cee 370 environmental engineering principlesc ao a 1 k q c v c ao a 1 1 q k v ao a e c c example:...

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CEE 370 Lecture #11 10/2/2019 Lecture #11 Dave Reckhow 1 David Reckhow CEE 370 L#11 1 CEE 370 Environmental Engineering Principles Lecture #11 Ecosystems I: Water & Element Cycling, Ecological Principles Reading: Mihelcic & Zimmerman, Chapter 4 Davis & Masten, Chapter 4 Updated: 2 October 2019 Print version Monday’s local paper David Reckhow CEE 370 L#11 2

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Page 1: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 1

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

David Reckhow CEE 370 L#11 2

Page 2: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 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

Page 3: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 3

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?

CA

VCA0

Q0

CA

Q0

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 = dt

dmAoutin

A

QC QC = dt

dCV outin

A

C CV

Q=

dt

dCoutin

A

inout CCy

Page 4: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 4

David Reckhow CEE 370 L#11 7

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(

)(

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 eCC

CC

tVQ

in

in eC

CC

tV

Q

inin eCCC

tV

Q

in eCC 1

and

Cin

C

t

Decreasing Q/V

Page 5: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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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 = dt

dmAoutin

A

outoutin kVC QC QC = dt

dCV

outoutin kC C CV

Q=

dt

dC

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

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

Page 6: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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David Reckhow CEE 370 L#11 11

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

David Reckhow CEE 370 L#11 12

Conclusion: PFR is more efficient for a 1st order reaction

Distance Across Reactor

CMFR

PFR

Page 7: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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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

Page 8: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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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

Page 9: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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David Reckhow CEE 370 L#11 17

Retention Time

Q

V

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

Page 10: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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David Reckhow CEE 370 L#10 19

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

David Reckhow CEE 370 L#10 20

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

Page 11: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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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 - b

aC 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

CA

VCA = ?CB = ?

Page 12: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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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 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

Page 13: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 13

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

Page 14: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

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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

Page 15: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 15

David Reckhow CEE 370 L#13 29

David Reckhow CEE 370 L#13 30

Page 16: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 16

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

Page 17: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 17

David Reckhow CEE 370 L#13 33

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

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

kV

Q

Page 18: CEE 370 Environmental Engineering PrinciplesC Ao A 1 k Q C V C Ao A 1 1 Q k V Ao A e C C Example: V=100L, Q=5.0 L/s, k=0.05 s-1 0.50 0.05 5 1 100 1 Ao A C C 0.37 5 0.05100 e C C Ao

CEE 370 Lecture #11 10/2/2019

Lecture #11 Dave Reckhow 18

David Reckhow CEE 370 L#13 35

From Chapra, 1997

David Reckhow CEE 370 L#11 36

To next lecture