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David Reckhow CEE 370 L#11 1
CEE 370Environmental Engineering Principles
Lecture #11Ecosystems I: Water & Element
Cycling, Ecological PrinciplesReading: Mihelcic & Zimmerman, Chapter 4
Davis & Masten, Chapter 4
Updated: 1 October 2019 Print version
http://www.ecs.umass.edu/cee/reckhow/courses/370/slides/370l11p.pdf
Monday’s local paper
David Reckhow CEE 370 L#11 2
Follow-up on Tuesday
dd
David Reckhow CEE 370 L#11 3
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)
David Reckhow CEE 370 L#11 4
David Reckhow CEE 370 L#11 5
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?
CAV
CA0Q0
CAQ0
Equalization tank
David Reckhow CEE 370 L#11 6
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 = dtdm
AoutinA −−
QC QC = dtdCV outinA −
( ) C CVQ=
dtdC
outinA −
inout CCy −=
David Reckhow CEE 370 L#11 7
Non SS CMFR without reaction So that:
Rearranging and integrating,
Which yields,
or
y VQ=
dtdy
−
∫∫ −=tty
y
dtVQ
ydy
0
)(
)0(
tVQ
yty
−=
)0()(ln
tVQ
ey
ty −=)0()(
David Reckhow CEE 370 L#11 8
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 eCCCC −=−−
tVQ
in
in eCCC −=
−− tVQ
inin eCCC
−
−=−
−=
− tV
Q
in eCC 1
and
Cin
C
t
Decreasing Q/V
David Reckhow CEE 370 L#11 9
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 = dtdm
AoutinA −−
outoutin kVC QC QC = dtdCV −−
( ) outoutin kC C CVQ=
dtdC
−−
David Reckhow CEE 370 L#11 10
Non SS CMFR; Cin=0 So that:
Rearranging, recognizing that in a CMFR, C=Cout, and integrating,
Which yields,
or
∫∫
+−=
ttC
C
dtkVQ
CdC
0
)(
)0(
tkVQ
CtC
+−=
)0()(ln
tkVQ
eC
tC
+
−
=)0()(
( ) outout kC CVQ=
dtdC
−−
dt kVQ=
CdC
+−
David Reckhow CEE 370 L#11 11
SS Comparison of PFR & CMFR CMFR PFR
QkV
AoA eCC−
=kQVCC AoA
+
=1
kQVC
C
Ao
A
+
=1
1
−
= QVk
Ao
A eCC
Example: V=100L, Q=5.0 L/s, k=0.05 s-1
( )50.0
05.051001
1
=
+=
Ao
A
CC ( )
37.0
510005.0
=
=−e
CC
Ao
A
1stor
der r
eact
ion
David Reckhow CEE 370 L#11 12
Conclusion: PFR is more efficient for a 1st order reaction
Distance Across Reactor
CMFR
PFR
David Reckhow CEE 370 L#11 13
Rate of reaction of A is given by VkCA
David Reckhow CEE 370 L#11 14
Response to Inlet Spikes
David Reckhow CEE 370 L#11 15
Selection of CMFR or PFR PFR
Requires smaller size for 1st order process CMFR
Less impacted by spikes or toxic inputs
David Reckhow CEE 370 L#11 16
Comparison Davis & Masten, Table 4-1, pg 157
David Reckhow CEE 370 L#11 17
Retention Time
QV
=θ
David Reckhow CEE 370 L#10 18
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
David Reckhow CEE 370 L#10 19
Conversion or Efficiency
k aA + bB pP + qQ→
X = (C - C )C
Ao AAo
A AoC = C (1 - X)
And the Conversion, X, is:
or
Some use “efficiency” (ɳ) to indicate the same concept
Stoichiometry
David Reckhow CEE 370 L#10 20
Conversion/efficiency (cont.)(N - N ) =
ba
(N - N )Bo B Ao Awhere,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
= ba
(C - C )C
= ba
XBo BAo
Ao AAo
David Reckhow CEE 370 L#10 21
Conversion/efficiency (cont.)This expression can then be solved for the concentration of B in terms of other known quantities:
B Bo AoC = C - baC
X
David Reckhow CEE 370 L#10 22
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
CAV
CA = ?CB = ?
David Reckhow CEE 370 L#10 23
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
David Reckhow CEE 370 L#10 24
Solution to Conversion Ex. (cont.)
BC = 0.5 M - 0.45 M = 0.05 M
Alternatively, using Eqn:
B Bo AoC = C ba C
X = 0.5 M 21
(0.3 M x 0.75)− −
BC = 0.05 MThe conversion of B is then:
BBo B
BoX =
(C - C )(C )
= 0.5 M - 0.05 M)0.5 M
BX = 0.9
B
Bo
Ao
C
=
C
b
a
C
X
=
0.5
M
2
1
(0.3
M x
0.75)
-
-
æ
è
ç
ö
ø
÷
Reactors in Series
David Reckhow CEE 370 L#11 25
Q
Q
w
w
w w w
Reactors in Series
David Reckhow CEE 370 L#11 26
David Reckhow CEE 370 L#13 27
Lake Volume (109 m3) Outflow (109 m3 y-1)Superior 12,000 67Michigan 4,900 36Huron 3,500 161Erie 468 182Ontario 1,634 211
David Reckhow CEE 370 L#13 28
Example: 90Sr fallout in Great Lakes
1
1
2
3
4
David Reckhow CEE 370 L#13 29
David Reckhow CEE 370 L#13 30
David Reckhow CEE 370 L#13 31
David Reckhow CEE 370 L#13 32
Loading Function Close to an
impulse load, centered around 1963
estimated value is: 70x10-9 Ci/m2
same for all lakes
David Reckhow CEE 370 L#13 33
Non Steady State Solution Impulse Load
1st lake
2nd lake 3rd lake
toecc 111
λ−=C11
( )ttoto eeVQcecc 212
)( 12212
122λλλ
λλ−−− −
−+=
C21C22
( )
−−
−−−
−+
−−
+=
−−−−
−−−
23131223
12231
233
23233
3231
323
)(
)(
λλλλλλ
λλλλλλ
λλλ
tttt
o
tto
to
eeeeVV
QQc
eeV
QceccC31
C33
C32
kVQ+=λ
David Reckhow CEE 370 L#13 34
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 ++++=
kVQ+=λ
Parameter
Units
Superior
Michigan
Huron
Erie
Ontario
Mean Depth
m
146
85
59
19
86
Surface Area
106 m2
82,100
57,750
59,750
25,212
18,960
Volume
109 m3
12,000
4,900
3,500
468
1,634
Outflow
109 m3/yr
67
36
161
182
212
David Reckhow CEE 370 L#13 35
From Chapra, 1997
David Reckhow CEE 370 L#11 36
To next lecture
http://www.ecs.umass.edu/cee/reckhow/courses/370/slides/370l12.pdf
CEE 370� Environmental Engineering PrinciplesMonday’s local paperFollow-up on TuesdayCMFR: non-SSNon-steady State CMFRNon SS CMFR (cont.)Non SS CMFR without reactionNon SS CMFR w/o reaction (cont.)Now add a reaction termNon SS CMFR; Cin=0SS Comparison of PFR & CMFRSlide Number 12Slide Number 13Response to Inlet SpikesSelection of CMFR or PFRComparisonRetention TimeAnalysis of Treatment ProcessesConversion or EfficiencyConversion/efficiency (cont.)Conversion/efficiency (cont.)Conversion/efficiency ExampleSolution to Conversion Ex.Solution to Conversion Ex. (cont.)Reactors in SeriesReactors in SeriesSlide Number 27Example: 90Sr fallout in Great LakesSlide Number 29Slide Number 30Slide Number 31Loading FunctionNon Steady State SolutionHydrologic ParametersSlide Number 35Slide Number 36