chapter 2-mass reactor model (102 p)
TRANSCRIPT
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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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Methods for Determining Reaction Order and Rate Constants
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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Methods for Determining Reaction Order and Rate Constants
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?
2) G hi l M th d I t ti M th d ( t )
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Kt
AA eCC t-
0=
2) Graphical Method + Integration Method (no computer program)
- 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
2) G hi l M th d I t ti M th d ( t )
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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
2) Graphical Method + Integration Method (no computer program)
- 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 ==
2) Graphical Method + Integration Method (no computer program)
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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67
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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69
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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71
PFR
b) Delta Input
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72
=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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73
= 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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75
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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76
=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
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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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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
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=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 =
+
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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)
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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
=+
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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
=
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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
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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
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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
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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 =
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92
One input
Input
CSTR
PFR
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93
Input
CSTR
PFR
Reactions in Cascade of CFSTRs (not tracer test)
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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
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95
1H
0e1 K1
C
C +=
=eC2
=eC3
steady state
CFSTRs 2nd Order steady state
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122AA
2AH
CCCK0 +=
solve for CA1solve above quadratic for CA2
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Multiple CFSTRs in Series (Cascade)- Model Example
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- 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
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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
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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
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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
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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