SP #1: Review

1. What is the chemical formula for carbonate, bicarbonate, sulfate, calcium hydroxide,

and potassium phosphate?

Solution:

Carbonate: CO32Bicarbonate: HCO3Sulfate: SO42Calcium hydroxide: Ca(OH)2

Potassium phosphate: K3(PO4)3

2. What are the radicals that cause alkalinity? (Problem 1.6 page 17 of Reynolds and

Richards)

Solution:

Radicals that cause alkalinity are carbonate (CO32-), bicarbonate (HCO3-), and hydroxide

(OH-).

3. If a water has a pH= 8.0, what will be the concentration of H+ in mg/l.

Solution:

pH= -log[H+], where [H+] is the concentration of H+ in mole/l.

Thus, [H+]= 10-8 mole/l

Since the atomic weight of H+ =1.0 g/mol, then the concentration of H+ = 10-8 g/l = 10-5

mg/l.

4. What is the molarity of a solution that contains 400 mg/l calcium?

Solution:

# moles (mass / molecular weight )

(mass / l )

=

=

l

l

molecular weight

400 mg / l

Thus, the molarity of the solution =

= 10 −2 mol / l

3

40 × 10 mg / mol

Note that the atomic weight of Ca is 40 g/mol.

Molarity =

5. A water has a hardness of 185 mg/l as CaCO3. What is the hardness expressed in

meq/l? (Problem 1.7 page 17 of Reynolds and Richards)

Solution:

Hardness (meq / l ) =

185 mg CaCO3

meq

×

= 3.7 meq / l

l

50 mg CaCO3

6. Covert a flux of 15 gallon/min/ft2 to m3/hr/m2.

Solution:

15 gallon 1

60 min 10.76 ft 2

m3

Flux (m / hr / m ) = (

⋅ 2 )×(

)(

)(

) = 36.65

min

264.2 gallon

hr

ft

m2

7. (Problem 4.5 page 92 of Reynolds and Richards). Note you need to solve problem 4.4

first to find the total solids concentration.

3

2

Solution

From problem 4.4, the total solids concentration is determined as

44.6484 g − 44.6420 g 1000 mg 1000 ml

TS (mg / l ) = (

)×

×

= 640 mg / l

10 ml

g

l

a. From the information given in problem 4.5, the total suspended solids concentration is

17.5504 g − 17.5482 g 1000 mg 1000 ml

TSS (mg / l ) = (

)×

×

= 22 mg / l

100 ml

g

l

b. Total dissolved solids is now calculated as:

TDS = TS- TSS = 640-22 = 618 mg/l

8. The mass flow rate of a contaminant entering a treatment plant is calculated as:

Mass flow rate = Volumetric flow rate × Concentration × conversion factor

If the unit of the volumetric flow rate is million gallon per day, the unit of concentration

is mg/l, and the unit of mass flow rate is lb/day, what is the value of the conversion

factor?

Solution:

lb million gallon mg 3785000l

lb

=

×

×

×

d

day

l

million gallon 453592.4mg

lb million gallon mg

=

×

× [ 8.344]

d

day

l

Thus, the value of the conversion factor is 8.344

9. Obtain a snap shot from Goggle Earth for a water and a wastewater treatment plant in

the UAE.

Solution:

The top aerial photo below is for Al Fujairah Desalination Plant which is a water

treatment plant) while the photo at the bottom is for Al Aweer Wastewater Treatment

Plant in Dubai.

SP#2: Water Quality

1. A solution has 46 mg/l of phosphate. What is the concentration of phosphate in mg/l as

P?

Solution:

Concentartion (mg / l as P) = concentration (mg / l as PO4 ) ×

= 46 ×

mass of P per mole of PO 4

mass of PO 4 per mole of PO 4

31

= 15

(31 + 4 ×16)

2. (Modified Problem 4.3 page 92 of Reynolds and Richards)

A city has a present population of 53,678, and the average water consumption is 162

gal/cap/day. A new water treatment plant is to be built to serve the city for the

coming 10 years. The expected population 10 years from now is 65,300

(a) What is the per capita water consumption 10 years from now?

(b) What is the design capacity of the water treatment plant?

Solution:

(a) Percent increase in population= [(65300-53678)/53678]*100% =21.6%

Per capita water consumption= 162 (100%+ 10%*21.6%) = 165.5 gal/cap/day

(b) Capacity= (165.5 gal/cap/day)* 65,300 = 10807150 gal/day = 10.81 million gal/day

3. The US Primary Drinking Water Standard requires that nitrate in drinking water should

not exceed 10 mg/l as N. A groundwater has a nitrate (NO3-) concentration of 15.5 mg/l.

If this water is to be used for drinking purposes, is there a need to reduce the nitrate

concentration to meet the US Primary Drinking Water Standard? Explain.

Solution:

Convert the concentration of nitrate in groundwater from mg/l to mg/l as N

concentartion (mg / l as N ) = concentration (mg / l as NO3 ) ×

= 15.5 ×

mass of N per mole of NO3

mass of NO3 per mole of NO3

14

= 3.5 < 10

(14 + 3 × 16)

Thus, there is no need to reduce the nitrate concentration as it is below the maximum

contaminant level.

4. Water samples from the effluent of a water treatment plant of a small town (18,000

people) were analyzed at regular intervals over a month period. The numbers of coliform

per 100 ml sample were as shown in the table below:

Sample

1

2

Coliform

per 100ml

0

0

Sample

9

10

Coliform per

100ml

1

0

Sample

17

18

Coliform per

100ml

1

0

3

4

5

6

7

8

1

0

1

2

0

0

11

12

13

14

15

16

1

0

2

5

2

1

19

20

21

22

23

24

0

0

0

1

1

2

According to the EPA regulations, are the number of samples and effluent bacterial

quality acceptable? Explain.

Solution

The population is 18,000, so the minimum number of samples should be 20. Since we

have 24 samples then the number of samples taken is OK.

Since the average number of coliforms is 0.875 which is not more than 1, then this

condition is OK.

Since the number of samples is 24>20, then 5% of the samples are allowed to have more

than 4 coliform/100 ml.

Number of samples allowed to have more than 4 coliform/100 ml= (5/100) (24)= 1.2.

Which means only 1 sample is allowed to have more than 4 coliform/100 ml. Inspection

of the table above shows that there is one sample (sample # 14) that has more than 4

coliform/100ml, which is Ok.

5. Treated wastewater samples from the effluent of a wastewater treatment plant were

analyzed at regular intervals. The influent of the plant receives waste that has a BOD of

300 mg/l and suspended solids of 220 mg/l. The characteristics of the effluent are shown

in the table below for 30 successive days. From the data in the table, determine if the

wastewater treatment plant violates the NPDES discharge requirements.

Sample

pH

1

2

3

4

5

6

7

8

9

10

11

12

13

7.8

7.4

7.6

7.4

7.7

7.8

7.9

7.8

7.7

7.8

7.4

7.5

7.4

Oil and

grease

(mg/l)

12

14

11

18

21

5

7

7

8

11

13

12

14

BOD

(mg/l)

SS

(mg/l)

Sample

pH

33

36

24

48

51

25

24

18

37

23

26

21

24

19

24

20

24

18

27

23

26

21

24

28

27

33

16

17

18

19

20

21

22

23

24

25

26

27

28

7.9

7.6

7.7

7.7

8.1

8.2

7.9

7.7

7.7

7.8

7.9

7.8

7.7

Oil and

grease

(mg/l)

17

5

6

7

7

8

11

13

12

14

11

12

21

BOD

(mg/l)

SS

(mg/l)

23

26

25

21

24

28

37

33

26

31

24

18

37

24

17

15

15

22

23

26

21

24

18

37

23

26

14

15

7.7

7.8

11

14

18

27

26

31

29

30

7.8

7.3

5

15

23

36

21

24

Solution

A spreadsheet is prepared to calculate average of 7 and 30-consuctive days for BOD, SS,

and oil and grease (O&G). Also, the percent removal of BOD and SS is shown in the

spreadsheet below.

Sample

pH

BOD

(mg/l)

SS

(mg/l)

O&G

(mg/l)

1

7.8

33

19

2

7.4

36

24

3

7.6

24

4

7.4

5

7.7

6

BOD(7)

BOD(

30)

SS(7)

SS(30)

O&G(7)

O&G(30

)

BOD

removal

(%)

SS

removal

(%)

12

89.0

91.4

14

88.0

89.1

20

11

92.0

90.9

48

24

18

84.0

89.1

51

18

21

83.0

91.8

7.8

25

27

5

91.7

87.7

7

7.9

24

23

7

34.4

22.1

12.6

92.0

89.5

8

7.8

18

26

7

32.3

23.1

11.9

94.0

88.2

9

7.7

37

21

8

32.4

22.7

11.0

87.7

90.5

10

7.8

23

24

11

32.3

23.3

11.0

92.3

89.1

11

7.4

26

28

13

29.1

23.9

10.3

91.3

87.3

12

7.5

21

27

12

24.9

25.1

9.0

93.0

87.7

13

7.4

24

33

14

24.7

26.0

10.3

92.0

85.0

14

7.7

18

26

11

23.9

26.4

10.9

94.0

88.2

15

7.8

27

31

14

25.1

27.1

11.9

91.0

85.9

16

7.9

23

24

17

23.1

27.6

13.1

92.3

89.1

17

7.6

26

17

5

23.6

26.6

12.3

91.3

92.3

18

7.7

25

15

6

23.4

24.7

11.3

91.7

93.2

19

7.7

21

15

7

23.4

23.0

10.6

93.0

93.2

20

8.1

24

22

7

23.4

21.4

9.6

92.0

90.0

21

8.2

28

23

8

24.9

21.0

9.1

90.7

89.5

22

7.9

37

26

11

26.3

20.3

8.7

87.7

88.2

23

7.7

33

21

13

27.7

19.9

8.1

89.0

90.5

24

7.7

26

24

12

27.7

20.9

9.1

91.3

89.1

25

7.8

31

18

14

28.6

21.3

10.3

89.7

91.8

26

7.9

24

37

11

29.0

24.4

10.9

92.0

83.2

27

7.8

18

23

12

28.1

24.6

11.6

94.0

89.5

28

7.7

37

26

21

29.4

25.0

13.4

87.7

88.2

29

7.8

23

21

5

27.4

24.3

12.6

92.3

90.5

30

7.3

36

24

15

27.9

28.2

24.7

23.6

12.9

11.4

88.0

89.1

Regulation

6-9

45

30

45

30

20

10

Status

OK

OK

OK

OK

OK

OK

Not OK

>85%

2 times

violatio

n

>85%

1 time

violatio

n

The table below summarizes the regulations and the existing violations

7-consuctive days

30-consuctive days

Oil and grease

BOD

Suspended solids

pH

BOD removal

Suspended solids removal

20 (OK)

10 (Exceeded)

45 (OK)

30 (OK)

45 (OK)

30 (OK)

6-9 (OK all values within this range)

>85% (violated once)

>85% (violated twice)

SP # 3: Reactors

1. A completely mixed reactor has a volume of 40 m3 and receives a pollutant at a

concentration of 40 mg/l with a flow rate of 10 m3/d. The pollutant leaves the reactor at a

concentration of 5 mg/l. Answer the following questions:

(i) What is the influent concentration?

(ii) What is the effluent concentration?

(iii) What is the concentration inside the reactor?

(iv) Is the pollutant conservative?

(v) What is the incoming mass flow rate of the pollutant

Solution:

(i) 40 mg/l

(ii) 5 mg/l

(iii) 5 mg/l because the reactor is completely mixed

(iv) No, because the concentration dropped which means the pollutant decayed.

(v) The incoming mass flow rate = (40 mg/l) (10 m3/d) (1000 l/m3)

= 400,000 mg/d

= 0.4 kg/day

2. In the schematic shown below two streams (1 and 2) are joined to form a single

stream (3) with a suspended solids concentration (SS3) of 2500 mg/l. What is the

value of Q2 in million gallons per day (mgd)? Assume complete mixing at the joint

point.

Q1=2 mgd

SS1= 200 mg/l

SS3= 2500 mg/l

Q2=? mgd

SS2= 12000 mg/l

Solution:

Apply mass balance at the joint point, assuming no decay:

Mass input rate = Mass output rate

Q1*SS1 + Q2*SS2 = (Q1 + Q2)*SS3

(2)(200) + Q2(12000) = (2 + Q2)*2500

Q2= 0.484 mgd

3. A treatment unit is to be designed to remove 90% of a contaminant by a reaction

with a rate constant of 3.0 per day. Calculate the detention times required in

completely-mixed and in a plug-flow reactor.

Solution:

For a completely mixed reactor:

Co

1 + k ( V / Q)

Since (Co-C)/Co= 0.9, then C/Co=0.1.

Now solve for the detention time (V/Q) using k =3

V/Q= 3 days

C=

For a plug-flow reactor

C

= e − k (V / Q )

Co

Using C/Co=0.10 and k = 3 per day, V/Q will be 0.77 day.

4. The data presented in the following table show the effluent concentration (C) of a

dispersed plug flow reactor resulting from injecting a pulse of an ideal tracer into the

influent and measured effluent concentrations versus time after injection. Determine the

actual retention time (min) and the dispersion number of the reactor.

t, min

0

15

30

45

60

75

90

105

120

135

150

180

210

240

270

300

330

360

C, ug/L

0

0

0

3.5

16.5

46.5

72

89

95

88

78.2

55.2

33.5

20

12.1

7.5

4.6

2.6

Solution

t, min

0

15

30

45

60

C, ug/L

0

0

0

3.5

16.5

tC

0

0

0

157.5

990

t2C

0

0

0

7087.5

59400

75

90

105

120

135

150

180

210

240

270

300

330

360

Sum

46.5

72

89

95

88

78.2

55.2

33.5

20

12.1

7.5

4.6

2.6

624.2

3487.5

6480

9345

11400

11880

11730

9936

7035

4800

3267

2250

1518

936

85212

261562.5

583200

981225

1368000

1603800

1759500

1788480

1477350

1152000

882090

675000

500940

336960

13436595

The actual retention time:

t=

∑t c

∑c

i

i

85212

= 136.5 min

624.2

=

i

The dispersion number (D/vxL) can be found as follows:

σ2 =

∑t C

∑C

2

i

i

− t 2 = 2890.05

i

σ θ2 =

σ2

= 0.15508

t2

Using

σ θ2 = 2

D

D 2

− 2(

) (1 − e −v x L / D )

vx L

vx L

Solve for D/vxL using the graph of σ θ2 versus D/vxL. Thus, D/vxL is about 0.0847.

5. What is the effluent concentration of a contaminant with a retention time of 6 hrs

injected into a dispersed plug-flow reactor at a concentration of 75 mg/l. The axial

velocity and length of the reactor are 2 cm/hr and 80 cm, respectively. The dispersion

coefficient is 140 cm2/hr. Assume a first-order reaction with a rate constant of 0.35 hr-1.

Solution:

C

4ae ( Pe / 2)

=

Co (1 + a ) 2 e ( aPe / 2 ) − (1 − a ) 2 e ( − aPe / 2)

and

The Peclet number = Pe= vxL/D= 1.142

4k t R

a = 1 +

Pe

0.5

a = (1 +

4 × 0.35 × 6 0.5

) = 2.89

1.142

C

= 0.26

Co

⇒ C = 19.7 mg/l

SP #4: Equalization and Filtration

1. The flow (every two hours) at an industrial wastewater treatment plant is as given in

the following table. The wastewater treatment plant does not have an equalization

tank. You are asked to design an equalization tank for the plant. What is the design

capacity of the tank? Solve the problem by the spreadsheet and graphical methods

and compare the answer obtained by the two methods.

Time

0-2

2-4

4-6

6-8

8-10

10-12

12-14

14-16

16-18

18-20

20-22

22-0

Q (m3/s)

0.5

2

4.5

3.5

2.5

2

1.5

1

0.75

0.5

1.5

4

Solution:

A. The spreadsheet method:

The table below shows that the capacity of the tank is 46200 m3. The design capacity is

20% more which is 55440 m3.

Time

Q, m3/s

Volume, m3

Volume-Average, m3

0-2

2-4

4-6

6-8

8-10

10-12

12-14

14-16

16-18

18-20

20-22

22-0

0.5

2

4.5

3.5

2.5

2

1.5

1

0.75

0.5

1.5

4

Average=

3600

14400

32400

25200

18000

14400

10800

7200

5400

3600

10800

28800

14550

-10950

-150

17850

10650

3450

-150

-3750

-7350

-9150

-10950

-3750

14250

Sum=

Positive, m3

Negative, m3

-10950

-150

17850

10650

3450

-150

-3750

-7350

-9150

-10950

-3750

14250

46200

-46200

B. The graphical method

The table below shows the cumulative volume versus time. The data are plotted as shown

in the figure below. Tangents are drawn parallel to the average cumulative pumping rate

line. The capacity of the tank is a + b+ c = 12,000+ 20,000 + 14,000 = 46,000 m3. So the

design capacity is 46,000×1.2= 55,200 m3

Time

0

2

4

6

8

10

12

14

16

18

20

22

24

Cumulative volume, m3

0

3600

18000

50400

75600

93600

108000

118800

126000

131400

135000

145800

174600

c

b

a

2. The surface area of a granular media filter is to be designed such that the filtration flux

is 5 gpm/ft2. If the received flow rate is 2 million gallon per day, what would be the

surface area of the filter?

Solution:

Flux = flow rate/ surface area

2 × 10 6 gal

Flow rate

day

day

Surface area =

=

×

= 278 ft 2

5 gal

Flux

24hr ⋅ 60 min

2

min⋅ ft

SP #5: Settling

1. A grit chamber to collect sand particles (>0.25mm) is to be designed. The flow rate

through the chamber is 0.8m3/s, its depth is 2.5m and its width is 3.0m. What is the design

length of the chamber?

Solution:

g ( ρ s − ρ )d 2 9.81(2650 − 1000)(0.25 × 10 −3 ) 2

vs =

=

= 0.056 m / s

18µ

18 × 0.001

Q

Q

0.8

v s = vo =

⇒ As =

=

= 14.23 m 2

As

v s 0.0562

14.23

As = L × W ⇒ L =

= 4.7 m

3

2. The shown curve was obtained from a settling test in a 0.4-m cylinder. The initial

solids concentration (Co) was 2400 mg/l. Determine the surface area required for an

underflow concentration (Cu) of 12,000 mg/l with an inflow rate of 1200 m3/day.

Solution:

Ac:

Determine the absolute value of the slope of the early part of the curve to find vo.

vo= (0.4-0.36)/(10-0)=0.014 m/min.

Thus the area needed for clarification = Ac= Q/vo= 59.5 m2.

At:

To determine the area for thickening:

Draw tangents to the early and late parts of the settling curve and follow the procedure in

the handouts to construct the bold line shown on the curve.

Determine Hu: HoCo=HuCu, thus Hu= 0.08 m

Determine tu using the value of Hu and the bold line, thus tu is about 35 minutes (see

figure).

At=Qtu/Ho, thus At= 72.9 m2.

The design surface area is therefore 72.9 m2.

3. This is Problem 9.3 page 277 of Richards and Reynolds after modification.

A batch settling test has been performed on an industrial wastewater having an initial

suspended solids of 597 mg/l to develop criteria for the design of a primary clarifier. The

test column was 5 in. in diameter and 8 ft high, and sampling ports were located 2, 4, 6,

and 8 ft from the water surface in the column. The suspended solids remaining after

various sampling times are given in the table below. The wastewater flow is 2.5 mgd.

Determine the surface area of the settling tank for a 60% removal efficiency.

Depth (ft)

2

4

6

8

10

394

460

512

1018

20

352

406

429

1142

Time (min)

30

243

337

376

1208

45

182

295

318

1315

Solution:

Find the percent removal of suspended solids as shown in the table below:

Depth (ft)

2

4

10

34.0

22.9

Time (min)

20

30

41.0

59.3

32.0

43.6

45

69.5

50.6

60

75.2

63.8

60

148

216

306

1405

6

14.2

28.1

37.0

46.7

8

a

a

a

a

a= data showed an increase in solids concentration

48.7

a

Plot lines of equal percent removal as shown in the curve below

0

2

Depth

4

100%

34

41

59

69.5

75.2

22.9

32

43.6

50.6

63.8

14.2

30

%

28.1

10

20

6

37

40

%

46.7

50

%

60

%

48.7

8

30

40

50

60

70

Time

Find the percent removal at four different times, say 25, 37, 50, and 60 minutes

0

100%

2

Depth

4

30

%

6

60

%

50

%

40

%

8

10

20

30

40

50

60

70

Time

5.3

2.4

1.8

0.8

) + (50 − 40)( ) + (60 − 50)( ) + (100 − 60)( ) = 45.9%

8

8

8

8

5.6

2.5

1.0

R37 = 40 + (50 − 40)( ) + (60 − 50)( ) + (100 − 60)( ) = 55.1%

8

8

8

R25 = 30 + (40 − 30)(

6.2

3.8

0.9

) + (60 − 50)( ) + (100 − 60)( ) = 65.7%

8

8

8

7.0

4.8

1.7

R60 = 46 + (50 − 46)( ) + (60 − 50)( ) + (100 − 60)( ) = 69.2%

8

8

8

R50 = 44 + (50 − 44)(

Plot the removal efficiency versus time and from the curve determine the time to achieve

60% removal. The time will be 43 min. For design multiply the time by 1.75. Thus, the

design detention time will be 75 min. Now, find the surface area using:

Q× t

( 2.5 × 10 6 × 0.134 /(24 × 60)(75)

As =

=

= 2181 ft 2

Depth

8

Note: 0.134 is a conversion factor from gallon to ft3.

SP # 6: Desalination

1. A reverse osmosis unit is to demineralize 1750 m3/d of tertiary treated effluent.

Pertinent data are as follows: mass transfer coefficient = 0.22 liter/{(d-m2) (kPa)},

pressure difference between the feed and the product water = 2100 kPa, osmotic pressure

difference between the feed and the product water = 300 kPa, and membrane area per unit

volume of equipment = 2000 m2/m3. Determine

a. The membrane area required.

b. The space required for the equipment, m3.

Solution:

a.

0.22l

(2100 − 300)kPa = 396 l / d .m 2

2

d .m kPa

1750 m 3 / d

Q

A=

=

= 4419 m 2

Fw 396 × 10 −3 m 3 / d .m 2

Fw = K (∆p − ∆ψ ) =

b. Space needed =

4419 m 2

= 2.2 m 3

2

3

2000 m / m

2. An electrodialysis stack is to be used to partially demineralize 500 m3/d of brackish

water so that it can be used by an industry. The raw water has a TDS of 4000 mg/l and

the product water must not have more than 1000 mg/l. Given the followings:

The normality of the raw water is 0.075 eq/l.

The membrane dimensions are 75×75 cm.

Stack resistance= 4.5 ohms.

Current efficiency = 84%

Ratio of current density to normality= 400 (with current density as milli ampere/cm 2)

Power cost= 2.5 cents/ kwh.

Brine TDS concentration= 70 g/l.

Determine

a. The salt removal efficiency.

b. Product water flow.

c. Number of membranes.

d. Power cost per m3.

Solution:

a. Salt removal efficiency =

4000 − 1000

× 100% = 75%

4000

b.

500 × 4000 = Q(1000) + (500 − Q)(70000)

⇒ Q = 478.26 m 3 / d

c.

current density

= 400 ⇒ current density = 400 × 0.075 = 30

N

⇒ current = (30) × (75 × 75) = 168750 milli ampere

I=

FQNE r

nE c

168.75 amp =

96,500 amp. sec/ eq × 500 m 3 / d × 0.075 eq / l × 0.75

d

10 3 l

×

× 3

n × 0.84

24 × 60 × 60 sec m

⇒ n ≅ 222

⇒ number of membranes = n − 1 = 221

d.

Power = RI 2 = (4.5)(168.75) 2 = 128145 watts = 128.1 kw

2.5 cents

day

24h

Cost =

× (128.1kw) ×

×

= 15.4 cents / m 3

3

kwh

day

500m

SP #7: Precipitation

Lime-soda ash is to be used to soften a water with the following composition: CO2=

8 mg/l, Ca= 110 mg/l, Mg= 24.4 mg/l, Na= 11.5 mg/l, K= 19.6 mg/l, HCO3= 200

mg/l as CaCO3, SO4= 86.4 mg/l, and Cl= 45.5 mg/l.

a. Draw a meq/l bar graph for the raw water.

b. Determine the amount of lime and soda ash needed for excess lime softening.

Assume the practical limits of hardness removal for Ca is 30 mg/l and for Mg is 10

mg/l as CaCO3.

c. Determine the pH of the water after softening but before recarbonation.

d. Determine the concentration of sodium in the softened water.

Solution:

a.

Constituent

CO2

Ca

Mg

Na

K

HCO3

SO4

Cl

0.36

Conc.

8 mg/l

110 mg/l

24.4 mg/l

11.5 mg/l

19.6 mg/l

200 mg/l as CaCO3

86.4 mg/l

45.5 mg/l

0

CO2

M.Wt.

44

40

24.4

23

39

100

96

35.5

5.5

7.5

Ca

Mg

SO4

HCO3

0

4.0

Eq. #

2

2

2

1

1

2

2

1

Na

Cl

5.8

…

…

8.0

Eq. Wt.

22

20

12.2

23

39

50

48

35.5

meq/L

0.36

5.5

2.0

0.5

0.5

4.0

1.8

1.28

8.5

K

…

7.08

Note that there are some negative ions that do not appear in the bar graph due to

incomplete water analysis.

b.

1 meq/l of

1 CO2

1 Ca(HCO3)2

1 Mg(HCO3)2

1 Ca-noncarbonate

1 Mg-noncarbonate

Excess

Total

meq/l needed from

lime

1

1

2

0

1

1.25

Soda ash

0

0

0

1

1

Actual

conc.

present

0.36

4.0

0

1.5*

2**

Actual meq/l needed

from

lime

0.36

4.0

0

0

2

1.25

7.61

Soda ash

0

0

0

1.5

2

3.5

*

1.5=5.5-4.0 (This is the concentration of Ca that does not have alkalinity)

2=7.5-5.5 (This is the concentration of Mg that does not have alkalinity).

**

Amount of lime needed in mg/l

= (7.61 meq/l)× (37mg/meq) = 281.6

Amount of soda ash needed in mg/l = (3.5 meq/l)× (53 mg/meq) = 185.5

c. The concentration of OH after treatment includes that from excess lime (1.25 eq/l) and

that from the equilibrium dissolution of Mg(OH)2 which is 0.2 meq/l. So the total OH is

1.45 meq/l which equals to 1.45 mmol/ = 1.45*10-3 mol/l.

10 −14

[H + ] =

= 6.9 × 10 −12

−3

1.45 × 10

pH= -log[H+]= 11.16

d. The concentration of Na= original concentration in water plus the contribution from

the addition of soda ash

Na= 0.5 + 3.5= 4 meq/l = 4meq/l × 23mg/meq= 92 mg/l

SP #8: Sorption

1. A wastewater that has an organic compound at 10 mg/l is to be treated by granular

activated carbon. A batch isotherm test has been performed in the laboratory and the

following results were obtained:

Bottle

Carbon mass

Solution volume

C equilibrium

(g)

(ml)

(mg/l)

1

0.12

100

5.5

2

0.22

100

2.7

3

0.33

100

1.25

4

0.38

100

0.98

5

0.43

100

0.56

6

0.54

100

0.3

7

0.65

100

0.15

8

0.82

100

0.11

Determine

(a) The most appropriate sorption isotherm model that describe these data.

(b) Use the model selected in part (a) to find the carbon mass required to treat a batch of

200 m3 of this wastewater in the field such that the treated water has a phenol

concentration of no more than 1 mg/l.

Solution:

(a) Calculate S for each bottle using CoV=CV+mS and then plot graphs of S versus C

(linear model), logS versus logC (Freundlich model), and 1/S versus 1/C (Langmuir

model).

Find the best fit line for each case along with the coefficient of determination (R2). Note

that for the case of the linear model, the data were fit to a line with zero-intercept since

the linear model takes the form S=KC.

From the above, the Freundlich model (logS vrs logC) has the highest R2 value. But since

the model contains two parameters and the linear model contains one parameter, then to

judge which model is more suitable we need to use the corrected Akaike Information

Criteria (AICc). The AICc is given by:

2( P + 1)( P + 2)

SSR

AICc = N ln

+ 2( P + 1) +

N−P

N

Find first the SSR values associated with the linear and Freundlich models

Linear

Bottle

1

2

3

4

5

6

7

8

C

5.5

2.7

1.25

0.98

0.56

0.3

0.15

0.11

S

S

3.75

3.32

2.65

2.37

2.20

1.80

1.52

1.21

Res

5.07

2.49

1.15

0.90

0.52

0.28

0.14

0.10

SSR

Freundlich

S

Res2

2

1.75

0.69

2.25

2.16

2.82

2.31

1.90

1.22

15.09

3.94

3.23

2.61

2.44

2.08

1.75

1.44

1.32

SSR

0.036

0.008

0.002

0.004

0.012

0.002

0.005

0.014

0.083

Now calculate the AICc value for the two models

Model

N

P

P+1

SSR

AICc

Linear

Freundlich

8

8

1

2

2

3

15.09

0.083

11.47

-24.55

Since the Freundlich model has a lower AICc value than the linear model, it is more

suitable to use. Therefore, logS= 0.2787logC + 0.3892

b. Find S for a concentration of 1 mg/l using the above Freundlich equation:

logS= 0.2787log1 + 0.3892= 0.3892. Thus, S=2.45 mg/g.

Apply CoV=CV+MS

(10mg/L) (200m3*1000 L/m3)= (1 mg/L) (200m3*1000 L/m3) + M(2.45 mg/g)

Thus, M=734,636.5 g = 734.6 kg

1. What is the chemical formula for carbonate, bicarbonate, sulfate, calcium hydroxide,

and potassium phosphate?

Solution:

Carbonate: CO32Bicarbonate: HCO3Sulfate: SO42Calcium hydroxide: Ca(OH)2

Potassium phosphate: K3(PO4)3

2. What are the radicals that cause alkalinity? (Problem 1.6 page 17 of Reynolds and

Richards)

Solution:

Radicals that cause alkalinity are carbonate (CO32-), bicarbonate (HCO3-), and hydroxide

(OH-).

3. If a water has a pH= 8.0, what will be the concentration of H+ in mg/l.

Solution:

pH= -log[H+], where [H+] is the concentration of H+ in mole/l.

Thus, [H+]= 10-8 mole/l

Since the atomic weight of H+ =1.0 g/mol, then the concentration of H+ = 10-8 g/l = 10-5

mg/l.

4. What is the molarity of a solution that contains 400 mg/l calcium?

Solution:

# moles (mass / molecular weight )

(mass / l )

=

=

l

l

molecular weight

400 mg / l

Thus, the molarity of the solution =

= 10 −2 mol / l

3

40 × 10 mg / mol

Note that the atomic weight of Ca is 40 g/mol.

Molarity =

5. A water has a hardness of 185 mg/l as CaCO3. What is the hardness expressed in

meq/l? (Problem 1.7 page 17 of Reynolds and Richards)

Solution:

Hardness (meq / l ) =

185 mg CaCO3

meq

×

= 3.7 meq / l

l

50 mg CaCO3

6. Covert a flux of 15 gallon/min/ft2 to m3/hr/m2.

Solution:

15 gallon 1

60 min 10.76 ft 2

m3

Flux (m / hr / m ) = (

⋅ 2 )×(

)(

)(

) = 36.65

min

264.2 gallon

hr

ft

m2

7. (Problem 4.5 page 92 of Reynolds and Richards). Note you need to solve problem 4.4

first to find the total solids concentration.

3

2

Solution

From problem 4.4, the total solids concentration is determined as

44.6484 g − 44.6420 g 1000 mg 1000 ml

TS (mg / l ) = (

)×

×

= 640 mg / l

10 ml

g

l

a. From the information given in problem 4.5, the total suspended solids concentration is

17.5504 g − 17.5482 g 1000 mg 1000 ml

TSS (mg / l ) = (

)×

×

= 22 mg / l

100 ml

g

l

b. Total dissolved solids is now calculated as:

TDS = TS- TSS = 640-22 = 618 mg/l

8. The mass flow rate of a contaminant entering a treatment plant is calculated as:

Mass flow rate = Volumetric flow rate × Concentration × conversion factor

If the unit of the volumetric flow rate is million gallon per day, the unit of concentration

is mg/l, and the unit of mass flow rate is lb/day, what is the value of the conversion

factor?

Solution:

lb million gallon mg 3785000l

lb

=

×

×

×

d

day

l

million gallon 453592.4mg

lb million gallon mg

=

×

× [ 8.344]

d

day

l

Thus, the value of the conversion factor is 8.344

9. Obtain a snap shot from Goggle Earth for a water and a wastewater treatment plant in

the UAE.

Solution:

The top aerial photo below is for Al Fujairah Desalination Plant which is a water

treatment plant) while the photo at the bottom is for Al Aweer Wastewater Treatment

Plant in Dubai.

SP#2: Water Quality

1. A solution has 46 mg/l of phosphate. What is the concentration of phosphate in mg/l as

P?

Solution:

Concentartion (mg / l as P) = concentration (mg / l as PO4 ) ×

= 46 ×

mass of P per mole of PO 4

mass of PO 4 per mole of PO 4

31

= 15

(31 + 4 ×16)

2. (Modified Problem 4.3 page 92 of Reynolds and Richards)

A city has a present population of 53,678, and the average water consumption is 162

gal/cap/day. A new water treatment plant is to be built to serve the city for the

coming 10 years. The expected population 10 years from now is 65,300

(a) What is the per capita water consumption 10 years from now?

(b) What is the design capacity of the water treatment plant?

Solution:

(a) Percent increase in population= [(65300-53678)/53678]*100% =21.6%

Per capita water consumption= 162 (100%+ 10%*21.6%) = 165.5 gal/cap/day

(b) Capacity= (165.5 gal/cap/day)* 65,300 = 10807150 gal/day = 10.81 million gal/day

3. The US Primary Drinking Water Standard requires that nitrate in drinking water should

not exceed 10 mg/l as N. A groundwater has a nitrate (NO3-) concentration of 15.5 mg/l.

If this water is to be used for drinking purposes, is there a need to reduce the nitrate

concentration to meet the US Primary Drinking Water Standard? Explain.

Solution:

Convert the concentration of nitrate in groundwater from mg/l to mg/l as N

concentartion (mg / l as N ) = concentration (mg / l as NO3 ) ×

= 15.5 ×

mass of N per mole of NO3

mass of NO3 per mole of NO3

14

= 3.5 < 10

(14 + 3 × 16)

Thus, there is no need to reduce the nitrate concentration as it is below the maximum

contaminant level.

4. Water samples from the effluent of a water treatment plant of a small town (18,000

people) were analyzed at regular intervals over a month period. The numbers of coliform

per 100 ml sample were as shown in the table below:

Sample

1

2

Coliform

per 100ml

0

0

Sample

9

10

Coliform per

100ml

1

0

Sample

17

18

Coliform per

100ml

1

0

3

4

5

6

7

8

1

0

1

2

0

0

11

12

13

14

15

16

1

0

2

5

2

1

19

20

21

22

23

24

0

0

0

1

1

2

According to the EPA regulations, are the number of samples and effluent bacterial

quality acceptable? Explain.

Solution

The population is 18,000, so the minimum number of samples should be 20. Since we

have 24 samples then the number of samples taken is OK.

Since the average number of coliforms is 0.875 which is not more than 1, then this

condition is OK.

Since the number of samples is 24>20, then 5% of the samples are allowed to have more

than 4 coliform/100 ml.

Number of samples allowed to have more than 4 coliform/100 ml= (5/100) (24)= 1.2.

Which means only 1 sample is allowed to have more than 4 coliform/100 ml. Inspection

of the table above shows that there is one sample (sample # 14) that has more than 4

coliform/100ml, which is Ok.

5. Treated wastewater samples from the effluent of a wastewater treatment plant were

analyzed at regular intervals. The influent of the plant receives waste that has a BOD of

300 mg/l and suspended solids of 220 mg/l. The characteristics of the effluent are shown

in the table below for 30 successive days. From the data in the table, determine if the

wastewater treatment plant violates the NPDES discharge requirements.

Sample

pH

1

2

3

4

5

6

7

8

9

10

11

12

13

7.8

7.4

7.6

7.4

7.7

7.8

7.9

7.8

7.7

7.8

7.4

7.5

7.4

Oil and

grease

(mg/l)

12

14

11

18

21

5

7

7

8

11

13

12

14

BOD

(mg/l)

SS

(mg/l)

Sample

pH

33

36

24

48

51

25

24

18

37

23

26

21

24

19

24

20

24

18

27

23

26

21

24

28

27

33

16

17

18

19

20

21

22

23

24

25

26

27

28

7.9

7.6

7.7

7.7

8.1

8.2

7.9

7.7

7.7

7.8

7.9

7.8

7.7

Oil and

grease

(mg/l)

17

5

6

7

7

8

11

13

12

14

11

12

21

BOD

(mg/l)

SS

(mg/l)

23

26

25

21

24

28

37

33

26

31

24

18

37

24

17

15

15

22

23

26

21

24

18

37

23

26

14

15

7.7

7.8

11

14

18

27

26

31

29

30

7.8

7.3

5

15

23

36

21

24

Solution

A spreadsheet is prepared to calculate average of 7 and 30-consuctive days for BOD, SS,

and oil and grease (O&G). Also, the percent removal of BOD and SS is shown in the

spreadsheet below.

Sample

pH

BOD

(mg/l)

SS

(mg/l)

O&G

(mg/l)

1

7.8

33

19

2

7.4

36

24

3

7.6

24

4

7.4

5

7.7

6

BOD(7)

BOD(

30)

SS(7)

SS(30)

O&G(7)

O&G(30

)

BOD

removal

(%)

SS

removal

(%)

12

89.0

91.4

14

88.0

89.1

20

11

92.0

90.9

48

24

18

84.0

89.1

51

18

21

83.0

91.8

7.8

25

27

5

91.7

87.7

7

7.9

24

23

7

34.4

22.1

12.6

92.0

89.5

8

7.8

18

26

7

32.3

23.1

11.9

94.0

88.2

9

7.7

37

21

8

32.4

22.7

11.0

87.7

90.5

10

7.8

23

24

11

32.3

23.3

11.0

92.3

89.1

11

7.4

26

28

13

29.1

23.9

10.3

91.3

87.3

12

7.5

21

27

12

24.9

25.1

9.0

93.0

87.7

13

7.4

24

33

14

24.7

26.0

10.3

92.0

85.0

14

7.7

18

26

11

23.9

26.4

10.9

94.0

88.2

15

7.8

27

31

14

25.1

27.1

11.9

91.0

85.9

16

7.9

23

24

17

23.1

27.6

13.1

92.3

89.1

17

7.6

26

17

5

23.6

26.6

12.3

91.3

92.3

18

7.7

25

15

6

23.4

24.7

11.3

91.7

93.2

19

7.7

21

15

7

23.4

23.0

10.6

93.0

93.2

20

8.1

24

22

7

23.4

21.4

9.6

92.0

90.0

21

8.2

28

23

8

24.9

21.0

9.1

90.7

89.5

22

7.9

37

26

11

26.3

20.3

8.7

87.7

88.2

23

7.7

33

21

13

27.7

19.9

8.1

89.0

90.5

24

7.7

26

24

12

27.7

20.9

9.1

91.3

89.1

25

7.8

31

18

14

28.6

21.3

10.3

89.7

91.8

26

7.9

24

37

11

29.0

24.4

10.9

92.0

83.2

27

7.8

18

23

12

28.1

24.6

11.6

94.0

89.5

28

7.7

37

26

21

29.4

25.0

13.4

87.7

88.2

29

7.8

23

21

5

27.4

24.3

12.6

92.3

90.5

30

7.3

36

24

15

27.9

28.2

24.7

23.6

12.9

11.4

88.0

89.1

Regulation

6-9

45

30

45

30

20

10

Status

OK

OK

OK

OK

OK

OK

Not OK

>85%

2 times

violatio

n

>85%

1 time

violatio

n

The table below summarizes the regulations and the existing violations

7-consuctive days

30-consuctive days

Oil and grease

BOD

Suspended solids

pH

BOD removal

Suspended solids removal

20 (OK)

10 (Exceeded)

45 (OK)

30 (OK)

45 (OK)

30 (OK)

6-9 (OK all values within this range)

>85% (violated once)

>85% (violated twice)

SP # 3: Reactors

1. A completely mixed reactor has a volume of 40 m3 and receives a pollutant at a

concentration of 40 mg/l with a flow rate of 10 m3/d. The pollutant leaves the reactor at a

concentration of 5 mg/l. Answer the following questions:

(i) What is the influent concentration?

(ii) What is the effluent concentration?

(iii) What is the concentration inside the reactor?

(iv) Is the pollutant conservative?

(v) What is the incoming mass flow rate of the pollutant

Solution:

(i) 40 mg/l

(ii) 5 mg/l

(iii) 5 mg/l because the reactor is completely mixed

(iv) No, because the concentration dropped which means the pollutant decayed.

(v) The incoming mass flow rate = (40 mg/l) (10 m3/d) (1000 l/m3)

= 400,000 mg/d

= 0.4 kg/day

2. In the schematic shown below two streams (1 and 2) are joined to form a single

stream (3) with a suspended solids concentration (SS3) of 2500 mg/l. What is the

value of Q2 in million gallons per day (mgd)? Assume complete mixing at the joint

point.

Q1=2 mgd

SS1= 200 mg/l

SS3= 2500 mg/l

Q2=? mgd

SS2= 12000 mg/l

Solution:

Apply mass balance at the joint point, assuming no decay:

Mass input rate = Mass output rate

Q1*SS1 + Q2*SS2 = (Q1 + Q2)*SS3

(2)(200) + Q2(12000) = (2 + Q2)*2500

Q2= 0.484 mgd

3. A treatment unit is to be designed to remove 90% of a contaminant by a reaction

with a rate constant of 3.0 per day. Calculate the detention times required in

completely-mixed and in a plug-flow reactor.

Solution:

For a completely mixed reactor:

Co

1 + k ( V / Q)

Since (Co-C)/Co= 0.9, then C/Co=0.1.

Now solve for the detention time (V/Q) using k =3

V/Q= 3 days

C=

For a plug-flow reactor

C

= e − k (V / Q )

Co

Using C/Co=0.10 and k = 3 per day, V/Q will be 0.77 day.

4. The data presented in the following table show the effluent concentration (C) of a

dispersed plug flow reactor resulting from injecting a pulse of an ideal tracer into the

influent and measured effluent concentrations versus time after injection. Determine the

actual retention time (min) and the dispersion number of the reactor.

t, min

0

15

30

45

60

75

90

105

120

135

150

180

210

240

270

300

330

360

C, ug/L

0

0

0

3.5

16.5

46.5

72

89

95

88

78.2

55.2

33.5

20

12.1

7.5

4.6

2.6

Solution

t, min

0

15

30

45

60

C, ug/L

0

0

0

3.5

16.5

tC

0

0

0

157.5

990

t2C

0

0

0

7087.5

59400

75

90

105

120

135

150

180

210

240

270

300

330

360

Sum

46.5

72

89

95

88

78.2

55.2

33.5

20

12.1

7.5

4.6

2.6

624.2

3487.5

6480

9345

11400

11880

11730

9936

7035

4800

3267

2250

1518

936

85212

261562.5

583200

981225

1368000

1603800

1759500

1788480

1477350

1152000

882090

675000

500940

336960

13436595

The actual retention time:

t=

∑t c

∑c

i

i

85212

= 136.5 min

624.2

=

i

The dispersion number (D/vxL) can be found as follows:

σ2 =

∑t C

∑C

2

i

i

− t 2 = 2890.05

i

σ θ2 =

σ2

= 0.15508

t2

Using

σ θ2 = 2

D

D 2

− 2(

) (1 − e −v x L / D )

vx L

vx L

Solve for D/vxL using the graph of σ θ2 versus D/vxL. Thus, D/vxL is about 0.0847.

5. What is the effluent concentration of a contaminant with a retention time of 6 hrs

injected into a dispersed plug-flow reactor at a concentration of 75 mg/l. The axial

velocity and length of the reactor are 2 cm/hr and 80 cm, respectively. The dispersion

coefficient is 140 cm2/hr. Assume a first-order reaction with a rate constant of 0.35 hr-1.

Solution:

C

4ae ( Pe / 2)

=

Co (1 + a ) 2 e ( aPe / 2 ) − (1 − a ) 2 e ( − aPe / 2)

and

The Peclet number = Pe= vxL/D= 1.142

4k t R

a = 1 +

Pe

0.5

a = (1 +

4 × 0.35 × 6 0.5

) = 2.89

1.142

C

= 0.26

Co

⇒ C = 19.7 mg/l

SP #4: Equalization and Filtration

1. The flow (every two hours) at an industrial wastewater treatment plant is as given in

the following table. The wastewater treatment plant does not have an equalization

tank. You are asked to design an equalization tank for the plant. What is the design

capacity of the tank? Solve the problem by the spreadsheet and graphical methods

and compare the answer obtained by the two methods.

Time

0-2

2-4

4-6

6-8

8-10

10-12

12-14

14-16

16-18

18-20

20-22

22-0

Q (m3/s)

0.5

2

4.5

3.5

2.5

2

1.5

1

0.75

0.5

1.5

4

Solution:

A. The spreadsheet method:

The table below shows that the capacity of the tank is 46200 m3. The design capacity is

20% more which is 55440 m3.

Time

Q, m3/s

Volume, m3

Volume-Average, m3

0-2

2-4

4-6

6-8

8-10

10-12

12-14

14-16

16-18

18-20

20-22

22-0

0.5

2

4.5

3.5

2.5

2

1.5

1

0.75

0.5

1.5

4

Average=

3600

14400

32400

25200

18000

14400

10800

7200

5400

3600

10800

28800

14550

-10950

-150

17850

10650

3450

-150

-3750

-7350

-9150

-10950

-3750

14250

Sum=

Positive, m3

Negative, m3

-10950

-150

17850

10650

3450

-150

-3750

-7350

-9150

-10950

-3750

14250

46200

-46200

B. The graphical method

The table below shows the cumulative volume versus time. The data are plotted as shown

in the figure below. Tangents are drawn parallel to the average cumulative pumping rate

line. The capacity of the tank is a + b+ c = 12,000+ 20,000 + 14,000 = 46,000 m3. So the

design capacity is 46,000×1.2= 55,200 m3

Time

0

2

4

6

8

10

12

14

16

18

20

22

24

Cumulative volume, m3

0

3600

18000

50400

75600

93600

108000

118800

126000

131400

135000

145800

174600

c

b

a

2. The surface area of a granular media filter is to be designed such that the filtration flux

is 5 gpm/ft2. If the received flow rate is 2 million gallon per day, what would be the

surface area of the filter?

Solution:

Flux = flow rate/ surface area

2 × 10 6 gal

Flow rate

day

day

Surface area =

=

×

= 278 ft 2

5 gal

Flux

24hr ⋅ 60 min

2

min⋅ ft

SP #5: Settling

1. A grit chamber to collect sand particles (>0.25mm) is to be designed. The flow rate

through the chamber is 0.8m3/s, its depth is 2.5m and its width is 3.0m. What is the design

length of the chamber?

Solution:

g ( ρ s − ρ )d 2 9.81(2650 − 1000)(0.25 × 10 −3 ) 2

vs =

=

= 0.056 m / s

18µ

18 × 0.001

Q

Q

0.8

v s = vo =

⇒ As =

=

= 14.23 m 2

As

v s 0.0562

14.23

As = L × W ⇒ L =

= 4.7 m

3

2. The shown curve was obtained from a settling test in a 0.4-m cylinder. The initial

solids concentration (Co) was 2400 mg/l. Determine the surface area required for an

underflow concentration (Cu) of 12,000 mg/l with an inflow rate of 1200 m3/day.

Solution:

Ac:

Determine the absolute value of the slope of the early part of the curve to find vo.

vo= (0.4-0.36)/(10-0)=0.014 m/min.

Thus the area needed for clarification = Ac= Q/vo= 59.5 m2.

At:

To determine the area for thickening:

Draw tangents to the early and late parts of the settling curve and follow the procedure in

the handouts to construct the bold line shown on the curve.

Determine Hu: HoCo=HuCu, thus Hu= 0.08 m

Determine tu using the value of Hu and the bold line, thus tu is about 35 minutes (see

figure).

At=Qtu/Ho, thus At= 72.9 m2.

The design surface area is therefore 72.9 m2.

3. This is Problem 9.3 page 277 of Richards and Reynolds after modification.

A batch settling test has been performed on an industrial wastewater having an initial

suspended solids of 597 mg/l to develop criteria for the design of a primary clarifier. The

test column was 5 in. in diameter and 8 ft high, and sampling ports were located 2, 4, 6,

and 8 ft from the water surface in the column. The suspended solids remaining after

various sampling times are given in the table below. The wastewater flow is 2.5 mgd.

Determine the surface area of the settling tank for a 60% removal efficiency.

Depth (ft)

2

4

6

8

10

394

460

512

1018

20

352

406

429

1142

Time (min)

30

243

337

376

1208

45

182

295

318

1315

Solution:

Find the percent removal of suspended solids as shown in the table below:

Depth (ft)

2

4

10

34.0

22.9

Time (min)

20

30

41.0

59.3

32.0

43.6

45

69.5

50.6

60

75.2

63.8

60

148

216

306

1405

6

14.2

28.1

37.0

46.7

8

a

a

a

a

a= data showed an increase in solids concentration

48.7

a

Plot lines of equal percent removal as shown in the curve below

0

2

Depth

4

100%

34

41

59

69.5

75.2

22.9

32

43.6

50.6

63.8

14.2

30

%

28.1

10

20

6

37

40

%

46.7

50

%

60

%

48.7

8

30

40

50

60

70

Time

Find the percent removal at four different times, say 25, 37, 50, and 60 minutes

0

100%

2

Depth

4

30

%

6

60

%

50

%

40

%

8

10

20

30

40

50

60

70

Time

5.3

2.4

1.8

0.8

) + (50 − 40)( ) + (60 − 50)( ) + (100 − 60)( ) = 45.9%

8

8

8

8

5.6

2.5

1.0

R37 = 40 + (50 − 40)( ) + (60 − 50)( ) + (100 − 60)( ) = 55.1%

8

8

8

R25 = 30 + (40 − 30)(

6.2

3.8

0.9

) + (60 − 50)( ) + (100 − 60)( ) = 65.7%

8

8

8

7.0

4.8

1.7

R60 = 46 + (50 − 46)( ) + (60 − 50)( ) + (100 − 60)( ) = 69.2%

8

8

8

R50 = 44 + (50 − 44)(

Plot the removal efficiency versus time and from the curve determine the time to achieve

60% removal. The time will be 43 min. For design multiply the time by 1.75. Thus, the

design detention time will be 75 min. Now, find the surface area using:

Q× t

( 2.5 × 10 6 × 0.134 /(24 × 60)(75)

As =

=

= 2181 ft 2

Depth

8

Note: 0.134 is a conversion factor from gallon to ft3.

SP # 6: Desalination

1. A reverse osmosis unit is to demineralize 1750 m3/d of tertiary treated effluent.

Pertinent data are as follows: mass transfer coefficient = 0.22 liter/{(d-m2) (kPa)},

pressure difference between the feed and the product water = 2100 kPa, osmotic pressure

difference between the feed and the product water = 300 kPa, and membrane area per unit

volume of equipment = 2000 m2/m3. Determine

a. The membrane area required.

b. The space required for the equipment, m3.

Solution:

a.

0.22l

(2100 − 300)kPa = 396 l / d .m 2

2

d .m kPa

1750 m 3 / d

Q

A=

=

= 4419 m 2

Fw 396 × 10 −3 m 3 / d .m 2

Fw = K (∆p − ∆ψ ) =

b. Space needed =

4419 m 2

= 2.2 m 3

2

3

2000 m / m

2. An electrodialysis stack is to be used to partially demineralize 500 m3/d of brackish

water so that it can be used by an industry. The raw water has a TDS of 4000 mg/l and

the product water must not have more than 1000 mg/l. Given the followings:

The normality of the raw water is 0.075 eq/l.

The membrane dimensions are 75×75 cm.

Stack resistance= 4.5 ohms.

Current efficiency = 84%

Ratio of current density to normality= 400 (with current density as milli ampere/cm 2)

Power cost= 2.5 cents/ kwh.

Brine TDS concentration= 70 g/l.

Determine

a. The salt removal efficiency.

b. Product water flow.

c. Number of membranes.

d. Power cost per m3.

Solution:

a. Salt removal efficiency =

4000 − 1000

× 100% = 75%

4000

b.

500 × 4000 = Q(1000) + (500 − Q)(70000)

⇒ Q = 478.26 m 3 / d

c.

current density

= 400 ⇒ current density = 400 × 0.075 = 30

N

⇒ current = (30) × (75 × 75) = 168750 milli ampere

I=

FQNE r

nE c

168.75 amp =

96,500 amp. sec/ eq × 500 m 3 / d × 0.075 eq / l × 0.75

d

10 3 l

×

× 3

n × 0.84

24 × 60 × 60 sec m

⇒ n ≅ 222

⇒ number of membranes = n − 1 = 221

d.

Power = RI 2 = (4.5)(168.75) 2 = 128145 watts = 128.1 kw

2.5 cents

day

24h

Cost =

× (128.1kw) ×

×

= 15.4 cents / m 3

3

kwh

day

500m

SP #7: Precipitation

Lime-soda ash is to be used to soften a water with the following composition: CO2=

8 mg/l, Ca= 110 mg/l, Mg= 24.4 mg/l, Na= 11.5 mg/l, K= 19.6 mg/l, HCO3= 200

mg/l as CaCO3, SO4= 86.4 mg/l, and Cl= 45.5 mg/l.

a. Draw a meq/l bar graph for the raw water.

b. Determine the amount of lime and soda ash needed for excess lime softening.

Assume the practical limits of hardness removal for Ca is 30 mg/l and for Mg is 10

mg/l as CaCO3.

c. Determine the pH of the water after softening but before recarbonation.

d. Determine the concentration of sodium in the softened water.

Solution:

a.

Constituent

CO2

Ca

Mg

Na

K

HCO3

SO4

Cl

0.36

Conc.

8 mg/l

110 mg/l

24.4 mg/l

11.5 mg/l

19.6 mg/l

200 mg/l as CaCO3

86.4 mg/l

45.5 mg/l

0

CO2

M.Wt.

44

40

24.4

23

39

100

96

35.5

5.5

7.5

Ca

Mg

SO4

HCO3

0

4.0

Eq. #

2

2

2

1

1

2

2

1

Na

Cl

5.8

…

…

8.0

Eq. Wt.

22

20

12.2

23

39

50

48

35.5

meq/L

0.36

5.5

2.0

0.5

0.5

4.0

1.8

1.28

8.5

K

…

7.08

Note that there are some negative ions that do not appear in the bar graph due to

incomplete water analysis.

b.

1 meq/l of

1 CO2

1 Ca(HCO3)2

1 Mg(HCO3)2

1 Ca-noncarbonate

1 Mg-noncarbonate

Excess

Total

meq/l needed from

lime

1

1

2

0

1

1.25

Soda ash

0

0

0

1

1

Actual

conc.

present

0.36

4.0

0

1.5*

2**

Actual meq/l needed

from

lime

0.36

4.0

0

0

2

1.25

7.61

Soda ash

0

0

0

1.5

2

3.5

*

1.5=5.5-4.0 (This is the concentration of Ca that does not have alkalinity)

2=7.5-5.5 (This is the concentration of Mg that does not have alkalinity).

**

Amount of lime needed in mg/l

= (7.61 meq/l)× (37mg/meq) = 281.6

Amount of soda ash needed in mg/l = (3.5 meq/l)× (53 mg/meq) = 185.5

c. The concentration of OH after treatment includes that from excess lime (1.25 eq/l) and

that from the equilibrium dissolution of Mg(OH)2 which is 0.2 meq/l. So the total OH is

1.45 meq/l which equals to 1.45 mmol/ = 1.45*10-3 mol/l.

10 −14

[H + ] =

= 6.9 × 10 −12

−3

1.45 × 10

pH= -log[H+]= 11.16

d. The concentration of Na= original concentration in water plus the contribution from

the addition of soda ash

Na= 0.5 + 3.5= 4 meq/l = 4meq/l × 23mg/meq= 92 mg/l

SP #8: Sorption

1. A wastewater that has an organic compound at 10 mg/l is to be treated by granular

activated carbon. A batch isotherm test has been performed in the laboratory and the

following results were obtained:

Bottle

Carbon mass

Solution volume

C equilibrium

(g)

(ml)

(mg/l)

1

0.12

100

5.5

2

0.22

100

2.7

3

0.33

100

1.25

4

0.38

100

0.98

5

0.43

100

0.56

6

0.54

100

0.3

7

0.65

100

0.15

8

0.82

100

0.11

Determine

(a) The most appropriate sorption isotherm model that describe these data.

(b) Use the model selected in part (a) to find the carbon mass required to treat a batch of

200 m3 of this wastewater in the field such that the treated water has a phenol

concentration of no more than 1 mg/l.

Solution:

(a) Calculate S for each bottle using CoV=CV+mS and then plot graphs of S versus C

(linear model), logS versus logC (Freundlich model), and 1/S versus 1/C (Langmuir

model).

Find the best fit line for each case along with the coefficient of determination (R2). Note

that for the case of the linear model, the data were fit to a line with zero-intercept since

the linear model takes the form S=KC.

From the above, the Freundlich model (logS vrs logC) has the highest R2 value. But since

the model contains two parameters and the linear model contains one parameter, then to

judge which model is more suitable we need to use the corrected Akaike Information

Criteria (AICc). The AICc is given by:

2( P + 1)( P + 2)

SSR

AICc = N ln

+ 2( P + 1) +

N−P

N

Find first the SSR values associated with the linear and Freundlich models

Linear

Bottle

1

2

3

4

5

6

7

8

C

5.5

2.7

1.25

0.98

0.56

0.3

0.15

0.11

S

S

3.75

3.32

2.65

2.37

2.20

1.80

1.52

1.21

Res

5.07

2.49

1.15

0.90

0.52

0.28

0.14

0.10

SSR

Freundlich

S

Res2

2

1.75

0.69

2.25

2.16

2.82

2.31

1.90

1.22

15.09

3.94

3.23

2.61

2.44

2.08

1.75

1.44

1.32

SSR

0.036

0.008

0.002

0.004

0.012

0.002

0.005

0.014

0.083

Now calculate the AICc value for the two models

Model

N

P

P+1

SSR

AICc

Linear

Freundlich

8

8

1

2

2

3

15.09

0.083

11.47

-24.55

Since the Freundlich model has a lower AICc value than the linear model, it is more

suitable to use. Therefore, logS= 0.2787logC + 0.3892

b. Find S for a concentration of 1 mg/l using the above Freundlich equation:

logS= 0.2787log1 + 0.3892= 0.3892. Thus, S=2.45 mg/g.

Apply CoV=CV+MS

(10mg/L) (200m3*1000 L/m3)= (1 mg/L) (200m3*1000 L/m3) + M(2.45 mg/g)

Thus, M=734,636.5 g = 734.6 kg

## One-Sample Estimation Problems (TOÁN)

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## Environmental Geology - Chapter 16: Water Pollution

## Pentachlorophenol: Toxicology and Environmental Fate

## Environmental chemistry - P1

## Environmental chemistry - P2

## Environmental chemistry - P3

## Environmental chemistry - P4

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