CEE 370 Lecture #11 10/2/2019
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David Reckhow CEE 370 L#11 1
CEE 370Environmental Engineering
PrinciplesLecture #11
Ecosystems I: Water & Element Cycling, Ecological PrinciplesReading: Mihelcic & Zimmerman, Chapter 4
Davis & Masten, Chapter 4
Updated: 2 October 2019 Print version
Monday’s local paper
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Follow-up on Tuesday
dd
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CMFR: non-SS Step loads are easier More complicated when reaction is
occurring Simpler for case without reaction
Example 4-5 (pg 128-129) With reaction
Example 4-4 (pg 125-127)
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Non-steady State CMFR Problem:
The CMFR is filled with clean water prior to being started. After start-up, a waste stream containing 100 mg/L of a conservative substance is added to the reactor at a flow rate of 50 m3/day.The volume of the reactor is 500 m3.What is the concentration exiting thereactor as a function of time afterit is started?
CA
VCA0
Q0
CA
Q0
Equalization tank
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Non SS CMFR (cont.) So the general reactor equation reduces to:
And because we’ve got a conservative substance, rA=0:
Now let:
Vr QC QC = dt
dmAoutin
A
QC QC = dt
dCV outin
A
C CV
Q=
dt
dCoutin
A
inout CCy
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Non SS CMFR without reaction So that:
Rearranging and integrating,
Which yields,
or
y V
Q=
dt
dy
tty
y
dtV
Q
y
dy
0
)(
)0(
tV
Q
y
ty
)0(
)(ln
tVQ
ey
ty
)0(
)(
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Non SS CMFR w/o reaction (cont.) And substituting back in for y:
Since we’re starting with clean water, Co=0
And finally,
tVQ
ino
in eCC
CC
tVQ
in
in eC
CC
tV
Q
inin eCCC
tV
Q
in eCC 1
and
Cin
C
t
Decreasing Q/V
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Now add a reaction term Returning to the general reactor equation:
And now we’ve got a 1st order reaction, rA=kC=kCout:
This is difficult to solve, but there is a particular case with an easy solution: where Cin = 0 This is the case where there is a step decrease in the influent
concentration to zero (M&Z, example 4.4)
Vr QC QC = dt
dmAoutin
A
outoutin kVC QC QC = dt
dCV
outoutin kC C CV
Q=
dt
dC
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Non SS CMFR; Cin=0 So that:
Rearranging, recognizing that in a CMFR, C=Cout, and integrating,
Which yields,
or
ttC
C
dtkV
Q
C
dC
0
)(
)0(
tkV
Q
C
tC
)0(
)(ln
tkVQ
eC
tC
)0(
)(
outout kC CV
Q=
dt
dC
dt kV
Q=
C
dC
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SS Comparison of PFR & CMFR CMFR PFR
QkV
AoA eCC
kQV
CC Ao
A
1
kQVC
C
Ao
A
1
1
QVk
Ao
A eC
C
Example: V=100L, Q=5.0 L/s, k=0.05 s-1
50.0
05.051001
1
Ao
A
C
C
37.0
510005.0
eC
C
Ao
A
1stor
der r
eact
ion
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Conclusion: PFR is more efficient for a 1st order reaction
Distance Across Reactor
CMFR
PFR
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Rate of reaction of A is given by VkCA
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Response to Inlet Spikes
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Selection of CMFR or PFR PFR
Requires smaller size for 1st order process CMFR
Less impacted by spikes or toxic inputs
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Comparison Davis & Masten, Table 4-1, pg 157
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Retention Time
Q
V
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Analysis of Treatment Processes Basic Fluid Principles
Volumetric Flow Rate Hydraulic Retention Time
Conversion Mass Balances Reaction Kinetics and Reactor Design
Chemical Reaction Rates Reactor Design
Sedimentation Principles
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Conversion or Efficiency
k
aA + bB pP + qQ
X = (C - C )
CAo A
AoA AoC = C (1 - X)
And the Conversion, X, is:
or
Some use “efficiency” (ɳ) to indicate the same concept
Stoichiometry
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Conversion/efficiency (cont.)(N - N ) =
b
a(N - N )Bo B Ao A
where,NAo = moles of A at t = 0, [moles]NA = moles of A at t = t, [moles]NBo = moles of B at t = 0, [moles]NB = moles of B at t = t, [moles]
If the volume of the reactor is assumed to remain constant, we can divide both sides of the expression by CAoV. The expression then becomes,
(C - C )
C =
b
a
(C - C )
C =
b
aXBo B
Ao
Ao A
Ao
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Conversion/efficiency (cont.)This expression can then be solved for the concentration of B in terms of other known quantities:
B Bo AoC = C - b
aC X
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Conversion/efficiency ExampleThe reactor shown in the Figure has an inflow of 750 L/hr. The concentration of A in the influent is 0.3 M and the concentration of B in the influent is 0.5 M. The conversion (of A) is 0.75. The reaction is:
A + 2B ProductsFind the conversion of B, XB, and the effluent concentration of A and B.
CAo = 0.3MCBo = 0.5MQ=750 L/hr
CA
VCA = ?CB = ?
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Solution to Conversion Ex.The first step in the solution is to determine the effluent concentration of A. This can be obtained as follows:
A AoC = C (1 - X) = 0.3M x (1 - 0.75)
AC = 0.075 M
For each mole of A converted to product, two moles of B are converted to product. Since we know the initial concentration of B we can calculate its final concentration:
Moles/ L of B converted = 2 x (0.3 M - 0.075 M) = 0.45 M
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Solution to Conversion Ex. (cont.)
BC = 0.5 M - 0.45 M = 0.05 M
Alternatively, using Eqn:
B B o A oC = C b
aC X = 0 .5 M
2
1 (0 .3 M x 0 .7 5 )
BC = 0.05 M
The conversion of B is then:
BBo B
BoX =
(C - C )
(C ) =
0.5 M - 0.05 M)
0.5 M
BX = 0.9
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Reactors in Series
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Q
Q
w
w
w w w
Reactors in Series
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Lake Volume (109 m3) Outflow (109 m3 y-1)
Superior 12,000 67Michigan 4,900 36Huron 3,500 161Erie 468 182Ontario 1,634 211
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Example: 90Sr fallout in Great Lakes
1
1
2
3
4
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Loading Function Close to an
impulse load, centered around 1963
estimated value is: 70x10-9 Ci/m2
same for all lakes
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Non Steady State Solution Impulse Load
1st lake
2nd lake 3rd lake
toecc 1
11
C11
tto
to ee
V
Qcecc 212
)( 122
12122
C21
C22
23131223
12231
233
23233
3231
323
)(
)(
tttt
o
tto
to
eeee
VV
QQc
eeV
Qcecc
C31
C33
C32
kV
Q
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Hydrologic ParametersParameter Units Superior Michigan Huron Erie OntarioMean Depth m 146 85 59 19 86SurfaceArea
106 m2 82,100 57,750 59,750 25,212 18,960
Volume 109 m3 12,000 4,900 3,500 468 1,634Outflow 109 m3/yr 67 36 161 182 212
Michigan Superior Huron Erie Ontario
mm cc 11
ss cc 11
mhshhh cccc 222221
mhesheheee ccccc 33333231
mheosheoheoeooo cccccc 4444434241
kV
Q
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From Chapra, 1997
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