Solutions manual for surface water-quality modeling 9780070113640, 0070113645, 9780070113657, 0070113653

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O4729 2468

Solutions Manual for

SURFACE WATER-QUALITY MODELING Stephen C. Chapra

University of Colorado

*

The McGraw-Hill Companies, Inc.

New York

St. Louis

San Francisco

Auckland

Bogota

Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto

sy

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umn TD

365 C43

1997

suppl. bKS

McGraw-Hill

A Division of The McGraw-Hill Companies

24

Solutions Manual for SURFACE WATER-QUALITY MODELING Copyright © 1997 by The McGraw-Hill Companies, Inc. Al rights reserved.

Printed in the United States of America. The contents or parts thereof may be reproduced for use with SURFACE WATER-QUALITY MODELING by Stephen C. Chapra

provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without permission of the publisher. ISBN 0-07-011365-3 1234567890 BKM BKM909876 http://www.mhcollege.com

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(3! fol a0

NOTE TO THE INSTRUCTOR AND/OR GRADER This manual has been developed to help you use Surface Water-Quality Modeling more effectively in the classroom. I have solved all the problems in the text and have reproduced my solutions on the following pages. In addition, a number of errors have come to my attention during classroom testing. I have listed these in the back of this manual. Some of the errata relate to homework

problems, and I have kept these errors separate from other types of errors. McGraw-Hill

should correct all these mistakes when the text goes into a second printing. In the meantime, you should consult the errafa in the back of this manual to minimize any confusion the errors may cause to your students. For example, I have reproduced the list and distributed it to my students so they can incorporate the corrections into their own books. Finally, if you find additional errors, or just have suggestions or comments you would like to send me, I’ve included both my current address and e-mail location below. Steve Chapra

November 3, 1996

Civil, Environmental and Architectural Engineering Campus Box 428 The University of Colorado

Boulder, CO 80309-0428

Phone: (303) 492-7573

Fax: (303) 492-1347

E-mail: [email protected]

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©

Digitized by Google

Original from

UNIVERSITY OF MICHIGAN

LECTURE 1: Ll m= eV =8500-&. 25L--o = 21258 nm’?

12 (@) Q = 100,000 capita(650 L / capita /d)-1

m?

SB

10° L 86400s

_9.7523_

W = 100,000 capita(35 grams/ capita /d). 2°54, KS () (6) 13

1000m_lhr

sec

8

1 mit _ 4007 5 ma

mgL | 20778

¢ = ha BSetewials 10

_3km

3

=041672

8

14

(@ a= W _ 6,950 mia 10° mg km 3 = 368,754 3 3 tonne 10° m°® yr () Ws ac= 868,75

_.§ ug/L = 4343.75 mta

pe/L

(2 % resin = $2504.20, so9y 375% 15 (a) The total flow is equal to the summation of the river flow, Q,, and the waste flow, Q,, The river flow can be determined as

Q, =UA, = 12.100 2?

5

3

3

1m _- 2930.

(3.281 Ry

5

The waste flow can be converted to the proper units,

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Ll

1m?/s = 10 MGD.——______. = 0.438 GD 29.8245 MGD 8

3

and the total flow calculated as

=O, +Q,, = 2.8314 0.438 = 3.269 cms _ 2p +QyCw _ 2.831 cms(0.2 mg/L)+ 0.438 cms(2 mg/L) = 0.441.208

©) 0-040,

3.269

L

16 (@ ¥,=1ga-— 1 _=3 78541 0.264172 gal on Mist Yate _ 3.7854 L(0.250 mg/L)+2 L(2 mg/L) _ 5 gg, me Yt,

3.7854 L+2L

(6) m, =eM, = 0.25078 (3.7854 L)

8 — = 0.9464x107 g

L

my == OV, Cx

mg

=2—-2 L ¢ L

oe

g

m= cV = 08552 (5.7854 L) L

“L

10° mg =4x10 x

-3

g

8 — = 4.9464 x10" g

10° mg

1.7 For the present case, the following balance must hold

Q,(0.5 g/L) +Qp(0.05 g/L) =4 cma(0.1 g/L) Because this equation has two unknowns, another equation is required. This additional formulation arises from the fact that

Q4 +Qp = 4 cms This equation can be solved for 2, and the result substituted into the first equation to give

Q,(05 g/L)+(4—Q, X0.05 g/L) = 4 cms(0.1 g/L) which can be solved for

*

0.2 = 0.444 cms =—= 4-045 and Qp = 4-0.444 = 3.556 cms

1.2

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18

@

c= 10. mL(100 mg/L) _ 3 3m 300 mL

L

(6) m= 33358. -300 mL19

g

1L

=1x107g

10° mL 10° mg

_8.3g/m'(12x10% m? )+1.0 g/m3(9x10° m?) =517&g

12x 10° m>+9x10° m3

m

1.10 The following balance should hold

55 - 10100) +2, 0) Q,+1

which can be solved for _

3

Q, = 1100) —55 = 17182—L..___

55

min

1000L

min = 2864x 10

605

—m?

s

1.11

(©) The rate at which the lake was being filled in would have to balance the flux; that is,

J=vye,

where v, = the rate at which the sediment is accumulating, and c, = the concentration of solid matter

in the sediments. This latter quantity can be calculated as

o,=(1- pp where ¢ = porosity and p = density. Therefore, the rate at which the lake fills in can be calculated

as

vy = a

(l-#p

5

(1-0.9)2.5 g/cm?

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3

10'

00m

13

vp, = 0.002—m 10° ym

mm

=2— yr

1.12 7 - AM. _ Yao _ He

~ Abt

Abt

At

_5m(100

20)

20)

28.

1 mo

1 mo

z

30.42d 10° mg

= 1315-8

md

1.13 The flow in the creek and the treatment plant should be equal to the flow below the mixing point, Q,+Q,

= 0494

The following mass balance should hold 639 = Q4 (170) + Qp(820) 0.494 These two equations can be combined to yield,

639 = 2a01 70+ (0.494-Q, 820)

7

0.494

which can be solved for Q,

_ 0.494(639 — 820) = 0.13756 cms 170-820

Qs = 0.494— 0.13756 = 0.356 cms

2-8 (©) M=2—7

23042d “Skg _ 6,084

(1 mo)10" m are mo

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10°,

1.4

*

LECTURE 2: 2.1 Applying equal-area differentiation t

c

_e

@

(mg/L)

a

10.5

0 2

5.1

4

31

6

28

8

22

10

19

ke

(mg L 1)

2.7

4 17

1

0.5

0.15

0.27

03

0.12

os

0.8

4

3 \ 2

\

1

° -deidt

N “LN

o

2

4

6

19%

8

5

10

a s

44 a

O14 0.01

Z + 10

1

c

H 100

Linear regression yields a least-squares fit of te{ -]

=—L6213+ 2.333loge

Therefore according to the differential approach, the parameters are estimated as n = 2.333 and k = 0.0239.

Go

gle

2.1

On the basis of this result, we can try the integral method with n = 2. yields a best-fit line of

A fit of the plot of 1/c versus f

LL 0.1135 + 0.04236¢ c q

Therefore, this fit yields k = 0.04236. Finally, a nonlinear regression can be used to directly fit the model (see Lec. 8 problems for

additional details on how this is done). The result is

6 «0.03266 dt The results can be summarized as

order

integral

integral/least-squares

differential

0.04236

0.0326

0.0239

2

rate

2.23

2.333

Thus, the integral/least-squares method falls between the integral and differential results. 2.2 de __pe3

dt

Safar ¢

-2

faite

2

The constant can be evaluated by applying the initial condition: c = c, at t= 0,

—2 Therefore, the solution is 4 aa+t c

&S

Consequently, if the reaction is third order, a plot of Vc? vesus t should yield a straight line with a

slope of 2k and an intercept of 1/c,?.

2.3 Plots of c versus ¢ and 1/c versus ¢ both yield nonlinear patterns. In contrast, a plot of In c versus t yielded a straight line (see plot below). A linear regression was employed to fit the following line to this data:

Inc = 1.609 - 0.035¢

(1? = 0.99993)

Therefore, k = 0.035 min“).

2.2

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24 Taking the logarithm of Eq. 2.44 yields log(k(t)) = log(k(20)) + (T — 20) log9 Thus, a plot of log[k(t)] versus (7 - 20) should yield a straight line with a slope of log@ and an intercept of log[k(20)].

Linear regression can be used to do this yielding

log(k(t)) = -0.60474 + 0.020722(T — 20) 8 = 10°°?™ — 10489 Keo = 10°54 = 0.248 2.5

6= Qf) = 1.9% = 1.066 kg = 1.6(1.066)?0-2° = 3.04 a!

2.6 (a) The glucose is decomposing, and using oxygen as the oxidizing agent. Equation 2.3 defines the stoichiometry of the reaction. Therefore, the glucose can be expressed in oxygen equivalents as

«g, ~ 10-mb100 mgC/L) 6(32)mgO _ 5 555 mg0 *

300 mL

o2)mg¢

SCL

This is precisely the amount of oxygen depletion that occurs in the bottle. decomposition is first-order, the following model could be hypothesized

Therefore,

if the

c=c, -e,(1-e*) where c, = the initial oxygen concentration in the bottle. This result can be manipulated to give

a= fe) °s

(b) Consequently, a plot of the right-hand-side versus ¢ should yield a straight line with a slope of k and a zero intercept. When this was done for the first 8 points (the last two seem to have already reached completion of the reaction) , the result was k = 0.09485 d"'.

2.3

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2.7 Because the model follows a first-order process, the natural log of concentration versus time should yield a straight line with a slope equal to the removal rate. When this is done, the least-squares fit is Ine = 7.94175— 0.01361t Therefore, the removal rate is 0.01361 d''. compute a settling velocity

This value can be used in conjunction with the depth to

v= KH = 0.01361(55) = oo7487 2.8 A plot of the natural logarithm of population versus time can be fit with a straight line. The resulting equation can be employed to compute,

In p = -29086+ 0.149985(1995) = 8.35525 P(1995) = * 3555 = 4252 2.9 The rate of growth can be computed as k

5

_ 14x 10°)—1n(05x 10°) 300 yrs

= 0.00693 yr?

Thus, in another hundred years (t = 400), the population would be

p= 05x 10° 49640) _ g x 10° 2.10 After overturn, the lake will reaerate and approach saturation. be modeled by the following mass balance

Therefore, the dissolved oxygen can

#4 (c,-0)

dc V—=kV(c,-

where k, = the reaeration rate [¢''] and c, = the saturation concentration [d"!]. If c =c, at t = 0, this equation can be integrated to yield, c=egt(c, ~e,(1-e™) which can be manipulated algebraically to give,

Therefore, a plot of the natural log of (c, - c\(c, - c,) versus ¢ should yield a straight line with a slope

of -k, and an intercept of zero. The values are shown in the table below. The line of best fit is developed (with a linear regression algorithm with a zero intercept) to yield an estimate of k, = 0.149 d!_

This number can be converted into a transfer coefficient by multiplying it by the mean depth of

the lake,

¥q = 0.14924(12) = Lat 2.4

Google

t

ec

c,-¢ c,—

0

4.6

3

6

0.000

63

-0.448

8.0 8.4 8.7 8.9

1.343 -1.791 -2.239 -2.686

13

9 12 1S 18

21

-0.895

9.0

-3.134

2.11 According to Eq. 2.46,

0 =

Qh = (2.2) = L082

which can be used to compute

kys = 0853 x 1082)*5 = 0.388 wk! 2.12. In this problem, we are treating the person as if he or she were a batch reactor with first-order decay. Thus, their level of anaesthetic will be governed by the model v=—m et

M

where v is the mass-specific concentration of the anaesthetic [mg/kg], m is the amount of anaesthetic administered [mg], / is the mass of the individual [kg], and & is the first-order decay rate. Thus, the

dose can be determined as

m= Mve™ = 50(10)e°%05) — 675 mg 2.13 The first-order decay model can be expressed in terms of half-life as

c=c,e

0.693 Sr

'

which can be solved for

tia

t=—42_In

,(%)_

—

0.693 at )

5730

|=——In

0.693

1

—

|= 30,

goa }~ 20700 ys

2.14 Both the differential and integral methods suggest that the rate is second order. If this is correct, the integral method yields a rate estimate of 0.06 L/(mole min). The plot of I/c versus ¢ is depicted

below:

2.5

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b @ @ ON

tic

12

2s © 2-43 dt

sS = -kfar

;

-2

—=-kt+C The constant can be evaluated by applying the initial condition: c = c, at t= 0,

c=

2

2

Therefore, solution is

p= dete c

6

Consequently, if the reaction is third order, a plot of 1/c? versus t should yield a straight line with a slope of 2k and an intercept of 1/c,?. For the present problem this yields a slope of 0.002. Therefore, the reaction rate would be 0.001. 2.16

Note that the value of @ is independent of the base temperature.

be computed as

Therefore, the value at 20°C can

(20) = 0.11.06)?"> = 0.074726 and the new formula would be

K(T) = 0.074726(1.06)" 2°

2.17 Simple divided differences can be used to develop slope estimates at = 0, t

1

c

1.00 0.95 -0.05

2.00 1.87 -0.13

Google

5.00 4.48 -0.52

2.6

10.00 8.59 -1.41

The log of the negative of the rate estimate can be plotted versus the log of c.. The result has an intercept of -1.30988 and a slope of 1.459. Thus, the order is 2 = 1.459 and the rate is

k= 10-3" — 0.049 2.18

A plot of the logarithm of half-life versus the logarithm of the initial concentration yields a straight line with a slope of -0.50353 and an intercept of 1.199.

These values can be employed to compute

n=1+ slope = 1— 0.50353 = 0.49647 gry

20.49647-1 _

(n-1y10™"""

(0.49647 - 110"

k =———____ = _*______- 0037 2.19

The integral least-squares approach (using Heun’s method as the numerical integrator) can be employed to estimate values of k = 0.053 andn = 1.46. The fit is shown below.

S

o

a

>

»

of

oN Fe OF

Of

10

2.20

Fo

=

1x31 mgP

pros “1x31 mgP+4x16mgO

= 0326 MePOs

"sa

mgP.

0326—P_ 19. MBPOs _ 326m mgPO,

m

Therefore, if the concentration is actually phosphate and you erroneously interpret it as phosphorus, you would be off by a factor of about 3. 2.21 (a)

7.=

10732 _, 680 106 x 12 gc

gc x 269 80 = 26980 () 10°F (c)

gC

10=>

m

16x14 gN 10° mgN_

x ot

106x12gC

mas

gN

2.7

Google

LECTURE 3: 3.1 (@) (Note that the nomenclature L is conventionally used to designate BOD)

gal m cap-d 264.172 gal

m’? cap-d

P, =150-___"___ = 0568 ___ P,,, 4

= 9?

Ib g cap-d 0.002205 Ib

g cap-d

=025—-_-—_ 8 ___ = 1134 _8—

1, = 1134 - 1997 0568

m

(b) V =3(2x10°)=6x10'm> =

= x10" 2(14)

m?

3

42657

3

dm?

AV =0.1(6x 10°) = 60000"

vA, =0.1(2 x 10°) = 200007

a= 42857 + 60000 + 20000 =122,857

3

As reflected by the individual terms of this equation, the most effective purging mechanism is the decay reaction (60000), followed by flushing (42857) and settling (20000).

(c) The total flow through the lake will be enhanced by the subdivision flow, mm?

Q,, = 05678(1000) = 568—-

Q = 42857+568= 434257,

3

43425 B =———————-= 43425 + 60000 + 20000

0352

(@) W, =42857(4) = 1714298 with:

L

W, =568(199.7) =1134008

= 7848284 —_ MB 43425 + 60000+20000 ~~ L

ithout: L=

W= 2gagze%

171429 __14m8 43425+60000+20000 LL

3.2 (a)

(()

= =W(t)+ Qe, -Qe-kVe

= AH =11x10°(5)=55x108 m? =

Vv

Tw

6

= 55x10 m 4.6 yrs

3

1196x108

3

yr

cx W+Qe,, _ 2000x106 g/yr+11.96x10° m? /yr(15 g/m?) _ 14.88

Q+kv

11.96 x 10° m3 / yr +0.1/yr(55x 10°)

Google

3.1

oL

()

W=Qc+kVe -Qe,,

=11.96x 10° m3 /yr(30 g/m?) + 0.1/ yr(55 x 10°)30 g/m? — 11.96x 10° m? /yr(15 g/m?) = 344.4108 g/yr

% reduction = 200010 =344.4%10° oo _ a 789 6 _

2000x 10°

@ _ 0,5-2000x 10 g/yr+11.96 x 10° m3 /yr(15 g/m?) _ 67.24 BB

@e

11.96 x 10° m? / yr +0.1/yr(55x 10°)

L

x 10° g/ yr +11.96 x 10° m?/yr(15 8/m°) _ 94 g MB (i) c= 2000 .

11.96 x 10° m? / yr +0.1/yr(2-55x 10°)

L

3 Gy c= 2000%108 g/ yr +1196 10! 6 m!/yt(15 g/m?)3 _ 74 47 BB 2-11.96 x 10° m3 / yr +0,1/yr(55x 10°) L

Therefore, judged solely on how much the concentration is reduced, option (i) is the most effective. (e) There are a variety of other factors that should be considered when making a decision in the "real world." These include: * economics (for example, how much does each option cost) ¢ — impacts on the biota (for example, dredging might impact bottom organisms which sometimes serve as important food sources for fish). ¢ impacts on existing uses (for example, dredging could have an immediate impact by decreasing water clarity during the dredging period. In addition, increasing the depth could have a longer term impact. For instance, a deeper system would undoubtedly affect the lake's temperature and possibly interfere with uses such as swimming and sport fishing. secondary benefits (for example, in addition to lowering the concentration, doubling the inflow

could affect other quality parameters such as water clarity).

Vv

OOS

©

.

(i)

ee

95-30

D

tgs

bss

TE

ee

55x 10°

(55x 10°)

= 3——_ ——@]| ——

1196x10® +0.1(2x 55x 10°)

=

1437

(iii) ty = 3 ———>*55x 10°—@______=5, 608 9 t9 = 3

= 9.45 yrs

Ti9exi08 +0.1(55x 10°)

i96x 108 +0.1(55x 10°)

ms yrs

3.2

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33

W _10500x 10° kg/ yr 10° mg, =e

© (n= 6-312

x10 mm/yr

ke

95

12

L

(0) The steady-state mass balance can be written as

0=W-Qc-vA,c which can be solved for

ya =e A,c 3.4

_ Hen =e) _ 212x10 9 095 21) 974 ™-00751™ Ac 10500 x 10°(21) yr d

V _ 1000000 m’ == = 1,333,333

@ O-=~O75 c

yO

W_ 1x10" mg === 15

133x10"

08

b)) B == 7g = 0.10667 e175 © i=

Qe _ 1x10 A,

7

6 1333 10°08) _ 5933 mg

1x10

m’yr

@ v, = 22 = 2933 | 111667 =0305™ ce 075 yr d 3.5 The values can be used to determine the consumption rate, and In(r) versus In(c) plotted and fit.

Q@sry) 01. 0.2 04 08 16

= =Q(mid) 0.0024 0.0048 0.0096 0.0192 0.0384

se (g/m’) 23 31 41 52 64

nd’) 184.8 331.2 566.4 921.6 1382.4

10000

1000

100

+

+—

+

t+——+-—+_+

10

+

++ 100

33

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The resulting, slope is 1.968 and the intercept is -0.9553.

is e°°° = 0.385 d’.

Therefore, the order is 1.968 (= 2) and the rate

3.6 The first-order model is

which can be manipulated (take natural log and algebraically rearrange) to yield t= me c)

@

The parameter ¢ can be expressed in terms of concentrations as

100- $= 100 G

which can be solved for

£ 2100-8 199 100

co

which can be substituted into Eq. i to give the final result (Eq. 3.40),

uf

100

__\l00-¢ ‘ A

}

3.7 The eigenvalue for this problem can be computed as

=3m(2x10° m?)=6x 10° m3 g=¥V 28x10 Ty

m

3

A week _ 49 957m

2weeks

3

7 days

Q v 42,857 m?/d A= S+k+ as Vv H 6x10° m?

orsdt

0.1m/d = 0.20476/d mn

The response times can be computed as 1.39 ths = —-—— = 6.79 d 75 0.20476

to = 90

2.3

"0.20476

=11234 3.4

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tos

95

3 0.20476

=

=14.65d

3.8 Recall from Lec. 2 that the solution for a second-order reaction in a batch reactor is

c=c,

1 1+ke,t

If the half-life is defined as the time where c/c, = 0.5,

1 1+ ke,ty

05=—_—_

which can be solved for

bso

~ ke,

This is a useful result, if you are strictly interested in determining a 50% reduction. However, it shows that a half-life is not a good general way to summarize the kinetics of a second-order reaction in the same neat fashion as for the first-order case. For the first-order reaction, this number tells you

how long it takes for the quantity to be halved at the set time interval of the half-life.

For the second-

order case, since the ts is concentration dependent, it would take a different time to go from c/c, = 1

to 0.5, than from 0.5 to 0.25. Consequently, it’s only good for the first halving. 3.9 The first-order reaction rates can be computed as: (@ k= 0.693

- 0.0231

30 yrs ®

yt

k= 0.693 - 0.086625 365 d = 3162 8d

() k=

d

0.693

yt

yt

= 0.05653

12.26 yrs

yr

3.10 The eigenvalue for this system can be written in terms of the residence time and half-life as

aa 2p at 28 tT

bso

which can be substituted into the response time formula (Eq. 3.42) to give tye —twhso

tsy + 0.6937,

100

100-¢

This formula can be used to develop the following plot,

3.5

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Response time (d)

1°00

Benner

100

9 Fo

10

tpg [ee

SS

ooo wei essecteeeaceeeeeseeeeeeeceeseessess Onset

14

— Cts

nee cen eee eee HH

1

10

Ht 100

Half-life (d)

3.6

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+ 1000

1 10000

year month week

LECTURE 4: 4.1 For simplicity, we will formally solve for the case where the forcing function is a decaying exponential. The growing exponential would employ a similar solution with appropriate sign changes. The differential equation to solve is

de + ac= We bt dt

Vv

subject to the initial condition that c = c, at ¢= 0. Applying Laplace transforms yields,

W,1V

sC-¢g Cot+ AC= othe

Simple algebraic manipulations can be employed to solve for

Coe

lo

WV

(s+ B, (s+ A)

As usual, the last term requires some algebraic manipulation to express it in a format that is amenable to an inverse transformation. One way to do this is with a partial fraction expansion,

WV

AS

(s+B, std)

st+B,

OB sta

The right-hand side can be consolidated to give

WIV

(s+ 8, (s+)

_ As+AA+Bs+BP, (s+By(s+A)

For the equality to hold, the numerators must be equal. following pair of equations

Therefore, equating like terms yields the

Ads BB, =e

(A+ B)s=0 which can be solved for

Azte_1

p=“

V a-B

_}

V A-B

Therefore, the total equation to be inverse transformed is

coe _ | 1 std

V(A-B,)|s+B,

std

Using the information from Table 5.1, this equation can be converted back to the time domain to give

41

Google

c= ce

Tame

|

Thus, the first term on the right of the equal sign is the general solution and the second is the particular solution. 42 The equation to be integrated is de

at

+c

= mo(t)

v

where c, = 0. Taking the Laplace transform gives

sC+Ac=™ Vv which can be solved for

c=™ 1 WVs+a The inverse transform yields m_

c=—e* V

43 The equation to be integrated is

©

dt

je =e sinax Vv

where c, = 0. Taking the Laplace transform gives

W, V +07

sC+AC=—

which can be solved for

Cc

aA

o

a

@

V st+@7* sta

The right side can be evaluated by a partial-fraction expansion

(W,/V)o___ A+B . aOD (+a std) #40" std

; “

The terms on the right side can be combined and then the similar terms in the numerators equated

42

Google

A+D=0

AA+B=0

we= BA+Da*

These three equations can be solved for 4, B and D and the results substituted back into Eqs. ii andi to give,

Wo

“VR +07)

(- stots] st+@*

st+m?

sta

The last term within the brackets can be dropped because it represents a transient part of the solution. The inverse transform yields,

c=

es (Asinax — @cosat)

(iii)

Now invoking the trigonometric identity,

sin(at — ¢) = cos¢sin at — singcosat algebraic manipulation can be employed to show that Eq. iii reduces to c=

WIV

Tera?)

vom

where

44

[NOTE TO INSTRUCTOR:

The following parameters should be used: spill mass = 5 metric

tonnes, v, = 0.1 m/d and outflow = 1x10° m’°/d.]

(a) After computing the lake’s volume [V’= 5(1x10°) = 500000 m°], the following equation can be developed to solve for the concentration as a function of time following the spill, _ 5x 108

Sy” ~ 5x10

1x10? 0.1

ie

a)

= 10¢e°2"

which can be used to compute values. The first few points are tabulated below, and the complete time series is displayed in the figure.

time (d)

0 1 2 3 4 5

c (mg/L)

10.00 8.03 6.44 5.17 4.15 3.33

43

Google

oN 2 OO

3 b) t, =—— = 13.636d

©)

bss

0.22

(©) t(e=0.1) =

In(10/0.1)

= 20.933 d

4.5 The lake’s eigenvalue can be computed as

Ov _ 5x10"

Az= VoH

=

4x10’

7

8 _ 10125 8

yt

The solution for the step loading is

c= HY i-e*)= a

hem)

= 12346{1—e Ho2H-19%9)

This equation can be used to compute the values for the lake over time. base concentration of 5 yg/L are shown in the following plot

e (ng/L)

These values, along with the

Or 16 $

10 $ 64 O $+ +t ttt ttt 1990

1995

2000

4.6 The lake’s eigenvalue can be computed as 1 = 1.0125/yr. =

200(05 x 10°)

4x 107 (101254 0.2)

(er20-19

—etoaxe-197))

2005

The solution for the exponential loading is

= 2,062(e9«-1997

This equation can be used to compute the values for the lake over time. base concentration of 5 g/L are shown in the following plot,

44

Google

2010

—

1012-197)

These values, along with the

4.7 The lake’s eigenvalue can be computed as A = 1.0125/yr. The solution for background concentration, along with the step and exponential loadings are c=5+ 12,346(1-e°19!251-19)) 4 2.062{e°2('-1997

=e 10124t-1997))

This equation can be used to compute the values in the following plot

4.8 (a) The eigenvalue can be computed as

pe Qpth ape 20X10gg 6 th0S=125/yr Eq. 4.32 can be employed to determine,

0= tan -t= 6280.75) — 1571 = 3.14159 These values, along with w =2/lyr=6.28/yr compute the concentration in the system as

ox

30x10 6 -(-e™) 1.25(20x 107)

Google

4.5

and ¢=tan'(6.28/125)=1374 can be used to

6

4 —_15x10" sing .283¢ — 31142 -1374]20x 107 v1.25? + 6283?

20x 107

15x10

6

¥L.25? + 6283?

sin{-3.142- 1374]e2*

The results are displayed below: 03

0.2

c(mg/L) O41

0

oO

2

4

6

8

10

time(yrs) n = Zeer 13T4 3654

ten mx

es epee 20)

6.283/ yr

—

4.9 The solution must be modified slightly to reflect that the lambda’s on either side of the equation differ. The mass balance is

de

a t+Ac=

10A.

.

20A,..

x 1 +5A, sin(2at) —74a sin 4

x

+2)

where A, = Q/V and A, = Q/V + k. The response can be determined as

5A, 20A 4 ¢= oh 4 —_741__sin(2at - 0,) - =

An

J +(2n)

4

:

sn 4

3afa2 + (42)?

a

+ 0, }

and the 6's are based on 4,. The result is shown below.

The major impact of including the reaction is on the mean level (drops from 3.2 to 2.9), whereas the peak and the amplitude swing are affected negligibly.

4.6

Google

4.10 Taking the derivative of Eq. 4.30 yields

< =¢,A(@)w coat - 4a) The cosine is zero at x/2. Therefore,

which can be solved for a s+ ate

t=+— @

4.11

This problem can be solved in a number of ways. First, it can be recognized that the function is linear with respect to the parameters since it can be reformulated using trigonometric identities as

I(t) = T+ Tagg SiN 04 0089 — Sgqy COSOOESIN O Therefore, general linear least-squares can be used to estimate the parameters (Chapra and Canale 1988). The approach is particularly efficient because the data is equispaced. In addition, nonlinear regression can also be employed. This can be done very conveniently with a spreadsheet solver. The tesult in either case is

J =296.5

Jeanp = 251.9 O=78.7

The data along with the fit is shown below:

0 0

61

4.12 The lake’s eigenvalue is 2 = 0.75/yr. from t= 0 to 0.5 yrs,

ox

122

183

243

304

365

The step solution can be employed to calculate concentration

10x10 6, (I-e*”) 0.75(1x 10°)

At t = 0.5, this equation can be employed to determine that c(0.5) = 41.695 pg/L.

This value can

then be used as the initial condition for the general solution over the next interval (¢ = 0.5 to 1 yr),

Google

47

© =41695e C7605) The application of the step and general solutions can be repeated out to ‘= 5 yrs. The results are displayed in the following plot:

200 150 c¢ (égiL) 100 50 of

0

+

+

+—

+

1

2

3

4

5

time (yrs) 4.13 The lake’s eigenvalue is 1 = 0.75/yr. The loading can be approximated by the Fourier series, weoy= 10+

40; ..

1.

1.

| sincom +4 sing) + 4 into) +]

. For three terms, the result looks like,

time (yrs) Each term in the series can be employed as a forcing function. The resulting solution is

o(t) =© 1x10 (re) 0.75(1x 10° +} 4

(400x10°)/7

1x 10°V0.75? +628?

(400x10°)/z sin( att — 1.45) - a

(400x10°)/Ga) 6,

1x 10°07? + 6.28?

_ssnréat — 153)

Ix 10° 0.75? +(3-6.28)?

4

(400x 10°) /(6n) 6,

6. (400X109)

siny 150°

1x 10° 0.73? +(3-6.28)?

sin(1 Ont — 155)

Ix 10°J0.75? + (5-628)

Google

si(—-1 450075

(40010 16, )/(Sa) Ix 10° 0.75? +(5-6.28)?

4.8

sin(—155)e-°™

which can be plotted versus time as 200 150

c (ug/L)

100 50 04

+ 1

0

+ 2

+ 3

+ 4

4 5

time (yrs) 4.14 A semi-log fit of Inc, versus ¢ can be used to evaluate the rate of the first reaction 3

oO

a

0

60

80

100

+@ The best fit line has a slope of -0.023 which is an estimate of the first substances decay rate. Therefore, the half-life can be computed as ten,

son

0.693

==—— = 30!

"0.023

The decay rate along with the ¢, = 35 can be substituted into Eq. 4.21 to give

5 = 1X0.023/ k,) 0.023—k, which can be solved numerically for k. = 0.035, which can be used to determine

0.693 ‘30a = 0.035

=198 yrs

49

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LECTURE 5:

5.1

(a) The solutions for the two lakes are

==

5x10 5 50.1 x 10°)

a=

1667(10)

?

'2.73-2833

Ixt0*

3.65

oS (e2#

= 10e*™* 2?)

© el)

O-NHWANBVN®OS

which can be used to compute the trends in the following plot:

oO

os

1

15

2

time (yrs) (b) Equation 4.21 can be used to compute the peak time as

1, = WM2.73/2833) _ 9359 yp 2.73— 2.833 which agrees with the plot. 5.2 (a) Steady-state mass balances for phosphorus can be written for each lake and solved for concentration

a=

4x10

1 67x 10° +16(82100x 10°)

.

6.95 x 10"

__

4x10

1.3806x102

~~

ow L

—695%10"_ 14 HB

a 36x 10° +16(57750x 10°) _ 096x107 —

L

c _ 458x 10 +67 x 10°(2.9) +36 x 10°(7.24) = 45/8 > 161x 10° +16(59750x 10°) L

~ 182x107 +161x10°(45) _ 45 448 4" 182 x 10° +16(25212 x 10°) L

Google

5.1

_ 6.65x 10%12 +182 x 10° 9 (32.24) =242948

os=

6 10°) 212 x 10° 9 + 16(18960 x

L

(5) Individual solution can be used to assess how much of each lake’s concentration is attributable to

its own loading as well as the loadings to upstream lakes. The result is loadings 8

m

s m h | 386% e 015% ° 0.07%

bh

10.00% 518% 020% 0.09%

90.96% 349% 164%

e

°

96.16% 45.08%

5312%

Consequently, for total phosphorus, only Lake Ontario is affected strongly by an upstream lake (45% of its concentration is attributable to Erie).

5.3 Using Eqs. 5.18 through 5.20, with lake 1 being Huron, the following time series can be computed for the three lakes,

27

¢(uCim)

5.4 The following plot is for the 64-reactor case. 14

°

a

a

o

2

10

of

The slope of this fine can be evaluated as -0.01969.

Interestingly, this is approximately equal to the

ratio k/U = 0.02. In fact, it will be shown in Lec.

9 (Eq. 9.9) that the solution as 1 —> © of the

cascade model is

c=c,e

i): 5.2

Google

5.5 (a) First, the flow can be converted into proper units, 3

Q= 29% 864008 _1m

sd

(6)

Pe

3

= 48,924

(3281)

3

The mass balance for two reactors in series can be written as

2

-£.[_2

on -(323)

which can be solved for

A - 20-¥A) = 489241 V0.4) = 1.42x10° m? Viv ¥0.4(0.2) Thus, the total area for the two tanks is 2.84x10° m?. © The two tanks seem to represent the best option because they require less area to effect the same temoval.

5.6 Pseudocode to implement Di Toro’s algorithm in the format of a function is FUNCTION

prodl

=

1

cn

(t,

n,

lambda(),

DO FOR j =1T0n-1 prod1

END

= prod1

*

lambda(j

c10) +

1,

j)

DO

sum = 0 DO FOR i=1TOn prod2 = 1

DO FOR j =1T0On IF j * i THEN prod2

END

END

sum

=

prod2

*

(lambda(j,

j)

-

lambda(i,

IF

DO

= sum

END DO cn = prodl * END FUNCTION

+ cl(t,

lambda(i,

i),

c10)

sum

5.3

Google

/ prod2

i))

FUNCTION

cl

END

= cl0

cl

(t,

lambda,

* EXP(-lambda

FUNCTION

*

c10)

t)

5.7 (a) The solutions for the three lakes in series for an impulse flux of Strontium-90 to Lake Huron are

Cy = 11862097" c=

€

24.5(L186)

2589—-0,0701

~e 25890)

(eoor™

0.355(24.5)1186|

e-0.07l" _ ¢-0aoae _ 72589 _ 90404

(2589—0.0701)|

(0.404—00701)

(0404-2589)

These equations can be used to compute the following time series, 12

Huron

1

5

os

+

08

St Clair

= aad

\sne

n2

°

°

10

2

time (yrs)

30

“

(6) For our problem, the mass balances are de,

de,

a = Amen

a.

Ann Ache

de,

7a = Mele

Aaele

The following dimensionless variables can be defined

t= Angt 7 = A HE Ane he rm

of = 16093 195697 55.74 3511

3

Employing, distances in meters, the following, equations define the concentration profiles in each Teach:

e(x) =Se

a 5267 Bf.

Cy (x) = 0, (8046)e

1 5267

0 280468) 5267

1,000,000] -035

ose) =| Haase

The results are displayed below using distance in kilometers for the abscissa:

2s 20

15

© (mg/L)

10

x (km)

The concentration falls below 5 mg/L at about 46.7km. 9.8

First solve for the velocity and convert the dispersion coefficient into units that are commensurate with the rest of the problem

u=-2.A,

200

=250%

d

Google

2

2

£=1° = _™m’ _ s

10¢cm?

86,400 s = 864,000

d

2

9.6 (a) =) =17" = 10545 kay5 = 0.2(1L0545)77>"*° = 0.298 (6) Solve for the velocity

= 2 _5x10*_ 1252 A e

400

d

Eq. 9.57 can be solved for

W =5x10* 3 (20

x10

0.298)1,000,000 0298).000,0

mg/m? fi

mel)

123?

- £2, 9.7 cy =10e 3600 (9000) “Bhi.

02

=n

kg

10° mg

=8.79-—2kg

d

= 5738453281 = soos

To find the level atB, we must determine how much the concentration has depleted by this point - =02 (25000-10000

(25000) = 59.0186" 34007%0-100%) = 25,64978

L

Then we have to perform a mass balance to determine the concentration,

ep

25.649(43200)+ 10 x 10° = 12948

86400

L

Finally, the concentration at the water intake can be calculated as ~ 22, cc = 12.94e aa00

(35000-25000)

) 9538T

The entire solution is shown below. 80

T

A

8B

© (mg/Lyo

c

0 oO

10

2

x (km)

9.3

Google

wv

«0

9.8

We will assume that all the additional flow derived from the source is reflected in an increase velocity

downstream.

Although this would not be strictly true (some would be reflected by an increase in

depth), it is a valid first approximation.

As in the development of Eqs. 9.58 and 9.59, the solution

will remain 6 = ce**

forx 0

However, the eigenvalues are modified to reflect that the velocities and flows are different above and

below the discharge

yet2E

agar fi

u2

2E

Uj

The mass balance at the outfall now becomes W+U,Ac, — EAA,c, —U,Ac, + EAA,¢, =0 which can be solved for Ww

Cy=

Q,-2, Qh.

fot]

2.

|

99 If Eq. 9.61 is used to determine the eigenvalues, the solution for the downstream segment becomes

1.

uv Wyn

ake

20

ope Thus, it can be seen, that as E approaches zero, this solution becomes

w

-4*s‘

c=—eU

x20

Q

which is the plug-flow result. Note that this approach does not help for x < 0 because a singularity occurs in that direction for the plug-flow case.

Google

9.4

LECTURE 10:

coven coaronaienonaaines soonEARGONNER NSN

10.1 (a) After converting the english units into commensurate metric units (B = 1000 ft = 304.79 m; H= 10 ft = 3.0479 m, Q = 500 cfs = 1,223,109 m’/d), the velocity and the loss rate can be computed as

1223109 m3 /d

m

= __ £13167 — 304.79 (3.0479 m) d

v

k(total) = k +— = 0.05 +0.3/3.0479 = 0.148 d? {coral H

which can then be used to calculate the estuary number

n= ‘KE_ 0.148(1x10°) = 0.0856 uv?

1316.7?

Therefore, the system can be classified as being mildly advective. (b) Eq. 10.24 can then be employed to compute the profile at the downstream station (x = 8 mi =

12874 m), (x, =

(12874-(1316.7)1)? aaa,

10x 10° /928.9

0.1487

ayn(ix 10°)t

The results are shown below.

ce

(ug/L)

03

10.2 The data can first be plotted:

120 +

100+

20

|

¢ (mg/L) 60

10.1

Next, the following, summations can be developed: Station 1

ml

XG

@=5

2,317,600

11,325,000

359,376,000

8,284,250,000

15520

+ Cr Mtin —4)

ind

ot

Yet + Crain Xl — 4) ind

nl

Yer

Station 2

(x= 1km)

+ Carlin Xa —4)

km)

15600

ind

These values can be used to compute:

~ 15520 1s a =~ = 3882 2(260- 60) L

-

3

M, =37x10° (388 8

}60- 60)min 20400

m

15600 ee 2(950- 550)

» =

d=

s

(19528 \950-ss0ymin9200* m

KS _ _ 1994 1g

10° mg

3

My =3.7x 10?

1s, 1S L

d=

ks

= 2.004 kg

10° mg

Thus, the mass of the conservative tracer is conserved.

j, = 2317600 _ 1493 min

jy = 11325000 _ 7960 min

3, = 359376000_ 1 49 3? = 8563 min? , 15520

a, = 8284250000 _ 99602 = 40215 min? 15520

15520

_

(5-1) km

(7260-1493) min 5

3

= 3x10 m1 9989 m/d

15600

10° m 1440 min _ gogq™

d

d

57 nt

H=2™

2(726.0— 149.3) min 10mg

2

60m

5 = 69989 m/d)?(40215-8563)? min? 10g

d

,e

10.3 (0) = —&_—_ “8-4 4) = 7500 m? g L

c, =4e%

d

_96173m

dd

1440 min 86400 sec

10000cm? _ 201s

m?

2000

A pood = =~ = 08/4 Pood “2500

1000

F etanool = 3599 ~ 054

10.2

Google

s

= ge 0-81-05) Cy =4e A mass balance at the end of the channel yields the following concentration at C. 2000

“e = 30000

4e8(t-03) _ 9. 42-08(1-05)

The results are shown below.

c 4 (mg/L) | * 2 0 10 ppb (b) m m=10 Bx 0.4 ppb = 250 g (0)

10g 10° mg yg (0)=—_* 8 4

10.4

= F500 mm

2000 A pond = —— = 08/d

L

pond

= 4e08

“pena Coad

g

of

steaee!

“2500

1000

=——=05d

“3.000

= 4e OX t-05)

Mass balance for lake (assuming f = 0 at 0.5 d): de

dt

40 = ORO

c=

The results are shown below.

10.3

8

(08-2)

(em

-e*#)

The time of the maximum concentration in the lake can be calculated as:

1n(0.8/2)

t, = 05+ 10.5

08-2

= 126 days

(a) Recall from Fig. 10.5 that the distance encompassing 99% of the area under the bell-shaped

curve due to pure diffusion can be defined as o = 52/2Dt. Therefore, the time required for one side of the distribution, defined by this bound, to move a distance x is

t

_ 2x/52) _ [2010)/52) 2D

210°)

= 7,396,450 8 (= 85.6d)

The distribution in the sediments at this time would look like (for the plot we assume an initial

condition of mp = 1 g/cm’).

0

0.05

0.1

which can be evaluated at the sediment-water interface as (expressing all times and rates on a day basis)

Jom, 10-2

4tynDt

_ (oes?

am

_

m, 10-0 =

4(856),/77(0.0864)85.6

_

(0-0)?

Xo0ssayass _ 0,000206m,

If the initial m, is assumed to be unity (i.e., 1 g/m”), then the flux would be 0.000206 g/cm7/d.

Google

10.4

LECTURE 11: 11.1 Inlet condition:

0=W,

+0,

a

+ Qore — Dre

2

+ By (q 4)

Vig,

W,+ Qyaco = Qyr /2+ By + hiK)e, + Qua /2- Ein er Outlet condition:

0= H+ Qnan

Chaatcy

—Qumeren t Entiat ~ en)

2

kVaSn

Wy = (Quin 12+ Ena naa + (Onset — Qn-in !2+ Epica + kV en 11.2 The matrix inverse for the case where n = 10 is 0.00068 0.000588 0.000508 0.00044 0.000381 0.000331 0.000289

0.000256

0.000231 0.000216

0.000392 0.000627 0.000542

0.000469

0.000226 0.000361 0.000601

0.00013 0.000208 0.000347

0.000391 0.000342 0.000302 0.000273 0.000256

0.000443 0.000387 0.000342 0.000309 0.000289

0.00052 0.00045

0.000406 0.000353 0.000309 0.000273

0.000246

0.000231

0.000588 0,00051

7.53E-05 0.00012 0.0002

4.36E-05 6.98E-05 0.000116

0.000507 0.000443 0.000391 0.000353 0.000331

0.000583 0.00051 0.00045 0.000406 0.000381

0.00034 0.000583

0.000197 0,000338

2.54E-05 4.06E-0S 6.76E-0S

1.5E-05 2.39E-05 3.98E-05 6.76E-0S 0.000116 0.0002 0.000347 0.000601 0.000542 0.000508

0,000115. 0.000197 0.00034 0.000588 0.00052 0.000469 0.00044

9E-06 1.44E-05 2.39E-05 4.06E-05 6.98E-0S 0.00012 0.000208 0.000361 0.000627 0.000588

5.62E-06 9E-06 1.5E-05 2.54E-05 4.36E-05 7.S3E-05 0.00013 0.000226 0.000392 0.00068

The results of the calculation are shown in the figure and table below. Halving the spatial interval results in approximately a halving of the percent relative error of the result. c

1

(mg/L) 2

0.5

n=10

n=5 0+

+

0

+

20

x 0

analytical 0.766

10 15

0.657 0.609

25 30 35

0.522 0.484 0.449

45 50

0.385 0.358

5

20

40

0.709

+

+

40

+

+

+

60

80

n=5 numerical error

0.609

7.28%

0.462

4.54%

0.564

0.416

Google

0.353

1.38%

+

4

100

n=10 numerical error

0.680

4.16%

0.588

3.46%

0.508

2.74%

0.440

1.98%

0.381

1.16%

55 60 65 70 75 80 85 90 95 100 11.3

0.332 0.308 0.287 0.268 0.250 0.235 0.223 0.213 0.207 0.204

0.275

2.68%

0.229

7.40%

0.331

0.24%

0.289

0.83%

0.256

2.09%

0.231

3.46%

0.216

4.57%

The matrix inverse for the problem is included in the solution for Prob. 11.2. Note that the key element needed to answer part (a) isaid = 0.000216. This result can be used to calculate, Cc. 01 ¢ =—— = —__= a;¢,(Q) _ 0.000216(1000)

0463

Thus, the result is higher than was originally computed

in Example

11.2a_

The

reason for the

discrepancy between the two results is illustrated by a graphical comparison among the two numerical

and the analytical

solutions below.

Note

that at the end of the tank,

the numerical

solutions

overestimate the analytical result, and the coarser scheme is higher than the finer one. Further, also notice that the last value for the coarse case corresponds to a broad region from x = 80 to 100 centered on x = 90. In contrast, the fine case corresponds to a region from 90 to 100 centered on 95. It is both these biases that lead to a higher reduction for the finer case.

n=10

11.4 Following the same general approach outlined in Box 11.2, the first derivative can be approximated with a centered difference, de _ Sin = S11 dx dx This approximation, along with the centered difference second derivative approximation, can be substituted into the steady-state advection-dispersion-reaction equation to give

O= FSH ae +e

Ax?

y

et

2dx

—ke,

Collecting terms and multiplying by V; = A Axgives

11.2

Google

~(Q+ EB’), + (2E'+kV, ey —(-Q+ E" E44, = 0 11.5 Backward differences for each of the regions can be written as

km 008.

Wy + Qin 64-1 — Disses — WV 0, + SV, =

km 8 to 16: Wy + Qh-s04-1 — Diss — MYye, = 0 16 km: W, + Q-rs6i-1 - Disrer — Vie, = 0 pl6 km:

Qh 40}-1- Dsarer -— Vie, = 9

(a) and (c) The numerical and analytical solutions are compared below.

20

10

O+

+ 20

0

+ 40

+ 60

(5) The numerical dispersion for the calculation can be estimated as

km Oto 16: E, -fu

=

on

2i” 1,080,000 m?

> 16km: E, = Fu = EB so B= 810,005 2

d

m?

a

d

These results can be put into perspective by using then to compute an estuary number for the

segments

kam Oto 16, p= AZ. = 101,080,000) = 0.00579 43207

> 16km: = ae. 21610.000) _ 599772 U 3240

These results are quite small and indicate that the model is still advection dominated. 11.6 The following general centered difference equation can be written for the interior segments, as well as the first segment. =

0O=M +O,

Gate S

a

—Qiss

G+, i 2

+ Bus

-4)+ Evga

— 0) - kc,

For the last segments, the zero derivative condition can be included by approximating the derivative between segments 18 and 19 as

Google

113

ds19 _ 9 ~ Sis de

Aisa

which can be solved for

or, because the derivative is zero, C19 = Cis. Therefore, the balance for the last segment is 7+,

,

2 2 18 — Dig rors + Ein aa(Cr7 — er) — kasMastis

O=Ms+ Qi

The solution to the resulting set of equations along with the analytical solution is shown below,

0.02

-20

-10

0

10

20

30

11.7 [NOTE TO INSTRUCTOR: The point source has units of g/d.] (a) The concentration at the mixing point can be computed as

co —

4kE

Oy or

12x10"

719938

fe MOTD

“ae

2507

(6) The estuary number is

_ KE _ 0075(1x 10°) u?

2507

=12

(c) The segment size required to attain a desired degree of accuracy can be determined as

fray) (1-005)

“5

0.075

Pp = 750.05 m

11.4

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(d) A backwards difference scheme can be employed to compute the profile using a Ax = 750 m. The concentration at the injection point is 19.12, which corresponds to a percent error of 4.1%.

a

oe

10,

-20000 11.8 (a) The system can be developed in matrix form as in (note that we have employed backward differences for the advective terms),

182x110"

-65x10°

-177x10" 0

199x100! 65x10?

0

co)

65x10!" 221x10"

{22x10

He, b=) 2x10? Ile, 3x10"

The matrix can be inverted to give

568x107? 517x107? 152x107

19x10") — 142 x10° 534x10" 399x107? 157x107?) 465x107"!

and multiplied by the right-hand-side forcing function to yield concentrations of c, = 12.91, c2 = 23.24 and c3 = 20.43.

(6) The element of the inverse needed to answer this part is a3; = 157 x 10"? , which can be used to

determine

Ac, = 157 x 10"?(0.5x 2x10!)= 1s748 11.9

Using the same approach as for Example

11.1, a backwards difference can be used to characterize

the tank with n= 5. The following equations result,

48 4 0 0 0

-2 0 68 -2 -4 68 0 -4 0

0 07%] (200 0 Olle” 0 -2 0 Ke,}=10 68 -21), 0

0 -4

438}it

‘5

Google

0

11.5

with the resulting solution: 60.92, 46.21, 35.26, 27.48, 22.90.

centered difference approach can be used to develop

38 3 0 0 0

-1 48 -3 0 0

0 -1 48 -3 O

0 0 -1 48

Alternatively, as in Example

11.2, a

OF" ] (200 Off” 0 0 fe,}$=4 0 -1l|, 0

-3

38 es

0

with the resulting solution: 65.34, 48.28, 35.75, 26.74, 21.11.

Both results are shown in the plot

below. Notice how the backwards difference exhibits numerical dispersion. be obtained by using a greater number of segments.

More precise results can

100 75

Backwards

50

Centered

25 Oo +—+—_+—__+—_+—_+_+_ +_ +_ +H

0

5

11.6

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10

LECTURE 12:

AGS NERRENORREERRHOEIE

25? 12.1 (a) At_ = 0.002083 hr (= 0.125 mi (@) At

2

@)

Se)

FR peat)

min

peck

‘mas

(6) The equations for f(x) up to the peak can be expressed as Say = LO)

— ¥min Xpeak

S Xpek Xmin SX

ea)

Eq. i gives Substituti ng f= '

20 © (Xa — Fania X% peak ~ Fain)

gy

in)

Xmin

=

=

S¥SX,

Gi)

pak

approach yields a similar From the peak to zer o fi)

(iii)

S Xmnax Xpeak S¥

-)

=, (max ~ ¥min X¥max ~ *peak)

(c) The formulas for F(x) can be developed by integrating Eqs. ii and iii (a Ja Eq. 18.10), F(x)=

2 ———_—_——_} ——-y x . qos =

B=

Rep)

1)

xi 2

y4—mn

2

x?

Et

max ~ min

\*max

Xnin S¥SY,

™

x

Posk

peak

Xpeak S¥S Xanax

Both f(x) and F(x) are displayed below for the values of the diffusion coefficient used in Problem 18.2.

Google

18.5

fe) 0.0002 + 0.0001+

0

0 x

a

.

Digitized by Google

6000

12000

x

18.6

igi Original from

UNIVERSITY OF MICHIGAN

LECTURE 19: |

sano

19.1 Separation of variable can be applied, fz

= ~reghof. edt

which can be integrated to give

0-0, = YogBole™™ - 1] or 0 = 0, -Mog8(1-e*")

(@) meson

7

gees)

6x32 gO

LJ

6

(6x12+12x1+6x16) giglucosey)

(@®) Inoy =-13934411+ 1575701x 10° 5 _ 6.642308 x10 283.15

28315

7

+ 1243800 x10

10

28315

_8621949x10!" 283:15¢

11

_ 5 44

Og =e? = 11288

0, = 11.288{1— 0.000035 x 11000]= 6.94 mg/L 19.3 (a) The diffusion coefficient can be estimated with a chloride balance 0= O5ig — Os, + Ein (5, -5) Ey = 480020210 = 38,400 m’ da? 10-5

(6) The BOD in the lake and the bay can be computed, without the subdivision, as

a=

QrinCrinlQe + Ein + (K+,

1=

[Q, +Ey =

7

/ Ha Wa) - Qo ia r,in(—Er2 - 22) 7

7

:

+k +, / hy W,MQ, + Ey2 +k +¥, / Hy W2)- Ey2 (Ey. - 2) 0— 4800(57)(38400 + 4800)

[60477.5 + 38400 + (0.1+ 0.1/8)128 x 107 [4800+ 38400 + (0.1+ 0.1/3)1.2 x 10°] 38400(-38400 — 4800)

= 0.038 mg L"

=

— 21in°2 inl D1 tEn tk +¥,/HWNih = Drintria a

:

(Q, + Ey, +k +¥,/ Hy Wis + Eyo +k +¥, / HW )- Ey(Era +2) _

4800(57)[54800+ 38400+ (0.1 + 0.1/8)1.28 x 107]-0

[604775+ 38400+ (0.1+ 0.1/8)128 x 107 ][4800+ 38400+ (0.1 + 0.1/3)L2 x 10°]- 38400(38400-+ 4800) 19.1

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= 1.35 mgL7 The loading and flow from the subdivision can be computed as

Quay = 0568(10000) = 5678 md Quay = 113.4(10000) = 1,134,000 gd The BOD in the lake and the bay can be computed, with the subdivision, as

_

O

0-[4800(57)+ 1134 x 10°](-38400- 4800) [60477.5+ 38400+ (0.1 + 0.1/8)128 x 10” }[4800+ 38400+ (0.1+.0.1/ 3)L2 x 10°]— 38400(-38400— 4800) = 0.22mgL!

?

=

[4800(57)+ 1134 x 10° ][54800+ 38400+ (0.1 + 0.1/8)128 x 107]-0 [604775+ 38400+ (0.1 + 0.1/8)1.28 x 10” ][4800+ 38400 + (0.1+ 0.1/3)12 x 10°] - 38400(-38400 — 4800)

= 6.78mgL" 19.4

The basic approach would be to regress the In of the particulate and dissolved components versus

distance.

This yields

InL, = 3.056- Ox

(adjusted r? = 0.997)

InLy = 2.768-0.0137x

(adjusted r? = 0937)

Because the fit of the particulate fraction looks O.K., the slope can be converted to a settling velocity,

v, = 0.1 km"(6600 m d''\(1 km/1000 m)/2 m = 0.33 m d. [A caveat to this manipulation is that the

disappearance of particulate BOD may be due to mechanisms other than settling. For example, some of it can be converted to dissolved form.)

100

10

0.1

19.2

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Although the 7’s appear good for the dissolved BOD, inspection of the fit indicates that the

dissolved component seems to systematically deviate from the straight line. [Aside: This is a good example of how a “good” or “high” 77 does not always tell the whole story. It also illustrates why plots must always be developed and scrutinized]. The fits suggest that part of the dissolved component seems to decompose readily, whereas some of it seems to be refractory. This hypothesis

can be expressed as an alternative model,

Ly =(1- F, Lye Ce

+ FL gee

where F, = the fraction of the dissolved component that is refractory, and kz, and ka are rates for the refractory and labile fractions, respectively. The model parameters can be estimated in a number of ways. For example, nonlinear regression can be used to estimate,

Ly = (1-0.49)L4,e 0 + 0.491 60°81 where the fit is shown in the following plot,

20

0

80

Thus, from the fit, it appears that the dissolved BOD is about 50% velocity, the exponents can be expressed as rates,

kegp

1km

= 0.00784/

km(6600 m d*!) ——— = 0,052 d?

* kent 7000m ka = 0.0618/kn(6600 m ay = 0.408 d"? m

19.5

(a) As was done for oxygen in Example 19.2,

on P= 9781 _ e

0.684

gig Ble m?>

c= 11418 BOle 28nitrogen 8_ 5197 ME m

mole

L

(b) As was done for oxygen in Example 19.2,

19.3

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+

T

total dissolved

4 1

100

refractory. Using the stream

Pp 0.000314 _ c=fs Hq, e 0.0272

mole

c= conseHeme Ae ee mole

050798

19.6 (a) For convenience in later calculations, the flows can be converted to daily units (Q,, = 172800, Q, = 432000, Q = 604800 m’ d"), Then the mass balance at the input along with the conversion factor of CBOD;/CBOD, = 0.31, can be used to compute Ly

=

172800(10) + 43200000) 1 _ 944 mp1 604800 031

(6) The distance downstream to attain 95% reduction can be computed as

kos = 0.2(L047)"*-* = 0.2888 d"!

wy =

= —_3_= 251300 m

k/U = 0.2888/24192

19.7 Using Egs. 19.32 and 19.33, the results are

distance

temperature, °C

salinity, ppt o,mgL" og, mg L" Ox, mg L? % saturation

19.8

30

20

5 5 8.263 8.031 62%

10 65 8.744 8.250 19%

25

22

10

18

20 15 9.467 8.398 89%

Using Eq. 19.32, the saturation concentration at sea level can be computed as 8.263 mg/L.

This

value can be corrected for salinity (Eq. 19.33) to give 7.81 mg/L. Finally, Eq. 19.39 can be used to adjust for the altitude effect to give 6.91 mg/L.

19.4

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LECTURE 20:

20.1 (a) The amount of CBOD spilled can be calculated as

} Mepop = 300001{ 1008-2 L

6 mole-32 gO, / mole

1 mole-(6-12 + 12-1+ 6-16) g- glucose / mole

=32x10° g-CBOD

(6) The mass balances for CBOD and oxygen can be written as

v=aLap =-QL-kwWL Eke

v2 = 00, -Qo-k4VL+k,V(0, 0) If L=L, and c =c, at t = 0, these balances can be solved for

1 L= Lelie"

ha} fe} 1

kha —ka

1

The remaining model parameters can be evaluated as: 0, = 6.942, k, = 0.08925, kg = 0.126.

can be substituted to give the following profiles:

t.=7.1d (c) The time of minimum DO can

These

time (days)

be calculated as

= 1n(0.15968 — 0.122583) _ 713 days 0.15968— 0.122583

20.2 First the key model parameters can be calculated as: V = 2.5x10°, Q = 3.57x10°, ka = 0.11, and ky= 0.158.

Next, the mass balances can be formulated in terms of population,

20.1

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0= Popx P, -(Q+ Popx Py) L—k,VL 0=Qo, -(Q+ Popx Py)o—kgVL + k,V (0, — 0) where Pop = population, P; = per capita BOD generation rate (= 125 g/cap/d), and Pg = per capita flow generation rate (= 0.57 m’/cap/d). The first equation can be solved for L=

Popx P, (Q+ Popx Po) tkv

This result can be substituted into the second equation which can be solved (with c = 6 mg/L) for Pop = 14,417 people. This can in turn be used to compute

W =Popx P, = 14417 «125 = 181082 Note that the corresponding waste flow is mn?

Q,, = Popx P, = 14417 x 057 = 8217——

Finally, the inflow concentration can be calculated as

= 1x10 __ yg, me 6.

357 x 10° +8217

L

20.3 Using the same approach as in Example 20.1,

K,

t

_ 2.09 x 107 om? /s_

O.lcm

1m_

100cm

86,400s

sd

= 01806 d

Thus, the use of a much thinner boundary layer (1 mm as opposed to the original 2.6 cm), raises the mass-transfer coefficient by a factor of 260. k

_ 9.1806 m/ d{7(0.0305)? ] m? . 10° mL *

300 mL

mm

Next, we can calculate the reaeration coefficient

= 01806 d"

Now the exchange is increased further because of the increased surface area for the open beaker. all leads toa much shorter response time,

This

ty = 3 = 171d 5176

which is more like the type of exchange actually observed in nature.

Although opening up the

constriction of the bottle neck made some difference, the major factor here was the much

thickness of the boundary layer.

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20.2

smaller

20.4 The saturation be calculated with Eqs. 19.32 and 19.39 to be 6.27 mg/L which corresponds to a deficit above the dam ofD = 6.27 — 6 = 4.27. Using coefficients from Table 20.2, Eq. 20.44 can be employed to calculate

r =1+40,38(0.65)0.7(2(1— 0.11

21+ 0.046 x 26) = 159

which can be used to determine the deficit below the dam as

427 = 2.68 mg L 4 Doetow =e! 159 mB which can be used to compute Opsew = 6.27 — 2.68= 3.6 mg/L.

20.5 Equation 20.56 along with the diffusivity of oxygen (2.09x10°) can be substituted into Eq. 20.53 to yield

12.919]°*

Ki Kol 5a

which can be manipulated to give

0.

36.74°7!

“Kio, Ser

which can be simplified to the same general form as Eq. 20.58

36.74

7°95

K= Kyo, [7

The two formulas are compared below,

KUK,02

2

Eq. 20.58

15

Eqs. 20.53 and 20.56

1 05 0

0

400

200

20.3

300

20.6 NOTE TO INSTRUCTOR:

The time units for Station 2 should be hours.

(a) The data can be integrated to determine the mass of ethylene passing each point,

M, =Of fyede =222-™ 222" m?(100-0 (100-0 min) min)._ \0+2(9 M, =of

fy

m?

¢ dt = 133205 (16-95 hr)

+69 +81+---+80) +0 mg

7

OE8 =_ 1 1,380,840 380,840 mg

0+2(14+2.9+---+0.6)+0 mg

203)

a] = 191,800 mg

which can be used, along, with the travel time and Eq. 20.60 to calculate,

k=

1

05d

In 1,380,840

191,808

= 3.948 d7

The correction for ethylene (Eq. 20.62) can then be used to estimate the reaeration

ky = 115(3.948) = 4.5447! (6) The velocity between the two points can be calculated as

y = 13500-6000 6000 mm _ 15000 ™ _625 ™ OSd d hr Inspecting the values at KP 6 indicates that the centroid of the dye patch passed at about 50 min. This can then be employed to determine the distance upstream at which the dye was introduced,

x= 625-™50 min— = 52083 m hr

60 min

Therefore, the two measurement points are at x = 520.83 and 8020.83 m.

ethylene discharged can be estimated as

Further, the initial mass of

1,380,840 = 1583,713 mg Mo = —ssaasoreoray This value can be normalized to the cross-sectional area (calculated from the ratio of Q/U as 21.3 m”) and the result, along with the other parameter values, substituted into Eq. 10.24 to compute the following,

20.4

20.7 The travel time between the two stations can be determined by

> _; _4000-500m_ 1d h-h= =0.2025d 02m/s 86400s This value, along with the concentrations can be substituted into Eq. 20.61 to yield

ke

0.2025

“| = 4842 7 \150

which can be converted to a reaeration rate by Eq. 20.62

ky = Rk = 138(4842) = 6.68 d 20.8

The reaeration rate can be determined in a number of ways. Nonlinear regression estimate ka = 0.09 d'. Alternatively, Eq. 20.49 can be manipulated,

be used to

0-0, =(0,-0,)e* o-90.

kt=n} —* 0; — 95 Thus, a plot of the right-hand-side versus ¢ should yield a straight line with an intercept of zero and a slope of -k,. As shown below, such an analysis yields an estimate of 0.088 d’.

0

10

20

30

0.5 1.5 se

s

-2.5 . 3.5 20.9 (a) According to Fig. 20.7, the Owens-Gibbs formula should hold, 0. 4°67

ky =532 oats 10247 = 28.68 d (b) For carbon dioxide, Eq. 20.58 provides a first estimate 32

kacon = 26(2)

0.25

=265d7

20.5

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20.10 Manning’s equation can be formulated for a rectangular channel as

15=—L20,

30)

0.03

2/3

20+2H

0.001'?

which can be solved for H = 0.206 m. 0.363 mps.

The velocity can then be computed as U = 1.5/(0.206x20) =

The Owens-Gibbs equation (which is chosen because the stream is shallow) can then be

employed to give,

4, =532.2383 0.67 206 02

04"

The average temperature below the discharge is T = (1x10+0.5x25)/1.5

corrected for temperature as in

= 15°C.

The rate can be

Kays = 50x 1.02415 = 44.4 dt 20.11 (@) The Banks-Herrera formula can be used to compute,

K, = 0.728(2)°* — 0.317(2) + 0.0372(2)? = 0544 md™ which can be converted to 16°C by

K, = 0544 x 1024"? = 0.495 md! Finally, the result can be expressed as a reaeration rate by

ky = Xt = 945 < on07 47 H 7 (6) For ammonia, we must also estimate the gas-film transfer rate. This can be done with Eq. 20.55 and 20.59

Kg s0 = \68U,, = 168(2) = 336 md K,= aq

18

|

025

= 3408md7?

Then Eq, 20.14 (along with the value of H, = 1.37x10° from Table 19.3) can be used to determine

1

v,

=

1

0495

+ 8.21x 107°(289.15)

137

10°(3408)

= 1844508 = 692

v= al. 0145 ma?

692

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20.6

20.12 (a) The velocity can be calculated as U = 800/(2.7x200) = 1.48 mps. the Churchill formula is used to compute

148

k= 525s ar =148d

For this velocity and depth,

4

The Wanninkhof formula can be used to compute

K, = 0.0986(L5)'™ = 0.1917 md” which can be combined with the Churchill result to obtain

k, = 148+

0.1917 =1555d" 27

which can be temperature corrected to give

Ka s (2,25 = 1555x 10247 = 17547 (b) The reaeration rate can be expressed as a mass transfer coefficient by multiplying it by depth, K, = 1,75(2.7) = 4.726 m/d. The gas-film exchange coefficient for water vapor can be estimated as

K, 5.1140

= 16815) = 252°m

Both these rates can be scaled to apply them to ammonia,

K,= and 2)

0.25

17

K = 25x 8 8

17

0.25

= 403s = 2556

d

Then, using the Henry’s constant from Table 19.3, Eq. 20.15 can be used to determine the ammonia mass-transfer coefficient,

137x 10> v, = 4035 —___~"____________z9, ——__, 137 x 1075 +8206x 107°(298.15)4.035/ 2556

s™

d

Note, that this shows that ammonia is predominantly gas-film controlled.

20.7

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LECTURE 21: 21.1 The amount of CBOD spilled can be calculated as

Megoo = 30000 L{ 100§-892°% ) L

6 mole-32 gO, / mole 1 mole -(6-12 +12-1+ 6-16) g- glucose / mole

= 32x 10° g-CBOD

Mass balances for CBOD and oxygen can be written and solved for

L=L,e(tethy

o=0,-

kal, (estoy

kaka

me Utetha )

The remaining model parameters can be evaluated as: 0, = 6.942, k, = 0.08925, ky = 0.126.

can be substituted to give the following profiles:

te=71d 21.2

These

time (days)

NOTE TO INSTRUCTOR: Part (c) requires that the depth of the reactor be specified. following solution assumes an H = 1 m.

The

(a and b) The BOD can be computed simply as

L=267ee™ where co = the initial organic carbon content and ky = 0.2(1.047'**) = 0.159. The oxygen can be

calculated in an identical fashion to Eq. 19.14

0= 0, -2.67¢(1-e"*") Together with a saturation value (calculated from the temperature and elevation data) of 7.79, the following plot of the progression in the bottle can be developed,

12

21.1

Notice that after about 8.2 days, the bottle runs out of oxygen, which terminates the BOD decoinposition. In point of fact, the system might switch to anaerobic decomposition at this point. Assuming that the reactor is 1 m deep, the Banks and Herrera formula (Eq. 20.52) can be used to compute a temperature-corrected reaeration rate of 0.4 d'. This value can be employed to compute the oxygen in the open reactor as

oxo. — a2 670 (6-4 -eh') *

kaa he

The results are shown below.

Oo.

4

8

12

21.3 The first step is to estimate k, by fitting a straight line to a plot of In L versus ¢, 4

0

5

10

The best fit yields a slope of -0.1281 d'. Assuming rate of

15

‘@

2

a theta of 1.047, this can be converted to a 20°C

kao = 01281x 10479)? = 0.2028 Next, the value of 0.1281 can be substituted into Eqs. 21.13 and 21.14 (along with D,= 11.288 - 10= 1.288, L, = 50, ,= 8, and D, = 9.988) formulated as a roots problem,

21.2

Google

Silkarkad

=

1

k, ee 1288(k, — 0.1281) ee

a iam

So aska) = kg(50)}

atai ka

k, | 0.1281

ka(50)

1- 1288(k, —- 0.1281)

k4(50)

}

it-s=0

0.1281

~ k-01281 -9988=0

These equations can be solved simultaneously to yield estimates of k, = 0.10992 and kz = 0.06118 which can be converted to 20°C rates of k, = 0.139 and kz = 0.097. These values can be used to simulate the sag. The results are compared with the data below. Note that the data for this problem was generated with values of k, = 0.126, ky = 0.063, and k, = 0.118, which correspond to 20°C rates of k, = 0.2, kg

= 0.1, and k, = 0.15.

were contaminated with random error generated between -1 and 1.

The simulation results

10

¢ (mg/L) O+

oO

+ 5

+ 10

+t 18

4 20

t (days) 21.4 Mass balances can be used to determine values below the mixing point: Q =9 m’/s, T,= 11.7778 °C,

So = 4.111 g/L, and o, = 10 mg/L. The saturation concentration [with temperature (10.86 mg/L), salinity (10.59 mg/L) and elevation corrections] can be calculated as 0, = 8.767. Therefore, the initial

deficit is D, = 8.767 — 10 =-1.236 mg/L.

21.5 Mass balances can be used to determine values below the mixing point: @ =122311 m’°/d, 7, = 20 °C, L, = 101 mg/L, and c, = 6 mg/L. The saturation concentration can be calculated as 0, = 7.34.

Therefore, the initial deficit is D, = 7.34 - 6 = 1.34 mg/L. The reaeration rate can be determined as k, = 216—,

1° 67 135

= 216d"

Both this rate and the deoxygenation rate (ky =

because the mixed temperature is exactly 20 °C. The deficit solution is therefore,

21.3

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1 d') do not need to be corrected for temperature

216

216

1

2635" _ 1340" =) D=134eso» MaDe 1-216 which can be used to determine the profile shown below:

8

x (m) The minimum DO is equal to 3.25 mg/L and occurs at about x, = 3.5 km. 21.6 The first step is to perform mass balances at the point source. The results are

Q

waste

1

river

4

T

25

15

L

50

5

o

2

mixed

5

7 4

7

Next, the reaction rates and hydraulic characteristics for each reach can be established. Note that Eq. 19.28 was used to set kz and Eq. 20.41 to set ka.

U, mps Aim

U, mpd T,°C karo, Id

kar, Id

kao, Id Kar, Id

o,, mg/L

1

2

0.5

0.2

3

5

3

0.4

7

43200 17 0.3

17280 17 0.3

34560 17.5 0.3

0.261385

0.261385

0.267457

9.694024

9.694024

-9.593855

0.534805 0.498076

0.1572 0.146404

0.134207 0.126481

The BOD and DO deficit equations can now be developed. The junction at KP 30 merely necessitates a reinitialization of the variables and a change in parameters. A formal mass balance must be

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21.4

developed at KP 60, because of the point source that enters there. Some results are listed below along, with a plot. 0 10 20

L 1400 13.18 1240

D 367 404 433

o 603 565 536

40 50 60 60 70 80

10.04 863 7.42 785 7.26 672

5.75 663 7.25 673 7.06 7.33

3.95 3.06 244 287 2.54 2.26

30

90

1168

455

6.22

5.15

7.56

16 > 7

2.03

o, 9.69 9.69 9.69

9.69

9.69 9.69 9.69 9.59 9.59 9.59 9.59

BOD

12 PN

DO saturation

s+

4+

D0

0

+

0

|

100

21.5

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+

+

200

{

LECTURE 22: 22.1 NOTE TO INSTRUCTOR:

This problem requires a cross-sectional area. The following

solution uses an A, = 10 m’.

The solution for a distributed source with no flow is

.

c = 5a

Ak

k

l-e

where S,” = the distributed source expressed as a line load (recall Fig. 22.2a) with units of mg md

‘. This equation can then be solved for

-

eA,k __ 2(100.693/5)

Sa=—-*, —2owas 028" 10000 BT l-e JU

1—e

01(86400)

mg

md

22.2 The depth of the stretch must first be determined with the Manning equation, which for rectangular channel is

v1

HB

n\

y's

B+2H

Substituting the parameters, the resulting equation can be expressed as a roots problem, 1

I= 555 which

(

50H

50+2H

2/3

0.0002" - 0.2

be solved for H = 0.279 m. Because the

stream is so shallow, the Owens-Gibbs formula

can be used to compute k, = 19.14 d'. This value can then be used in conjunction with Eq. 22.2 and 22.4 to solve for the deficit over the stretch,

, D= Sa

Hk,

k [-" | = 067 mm

(amt

1 }

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25.2

(6) A numerical approach can be developed by writing balances for the aerobic layer as

dn, _D, H,—l=—(n, ht

Dn Ay (m2 -n,)-—2n, -m) Ay My —-k,,H, Ky lm

_ D, H,—dm ht =—"(m OH (m, —m)—-—*m —m) TM —-k,,.Hym K py lt ymy For the 4 anaerobic layers of equal size, balances can be written as

“dey

Hy

Hy —* = —* Je -keH: a@oH° * 202 D,

Ay oa = Anke Hey a D,

1,

des_

Hs He3 Jo CH

s -n)

D,

keHcy +7

Hs— sat > =

D,

oma)

~ma)+ a mma)

-k,H. Ret ses

dn.

D,

Hs 7 = Ack, Hscs +7, dm

D,

Hse =k Hes +,

Ns) —ms)

where H, = the total thickness of the active anaerobic sediment. Because the aerobic layer is so thin, the aerobic equations can be solved at steady state for mn

Dy! Fy

“2D, / Ay +kyFh

6

D/H,

@

"\*2D,1H, +kgHh,

Thus, only the anaerobic equations need to be integrated and Egs. i and ii provide the means to determine m, and m,. Now the next major step in solving these equations involves determining H,. This can be done by recognizing that

SOD = rk _ fT ym + Fon! ky Hy and that

25.3

Given values of m, and 7, these equations can be solved simultaneously for 7, and SOD. A nice way to solve the system is to integrate the anaerobic equations until they reach a steady state. At each time step, the methane in the anaerobic layers can be compared with the methane

saturation value. If it is exceeded, the methane value in the layer would be set at saturation.

At each time step, Eqs. i through iv are solved for the aerobic layer characteristics. The final results for the present context are an SOD = 1.23 g/m’/d and a thickness, H, = 0.589 mm. The other variable values and transfers are shown below.

0834

0.834

Poc

loss

tipattiom'°° bubble

input Input from wom O22 Poc

METHANE Sediment

methane

and

ammonium

AMMONIUM budgets.

All

arrows

are

expressed

as

equivalents (gO m* d*). Both the methane and ammonium have units of mg/L.

flux

of

oxygen

25.5 The addition of the boundary layer necessitates writing extra equations for that layer which should be solved dynamically. In addition, oxygen balances need to be written for both the boundary and the aerobic sediment layer. Although this approach would be necessary for most model applications, a simpler approach is possible for the present problem. Because we assume that the characteristics of the two layers are identical, we can approximate them as a single large segment. If this is done, we rapidly recognize that the increased residence time (because of the added thickness of the boundary layer) results in negative oxygen concentrations. Thus, to attain results comparable to those of Prob. 25.4, we must employ much smaller decomposition rates. These can be determined by trial and error

as k, = 0.0448 and k, = 0.064 d' which result in a total layer thickness of 10.911 mm. Interestingly, the associated SOD is 0.067 g/m’/d, which is lower than the value of 1.23 calculated in Prob. 25.4. A higher value could be induced by using a smaller layer thickness. However, that would mean that the boundary layer would have to be less than 1 cm.

This problem has been intended to illustrate the potential importance of sediment-boundary layers on the computation of SOD. These should become increasingly important as SOD calculations are coupled with detailed hydrodynamic computations

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25.4

LECTURE 26: 26.1

NOTE TO INSTRUCTOR:

There are several incorrect parameter values and omitted values

in this problem statement. The volumes of segment 1 and 2 should be set at 5x10° m° and the surface area of the sediments should be 1x10° m?, The SOD should be changed to 1 g/m’/d and

the flows and bulk dispersion coefficient units should be m°/d.

Mass balances for BOD can be written for segments 1 and 2 (using backward differences),

O= Wy +Doro + Ona + Bay (La ~ )+ Ej3( La Ly) O34 —kahily

O= Wy + Oya Ly — Opi Ly + Eay(Ly — Ly) + Eas (Ls - La) Onila — heal Collecting terms gives

(Qs + Ei3+ En +kaNi dy

(En + Qu Ly = 11+ Qoilo + Eisls

~ Exyly + (Qa + Eni + Baa + kanVa Ua = Win + Oral + Exals

or in matrix format,

[22+

into Ey

~En-Qn Qay + En) + Es +k aa

fae [fersoate seoo| |\2) (Wp. +Onls+ Exls

Substituting parameter values for the [4] matrix yields [A]=

10700000 1200000

-—3700000 6700000

which can be inverted to give

pay!=| 996x108

55x 10%

178x107

159x107

Mass balances for DO can be written for segments 1 and 2 (using backward differences),

= Wo1 + Doro + D102 + E21(0 ~04)+ E13(03 ~ 01) - 230 — kala + kaMi(s ~ 04)

O= Won + Qyn03 ~Qay09 + Ea (04 — 03) + E3303 ~ 02) O21 ~ kaaVala - Seva

These equations can also be expressed in matrix format as,

[: +E 3+ En tka, -E

-Ey ~ Oa fe } = {ye +210 + E1303 + kaVion +kaMily Qa + Eq, + Ep3 JL) Wop + 05203 + E3303 + kaaV2ly + SpAn

Substituting parameter values for the [B] matrix yields [B]= 11200000 ~ {-1200000

-3700000 6200000

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26.1

which can be inverted to give

tay! =| 954x109

69x 107%

185x10%

172x107

The [C] matrix can then be generated as

-1 _ppy-1[500000

(v4

[ 0

0

500000 4]

-1 | 526x107 = 7.15x10°

-[s2ecies

rs

|

The Excel solver was then used to vary W,; until the resulting concentrations of oxygen were above 4

mg/L. The resulting loading was 1.13x10° g/d. The final steady-state calculation was

to} = [B1"{,}-1cyfw;} oy =| 954x10°S 185x10%

56910°§ || Hoi + Qo100 + £is0s + kaiVion | _ 526x109

1.72x 10° |] W,4 +Qy905 + £9303+SpAqq |

[24610

715x107 Wii + Qorlo+ Eisls

tor={7o3} 26.2 NOTE TO INSTRUCTOR:

Change the BOD of the waste source to 200 mg/L.

The QUALZ2E result is shown along with the analytical solution below.

mg/L

20

10

ANALYTICAL SOLUTION: x 05 45

L 19.69 1848

o 902 823

x 395 445

145 19.5 245 295 34.5

15.78 1458 13.47 1245 11.50

7.24 7.07 7.02 706 7.15

545 595 645 695

9.5

17.08

7.60

49.5

L 1063 982

o 7.27 7.41

839 7.75 716 662

7.70 7.85 7.99 812

9.08

7.55

26.2

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14210" ||W,4 +Q59L5 + Eq3ls

26.3 The analytical solution to this problem is displayed below. x 0 10 20 30 40 50

L 14.00 13.18 12.40 11.68 10.04 8.63

D 3.67 4.04 4.33 4.55 5.75 6.63

o 6.03 5.65 5.36 5.15 3.95 3.06

Oo

9.69 9.69 9.69 9.69 9.69 9.69

x 60 60 70 80 90

L 742 7.85 7.26 6.72 6.22

D

7.25 673 7.06 733 7.56

o

(2.44 2.87 2.54 2.26 2.03

oO,

9.69 9.59 9.59 9.59 9.59

16 7

12 pe

DO saturation

26.4 The QUALZ2E result should correspond approximately to the analytical solution listed below. The actual plot generated with QUALZ2E is also shown. x 5

345600 734400 734400 734400 734400 734400 734400 734400 734400 734400 734400

L 2.00 27.41 ° 26.88 26.36 22.56 18.56 15.27 12.57 11.03 9.67 8.49

o 7.50 5.65 5.56 5.47 5.00 4.77 4.80 4.98 5.81 6.38 6.80

26.3

Q 734400

120 130 140 150

734400 475200 475200 475200 475200 475200 475200 475200 475200

L 7.45 6.55 5.65 4.85 4.17 3.58 3.08 2.64 2.27 1.95

713 744 7.93 8.19 8.36 8.47 8.56 8.64 8.70 8.76

LECTURE 27: 271

3

(@) Q,, = 100,000 capita (0.5 m? (eaita 8) = 50,000 77 = 50000(25)+100000(15) = 18,333 °C 7 150000 15x10!! number / capita/d m? ) N, . = . 05 m? /capita/d 10*(100 yy, = 30000(30% 10°)+ 100000(0) ° 150000 (c)

¢ number = 30x 10' 100 mL

mL)

¢ fumber = 10x 10° ——

3

m

Q@=100,000+ 50,000= 150,000 — 8

Q 150,000 m == =—— =15,000— HB 05(20) d (@) Kp, = 08-1.07'83*- - 9.7147/d

(©) 16 =Isnee 0

2s cal 213) 1d —— = 650—-——— = 9. oT cm’d (24) 24hr

cal om?hr

9339 (1 (0505) e ) =_ 8263.4 a 0 _= F505 ()

_0.715+8263,

N=10x10% — Lsx10*

= 25,150 number

27.2 (a) The bacterial die-off rate can be calculated as

' 200/24 ky » == [08+.0.006(50)} [08+-0.006(50)}1.0727-20 4