeBB 212

12

010

+=

])[()(

12

01 122 KeKK

AKeB

tKKtK +

=

Need to determine constant of

integration


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( )( )tKtK

12

01t

21 eeKK

AKB

=

[ ] [ ] [ ] [ ]0ttt ACBA =++

[ ] [ ] [ ] [ ] *BAAC tt0t =

( )tKtK12

01tK00t

211 eeKK

AKeAAC

=

Now, in case of B0 = 0

Lets calculate Ct

for all [C]0 = 0

[B]0 = 0

tKot

1eAA=


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+

=

12

1

12

10

21

11KK

eK

KK

eKeAC

tKtKtK

t

*This can be done for all types of reactions.

Remember we have discussed so far only regarding batch reaction


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Methods for Determining Reaction Order and Rate Constants

1.Integral Method with Graphical Determination

- It consists of guessing n and integrating the equation to obtain a function

- Graphical methods are then employed to determine whether the model fits the data

adequately.

- The graphical approaches are based on linearized versions of the underlying

models.

nkC

dt

dc=

Order Rate units Dependent (y) Independent (x) Intercept Slope

Zero, n=o

First, n=1

Second, n=2

General, n1


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2. Integration with Tabular Averaging of K

E.g., First Order

t

C

Cln

K 0

t

A

A

=

t CA -ln CAt /(CAO) K

_ _ _ _

_ _ _ _


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11

11

+

+

=

ii

iii

tt

CC

t

C

dt

dC

3. Differential Method

4. The Method of Initial Rate

5. The Method of Half-Lives

6. The Method of Excess

7. Numerical Method, etc


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Simple Example of Determination of the Reaction Order and the Reaction Rate

Constant

Batch Reactor Test

Determine the reaction order and reaction rate constant

Time, d Concentration, CA, mole/L

0 250

1 70

2 42

3 30

4 23

5 18

6 16

7 13

8 12

A B


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Batch Reactor Test

Vr00Vdt

dCAA += AA rdt

dC

=

Accumulation = In Out + Gen

krdt

dCA

A ==

][Akrdt

dCA

A ==

2][Akrdt

dCA

A ==

Time, d Concentrat ion, C, mole/L

0 250

1 70

2 42

3 30

4 235 18

6 16

7 13

8 12


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- First, in case of first order.

- By Computer program (least square)

][Akrdt

dCA

A == KtAAeCC

t

-

0=

Time, day

0 2 4 6 8

CA

0

50

100

150

200

250

300

1) By Computer program (least square)

Time, day

0 2 4 6 8

CA

0

50

100

150

200

250

300

Kt

A eC t-250=

K=1.02

R2=0.957

- Does not fit-well.

- However, the program gives a good number for R2.(this is a problem of

computer program).

1) B C (l )


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- In case of second order.

- By Computer program (least square)

2][Akr

dt

dCA

A ==

Time, day

0 2 4 6 8

CA

0

50

100

150

200

250

300

1) By Computer program (least square)

- Perfect fit, excellent number for R2.

- Now we say the reaction follows second order and k = 0.01

KtC

1

C

1

0t AA

+=

K=0.01

R2=0.998

ktC

CC

A

AA

0

0

1+=

ktCA

2501

250

+=

2) G hi l M th d I t ti M th d ( t )


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- First, in case of first order.

][Akrdt

dCA

A == KtAAeCC

t

-

0=

Time, day

0 2 4 6 8

CA

0

50

100

150

200

250

300

2) Graphical Method + Integration Method (no computer program)

- How can you determine k?



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Kt

AA eCC t-

0=


- We need linear form of the equation.

KtC

C

A

At =0

ln

Time, d C -Ln (C/Co)

0 250 0

1 70 0.533

2 42 0.775

3 30 0.921

4 23 1.036

5 18 1.143

6 16 1.194

7 13 1.284

8 12 1.319

Time, day

0 2 4 6 8

-L

n

(C/C0)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

K?????

Not linear.

So, not first order



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- In case of second order.

2][Akr

dt

dCA

A ==

Time, day

0 2 4 6 8

CA

0

50

100

150

200

250

300


- How can you determine k?

ktC

CC

A

AA

0

0

1+=

2) G hi l M th d + I t ti M th d ( t )


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- In case of second order, we also linear form

2][Akr

dt

dCA

A ==


ktC

CC

A

AA

0

0

1+=

Kt

C

1

C

1

0t AA

+=

Time, d C 1/C

0 250 0.004

1 70 0.014

2 42 0.024

3 30 0.033

4 23 0.044

5 18 0.056

6 16 0.063

7 13 0.0778 12 0.083

Time, day

0 2 4 6 8

1/C

0.00

0.02

0.04

0.06

0.08

0.10

R2=0.998

0.004

01.028

024.0084.0=

=

ddk

- Reaction follows 2-order or (pseudo 2 order).

3) Tab lar A erage + Integration Method (no comp ter program)


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3) Tabular Average + Integration Method (no computer program)

t CA -ln CAt /(CAO) K

_ _ _ _

_ _ _ _

First orderKt

AA eCC t-

0= Kt

C

C

A

At =0

ln

Second order

t CA

1/CA K

_ _ _ _

_ _ _ _

ktC

CC

A

AA

0

0

1+= Kt

C

1

C

1

0t AA

+=

4) Diff ti l th d


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4) Differential method

N=2.06 (pseudo 2 order)

* If the reaction is more complicated, you need your analytical skills tosolve the problem.

A B* This is a just simple case.


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Temperature Effect on Reaction Rate

Possible temperature affects:

-The rates of most reactions in natural waters increase with temperature.

- A general rule of thumb is that the rate will approximately double for a

temperature rise of 10 C.

- A more rigorous quantification of the temperature dependence is provided by an


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A more rigorous quantification of the temperature dependence is provided by an

empirical equation, Arrhenius equation:

aRTE

a AeTK/

)(=

A = Arrhenius constant, a pre-exponential or frequency factor

E = Activation energy for a reaction (J mole-1)

R = the gas constant (8.314 J mole-1 K-1)

Ta = Absolute temperature (K)

-The equation is often used to compare the reaction rate at two different

temperature

12

12 )(

1

2

)(

)(aa

aa

TRT

TTE

a

a eTK

TK

=


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- Temperature in most water bodies vary over a rather narrow range, so the

product of Ta1 and Ta2 is relatively constant.

- The difference in temperature (Ta2 Ta1) is identical whether an absolute or a

centigrade scale is used.

- Consequently, the following can be defined as a constant

12

12 )(

1

2

)(

)(aa

aa

TRT

TTE

a

a eTK

TK

=

12 aa TRT

E

e

=)(

)(

1

2

TK

TK

)(TK


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- In water quality modeling, many reactions are reported at 20 C.

- The temperature dependency of biologically mediated reactions is often

expressed as the quantity Q10, which is defined as the ratio

- Meaning of Q10?

- Some typical values of used in water quality modeling

12

)(

)(

1

2 TT

TK

TK =

202)20()( = TKTK

10

10)10(

)20(

== K

K

Q

Q10 Reaction1.27 Oxygen reaeration

1.58 BOD decomposition

1.89 Phytoplankton growth

2.16 Sediment oxygen demand (SOD)


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=HRT

Hydraulic Residence Time in Reactors

HRT =

- This is the theoretical HRT

- Actual HRTs can be less than the theoretical due to dead volume. We can determine

the actual HRT using a tracer.

Input a slug of tracer and

measure the concentration in

the effluent.

It is assumed that the average HRT for the tracer molecules is equal to the


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g q

average HRT for the water molecules, i.e. the tracer is completely soluble.

Hact

= mass of tracer added

= center of mass time to center of mass

or

=ii

iii

actH tC

tCt= x axis center of the whole area

= oHact timeresidencefinitemasselementaleachaddedmasstotal ][

addedmasstotal

timeresisdencefinitemasselementaleachHact

= 0

][

=

o

out

o

out

dtQC

dtCtQ

=

o

out

o

out

dtQC

dtQCt

Steady State Materials Balance for Conservative (Non - reactive) Substances


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Steady State Materials Balance for Conservative (Non reactive) Substances

Assume that two streams meet and completely mix instantaneously

Acc (zero) = In - Out + Generation (zero)

Mass balance on flow

Mass balance on conservative substances.

=3C

Mass flux


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Plug Flow Reactors

How to recognize a PFR


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Look at the response of the system to the input of a non-reactive tracer.


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PFR Behavior


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Accumulation = In - Out + Generation

=

V

t

CA

+=+=+ X

X

CCQCCQQC AXAAXAXXA )(


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VrXX

CCQQCV

t

CA

A

XAXA

A +

+=

XArXX

CQ

t

XAC

A

AA +

=

AAA r

X

C

A

Q

t

C+

=

VrQCQCVt

CAXXAXA

A +=

+

AXV =


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AAA r

X

C

A

Q

t

C+

=

HV

Q

XA

Q

=

=

1

A

H

AA rC

t

C+

=

Taking limit as X 0

CC


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0t

CA =

H

A

H

AA

d

dCor

Cr

=

dt

dCr AA =

At steady state:

Then

The response of a PFR as a function ofH at steady state is the same as the

response of a batch reactor, and solutions are exactly the same.

Compare with batch

AH

AA rC

dt

C+

=

PFR

What does that mean? Compare reactors in PFR vs. Batch.

Compare with batch


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H

A

H

AA

d

dCor

Cr

=

dt

dCr AA =

PFR

p

AA dCC


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=H

A

d

dC

0 Order rA =

CAe = Effluent value of CA

CAi = Influent value of CA

H

A

H

AA

d

dCor

Cr

=PFR

=AeC

AA dCC


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=H

A

d

dC

=AeC

1st Order rA

=

=H

A

d

dC

2nd OrderrA =

*The conversion is the same for all of the reaction rates we covered in batch

reactors. Substitute CAe for CAt, CAi for CA0, and H for t

H

A

H

AA

dorr

=PFR

Complete Mix Reactors


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1) CFSTR Continuous Flow Stirred Tank Reactor

2) CSTR Continuous(ly) Stirred Tank Reactor

How to recognize a CFSTR


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How to recognize a CFSTR

a) Tracer Response- Slug input

Non-reactive


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GenOutInAcc +==V

dt

dCT

rT = 0 because non reactive tracer (just for this recognition experiment).

TT C

V

Q

dt

dC=

=tT

0T

C

C

t

0T

T dtV

Q

C

dC

tT

C tT dt

QdC


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CT0 = Concentration in the reactor at t = 0

=0T

C 0T

T dtV

Q

C

tt

C

CC

HT

T

T

t

0

1ln

0

=

HT

T tC

Ct

=0

ln Ht

t

TT eCC=

0

CSTR


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PFR

b) Delta Input


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=Acc

VrQCQCV

dt

dCTTT

T

i+=

=Vdt

dCT

=tC

TT

T dtV

Q

CC

dCtT

i 00

tC

T dQdCt

T


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= TTT dt

V

Q

CCi 00

HT

TT t

0C

CCln

i

ti

=

( )H

C

0TT

tCCln

tT

i =

H

i

ti

t

T

TT

eC

CC

=

Hii

t

TtTT eCCC

=

=tT

C

Here it is assumed CT0 = 0 in the reactor at t = 0.

CSTR


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PFR

CSTR Behavior


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GenOutInAcc +=

=Vdt

dCA

KVQCQC0ei AA

=

Q

VKCC0

ei AA=

=eA

C

0 Order, Steady state; rA = -K

Now, we finished tracer test just to recognize CSTR

Of course, we do not have CA over time.

0 Order Non Steady state; rA = -K


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=Vdt

dCA

=dt

dCA ( ) KCC1

dt

dCAA

H

Ai

=

=

tA

0Ai

C

C

t

0

H

A

H

A

A dtCKC

dC

( ) +=+ bXabbXadX

ln1

0 Order Non Steady state; rA K

tC

KC tA

A

i

C

CH

A

H

A

H=

0

ln

GenOutInAcc +=

C

AA tCKC tA

i =

lntC

KC tA

i

C

AA =

ln


77/102

77

HCHH

K

A

=

0

ln

H

H

A

H

A

H

A

H

A

t

CK

C

C

K

C

i

t

i

=

0

ln

Hiti

t

H

A

H

A

H

A

H

A eCKCCKC

=

0

( ) Hiit

t

HAAHAAeKCCKCC

=0

At t , =tA

C

At t 0, =tA

C

tK

ACHH

H=

0

ln

The same as steady-state

Of course, we have CA over time.

1 t O d t d t t KC


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78

VKCQCQCVdt

dC

AAA

A

i =

Q

VKCCC0

eei AAA=

=iA

C

=eA

C

1st Order steady state, rA = -KCA

GenOutInAcc +=

Of course, we do not have CA over time.

1st Order-Non-Steady State CFSTR


79/102

79

=dt

VdCA

=dt

dC

Q

V A

( )

=

+

t

0H

C

C HAA

A dt1

K1CC

dCtA

0A i

y

( )( ) HAHA

AHA

H

t

CK1C

CK1Cln

1K

1

0i

ti

=

+

+

+

GenOutInAcc +=

( )

+=

+

bXa

bbXa

dXln

1

( )( )

AHA tCK1Cln

1 ti =

+


80/102

80

( ) HAHAH CK1C1K 0i

++

( ) ( )H

H

H

H

iti

tK

HA

tK

AHAA eKCeCKCC

+

+

+=+

)1()1(

11 0

H

H

H

H

i

t

tK

A

H

tK

A

A eC

K

eC

C

+

+

+

+

=)1(

)1(

0

1

1

=tA

C

=tA

C

At t ,

At t = 0,

Of course, we have CA over time.

The same as steady-state

2nd Order Steady state rA = -KCA2

GenOutInAcc +=


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81

=Vdt

dCA

Q

VKCCC

eei AAA

2=

0CCCKiee AA

2AH =+

a2

ac4bb

X

2 =

Solve using quadratic formula,

Saturation Reaction: 1 0 order, steady state


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82

VrQCQC0 AAA ei +=

e

e

eiAS

AHAA

CK

CkCC0

+

=

e

e

AS

A

A CK

kC

r +=

ei

e

eAA

AS

AHCC

CKCk =

+

2eASAAASAAH

CKCCCKCCkeeiie

+=

( ) 0CKCKkCCiiee ASASHA

2A =++

Solve using quadratic formula,a2

ac4bbX

2 =

Types of Reactors: (1) Batch, (2) Complete Mix, CFSTR, (3) Plug flow,

(4) Cascade of CFSTR and (5) Packed Bed


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83

(4) Cascade of CFSTR, and (5) Packed Bed

PFR Batch

Comparison of PFR vs CFSTR


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84

O Order CFSTR and PFR

1st Order CFSTR 1st Order PFR

2nd Order CFSTR 2nd Order PFR

Mult iple CFSTRs in Series (Cascade)


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Instead of th is big reactor,

Mult iple CFSTRs in Series (Cascade)


86/102

86

=)(1 tC

How to recognize a series of CFSTRs . Look at response to slug input of tracer.From previous analysis.

=dt

VdC2

)t(2)t(1H)t(2

CC

dt

dC=

2

t

)0(1H2 CeC

dt

dC1H =

Write Mass Balance for Reactor 2

H

t

)0(122 eC

1CdC

=+


87/102

87

22

)0(1HH

eCdt

=

+

( ) ( )tQYtPdt

dY=+

= Qdt1

Y

2H

t

0 2H

tdt1

ee

==

= Pdte

dteC1

eeY 1H

2

2H2H

t

0

H

tt

=

H

22

t

0HH

22 eC1

)t(Q,C

y)t(P,dt

dC

dt

dY

=

==

Linear First Order D.E.

21 HH=If

dteC1

eeY 1H2H2H

t

0

tt

=


88/102

88

dtC

eYH

t

H =

2

2 0

+

=

KtC

eYH

t

H 0

H

t

H0 etC

Y

=

( ) Ht

H

)0(1t2 e

tCC

=

Do not take constants outside the integral.

At t = 0, Y= C2( int ) = 0

K = 0

dteCeeY

2

0H

Need to determine constant of

integration

Other approach to solve the D.E


89/102

89

H

t

HH

eCC

dt

dC

=

+ 022

dtdC'C 2=

H

1

=

( ) tt

H

t eeCCCe H

=+0'

( ) tt

H

tttttt eeC

CCeCeCeCeCeCe H

=+=+=+= 0)'('')'('

dteeC

Ce HHHtt

H

t =

0

Integrate with inserted

=

dtC

Cet

H 0dteeC

Ce HHHttt =

0


90/102

90

H

KtC

CeH

tH +

= 0

HH

tt

H

0 eKetC

C +

=

( ) Ht

H

0t2 e

tCCC

==

H

t2

H

)0(13 et

2CC

=

( )H

t1n

H

)0(1n e

t

!1n

CC

=

At t = 0, Y = C2( int ) = 0

K = 0

H

H

t

t eCC

= 0)(1

Note H is for 1 CFSTR


91/102

91

368.0

C

C

)0(1

max2 =

271.0C

C

)0(1

max3 =

( )( )

( )1n1n

)0(1

maxn e!1n

1n

C

C

=

Slug input of nonreactive tracer

For 3 CFSTRs, C3 max is at t =

For n CFSTRs, Cn max is at t =

H

For 2 CFSTRs, C2 max

is at t =


92/102

92

One input

Input

CSTR

PFR


93/102

93

Input

CSTR

PFR

Reactions in Cascade of CFSTRs (not tracer test)


94/102

94

H

01K1CC

+=

H

e1e2

K1

CC

+

=H

e2e3

K1

CC

+

=

1st Order, assuming V1 = V2 = V3, steady state

Reactor 1

Reactor 2 Reactor 3

only for V1 = V2 = V3( )nH0

neK1

CC

+=

nth Reactor

321 VVV 1st Order


95/102

95

1H

0e1 K1

C

C +=

=eC2

=eC3

steady state

CFSTRs 2nd Order steady state


96/102

122AA

2AH

CCCK0 +=

solve for CA1solve above quadratic for CA2


97/102

Multiple CFSTRs in Series (Cascade)- Model Example


98/102

- Incompletely mixed systems such as estuaries, complicated rive..

- This can be represented as a series of coupled completely mixed systems.

Some Useful Tips for PFR vs. CSTR in Environmental Systems


99/102

1) Shock charge.

2) Microbial or catalytic activity, in case.

3) Reaction rate, if first order.

4) Treatment time.

5) Effluent quality.

PFR

CSTR

Multiful CSTR

PART I. REACTION KINETICS AND REACTOR MODELS


100/102

100

Summary through Example Demonstration

Chapter 2. Mass Balance, Reactor and Flow Models

PFR

CFSTR or CSTR

Vs.

(Similar to Batch)

Problem or Question Statement


101/102

101

The elementary, liquid-phase, irreversible reaction below is to be carried out in a

flow reactor.

CBA +Two reactors are available, a PFR and a CSTR with 200 L volume at 300 K.

The two feed streams to the reactor mix to form a single feed stream that is equal

molar in A and B, with a total volumetric flow rate of 10 L/min.

Additional information is

K = 0.07 L/mol-min

CA,o = 2 mol/L, QA, o= 5 L/min

CB, o = 2 mol/L, QB, o= 5 L/min

Which reactor between CSTR and PFR are you going to use to best perform

the reaction in terms of transformation of chemical A (or B)?

Result


102/102

CSTR

PFR

300 K

Complete mixing

200 L

CA,o=2 mol/L,

QA,o = 5 L/min

CB,o=2 mol/L,

QB,o = 5 L/minCACBCC CA,eff=0.56 mole/L

CB,eff=0.56 mole/L

CC,eff=0.44 mole/L

CA,o=2 mol/L,

QA,o = 5 L/min

CB,o=2 mol/L,

QB,o = 5 L/min

300 K

200 L

CA

CBCC

CA,eff=0.42 mole/L

CB,eff=0.42 mole/LCC,eff=0.58 mole/L

K=0.07

K=0.07

58% Conversion of Chemical A

44% Conversion of Chemical A

Chapter 2-Mass Reactor Model (102 P)

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