66865-Chptr6PrblmSolns

7/29/2019 66865-Chptr6PrblmSolns

1/27

6.1 a. A primary standard is an enforceable limit on the concentration of acontaminant in water or an enforceable requirement that a particular treatmenttechnique be implemented. Primary standards apply only to contaminants that

impact human health. A secondary standard is a recommended limit on the

concentration of a water constituent or on the measured value of a water quality

parameter (e.g., turbidity). Secondary standards apply to factors that affect adrinking waters aesthetic but not human health attributes.

b. An MCL is a primary standard, whereas an MCLG is a maximumconcentration goal for a drinking water contaminant, which would be desirable

based on human health concerns and assuming all feasibility issues such as cost

and technological capability are not considered. An MCLG is not an enforceablelimit, but does provide the health-based concentration, which the MCL should

seek to approach as closely as possible within the constraints of practical

feasibility. There are many contaminants with MCLGs, but no numeric MCL.These include acrylamide, copper and lead.

6.2 The CWA sets up a system by which the maximum concentration of contaminantsin discharges to surface waters and in the surface waters themselves are set andenforced, while the SDWA provides the legislative mechanism necessary to set

and enforce the maximum contaminant concentrations in drinking water. The

CWA is designed to ensure that the quality of surface waters in the U.S. is, at aminimum, appropriate for the beneficial uses for which the water is designated.

The SDWA is designed to ensure that water supplied by public water systems forhuman consumption meets acceptable health standards at the point of use.


2/27

6.3 a. The hydraulic detention time is:( ) ( )

( )hr4.17d0.174

gal7.4805

ftgal/d102.5

ft0.10ft43.0

36

2

==

==

Q

V

b. The critical velocity is:

( )( ) hrft

ft2.40

ft43.0

hr24d

gal7.4804ft

dgal102.5

2

3

2

36

b

o =

==A

Qv

c. The weir loading rate is:

( )( ) hrft

ft51.5ft43.02

hr24

d

gal7.4804

ft

d

gal102.5

rateloadingweir

3

36

w =

== LQ


3/27

6.4 From Appendix C: At 20C, = 998.2 kgm-3 and = 0.00100 kgm-1s-1a. The particle settling velocity is:

( ) ( )( )( )( )s

m102.18

sm

kg00100.018

m100.1kg/m2.9984.0m/s9.807

18

-

5-

2-5322

pp

s =

==

dg

v

b.

( )

( )( ) sm102.89

m30m10

s606024

dd/m7,500

4-

3

b

o =

==A

Qv

Since vo > vs, less than 100% of the particles will be removed.

c. The settling velocity of the smallest particle which is 100% removed is equalto vo. So,

( )( ) s

m36.5

s

m103.65

m

kg2.9984.0

s

m9.807

s

m1089.2

sm

kg0.0010018

-

18 5-

21

32

4-2

1

p

sp ==

=

=

g

vd


4/27

6.5 From Appendix C: At 20C, = 998.2 kgm-3 and = 0.00100 kgm-1s-1The length to width ratio is 5, soAb = 5w

2

Set vo = vs and solve forw:

( ) ( )m42.7

m10m

kg998.2-200,1

s

m9.8075

sm

kg00100.0

s

m0.10018

d-5

18

21

25

32

3

21

2

p

=

=

=

g

Qw


5/27

6.6 From Appendix C: At 15C, = 0.00114 kgm-1s-1a.

b

V

PG

=

( )

( ) ( )

1-

21

21 s50.0

m75.3m17.4sm

kg0.00114

W186 =

=G

Similarly, G2 = 20.1 s-1

and G3 = 10.0 s-1

.

b.

( ) ( )4-

36-153-6

0

3

p

107.546 m

mL10mL108.1m100.20

6 =

==

Nd

Q

GV

N

N

b0 41

+=

( ) ( )( )( ) ( )17.9

s606024

d

d

m16,000

m75.3m17.4s0.501054.741.01

3

2-1-4

0 =

+=

N

N

Therefore, there will be on average 18 singlets per aggregate.


6/27

6.7 From Appendix C: At 15C, = 0.00114 kgm-1s-1a.

Accumulation = Reaction: )r(NVdt

dNV =

and

GNN

=

4-)r(

so,

=tN

N

dtG

N

dN

0

4-

0

which yields:

= Gt

NN

4-

0e

b.

( )1-

21

3

3

25-

s3.78

m004.0sm

kg0.00114

s

mkg105.6

=

==V

PG

c.

( )4-

3

735-

0

3

p101.64

6

m0040.0

100.1m100.5

6

=

==

Nd

( ) ( )( )( )( )

==

min30min

s60s78.31064.141.0-exp

L4.0

101.0e

1-4-74-

0

Gt

NN

L

particles106.07 5=N

d.

aggregatesinglets

4.12L1007.6

L0.4

101.0

1-5

7

0 =

=N

N


7/27

e.

( ) ( )( )( )( )

==

min30min

s60s78.31064.140.2-exp

L4.0

101.0e

1-4-74-

0

Gt

NN

L

particles101.88 6=N


8/27

6.8 a. ( )hr

m640m80

hr

m8.0

32

fa =

== AvQ

b. From the definition of filter efficiency: Vf (Vb + Vr) = fVf (6.8.1)

and the definition of the effective filtration rate gives:Vf (Vb + Vr) = refAftc (6.8.2)

Combining (6.8.1) and (6.8.2) and rearranging yields:

)( )( )cycle)(perm33,370

0.96

hr52m80hr

m7.7 3

2

f

cfeff ===

tArv

From (6.8.1): ( ) ( )( ) cycle)(perm1,3350.96-1m370,33-1 33ffrb ===+ VVV

Thus, the volume of water lost per cycle (Vb

+ Vr) is 1,340 m

3.


9/27

6.9 Note that this problem incorporates a practically realistic, yet possibly confusing,nuance in that the total backwash time per cycle (tb) is double the total time thatbackwash water is flowing through the filter (tb). Thus, calculation of the water

volume used in backwash is based on 8 minutes of flow per cycle, whereas the

time the filter is off-line for backwashing is 16 minutes.

a.

( )( )( ) L307,200min

s60min0.8m64sm

L10.0 2

2

'

bf

f

bb =

== tAA

QV

( )( )( ) L316,800min

s60min15m64sm

L5.50 2

2rfar=

== tAvV

( ) ( ) ( ) min1,409min15-min16min6024rbcf === tttt

( )( )( ) L102.98mins60min,4091m64smL5.50 722f

f

ff '

f

=

== tA

AQV

%980.979L102.98

L316,800-307,200-102.98

7

7

f

rbff ==

=

=

V

VVV

b. The fraction of a filtration cycle that is not backwashing is:( )

0.989hr24

min60

hrmin16-hr24

c

brf

=

=

+

t

ttt

so the number of filters required is:

( ) ( ) ( )6.89

989.0L1000

mm64

sm

L5.50

sm2.40

989.0

3

2

2

3

fa

plant =

==Av

Qn

Therefore, at least 7 filters are required.


10/27

6.10 a. For a steady state PFR, the Giardia concentration in the contactor effluent is:

*-

0ekNN=

so, ( ) ( )3

1-

3

0*b

m156000,1lnmin0.53

min

s

60sm0.20ln ===

NN

kQV

b. For a steady state CSTR, the mass balance is:QN0 = QN+ k

*NVb

so,( )

( )( ) 3

1-

3

0

*bm22,6001-000,1

min0.53

mins60

sm0.20

1- ==

=

N

N

k

QV

c. For a steady state series of CSTRs, the mass balance is:

( )m*0m

1

1

kN

N

+=

so,( )

( )( )( ) m67.51-000,1

min0.53

mins60

sm0.20

1- 351

1-

31

m

0

*b,1==

=

m

N

N

k

QV

and the total contactor volume is: mVb,1 = 5(67.50 m3) = 337 m3

d.

minmg

L0.265

Lmg2.0

min0.53

0.1

-1*

=

==nC

kk


11/27

6.11 For a steady state CSTR, the mass balance is:QN0 = QN+ k

*NVb

so,( )( )

( ) sL0.0156

1-1,000

s80.7L2.00

1-

-1

0

*

b ==

=

NN

kVQ

and the volume produced during ten hours of operation would be:

( ) L562hr

s6060hr10

s

L0.0156 =

=Qt


12/27

6.12 Hardness in meq/L:

meq/L7.50mg/meq20.0

mg/L150Ca 2 ==+

meq/L4.92mg/meq12.2mg/L60Mg2 ==+

Total hardness = 7.50 + 4.92 = 12.4 meq/L

Hardness as CaCO3:

33 CaCOasmg/L621

meq

CaCOasmg50.0

L

meq12.4hardnessTotal ==

Table 6.4 would classify this as very hardwater.


13/27

6.13 6.16 The solutions for these problems are the solutions for problems 6.2 - 6.5 inthe 2nd edition Solutions Manual.

6.17 Component Concentration

(mg/L)

Equiv. Weight

(mg/meq)

Concentration

(meq/L)

CO2 14.5 22 0.6591

Ca2+

110.0 20 5.500

Mg2+

50.7 12.2 4.156

Na+

75 23.0 3.261

HCO3-

350 61.0 5.738

SO42- 85.5 48.0 1.781

Cl-

16.2 35.5 0.4563

pH 8.2 - -

9.

656

5.

738

0 5.

500

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

Mg-NCH

9.

656

5.

738

0 5.

500

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

Mg-NCH

a. Mg-CH = CH Ca-CH = 5.738 5.500 = 0.238 meq/Lb. Mg-NCH = TH CH = 9.656 5.738 = 3.918 meq/Lc. 1 meq/L Ca(OH)2 neutralizes 1 meq/L CO2(aq)

so, Ca(OH)2 required = 0.6591 meq/L


14/27

6.18

Component Concentration Equiv. Weight

(mg/meq)

Concentration

(meq/L)

CO2 6.0 mg/L 22 0.2727

Ca2+ 50.0 mg/L 20 2.500

Mg2+

20.0 mg/L 12.2 1.639HCO3

- 3.1 mmole/L 61.0 3.100

pH 7.6 - -

4.

139

3.1

00

0 2.

500

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

Mg-NCH

4.

139

3.1

00

0 2.

500

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

Mg-NCH

Table P6.18 Components, lime and soda ash dosage, and solids generatedComponent (eqn.) Concentration

(meq/L)

Lime

(meq/L)

Soda ash

(meq/L)

CaCO3(s)(meq/L)

Mg(OH)2(s)(meq/L)

CO2(aq) (6.37) 0.2727 0.2727 0 0.2727 0

Ca-CH (6.38) 2.500 2.500 0 5.000 0

Mg-CH (6.39) 0.600 1.200 0 1.200 0.600

Ca-NCH (6.40) 0 0 0 0 0

Mg-NCH (6.41) 1.039 1.039 1.039 1.039 1.039Excess 0.400

Totals 5.412 1.039 7.512 1.639

a. Lime required = (5.412 meq/L)(37.07 mg/meq) = 200.5 mg/L

Soda ash required = (1.039 meq/L)(53.0 mg/meq) = 55.1 mg/L

b. Sludge generated:

+

mg10

kg

gal

L785.3

d

gal1015

meq

mg9.22

L

meq.6391

meq

mg0.50

L

meq512.7

6

6

kg/d24,040=


15/27

6.19

Component Concentration Equiv. Weight

(mg/meq)

Concentration

(meq/L)

Concentration

(mg/L CaCO3)

Ca2+ 90 mg/L 20 4.500 225.0

Mg2+ 30 mg/L 12.2 2.4596 123.0

HCO3-

165 mg/L 61.0 2.705 135.3pH 7.5 - - -

348

.0

135

.3

0 225

.0

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

Mg-NCHCa-NCH

348

.0

135

.3

0 225

.0

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

Mg-NCHCa-NCH

( )[ ][ ][ ]

3

4-

7-

5.7

1

-

3aq2

CaCOmg/L19.14M101.914M104.47

L

mol10

g61.0

mol

L

g0.165

HHCO

CO ==

==

+

K

Table P6.19 Components, lime and soda ash dosage, and solids generated

Component (eqn.) Concentration

(mg/L

CaCO3)

Lime

(mg/L

CaCO3)

Soda ash

(mg/L

CaCO3)

CaCO3(s)(mg/L

CaCO3)

Mg(OH)2(s)(mg/L

CaCO3)

CO2(aq) (6.37) 19.14 19.14 0 19.14 0

Ca-CH (6.38) 135.3 135.3 0 270.6 0

Mg-CH (6.39) 0 0 0 0 0

Ca-NCH (6.40) 90.00 0 90.00 90.00

Mg-NCH (6.41) 123.0 123.0 123.0 123.0 123.0

Excess 20

Totals 297.4 213.0

a.L

mg220.1

CaCOmg100

Ca(OH)mg74

L

CaCOmg297.4Lime

3

23required =

=

b.L

mg225.8

CaCOmg100

Ca(OH)mg061

L

CaCOmg213.0AshSoda

3

23required =

=


16/27

6.20 The solution for this problem is the solution for problem 6.6 in the 2nd

edition

Solutions Manual.

6.21Component Concentration Equiv. Weight

(mg/meq)

Concentration

(meq/L)

Concentration

(mg/L CaCO3)

Ca2+ 95 mg/L 20 4.75 238

Mg2+ 26 mg/L 12.2 2.13 107

HCO3- 160 mg/L 61.0 2.62 131

pH 7.0 - - -

6.

88

2.

62

0 4.

75

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

Mg-NCHCa-NCH

6.

88

2.

62

0 4.

75

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

Mg-NCHCa-NCH

( )[ ][ ][ ]

meq/L1.17M105.868M104.47

L

mol10

g61.0

mol

L

g0.160

HHCO

CO 4-7-

7

1

-

3aq2 ==

==

+

K


(meq/L)

Lime

(meq/L)

Soda ash

(meq/L)

CaCO3(s)(meq/L)

Mg(OH)2(s)(meq/L)

CO2(aq) (6.37) 1.17 1.17 0 1.17 0

Ca-CH (6.38) 2.62 2.62 0 5.24 0

Mg-CH (6.39) 0 0 0 0 0

Ca-NCH (6.40) 2.13 0 2.13 2.13 0

Mg-NCH (6.41) 2.13 2.13 2.13 2.13 2.13

Excess 0.400

Totals 6.32 4.26 10.67 2.13

a.L

mg234

meq

mg7.073

L

meq6.32Limerequired =

=

b.L

mg226

meq

mg35

L

meq4.26AshSoda required =

=


17/27

6.22 The solution for this problem is the solution for problem 6.7 in the 2nd

edition

Solutions Manual.

6.23Component Concentration Equiv. Weight

(mg/meq)

Concentration

(meq/L)

Concentration

(mg/L CaCO3)

Ca2+ 40.0 mg/L 20 2.00 100

Mg2+ 10.0 mg/L 12.2 0.820 41.0

HCO3- 110 mg/L 61.0 1.80 90.2

pH 7.0 (assumed) - - -

2.

82

1.

80

0 2.

00

0

Ca2+

HCO3-

CH

Mg2+

Ca-CH

NC H

Mg-NCH

Ca-NCH

2.

82

1.

80

0 2.

00

0

Ca2+

HCO3-

CH

Mg2+

Ca-CH

NC H

Mg-NCH

Ca-NCH

( )[ ][ ][ ]

meq/L0.807M104.034M104.47

L

mol10

g61.0

mol

L

g0.110

HHCO

CO 4-7-

7

1

-

3aq2 ==

==

+

K


(meq/L)

Lime

(meq/L)

Soda ash

(meq/L)

CaCO3(s)(meq/L)

Mg(OH)2(s)(meq/L)

CO2(aq) (6.37) 0.807 0.807 0 0.807 0

Ca-CH (6.38) 1.80 1.80 0 3.60 0

Mg-CH (6.39) 0 0 0 0 0

Ca-NCH (6.40) 0.200 0 0.200 0.200 0

Mg-NCH (6.41) 0.820 0.820 0.820 0.820 0.820

Excess 0.400

Totals 3.83 1.02 5.43 0.82

a.L

mg142

meq

mg7.073

L

meq3.83Limerequired =

=

b.L

mg54.1

meq

mg35

L

meq1.02AshSoda required =

=


18/27

c. Sludge generated:

+

mg10

kg

gal

L785.3

d

gal1073

meq

mg9.22

L

meq.820

meq

mg0.50

L

meq.435

6

6

kg/d41,400=


19/27

6.24Qf= 5 10

6L/d Qp = 3 10

6L/d

Cf= 1,500 mg/L Cp = 75 mg/L

Qc, Cc

QfCf= QcCc + QpCp and Qf= Qc + Qp

So,

( )( )( ) ( )( )[ ]

( )mg/L3,640

L/d103-105

mg/L57L/d103-mg/L500,1L/d105

-

-

66

66

pf

ppff

c =

==QQ

CQCQC


20/27

6.25Qf= 5 10

6L/d Qp = 3 10

6L/d

Cf= 1,500 mg/L Cp = 75 mg/L

Cc = 3,640 mg/L

Qc

= 2 106

L/d

a. Water recovery:

60%or0.60

dL105

dL103

6

6

F

P =

==

Q

Qr

b. Salt rejection:

95%or0.95

Lmg1,500

L

mg75

-1-1F

P === CCR

c. Log salt rejection:

( ) ( ) 1.300.95-1log--1log-log

=== RR


21/27

6.26 a. The main constituent of concern from the perspective of configuring the

POTW treatment train is biodegradable organic matter. It is, however, arguablefrom the perspective of human health that the main constituents of concern are

human pathogens.

b.Unit operation function based on an overall aim of removing BOD:i. The grit chamber removes the very largest and most settleable

particles, which may contain a modest fraction of organic matter.

ii. The primary sedimentation basin removes most of thegravitationally settleable organic matter (as well as inorganic

matter). Typically about 35% of the BOD can be removed by

primary sedimentation.

iii. The bioreactor converts dissolved and fine particulatebiodegradable organic matter into microbial cell mass and energy

for microbial metabolism.

iv. The secondary clarifier physically removes the cell mass generatedin the bioreactor by gravitational settling.v. Digestors further degrade the organic particles in the primaryclarifier sludge and/or the microbial cell mass separated into the

secondary clarifier sludge by exposing it to more prolonged

biodegradation.


22/27

6.27 Both primary and secondary wastewater treatment are designed to remove

biodegradable organic matter (BOM) and the superset of total solids. Primary treatmentonly removes that BOM which can be physically separated from the raw sewage by

floatation, gravitational settling or screening. On the other hand, secondary treatment

removes BOM that may be biodegraded by microbes within a relatively short duration

(typically the hydraulic retention time is less than 1 day). Much of the BOM degraded insecondary treatment is dissolved or colloidal and it is converted into microbial cell mass.

The cells are removed from the secondary effluent by settling in a secondary clarifier (orby exclusion by a membrane).

6.28 6.29 The solutions for these problems are the solutions for problems 6.9 - 6.10 in

the 2nd

edition Solutions Manual.


23/27

6.30

well-mixed

aeration pond

Q, S0,X0Q, S,X

V, S,X, rg, rsu

Q = 30 m3/d

S0 = 350 mg/L BOD5S= 20 mg/L BOD5Ks = 100 mg/L BOD5

kd = 0.10 d

-1

m = 1.6 d-1

Y= 0.60 mg VSS/mg BOD5

a. AssumingX0 = 0, the microbe mass balance yields equation 6.57:( )

( )dmds

-1-

1

k

kKS

+=

and solving for the hydraulic detention time, ,

( )[ ]( )( ) ( )[ ]

d6.0d1.0BODmg/L100-d0.1-6.1BODmg/L20

BODmg/L20100

--

1-

5

1-

5

5

sdm

s =+

=+

=dkKkS

SK

b. From equation 6.60 and 6.51,

( )( )

( )( ) LVSSmg

7.07

L

BODmg350d6.1d6.0

L

BODmg20-350

L

BODmg20100

BODmg

VSSmg0.60

-

51-

55

50s

=

+

=

+

= KS

SSSKY

X

and at steady state withX0 = 0, the effluent flux of microbial mass must equal therate of microbial mass production in the pond:

'

grVQX=

and,d

VSSkg0.212

mg01

kg

L

VSSmg07.7

m

L000,1

d

m30

63

3

=

=QX


24/27

6.31

well-mixed

aeration lagoon

Q, S0,X0Q, S,X

V, S,X, rg, rsu

l= 60m

w =5 m

h = 2 m

Q = 30 m3/d

S0 = 350 mg/L BOD5

S= 20 mg/L BOD5Ks = 100 mg/L BOD5kd = 0.10 d

-1

m = 1.6 d-1

a. The lagoons hydraulic detention time is:

( )( )( )d1.50

dm400

m2m5m60

3===

Q

V

and the steady state substrate concentration in the lagoon is:

( )( )

( )( )( )

( )( ) ( )( )[ ] LBODmg

161d08.0d1.5-1-d10.1d1.5

d08.0d1.51L

BODmg76.0

-1-

1 5

1-1-

1-5

dm

ds =+

=+

=k

kKS

Therefore, the lagoons BOD5 removal efficiency is:

( ) ( ) 52.2%100

L

BODmg336

L

BODmg161-336

100-5

5

0

0 =

== S SSEff

c. The lagoons daily oxygen demand is:

( )d

Okg70.2

mg10

kg

m

L000,1

L

BODmg161-336

d

m400- 2

63

5

3

0 =

=SSQ


25/27

6.32

Q, S0 Qs Qe,Xe, S0

WAS

Qw, Sw,Xw

RAS

Qr, Sr,Xr

V,X, SP.C.

Act. Sludge Basin

S.C.

Q = 0.300 m3/s X= 2,100 mg VSS/L Xr= 10,000 mg VSS/L

c = 9.0 d S0 = 220 mg BOD5/L

dVSSmg

BODmg0.52 5

=

MF

a. Based on the definition of the food to microbe ratio,

( )3

5

5

3

0 m5,220

L

VSSmg100,2

dVSSmg

BODmg0.52

d

s606024

L

BODmg220

s

m0.300

=

=X

MF

QSV

b. Noting thatXr=Xw, the cell retention time can be used to calculate the WASflow rate.

( )

( ) dm

122

L

VSSmg000,10d9.0

L

VSSmg100,2m5220

3

3

wc

w =

==X

VXQ

c. From a hydraulic flow balance around the activated sludge basin, recycle lineand secondary clarifier,

s

m0.299

s606024

d

d

m122-

s

m0.300-

333

we =

== QQQ


26/27

d. First, write a microbial mass balance around the secondary clarifier (S.C.).

QsX= QeXe + (Qr+ Qw)Xw

Noting thatXe 0 and Qr= Qs Qe - Qw, simplify and solve forQs.

( ) sm

0.378

L

VSSmg2,100-10,000

L

VSSmg000,10

s

m0.299

-

3

3

w

wes =

==XX

XQQ

and the hydraulic detention time is then:

( )d0.160

d

s606024

s

m

0.378

m5220

3

3

s

=

==

Q

V


27/27

6.33 Q, C0 Qe, Ce

Qs, Cs

Q = 20 MGDQs = 0.070 MGD

C0 = 800 mg TS/L

Ce = (1 0.18)C0 = 656 mg TS/L

a. A mass balance around the primary clarifier yields:

( )( )

( )( ) ( )

+

=+=L

TSmg8000.82

L

TSmg8000.82-800

MGD0.070

GD20- ee0

s

s CCCQ

QC

L

TSmg41,800s =C

b. The mass of solids removed annually is:

( )

==

yr

s365606024

mg10

kg

L

mg800,41

m

L1,000

MGD

sm0.04381

MGD0.07063

3

ss tCQm

yearpertonne4,040

yr

TSkg104.04 6 ==m

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66865-Chptr6PrblmSolns