CEE 370 Environmental Engineering Principles Lecture #11 - - PowerPoint PPT Presentation

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: 1 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

• ccurring

 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 the reactor as a function of time after it is started?

CA V

CA0 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:

V r Q C Q C = dt dm

A

• ut

in A

− −

Q C Q C = dt dC V

• ut

in A

−

( )

C C V Q = dt dC

• ut

in A

−

in

• ut

C C y − =

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Non SS CMFR without reaction

 So that:  Rearranging and integrating,  Which yields,  or

y V Q = dt dy −

∫ ∫

− =

t t y y

dt V Q y dy

) ( ) (

t V Q y t y − =         ) ( ) ( ln

t V Q

e y t y

      −

= ) ( ) (

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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,

t V Q in

• in

e C C C C

      −

= − −

t V Q in in

e C C C

      −

= − −

t V Q in in

e C C C

      −

− = −         − =

      − t V Q in

e C C 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)

V r Q C Q C = dt dm

A

• ut

in A

− −

• ut
• ut

in

kVC Q C Q C = dt dC V − −

( )

• ut
• ut

in

kC C C V 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

∫ ∫

      + − =

t t C C

dt k V Q C dC

) ( ) (

t k V Q C t C       + − =         ) ( ) ( ln

t k V Q

e C t C

      +       −

= ) ( ) ( ( )

• ut
• ut

kC C V Q = dt dC − −

dt k V Q = C dC       + −

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SS Comparison of PFR & CMFR

 CMFR  PFR Q kV Ao A

e C C

−

=

k Q V C C

Ao A

      + = 1 k Q V C C

Ao A

      + = 1 1

      −

=

Q V k Ao A

e C C

Example: V=100L, Q=5.0 L/s, k=0.05 s-1

( )

50 . 05 . 5 100 1 1 = + =

Ao A

C C

( )

37 .

5 100 05 .

= =

−

e C C

Ao A

1st order reaction

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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

• f 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 = θ

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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 )

C

Ao A Ao

A Ao

C = C (1 - X)

And the Conversion, X, is:

• r

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 aX

Bo B Ao Ao A Ao

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Conversion/efficiency (cont.)

This expression can then be solved for the concentration

• f B in terms of other known quantities:

B Bo Ao

C = C

• b

aC X

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Conversion/efficiency Example

The 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 → Products Find the conversion of B, XB, and the effluent concentration of A and B.

CAo = 0.3M CBo = 0.5M Q=750 L/hr

CA V

CA = ? 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 Ao

C = C (1 - X) = 0.3M x (1 - 0.75)

A

C = 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

• f 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.)

B

C = 0.5 M - 0.45 M = 0.05 M

Alternatively, using Eqn:

B Bo Ao

C = C b a C X = 0.5 M 2 1 (0.3 M x 0.75) − −      

B

C = 0.05 M

The conversion of B is then:

B Bo B Bo

X = (C

• C )

(C ) = 0.5 M - 0.05 M) 0.5 M

B

X = 0.9

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 67 Michigan 4,900 36 Huron 3,500 161 Erie 468 182 Ontario 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

t

• e

c c

1

1 1 λ −

=

C11

( )

t t

• t
• e

e V Q c e c c

2 1 2

) (

1 2 2 12 1 2 2 λ λ λ

λ λ

− − −

− − + =

C21 C22

( )

        − − − − − − + − − + =

− − − − − − − 2 3 1 3 1 2 2 3 12 23 1 2 3 3 23 2 3 3

3 2 3 1 3 2 3

) ( ) ( λ λ λ λ λ λ λ λ

λ λ λ λ λ λ λ t t t t

• t

t

• t
• e

e e e V V Q Q c e e V Q c e c c

C31 C33 C32

k V Q + = λ

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Hydrologic Parameters

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

 Michigan  Superior  Huron  Erie  Ontario

m m

c c

11

=

s s

c c

11

=

mh sh h h

c c c c

22 22 21

+ + =

mhe she he e e

c c c c c

33 33 32 31

+ + + =

mheo sheo heo eo

• c

c c c c c

44 44 43 42 41

+ + + + = k V Q + = λ

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 From Chapra,

1997

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 To next lecture