Environmental treatment Solved problems

The US Primary Drinking Water Standard requires that nitrate in drinking water should not exceed 10 mg/l as N.. A completely mixed reactor has a volume of 40 m3 and receives a pollutant

2-Calcium 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)

mg

l mg solution

the of molarity

the

Thus

weight molecular

l mass l

weight molecular

mass l

moles

Molarity

/10

/1040

/400,

)/()

/(

#

2 3

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

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:

l meq CaCO

mg

meq l

CaCO mg

l meq

50

185)/(

65.36)76.10)(

min60)(

2.264()

1min

15()/

/

2 3

2 2

m

ft hr

gallon

m ft

gallon m

ml g

mg ml

g g

.44(

ml g

mg ml

g g

ion Concentrat rate

flow Volumetric rate

flow

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?

million

l l

mg day

gallon

million

d

lb

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

31 46

) /

( )

/ (

4 4

4 4

=

× +

PO of mole per P of mass PO

as l mg ion concentrat P

as l mg ion

Concentart

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

10 5 3 ) 16 3 14 (

14 5

15

) /

( )

/ (

3 3

3 3

<

=

× +

NO of mole per N of mass NO

as l mg ion concentrat N

as l mg

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 Oil and

grease (mg/l)

BOD (mg/l) (mg/l)SS Sample pH Oil and grease

(mg/l)

BOD (mg/l) (mg/l)SS

14 7.7 11 18 26 29 7.8 5 23 21

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

(mg/l)

SS (mg/l)

O&G (mg/l)

SS removal (%)

1 time violatio n

The table below summarizes the regulations and the existing violations

7-consuctive days 30-consuctive days

Oil and grease 20 (OK) 10 (Exceeded)

Suspended solids 45 (OK) 30 (OK)

pH 6-9 (OK all values within this range)

BOD removal >85% (violated once)

Suspended solids removal >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

Solution:

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

Mass input rate = Mass output rate

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

σ 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

2 / ( 2

) 2 / (

)1()

1

(

4

aPe aPe

Pe

ae C

C

−

−

−+

)1

()(2

x x

x

e L

v

D L

1

6 35

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

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, m 3 /s Volume, m 3 Volume-Average, m 3 Positive, m 3 Negative, m 3

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 Cumulative volume, m 3

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?

a

b

c

Flux = flow rate/ surface area

2 2

6

278min6024min

5

102

ft hr

day ft

gal day gal Flux

rate Flow area

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:

s/m056.0001

.018

)1025.0)(

10002650

(81.918

d)(

g

2 s

8.0

m v

Q A A

o

m7.43

23.14LW

L

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

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

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

6 14.2 28.1 37.0 46.7 48.7

a= data showed an increase in solids concentration

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

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

%9.45)8

8.0)(

60100()8

8.1)(

5060()8

4.2)(

4050()8

3.5)(

3040(30

R

%1.55)8

0.1)(

60100()8

5.2)(

5060()8

6.5)(

4050(40

34

50.6

37 43.6

9.0)(

60100()8

8.3)(

5060()8

2.6)(

4450

7.1)(

60100()8

8.4)(

5060()8

0.7)(

4650

2

6

21818

)75)(

6024/(

134.0105.2(

ft Depth

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

2000

4419

m m

m

m needed

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

a The salt removal efficiency

b Product water flow

Q Q

/26

478

)70000)(

500()1000(4000

=

×

2 2

3 3 3

2 2

4419

/10396

/1750

./396)

3002100(

22.0)(

m m

d m

d m Fw

Q

A

m d l kPa

kPa m d

l p

ampere milli

current

density current

N

density

current

168750)

7575()30(

30075.0400400

10sec60602484

.0

75.0/075.0/500sec/

.500,9675

number

n

m

l d

n

l eq d

m eq

amp amp

2 2

/4.15

24500

)1.128(5

2

1.128128145

)75.168)(

5.4(

m cents day

h m

day kw

kwh

cents

Cost

kw watts

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

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

incomplete water analysis

*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

12 3

14

109.61045

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

(g)

Solution volume (ml)

C equilibrium(mg/l)

(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:

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

P N

P P P

N

SSR N

AIC c

−

+++++

Model

11.47

15.092

18

Linear

-24.55

0.0833

28

Freundlich

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

2 A wastewater having an organic contaminant with a concentration of 250 mg/l is to be treated by a fixed bed activated carbon at a flow rate of 120 m3/d The allowable effluent concentration is not to exceed 30 mg/l Breakthrough data shown in the table below have been obtained from a laboratory column

The laboratory column has the following characteristics: diameter= 10 cm, length= 1.20

m, weight of activated carbon = 3.0 and an applied flow rate = 15 liter/hr

a What is the volume and mass of carbon that can be used in a field column?

b Use the scale-up approach to determine the volume of wastewater that can be treated in the field with the mass of activated carbon determined in part a

column

Q

V Q

V

)(

hr d

d m

V hr

l

l

))24/)(

/120(()

So, Vcolumn in the field= 3.14 m3

Since the bulk density in the laboratory and the field are the same, the mass of activated carbon in the field is calculated from:

column

carbon lab

column

carbon

V

M V

M

)(

)

field

carbon lab

m

M m

kg

)14.3()10

b Plot effluent concentration (C) versus volume of water out of the laboratory column as shown in the figure below From the figure, the volume of water treated in the laboratory until the point of an allowable concentration of 30 mg/l is about 2500 Liters

Using treated lab treated field

M

V M

V

)(

)

Given that M lab=3 kg, Vtreated in the lab = 2500 Liters (from graph), and M filed=1000

kg, then Vtreated in the field will be 833,333 liters or 833 m3

SP #9: Disinfection

1 Results of a chlorine demand test on a raw water are as follows:

Chlorine dosage (mg/l) NH3, mg/l Residual chlorine after

a Sketch a chlorine demand curve showing also the changes in NH3 level

b What is the breakpoint chlorine dosage?

c What is the chlorine demand at a dosage of 1.5 mg/l?

d What chlorine dose is necessary to provide free available residual chlorine of 0.6 mg/l?

Solution:

a The chlorine demand curve is shown below

b Breakpoint chlorine dosage = 1.2 mg/l (see figure above)

c Chlorine demand at an applied dose of 1.5 is = 1.5-0.45=1.05 mg/l (see figure above)

d Chlorine dose to provide a free residual of 0.6 mg/l is 1.7 mg/l (see figure above)

2 Inactivation of coliforms using chlorination is affected by chlorine dose and contact time as shown in the following equation

b

c

d NH3 depletion curve