department of civil engineering-i.i.t. delhi cvl212...

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Department of Civil Engineering-I.I.T. Delhi

CVL212: Environmental Engineering

Additional Questions

Dr. Arun Kumar ([email protected])

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(Home Work 1 Solution

Q1. List three contaminants- of- your- concern in water, air and solid media. (1+1+1 =3 points)

Q2. Define physical, chemical and biological parameters? List names of two parameters in water and indicate their

significance. (2*3+2=8 points)

Q3. From your I.I.T. Delhi library, find out three journals in the area of environmental engineering? (3 points)

Q4. Provide estimate of your daily water consumption in different forms (activity-water consumed-water type-

water source) and provide three options to optimize your daily water consumption pattern. (3 +3 =6 points)

Q5. In 100 g/L C2H12O6 solution, do following:

1. Write a balanced reaction between organic compound and oxygen to produce CO2 and H2O.

2. Calculate number of moles of oxygen required to completely react with organic compound?

3. Calculate concentration (g/L) of oxygen required to completely react with organic compound? (This is

called theoretical oxygen demand) (2+1+1 points)

Q6. Say a raw wastewater sample from WWTP1 has 2000 mg/L 5-day BOD (due to carbon only) with reaction

constant (k=0.23/day at 20°C). Calculate ultimate BOD; amounts of BOD exerted on 1- and 15-day and comment

on observed differences between 1- and 15-day BOD values (1.5 +1+1+1.5 = 5 points)

Answer 5-day BOD = 2000 mg/L (due to carbon only); k=0.23/day

L(5)= L(0) [1-e(-k*5)

] = 2000 mg/L= L(0) [1-e(-0.23*5)

]

=> Ultimate BOD L (0) =2926 mg/L (answer)

time(day) t-day BOD

(mg/L) 1 601 15 2834 Here BOD (1day) is smaller than BOD (15 days) and BOD (15 days) is reaching towards ultimate BOD.

Q7. Balance these equations, which are generally used to determine nitrogen-based BOD, to write the final

balanced equation and also calculate number of moles of oxygen required for reacting with one mole of ammonia.

(1+1+0.5*2 points) NH3 +O2 � NO2

- + H

+ + H2O (1a)

NO2- +O2 + H

+ � NO3

- + H

+ (1b)

Answer: Balanced equation: NH3 +2O2 � HNO3 + H2O

1 mole of ammonia (i.e., 17 g) reacts with 2 moles of oxygen (i.e., 2*32=64 g). So, two moles of oxygen required

for reacting with one mole of ammonia.

Q8. In addition of using BOD, organic matter in wastewater is also expressed using chemical oxygen demand

(COD; unit: mg/L) and total organic carbon (TOC; unit: mg/L). Given that in an untreated domestic

wastewater, the BOD5/COD ratio varies between 0.4 and 0.8 and the BOD5/TOC ratio varies between 1 and

0

500

1000

1500

2000

2500

3000

3500

0 50 100 150

time(days)

BO

D(m

g/L

)

3

1.6. Using the BOD5 (20°C) value from Q6, calculate ranges of COD and TOC for the WWTP1 raw

wastewater?

Answer

5-day BOD = 2000 mg/L (due to carbon only)

COD (2000/0.8) to (2000/0.4 ) mg/L TOC (2000/1.6) to (2000/1.0 ) mg/L

Q9. What is the numerical relationship between 5-day BOD values and reaction rate constants? Qualitatively,

arrange 5-day BOD values calculated using reaction rate constants (k) = 0.50 (say k1), 0.23 (say k2), and 0.30 (say

k3) (unit: 1/day).

Answer: Assume ultimate BOD= 1000 mg/L (For calculation purposes)

k value (1/day) 5-day BOD(mg/L) (ranked from low to high values) 0.23 683 0.3 775 0.5 918

Q10. If the wastewater effluent given in Q6 has 50 mg/L ammonia also, calculate total amount of oxygen required

to oxidize both nitrogen- and carbon- based compounds biologically? (Hint: Ammonia reacts with oxygen to make

nitric acid and water.)

Answer As 1 mole of ammonia (i.e., 17 g) reacts with 2 moles of oxygen (i.e., 2*32=64 g), so 50 mg/L ammonia (i.e., 50

*0.001 g/L/(17 g/mole)=0.00294 moles/L) requires = 2/1*0.00294=0.00588 moles/L oxygen (or 0.00588*32

g/L=188 mg/L).

Total amount of oxygen required to oxidize both carbon and nitrogen

= (2926 mg/L) + (188 mg/L)

=3114 mg/L oxygen (answer)

++++++++++++++Additional Questions (Not for submission) +++++++++++++

QA1. How does carbon dioxide dissolution in lake water affect algal bloom?

Answer: When carbon dioxide dissolves in lake water it would increase carbon dioxide level in water (if it were

undersaturated). Correspondingly, it would increase algal growth, given sufficient nutrients are available. As CO2

dissolution influences solution pH as well, it might affect algal growth if pH change does not provide growing

environment to algae.

0

200

400

600

800

1000

0 0.2 0.4 0.6

Reaction constant (1/day)

5-d

ay B

OD

(mg

/L)

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CEL 212: Environmental Engineering

Second Semester 2015-2016

Home Work 2 Solution

Always write your name and entry number in all submissions. Please mention your assumptions explicitly.

Q1. Problem # 3-19 (Peavy et al. Text Book). A wastewater treatment plant disposes of its effluent in a surface

stream. Characteristics of the stream and effluent are shown below. [5+5+5 points]

Parameter wastewater stream flow (m

3/s) 0.2 5

Dissolved oxygen, mg/L 1 8 Temperature, °C 15 20.2 BOD5 at 20°C, mg/L 100 2 Oxygen consumption rate (K1 at 20°C) (1/day) 0.2 - Oxygen reaeration rate (K2 at 20°C) (1/day) - 0.3

(a) What will be the dissolved oxygen conc. in the stream after 2 days?

(b) What will be the lowest dissolved oxygen concentration as a result of the waste discharge?

(c) Also calculate the maximum BOD5 (20°C) that can be discharged if a minimum of 4.0 mg/L of oxygen

must be maintained in the stream?

Answer:

Parameter wastewater

(given)

stream (given) Wastewater and stream water

mixture

flow (m3/s) 0.2 5 =Qmixture=5+0.2=5.2 m/s

Dissolved oxygen, mg/L 1 8 DOmixture=(0.2*1+8*5)/(5+0.2)

=7.73 mg/L

Temperature, °C 15 20.2 Tempmixture=(0.2*15+20.2*5)/(5+0.2)

=20 deg C (No temp. correction

required)

BOD5 at 20°C, mg/L 100 2 BODmixture=(0.2*100+2*5)/(5+0.2)

=5.77 mg/L

Oxygen consumption rate (K1 at

°C) (1/day)

0.2 0.23 (No temp. correction required)

(assumed for stream water)

Oxygen reaeration rate (K2 at °C)

(1/day)

- 0.3 0.3 (No temp. correction required)

Ultimate BOD= Yultimate=L0 = (5-day BOD in mixture water)/[1-exp(-K1mixture*5)]

=(5.77 mg/L)/ [1-exp(-0.23*5)] =8.44 mg/L

Initial DO deficit (D0)

For 20°C stream water temperature, equilibrium concentration of oxygen =9.17 mg/L

D0 = 9.17 mg/L-7.73 mg/L = 1.44 mg/L

To get DO after 2 days in stream water after mixing, we need to calculate DO deficit after 2 days first and then

calculate DO (at 2days). DO deficit at 2 days is given by

D (t=2 days) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + D0 exp (-K2*t)

= [0.2*8.44]*[exp (-0.2*2)-exp (-0.3*2)]/ (0.3-0.2) + 1.44 exp (-0.3*2)

= [1.94]*[0.6703-0.5488]/ (0.07) + 0.7903 =3.07 mg/L

D (t=2 days) = DO saturated-DO (2day) = 3.07 mg/L

DO (2day) = 9.17-3.07=6.10 mg/L (answer for part i)

Time for critical DO deficit (tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))]

= 1/ (0.3-0.23)*ln [(0.3/0.23)*(1-1.44 *(0.3-0.23)/ (0.23*8.44))]

= 14.29*ln [1.3*(1-1.44 *0.036)] = 14.29*ln [1.23] =2.95 days

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Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc)

= (0.23/0.3)*8.44 exp (-0.23*2.95) = 6.47 *0.507 =3.28 mg/L

Dc = DOsaturated-DOcritical = 3.28 mg/L

DOcritical= 9.17-3.28=5.89 mg/L (answer for part ii)

Required minimum DO = 4.0 mg/L in stream water. As DO at critical location is 5.89 mg/L, greater than the

recommended DO level, no modification in wastewater effluent characteristics is required.

To calculate maximum BOD5 in effluent water, calculate allowable DO deficit (i.e., Dallowable)

= DOsaturated-DOminimum = 9.17-4.0 = 5.17 mg/L [Note that 5.17 mg/L DO deficit is allowable and we are having 5.89

mg/L critical DO deficit.]

Now with calculated allowable DO deficit (this is assumed to be the critical deficit now) and calculated tcritical

(assumed to be similar to previous case, i.e., 2.95 days), calculate ultimate BOD in this case. Then calculate 5-day

BOD of the mixture stream water and then calculate 5-day BOD of the effluent which will be the desired

maximum 5-day BOD value.

Dallowable (t=tcritical) =Dcritical,new

=> 5.17 = (0.23/0.30)*L0 [exp (-0.23*2.95)] = 0.77*0.51L0=0.3927L0

=> Ultimate BOD of the mixture water = L0=5.19 mg/L/ (0.3927) =13.17 mg/L

Now 5-day BOD in mixture water is calculated.

5-day BODmixture = L0 *[1-exp (-K1mixture*5)]

= (13.17 mg/L)*[1-exp (-0.23*5)] =9.0 mg/L

5-day BOD in effluent water is calculated now.

BODmixture= (5-day BODeffQeff+5-day BODstreamQstream)/ (5+0.2)

9.0 mg/L = (5-day BODeff*0.2+2*5)/ (5+0.2)

5-day BODeff *0.2+10= 9.0*5.2 = 46.8

=> 5-day BODeff = (46.8-10)/0.2= 184 mg/L (answer for part iii). This is maximum value of 5-day BOD in

wastewater effluent which can be discharged in the stream water without exceeding the minimum required

DO value of 4 mg/L.

Q2. For a given effluent-stream combination, say values of BOD reaction rate and stream reaeration rate are 0.26

and 0.42 per day, respectively and given that initial dissolved oxygen (DO) deficit is 2 mg/L with ultimate BOD of

the mixture equals to 18 mg/L, discuss the approach for calculating time (say t*) since mixing of effluent with

stream water after which DO deficit becomes 1% of the initial DO deficit? [5 points]

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Q3. The WWTP in the “AA” community discharges 10 million gallons/day of secondary effluent into a stream

“Red Cedar” whose minimum flow rate is 100 m3/s.

WWTP effluent Stream water

Temperature (°C) 20°C 15°C

BOD5 (mg/L) (at 20 °C) 200 1

Dissolved oxygen (mg/L) 2 80% of saturation level

Oxygen consumption rate (k1) (1/day) 0.3

0.7 Oxygen re-aeration rate (k2)(1/day)

Using above information, calculate the following:

1) Temperature, dissolved oxygen (DO) and BOD of the mixture.

2) Initial dissolved oxygen deficit at the place of mixing.

3) Critical oxygen deficit (Dc), time for maximum dissolved oxygen deficit (tmax) and its location from

discharge point (Xc).

4) The dissolved oxygen level and 20°C BOD5 of a sample taken at the critical point.

(Saturated oxygen concentration in stream before discharge is 10.07 mg/L at 15°C. Saturated DO at 15.7°C is 9.9

mg/L. Use temperature coefficients of 1.135 for k1 and 1.024 for k2 for applying temperature corrections for these

two constants.)

Answer: Qeffluent= 10 million gallons/day = 10*10

6 gallons/day* (3.78 liters/gallon)*(1/1000 m

3/liters)

= 3.78*104 m

3/day = (3.78*10

4)/(24*3600)=0.4375 m

3/s

DOstream = 80% of saturation level = 80/100*(10.07 mg/L) =8.056 mg/L

Using the mass balance approach for dissolved oxygen:

Dissolved oxygen in the mixture = [DOeffluentQeffluent+DOstreamQstream]/(Qeffluent+Qstream)

= [(2mg/L*0.4375m3/s)+(8.056 mg/L*100 m

3/s)]/( 0.4375m

3/s + 100 m

3/s)

= [(0.875)+(805.6)]/(100.4375) =806.475/100.4375=8.030 mg/L

Initial DO deficit (D0) at the place of mixing = 10.07 mg/L-8.03 mg/L = 2.04 mg/L

Using the mass balance approach for temperature :

Temperature in the mixture = [TeffluentQeffluent+TstreamQstream]/(Qeffluent+Qstream)

= [(20*0.4375m3/s)+(15 *100 m

3/s)]/( 0.4375m

3/s + 100 m

3/s)

= [1508.75]/(100.4375) =15.02 °C

Using the mass balance approach for 5-day BOD

5-day BOD in the mixture = [BODeffluentQeffluent+BODstreamQstream]/(Qeffluent+Qstream)

= [(200mg/L*0.4375m3/s)+(1 mg/L*100 m

3/s)]/( 0.4375m

3/s + 100 m

3/s)

= [(87.5)+(100)]/(100.4375) =187.5/100.4375=1.87 mg/L

Ultimate BOD (i.e., Yultimate or L0) = (5-day BOD in mixture water)/[1-exp(-K1mixture*5)]

=(1.87 mg/L)/ [1-exp(-0.3*5)] (assuming river water and mixture have same oxygen consumption rate)

=2.41 mg/L

K values after temperature corrections:

K1 (at 15.02 °C ) = (0.3/day) (1.135)(15.02-20)

=0.1597/day

K2 (at 15.02 °C ) = (0.7/day) (1.024)(15.02-20)

=0.6220/day

Time for critical DO deficit (tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))]

= 1/ (0.6220-0.1597)*ln [(0.6220/0.1597)*(1-2.04 *(0.6220-0.1597)/ (0.1597*2.41))]

= 2.16*ln [3.89*(1-2.04 *1.20)] = negative value (i.e., critical deficit will not happen)

Thus, Critical DO deficit (Dc)= 0

Dc = DOsaturated-DOcritical = 0

DOcritical= 9.9-0=9.9 mg/L (assuming that Saturated DO at 15.2°C is equal to that at 15.7°C, i.e., 9.9 mg/L)

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Q4. Assuming that reaeration rate (k2) given in Q1 varies between 0.23/day and 0.60/day with mean equals to

0.42/day, re-do question Q2 using these two revised estimates of k2 and comment on effect of variability in k2

values on estimate of t*.

Q5. Re-do Q1 and comment on DO values at t=0.5tc and t=2tc. For policy making, which DO value should be

considered? After mixing, what is the setback distance (i.e., downstream distance along the bank from the point of

mixing) where DO level is at least 70% of saturated DO level in river water?

Q6. Using Q1 data, if effluent water is not mixed in stream water, what is the maximum BOD concentration in

effluent to meet the stream water DO standard?

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Q1. Draw the dissolved oxygen (DO) sag curve and answer the following:

(i) Why do we get sag curve for DO in stream water?

(ii) Show equilibrium DO concentration, initial DO deficit, critical DO deficit, and time for critical

DO deficit in the oxygen sag curve.

Hint: Refer lecture notes

Q2. A water has the following analysis:

Ion type Concentration (mg/L) Ion type Concentration (mg/L)

Na+ 20 Cl

- 40

Ca2+

5 CO32-

10

Mg2+

10 SO42-

5

Calculate values of total hardness and non-carbonate hardness in mg/L as CaCO3? (1+1.5 = 2.5 points)

Answer: Only cations with bivalency contribute to hardness.

Hardness (in mg/L) as CaCO3 = M2+

(in mg/L) *[(50)/(Equivalent weight of M2+

)]

Cation Concentration

(mg/L) Equivalent weight

(g/mole) Hardness (in mg/L) as CaCO3

Ca2+ 5 20 =(5) *[(50)/(20)]=12.5

Mg2+ 10 12.2 =(10) *[(50)/(12.2)]=41

Total hardness = Hardness due to Ca2+

ions and Mg

2+ ions. = 12.5+41=53.5 mg/L as CaCO3 (answer)

For calculating non-carbonate hardness:

Anion Concentration

(mg/L) Equivalent weight

(g/mole) Hardness (in mg/L) as CaCO3

CO32-

10 =(12+3*16)/2= 30 =(10) *[(50)/(30)]=16.67

As the maximum value of carbonate hardness could be 16.67 mg/L as CaCO3 and given that total hardness is

53.5 mg/L as CaCO3, this non-carbonate hardness (minimum value)

= 53.5-16.67 =36.83 mg/L as CaCO3

Q3. Draw a schematic of water treatment plant treating IIT Delhi surface water obtained from a channel,

label different unit processes with appropriate sequence. Mention what types of parameters you would

like to measure, unit processes you would like to use and why? (2+2+1+1=6 points)


Q4. What is the basic difference between discrete and flocculent settling? Determine the settling velocity

of a spherical particle with a diameter of 100 µm and a specific gravity of 2.3 in water at 25°C? How

about settling velocity for 0.100 µm particles? (2+2+2=6 points)


Q5. Calculate different types of water demand for your hostel building floor on which you live?

Hint: Water demands for drinking water, bathing and washing clothes, cleaning water, flushing toilets,

cleaning floors, etc. (record your water usage and see how much you consume everyday and thus

produce wastewater with different water quality characteristics!)

Q6. Draw a schematic of water treatment plant treating IIT Delhi groundwater (or Laboratory

wastewater); label different unit processes with appropriate sequence. Mention what types of

parameters you would like to measure, unit processes you would like to use and why?


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Q7. Suppose IIT Delhi water has suspended solids, organic matter, microorganisms, hardness, and

dissolved CO2, which are problematic constituents. Other dissolved constituents are below problem

levels. Draw a schematic diagram of a treatment plant that will render this water potable. Identify each

unit and show points of chemical addition with their names.


Q8. Say a wastewater has calcium ions (50 mg/L), sodium ions (14.8 mg/L), 100 mg/L carbonate ions,

2000 mg/L 5-day BOD (due to carbon only with reaction constant, k=0.23/day at 20°C) and 50 mg/L

ammonia. Answer the following:

(i) Define carbonate hardness and calculate its value?

Calculate total amount of oxygen required to oxidize both nitrogen- and carbon- based

compounds biologically? [10+10=20 points]

(ii) Comment on sequence of exertion of both types of BOD. [10 points]

Hint: See solution from Q2 and refer lecture notes and HW1

Q9. A stream (flow rate=3.34 Km/h) contains no BOD, but has an oxygen deficit of 6 mg/L. Seventeen

kilometers downstream this deficit has been reduced to 4 mg/L. The stream conditions are uniform

throughout the stretch. What information do you need to calculate DO deficit 59 Km downstream from

the original point? Provide reasons. [10 points]


Q10. Say I am considering volunteering for a one-way trip to Mars [I can send weekly updates, though!],

which requires my commitment on optimizing/reducing my daily water consumption. One option is to

use treated wastewater for my non-drinking activities. Which two parameters do you suggest me to

monitor before I accept the condition and why? [10 points]


Q11. Look at the following BOD plot for three different water samples (A,B, and C) with identical

ultimate BOD = 2926.70 mg/L. Comment on their oxygen consumption rates and order these samples in

increasing order of oxygen consumption rate values. [20 points] Answer:

Note effect of k values on BOD values for different times. For example, for 5-day BOD: 5-day BOD(k1=0.10/day)

> 5-day BOD(k2=0.23/day) > 5-day BOD(k3=0.3/day). So, the more the reaction rate constant, the more oxygen

will be consumed in a given time, here the duration is 5-days. Remember that BOD vs. time curve where high

oxygen consumption rate will have higher curve compared to others (see curve for k=0.3/day compared to

k=0.1/day curve).

0

500

1000

1500

2000

2500

3000

3500

0 2 4 6 8 10 12 14 16

Time(days)

BO

D(t

) (m

g/L

)

10

Q12. A 200-mL sample of water has initial pH 10. Thirty milliliters of 0.025 N sulfuric acid is required

to titrate the sample to pH 6. What are total alkalinity, carbonate, bicarbonate and hydroxide alkalinity?

Define carbonate hardness? Using data from Q1, how much is carbonate hardness and why?

Answer:

pH 10 => All three types of alkalinity; pOH=14-pH = 14-10=4; [OH-] =10-pOH

= 10-4

moles/L

Number of hydroxide equivalents = (17g/mole*10-4

mole/L)/(17g/equivalent) = 10-4

equivalents/L =

number of equivalents of CaCO3

Hydroxide alkalinity = (equivalent weight of CaCO3 =50 g/equivalent)*( 10-5

equivalents of CaCO3/L)

=50*10-5

*103 = 0.5 mg/L CaCO3

As titration was done using 0.025N H2SO4 and titration was done till pH 6 where 30 mL sulfuric acid is

consumed. It indicates that sample has hydroxide alkalinity, carbonate and bicarbonate alkalinities, if

we assume that titration

The distribution of carbonate and bicarbonate alkalinities depend on concentration of carbonate and

bicarbonate concentrations at pH 6 and it depends on equilibrium between carbonate and bicarbonate

species.

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Q13. An environmental engineer wants to use DO sag curve for determining maximum permissible

BOD which can be discharged into river from an industry. Write down steps and formulate a model to

calculate this value. [20 points]


Q14. Define BOD and its relationship with COD? If the 5-day BOD of a wastewater is 200 mg/L at

20°C, calculate the ultimate BOD at 30°C (k1=0.23/day)? [5+5+10=20 points]


Q15. How does algal growth in lake helps in maintaining buffering capacity of a lake system? What

would happen if I inject carbon dioxide captured from coal-powered plants into this lake?

[15+15=30 points]

Answer:

Algae consume carbon dioxide in presence of sunlight and produce new algal cells and oxygen.However

in the absence of light, algae consumes oxygen to produce carbon dioxide. It results in diurnal variation

in CO2 concentration in water and thus it affects alkalinity of the system (and thus the buffering capacity

of the system).

For high algae growth pH changes to 10 due to above consideration. With increase in pH, forms of

alkalinity changes and CO2 can also be extracted for algal growth both from carbonates and

biocarbonates as per following equation:

2HCO3- <=>CO3

2- +H2O +CO2

CO32-

+H2O <=> 2OH-+CO2

thus the removal of CO2 by algae tends to cause a shift in forms of alkalinity present from bicarbonate to

carbonate and form carbonate to hydroxide. However in this process, alkalinity remains constant.

(see notes below also)

13





Always write your name and entry number in all submissions. Please mention your assumptions

explicitly.

Q1. Determine the settling velocity of a spherical particle with diameter of 100 micron and a specific

gravity of 2.3 in water at 25degC?

Answer: Use approach discussed in the class

Q2. Define following: (i) Type 1 suspension; (ii) flocculating particles (iii) dilute suspensions, and (iv)

concentrated suspensions.

Answer: Refer notes.

Q3. What is the basic difference between settling of inert suspended particles and biological solids (i.e.,

biomass)? How would the settling pattern change if I have a sample with 100 mg/L inert suspended

particles and 800 mg/L biological solids?

Answer: For inert SS, type 1 settling is appropriate compared to type -2 settling for biological

solids. So sample with 100 mg/L SS will have type 1 settling and sample with 800 mg/L biological

solids will have type 2 settling.

Q4. A settling analysis is run on a type 1 suspension. The column is 1.8 m deep and data are presented

below. What will be the theoretical removal efficiency in a settling basin with a loading rate of 25

m3/d/(m

2)?

Time (min) 0 60 80 100 130 200 240 420

Conc. (mg/L) 300 189 180 168 156 111 78 27

Mass fraction

remaining

Velocity (m/min)

Answer: After Minor 1

Q5. Revisit Q4. Consider a scenario where we have four discrete particles in beginning (settling

velocity=0.00017 m/min.). After settling for Z1 depth, two particles come closer and start settling

together (now we have only two discrete particles settling from depth Z1). Further settling of two

discrete particles (approximated for modelling purposes) for Z2 (Z2>Z1) depth from the top of the

column, all particles come closer and started settling together as a single particle only (now we have

only one discrete particle settling starting from depth Z2). Model the settling process of these

particles. Is it possible to model removal of these particles in the settling column and calculate overall

removal? What additional information you would like to have, if any?


Q6. Revisit Q4. Determine the theoretical efficiency of the selling basin if vt=0.04x.


Q7. Revisit Q4. Determine the theoretical efficiency of the selling basin if loading rate = 35 m3/d/(m

2)?


14

Additional Questions

QA1. For the following samples, calculate hydroxide, carbonate, and bicarbonate alkalinity by the

procedure (Alkalinity and pH measurements). The sample size is 100 mL, N/50 sulfuric acid is used

as the titrant and the water temperature is 25°C.

Sample pH Total mL titrant to reach end point

Phenolphthalein Bromcresol green

A 11 10 15.5

D 7 0 12.7

C 11.2 8.2 8.3

Answer: Refer previous HW solution (at pH 7, no hydroxide alkalinity and only carbonate and

bicarbonate alkalinities

QA2. A city water supply is obtained from a deep aquifer. The water has uniform quality, which is clear

and free of organics; hardness is in excess of 300 mg/L and consists of both calcium and magnesium.

Dissolved CO2 is approximately 15 mg/L and iron (Fe2+

) is about 1 mg/L. Other dissolved constituents

are below problem levels. Draw a schematic diagram of a treatment plant that will render this water

potable. Identify each unit and briefly state its purpose. Show points of chemical addition and identify the

chemicals.

Answer: Refer previous HW solution

QA3. Find out the flow in stream required per 1000 population in disposing of sewage from a residential

town in India with following data: average stream temperature= 300C; 5d-BOD of sewage at 30degC=

300 ppm (i.e., 300 mg/L); average sewage flow=135 lpcd (i.e., liters per capita per day or liters per

person per day); k1=0.15/day and k2=0.27/day at 30degC; minimum DO concentration to be maintained

in stream = 4 ppm.

Answer: Refer previous HW solution

15





Always write your name and entry number in all submissions. Please mention your assumptions

explicitly.

Q1. A settling analysis is run on a type 1 suspension. The column is 1.8 m deep and data are

presented below. What will be the theoretical removal efficiency in a settling basin with a loading

rate of 25 m3/d/ (m

2)? What parameters need to be changed to increase overall removal?

Time (min) 0 60 80 100 130 200 240 420 Conc. (mg/L) 300 189 180 168 156 111 78 27 Mass fraction

remaining =189/300 =180/300 =168/300 =156/300 =111/300 =78/300 =27/300

Velocity (m/min)

Hint: As discussed in the class and example given in the text book.

Q2. Revisit Q1. Consider a scenario where we have four discrete particles in beginning (settling

velocity=0.00017 m/min.). After settling for Z1 depth, two particles come closer and start settling

together (now we have only two discrete particles settling from depth Z1). Further settling of two

discrete particles (approximated for modeling purposes) for Z2 (Z2>Z1) depth from the top of the

column, all particles come closer and started settling together as a single particle only (now we have

only one discrete particle settling starting from depth Z2). Model the settling process of these

particles. Is it possible to model removal of these particles in the settling column and calculate overall

removal? What additional information you would like to have, if any?

Q3. Revisit Q1. Determine the theoretical efficiency of the selling basin if vt=0.04x.

Q4. Revisit Q1. Determine the theoretical efficiency of the selling basin if loading rate = 35 m3/d/(m

2)?

Q5. What is the basic difference between type-I and type-2 settling?

Q6. Using the following settling test data, determine the efficiency of a settling tank in removing flocculating

particles if the depth is 8 ft (i.e., 8/3.3 m) and the detention time is 30 minutes? What parameters need to be

changed to increase overall removal?

17

Q7. For a flocculants suspension, determine the removal efficiency for a basin 10 ft deep with an overflow rate V0

equal to 10 ft/h, using the laboratory settling data present in the following table:

18





Q1. What do you understand by coagulation process? Name two generally used coagulant agents. [5 points]

Q2. Why solution pH is an important parameter in the coagulation-flocculation process and need to be

maintained? Explain using alum water chemistry. [5

points]

Answer: pH determines the form of coagulant (i.e., ions or cationic or anionic species) added in solution. For

example, when alum is added in solution, at low pH only Al3+

ions are present and when pH is increased, Al

(OH)xy+

species are present and finally Al (OH)3 precipitate is formed. For charge neutralization and ionic layer

compression, neutralization and charge reversal: Al3+

ions and Al (OH)xy+

species are responsible which happen at

low solution pH. For floc formation during sweep coagulation, Al (OH)3 precipitate is required which happen at

high pH.

Further, solution pH is also maintained to ensure alkalinity of the solution as during alum dissolution, hydroxyl

ions are produced which could make solution basic.

Q3. During coagulation using alum, I forget to adjust pH. Discuss its possible implications on remaining

turbidity in water. [10 points] Answer: If I forget to adjust pH, ionic species and flocs might not form and thus they might not participate in

removing suspended solids of solution. Thus it increases turbidity of solution. More information can be obtained

from Q2 solution.

Q4. To aid sedimentation in the primary settling tank, 25 mg/L of ferrous sulfate (FeSO4.7H2O) is added to

the wastewater. Determine the minimum alkalinity required to react initially with the ferrous sulfate. How

many grams of lime should be added as CaO to react with Fe (HCO3)2?

FeSO4.7H2O + Ca(HCO3)2 <=> Fe (HCO3)2 + CaSO4 + 7H2O [10+10=20 points] Answer: Amount of CaO is required which produces Ca (HCO3)2 is solution. One mole of CaO produces one mole

of Ca (HCO3)2.

CaO+H2O+2CO2 <=> Ca (HCO3)2 balance the given equation:

FeSO4.7H2O + Ca(HCO3)2 <=> Fe (HCO3)2 + CaSO4 + 7H2O

m.wt of ferrous sulfate (FeSO4.7H2O) = 56+32+16*4+7*18=278 g/mole. So, 25 mg/L ferrous sulfate

(FeSO4.7H2O) = (25*0.001g/L)/( 278 g/mole) = 8.99 * 10-5

moles/L

One mole of ferrous sulfate requires one mole of Ca (HCO3)2 or one mole of CaO. So mole of CaO required for

reacting with 25 mg/L ferrous sulfate = 8.99 * 10-5

moles/L. Mass of CaO required = 8.99 * 10-5

moles/L * (40+16

=56 g/mole) = 5.04 mg/L (answer)

Q5. How does interparticle bridging with polymers and sweep coagulation using alum differ? Explain using

examples and reactions. [10 points] Answer: Interparticle bridging with polymers binds two particles with polymeric chains in solution whereas during

alum coagulation, flocs collects two particles together. For both cases, once they become heavy they settle.

Q6. What happens during addition of alum in mixture of primary solids (i.e., solids from primary

sedimentation tank) and secondary solids (i.e., solids from secondary sedimentation tank)? Explain using

alum chemistry? [10 points] Answer: Both primary solids and secondary solids are removed when alum is added in solution. Depending on pH,

alum ions or alum hydroxide precipitate is formed and correspondingly both types of solids are removed. For

biological solids, floc formation is essential otherwise it would not be removed from solution.

19

Q7. [The Mars Trip] I want to treat secondary wastewater effluent for producing drinking water using

coagulation-flocculation only. Is this a viable option? Justify. [2 points]

Answer: Coagulation-flocculation can only remove particles or some hardness-related ions, but its not very much

effective in removing small amounts of residual organic matter or remaining pathogens in wastewater effluent.

Thus coagulation-flocculation process can not effectively remove all residual parameters of secondary wastewater

effluent. Another unit process such as membrane filtration as polishing unit is required to remove low amounts of

residual organic matter and pathogens.

Q8. Plot different forces acting on a colloid particle as a function of distance between two colloids and label

them. Indicate changes in forces acting on a colloid after ionic layer compression and explain the

observation. [10+10=20 points] Answer: This plot is given in notes and different forces are shown in Figure.

20





Q1. For a single-stage softening, following water (see characteristics below) is softening using the Lime-soda ash

process (pH =9). No magnesium removal is required here. [5+10 =15 points]

Species Concentrations (milli-equivalents/L)

Carbon dioxide 1.0

Ca2+

4.0

Mg2+

2.0

Na+ 3.0

HCO3- 2.5

SO42-

5.0

Answer the following:

(i) Write balanced equations for removing CO2, calcium bicarbonate and calcium sulfate using the Lime-soda

process.

(ii) Calculate amount of CaO and Na2CO3 required? (unit: milli-equivalents/L)? For a flow of 10,000 m3/day,

calculate the daily chemical requirement and the mass of solids produced? Assume that the lime used is

90% pure and the soda ash is 80% pure.

Q2. In a WWTP, residuals from the sedimentation basin consist of solids carried by the raw water and the

precipitates formed by adding the chemicals. Summary of the chemical dosages and raw water constituents

producing sludge solids are given below.

Parameter Value

Design maximum day flow 113, 500 m3/day

Design average day flow 57,900 m3/day

Maximum turbidity 17 NTU

Maximum seasonal iron 0.7 mg/L

Maximum seasonal manganese concentration 0.4 mg/L

Optimum coagulant aid, cationic polymer 0.05 mg/L

Optimum coagulant ferric sulfate dosage 25 mg/L

Hydrated lime (Ca(OH)2 for pH adjustment 15 mg/L

Seasonal potassium permanganate (KMnO4)

dosage

4 mg/L

Seasonal PAC for taste and odor control 5 mg/L

* Assume 1 mg-total suspended solids = 1 NTU

Under maximum flow conditions, calculate the following:

(i) Amount of solids from raw water?

(ii) Amount of solids produced due to precipitation of iron as ferric hydroxide?

(iii) Assuming 20% of Ca (OH)2 precipitate as CaCO3, amount of lime solids produced during pH

adjustment?

Q3. The AXX well water has bicarbonate hardness equal to 100 mg/L as CaCO3. Calculate amounts of lime and

soda ash required during the lime-soda ash process (Given: Daily flow rate: 10,000 m3/day). [10 points]

Q4. Look at the following water quality data. Say one needs to treat this water using ion-exchange process

(capacity: 90 Kg hardness/m3 material at 0.4m/minute flow rate). For this material, regeneration is conducted using

10% NaCl solution which requires 100 Kg sodium chloride/m3 resin. Answer the following: [7+3=10 points]

(i) Calculate mass of resin required and chemical required for regeneration process?

(ii) Can I use this ion-exchanger for removing anions from water? Provide reasons.

Species Free CO2 Ca2+ Mg

2+ Na+ HCO3

- SO42-

Conc. (mequiv/L) 1.0 4.0 1.0 2.0 2.5 4.5

21

Q5. Calculate the amount of lime-soda ash required for treating 1 MLD of water? How much solid waste is

produced from this process? [5+5 =10 points]

Free carbon dioxide 3 mg/L Mg2+ 18 mg/L

Ca2+ 44 mg/L Na

+ 16 mg/L Alkalinity (HCO3

-) 122 mg/L

Purity in lime 85% Purity in Soda 100%

Q6. Calculate daily lime-soda ash requirement during softening of water (100 mg/L carbonate hardness (as CaCO3)

and 100 mg/L MgSO4) (Given: Daily water production: 1000 m3/day; lime is 90% pure and soda ash is 80% pure).

Q7. Draw a typical breakthrough curve for ion-exchanger unit and show different parameters on this curve.

Discuss its importance in deciding regeneration frequency of ion-exchanger unit. [2+2+6= 10 points]

22

Q8. Excess sodium intake can results in high blood pressure and inner ear problems for some people. The

regulatory body recommends maximum allowable concentration to be 20 mg/L sodium in drinking water. Now

sodium ions are used in ion exchange process. For reducing hardness from 6 to 1.5 meq/L in water, how much

sodium ion is produced (mg/L) and if it poses any health risk based on given maximum concentration guideline.

Solution:

Reduction in hardness value = 6-1.5 meq/L =4.5 meq/L (during the ion exchange process).

Sodium ions produced during hardness reduction = 4.5 meq/L

Concentration of sodium ions produced = 4.5 × (0.001) × (23g/mole/1)=0.1035 g/L= 103.5 mg/L

23

As regulatory body recommends maximum allowable concentration to be 20 mg/L sodium in drinking water, concentration of

sodium ions produced during the ion exchange process violates this requirement and thus this process poses health risk.

24





Q1. Groundwater contaminated with TCE is treated with powdered activated carbon by sorption.

The adsorption of TCE on PAC follows a Freundlich isotherm model as per following equation

Qeq = 129(Ceq)0.73

where Qeq is mass of TCE adsorbed on PAC (mg/g PAC); Ceq is concentration of

dissolved TCE in water (mg/L).

(i) If TCE concentration in water is 1 mg/L, what mass of PAC must be used per volume of

water (mg PAC/L water) to reduce TCE concentration to the allowed maximum

contaminant level of 0.005 mg/L?

(ii) if the amount of PAC applied is only half of what is calculated in part (i), what is the

equilibrium aqueous concentration of TCE?

Answer:

Qeq = 129(Ceq) 0.73

where Qeq is mass of TCE adsorbed on PAC (mg/g PAC); Ceq is concentration of

dissolved TCE in water (mg/L).

This represents a Freundlich model. Model parameters are K=129 [(mg/g)/ (mg/L)0.73

], n=1/0.73=1.37

Part (i)

For TCE removal: C0=1 mg/L, maximum allowable concentration = 0.005 mg/L. Say X (g) PAC is used.

Removal required = (1-0.005)mg/L=0.995 mg/L

Adsorption capacity of PAC = (0.995 mg/L)*(1 liter)/[X (g)] = 0.995 mg/X (this is Qeq)

Ceq=0.005 mg/L

(0.995/X) = 129(0.005)0.73

= 2.70

X (g) = 0.9975/2.70=0.37 (g) (so 0.37g/L PAC is required to treat one liter volume of water)

Part (ii)

If X* = (0.37g/L PAC)/2=0.185 g/L, then what is Ceq (the new equilibrium aqueous concentration of

TCE)?

Q* (new) = [(1-Ceq)*1/ (0.185g/L*1 liter)] = (1-Ceq)/ (0.185) (mg/g)

(1-Ceq)/ (0.185) (mg/g) = 129(Ceq) 0.73

(1-Ceq) =23.87(Ceq)0.73

=> 23.87(Ceq)0.73

+Ceq-1=0

say f=23.87(Ceq)0.73

+Ceq-1 and it needs to be zero to get Ceq value.

Ceq (mg/L) f

0.005 -0.496

0.001 -0.84488

0.01 -0.16234

0.02 0.392791

0.015 0.127754

0.011 -0.1017

0.009 -0.22461

0.013 0.015377

0.012 -0.04251

0.0125 -0.01342

0.0129 0.009642

0.013 0.015377

25

The Ceq comes out between 0.0125 and 0.0129 mg/L as

function (f) equals to -0.01342 and 0.00962. So Ceq=0.0127 mg/L (answer)

Q2. What is the difference between Langmuir and Freundlich adsorption isotherm mechanisms?

What do these models indicate about adsorption capacities of adsorbent? Can these models provide

information on homogeneity or heterogeneity of adsorption sites?

Answer:

See notes for difference in these models. Langmuir model provides information about homogeneity of

adsorption sites and Freundlich model provides information about heterogeneity of adsorption sites.

Q3. Untreated drinking water has 0.02 mg/L of geosmin, which gives earthy odor. How much

removal can be achieved by adding 10 mg/L powdered activated carbon (PAC) in water? Assume

geosmin adsorption on PAC is defined by Freundlich adsorption

Qeq = Kf (Ceq)1/n

where Qeq is mass of geosmin adsorbed on PAC (µg) per mg of PAC; Ceq is

concentration of dissolved geosmin in water (µg/L), and Kf= 0.5 (µg/mg)(L/ µg)(1/n)

and (1/n)=1.08.

Answer: Follow part (ii) of Q1 to determine Ceq and then to determine percentage removal.

Q4. How does drawing a breakthrough curve help in deciding time for changing of activated

carbon packed in activated carbon-based absorber? or

Q5. Draw a breakthrough curve and label all important information one can obtain from this

curve. Can this curve be used in setting time for changing adsorbent in absorber?

Answer: Breakthrough curve (BTC) provides information about remaining contaminant concentration as

function of amount of treated water passed (or time spent since start of reactor). The breakthrough point

is decided based on maximum allowable contaminant concentration (Cmax) in water and thus using BTC,

one can determine volume of treated water passed since start of Ceq exceeding Cmax. This stage

indicates breakthrough of contaminant from column reactor. This also indicates time for regenerating

activated carbon in adsorber and the after further monitoring one can decide about complete discarding of

adsorbent.

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 0.005 0.01 0.015 0.02 0.025

Ceq (mg/L)

fun

cti

on

26

Q6 Comment on adsorption kinetics of following six adsorbents.

Answer: Adsorption kinetics of six adsorbents is presented here. Out of these 6 adsorbents, IOCCEl25

and IOCFGI110 achieved approximately 90-95% removal compared to other adsorbents within 12 h

contact duration. Further, there is no further improvement in removal of arsenic from these adsorbents

with time, indicating attainment of saturation of adsorption capacity of these adsorbents. Other adsorbents

give removal between 85-90% after 12 h contact. More data in initial sampling period are required to

properly understand adsorption behaviour of these adsorbents during initial phase of adsorption kinetics.

IOCCEl25 and IOCFGI110 can be explored further for their equilibrium adsorption capacity after this

kinetic study.

Q7. Comment on adsorption kinetics of this adsorbent.

Answer: Here no kinetic data is given between 0 and 12 h and then approximately 95-100% removal can

be seen after 12 h of exposure. Adsorption capacity data is also given which indicates that this adsorbent

can adsorb in the range of 0.008 to 0.010 mg As/g media during 95-100% removal of As from water. It

indicates this material cannot adsorb As beyond this limit.

27

Q8. Compare equilibrium behaviour of two adsorbents.

Answer: This represents isotherm data of Sand110 and FGI110. Sand110 gives adsorption capacity in

the range of 0.0 to 0.005 (or so) (mg/g) and FGI110 gives adsorption capacity in the range of 0.0 to 0.8

mg/g, indicating that FGI110 can provide higher adsorption capacity for given mass of adsorbent. These

isotherms also indicate that for low remaining aqueous As concentration in water, FGI110 can be chosen

over Sand110 for maximizing removal of As from water.

Note that in the inset, USEPA 10 µg/L As guideline is given and this is used to determine which

adsorbent is better to give this As level in water with fewer amounts of media used. In this case, FGI110

is better than Sand110 for this goal.

Q9. Compare equilibrium behaviour of different comments.

Answer: Describe different isotherm models using approach given in Q8.

28

Q10. Compare breakthrough behaviour of two adsorbents, how much volume of water could be

treated without exceeding As guideline?

Answer: This represents presentation of column data as BTC. Here FGI110 breakthrough early

compared to FGI110_7d as can be seen by noting bed volumes passed (here it indicates volume of water

passed in terms of volume of column). FGI110_7d can provide As below the guideline for longer period

compared to FGI110 media.

For a given guideline (say for As: 10 µg/L As guideline), volume of water treated

=no. of bed volumes passed at breakthrough point (i.e, when As>=10 µg/L) *(column or absorber

volume).

29




Home Work 9 (Disinfection)

Q1. The Chick’s Law: For disinfection, assume N (0) is initial number of pathogens and N (t) is remaining

number of pathogens at time and given by: N (t) =N (0) × exp (-K×t); where K is disinfection rate

(unit=1/unit of time) and depends on disinfectant-pathogen interaction and solution characteristics. Here R-

log removal: R=-log10 [N (t)/N (0)].

If disinfectant concentration (Cdisinfectant) and contact time (tc) are related to each other by following equation

(the Watson’s Law):

(Cdisinfectant)n × (tc) = constant (standard unit: C in mg/L and tc in minute)

Calculate R-log removal value for 99.9% removal? What is the remaining pathogen concentration at this

removal after 1 minutes of contact time? (Assume K=0.046/min) [3+3=6 points]

Solution: R-log removal = -log10 [1-Nt/N0]

Given removal = 99.9%, so Nt/N0= (1-99.9/100) =0.001

R-log removal = -log10 [0.001] =3 (answer)

For calculating remaining fraction of pathogens after 1 minute of disinfection with K=0.046/min, use the Chick’s

Law: Nt/N0 = exp (-k ×t) = exp (-0.046/min×1 min) =0.9550 (i.e., 95.50%) (answer)

Q2. An experiment shows that a concentration of 0.1g/m3 of free available chlorine yield a 99% kill of

bacteria in 8 minutes. What contact time is required to achieve a 99.9% kill at a free available chlorine

concentration of 0.05 g/m3? Assume that Chick’s Law and Watson’s Law hold with n=1. [2+2=4 points]

Solution: Given: For 99% kill: C= 0.1 g/m

3 and time (t) =8 minutes

Chick’s Law: Nt=N0×exp (-k ×t)

Calculation of disinfection rate constant:

Nt/N0= (1-99/100) =0.01 in 8 minutes

From Chick’s Law: 0.01 = exp (-k ×8)

=> k = - (1/8) ln (0.01) = 0.5756/min (answer)

Using calculated k value, calculate time for getting 99.9% kill:

Nt/N0= (1-99.9/100)=0.001

Using Chick’s Law: 0.001 = exp (-0.5756×t) =>t = - (1/0.5756) ln (0.001) =12 min (answer)

Note: Watson’s Law: Cn×t=constant = > C×t=constant (as n=1)

For 99.9% kill: C= 0.1 g/m3 and time (t) =8 minutes. So, C×t value

=(0.1 *1000mg/1000L)*(8 minutes) = 0.8 (mg/L)(min.)

For 99.9% removal: Ct value= (0.05 ×1000mg/1000L)*(12 minutes) = 0.6 (mg/L) (min.)

30

Q3. For wastewater consists of ammonia, organic matter and microorganisms), draw breakthrough curve

using following information and answer following questions:

Chlorine dosage (mg/L) 0.1 0.5 1.0 1.5 2.0 2.5 3.0

Chlorine residual (mg/L) 0.0 0.4 0.8 0.4 0.4 0.9 1.4

(i) Discuss the significance of different regions. [8 points]

(ii) Calculate chlorine dose to achieve 0.75 mg/L free available chlorine? [4 points]

Solution:

31

Q4. Look at the following relationship between concentration of free residual chlorine and contact time

required for 99% kill (Watson’s Law: C0.86

tp= λ (constant) for different pathogens).

Pathogen type Adenovirus 3 E.coli Coxsackievirus A2

λ (constant) 0.098 0.24 6.3

For given chlorine dose, how long would you like to disinfect to achieve maximum removal of all pathogens?

Explain the result. [6+4=10 points]

Solution: For given chlorine dose: high contact time is required for high λ value. So we need longest contact time

for Coxsackievirus A2 than that for other pathogens. Thus to achieve maximum removal of all pathogens we need

contact time equal to that of Coxsackievirus A2.

Q5. Comment on decay of adenovirus (survival = Nt/N0) using low pressure (LP) and medium pressure

(MP) UV rays. Note UV dose is given in milli joules/cm2.

Solution: Example: Log (survival)= (-4) (i.e., 4-log removal)

Log (Nt/N0)=-4; Nt=N0 (10-4

) or Removal = (1-0.0001)100= 99.99%

LP UV rays produce high decay of adenovirus compared to MP UV rays. In initial regions, LP UV rays give high

disinfection rate than MP UV rays, but for high dose, both produces similar extent of virus decay.

Q6. Comment on effect of light and dark on concentration of five fecal indicators. Concentration values are

shown in box plots where middle line show median value, below 25th

percentile value (i.e., 25% of values

lower than this) and above 75th

percentile.

Solution:

Dark conditions give higher CFU values, i.e., lesser removal of pathogens than that due to light conditions.

Highest removal was observed for Enterococci than other microorganisms as lowest remaining concentration.

Lowest removal was observed for F-RNA phages than other microorganisms.

Higher removal for bacterial indicator (i.e., for fecal coliforms) was observed than viral indicators (i.e., for somatic

coliphage and F-RNA phages).

32

Q7. Comment on effect of UV dose on virus survival for two sets of data. Note that two linear models are fit

to data (this is first order data).

Solution: Adeno 2 virus is disinfected at slower rate than other pathogens with increasing UV dose. For high log-removal,

high UV dose is required for adeno 2 virus than other pathogens.

Both linear models appear to fit the observed decay data, however, it is difficult to comment on model fitting due

to difficulty in determining goodness-of-fit of fit models to data.

Q8. Comment on dose requirements for inactivation of viruses at different levels by UV light for each of

viruses studied.

Solution:

With high UV dose, high log-removal can be achieved.

For 99.99% removal (i.e., for 4-log removal), highest UV dose is required for adenovirus type 2 than other

pathogens with lowest dose required is for echovirus 2.

These data also indicate that adenovirus type 2 is most resistant to UV dose (i.e., hard to kill). Further, adenovirus

2 requires relatively higher UV dose than other pathogens (at least 4-5 times higher).

33




Home Work 10

Q1. An organic waste with 250 mg/L BOD5 (S0) needs to be treated using a completely-mixed activated sludge

process. The effluent BOD is to <=10 mg/L (Seff). Assume that the temperature is 20°C, the flow rate is 5 million

gallons/day and following information are available:

(i) Influent volatile suspended solids to reactor are negligible.

(ii) Return sludge concentration = 10000 mg/L of suspended solids = 8000 volatile suspended solids

(iii) Mixed-liquor volatile suspended solids (MLVSS)=X=3500 mg/L

(iv) Mean cell-residence time (θc)= 10 days

(v) Microbial parameters: kd =0.06/day; Y=0.65 mg cells/mg BOD5 utilized.

(vi) Effluent has 20 mg/L biological solids (here 80% is volatile and 65% is biodegradable). Assume that the

biodegradable biological solids can be converted from ultimate BOD demand to a BOD5 demand using the

factor 0.68 [i.e., BOD K value =0.1/day (base 10)].

Calculate (i) Biomass production rate; (ii) Observed yield (i.e., Y/ [1+kdθc]); (iii) Food to microorganisms ratio

(i.e., S0/ [Xθ]). [2×3=6 points]

Q2. Assuming that the endogenous coefficient (kd) can be neglected, develop expressions for determining substrate

and cell concentration as a function of time for a batch reactor. If the initial concentration of substrate and cell is

100 and 200 mg/L, respectively, calculate the amount of substrate remaining after 1 h. If the endogenous

coefficient is equal to 0.04/day, estimate the error made be neglecting this factor. Assume: k=2/h; Ks = 80 mg/L;

Y=0.4 mg/mg. [4+4=8 points]

Q3. An anaerobic digester is designed to remove 85% of BOD5 of an industrial organic waste with an ultimate

BOD =2000 mg/L. If (θc) = 10 days, estimate the amount of sludge to be wasted daily and the quantity of gas

produced each day. Assume that the flow =0.1 million gallons/day; Y=0.1; kd=0.01/day.

[3+3=6 points]

Q4. In the activated sludge reactor, bacterial oxidation and synthesis can be described using Eq. (1) and its

endogenous respiration can be described using Eq. (2). Calculate amounts of oxygen required for oxidation and

synthesis of 3500 mg/L MLVSS (i.e., bacterial concentration)?

Oxidation and synthesis (in presence of bacteria):

COHNS (organic matter) +O2 + nutrients �� CO2 +NH3 + C5H7NO2 (new cells) + other end products (1) Endogenous respiration (in presence of bacteria):

C5H7NO2 (cells) + 5O2 �� 5CO2 +2H2O+ NH3 + energy …… ……………(2)

Q5. From Q3, how much gas and sludge will be produced if the θc is increased to 20 days?

Q6. Calculate amount of methane produced per mg of ultimate BOD5 stabilized during anaerobic digestion

process? Assume the starting compound is tricarboxylic acid (C3H8O6).

34

Q7. Compare organic matter degradation of distillery industry wastewater (i.e., spent wash) using eight different

anaerobic reactors. Also comment on methane gas production from each of these eight reactors.

Q8. Compare COD removal effectiveness of different treatment types.

Q9. Comment on removal of Salmonella over time at different locations of wastewater treatment plant (anaerobic

sludge blanket).

35

Q10. Compare concentrations of fecal indicators and pathogens at different locations of wastewater treatment plant

(anaerobic sludge blanket).

36




Home Work 11 (Courtesy: Dr. Arvind K Nema)

Q1. An anaerobic reactor, operated at 35oC, treats wastewater with a flow of 2000m

3/d and a

biological soluble COD (bsCOD) concentration of 500g/m3. At 90% bsCOD removal and a biomass

synthesis yield of 0.04 g Volatile Suspended Solids/ g bsCOD used, estimate the amount of methane

produced in m3/d.

Step 1: Prepare a steady state mass balance for COD to determine the amount of influent COD

converted to methane.

CODin = CODeff + CODVSS(biomass) + CODmethane

CODin = COD concentration in influent x Wastewater Inflow

CODeff = COD concentration in effluent x Wastewater Inflow

CODVSS = 1.42 g COD/g VSS x Yield coeff g VSS/g COD x Efficiency of system

CODmethane= ?

Step 2: Determine the volume of gas occupied by 1 mole of gas at 35oC.

Step 3: The CH4 equivalent of COD converted under anaerobic conditions =

(L/ mole)/ (64 g COD/ mole CH4)

Step 4: CH4 production = CODmethane x CH4 equivalent of COD converted

Solution:

CODin = COD concentration in influent x Wastewater Inflow = ( 500g/m3) *(2000m

3/d)=1000 Kg/d

Biological soluble COD removal = 90%

CODeff = 0.10* 1000 Kg/d = 100 Kg/d

CODused = 0.90* 1000 Kg/d = 900 Kg/d

Biomass synthesis yield = 0.04 g Volatile Suspended Solids/ g bsCOD used

Biomass synthesis yield = (900*1000g/day)*(0.04 g VSS)=36000 g/day

COD corresponding to biomass = (36000 g VSS/day)*(1.42 g COD/g VSS)=51.120 Kgday

CODin = CODeff + CODVSS(biomass) + CODmethane

CODmethane = CODin - {CODeff+CODVSS(biomass)}

CODmethane = (1000 Kg/d)- {100 Kg/d+ 51.120} =848.88 Kg/d (answer)

CODmethane = 848.88 Kg/d (answer)

As 64 g COD comes from 1mole of CH4. So 848.88 Kg/d COD would result from

= (1 mole CH4 /64)*(848.88 Kg/d) = 13.26 mole CH4/day

Now we know that for STP conditions (i.e., 273 K and 1 atm pressure, 1 mole gas occupies 22.4 Liter

volume). Now we need to calculate volume of methane gas at 35°C (i.e., 308K) and 1 atm pressure:

V/T=constant (as P is constant here)

[22.4/273] = [V2/308] ==> V2=25.27 Liters/mole (this is true for methane gas also)

Now as biological process is producing 13.26 moles methane gas/day, it indicates that 335.1 liters/day (or

0.335 m3/day).

37

Q2. The water content of solids slurry (WW sludge) is reduced from 98 to 95 %. What is the

percent reduction in volume assuming that solids contain 70% organic matter of specific gravity

1.0 and 30% mineral matters of specific gravity 2.0? What is the specific gravity of 98 and 95%

slurry?

Solution:

w1=98% (i.e., 2% solids)

w2=95% (i.e., 5% solids) (this solid content is increased due to dewatering process)

Wsolids/(Ssolids*ρw) = Wmineral/(Smineral *ρw) + Worganic matter/(Sorganic matter* ρw)

Here S: specific gravity

W: weight

ρw = density of water (i.e., 1 Kg/liter)

Case 1:

Water content =98% (i.e., 2% solids);

Organic matter content =70% =0.7 Wsolids

Mineral matter =30% =0.3 Wsolids

ρw = 1 Kg/liter

Smineral=specific density of mineral = 2

Sorganic matter=specific density of organic matter =1

For determining specific gravity of all solids (Ssolids):

Wsolids/(Ssolids) = 0.3Wsolids/(2) +0.7 Wsolids/(1)

1/Ssolids = 0.15 +0.7 = 0.85

=> Ssolids =1/0.85 =1.18 (this is specific density of solids)

Now if specific density of water = 1 and water content =98% (i.e., solid content =2%), we can calculate

specific gravity of sludge =

1/Ssludge = (0.98)/1+(0.02)/1.18 =0.9969

� Ssludge = 1/0.9969 =1.003 (or Density of sludge = 1.003*1=1.003 Kg/L)

Density of sludge = Mass of sludge /Volume of sludge

=>Volume of sludge = Mass of sludge/Density of sludge

For 1 Kg mass of sludge, volume of sludge = 1/(solid content*density of sludge)

= 1/(0.02*1.003) = 49.85 liters (initial volume of sludge per Kg mass of sludge)

After solid waste treatment, water content is 95% (i.e., 5% solids).

V1=49.85 liters/Kg

Assuming specific gravity of sludge remains same (i.e., Ssludge = 1.003) or Density =1.003 Kg/L

New volume of sludge after reduction in water content per Kg of sludge

=V2= 1/(0.05*1.003) = 19.94 liters

So volume reduction = [1-(19.94/49.85)] *100 =60%

38

Q3. Assume that the data given in Q2 belongs to primary sludge. What will be the volume of

digested slurry in case 60% of the volatile solids are destroyed and water content is reduced to

90%?

Solution:

Say 60% volatile solids (or organic matter) is destroyed. Initially 70% solid was organic matter and now

remaining organic matter = (70%)*(0.40) = 28% (or mineral content =1-0.28=0.72 (i.e., 72%). Revised

water content =90% (and thus solid content =10%).

1/(Ssludge) = 0.72/(2) +0.28/(1) =0.64 �Ssludge =1/0.64 =1.56

Density of sludge = 1.56*1=1.56 Kg/L

Volume of sludge = Mass of sludge/Density of sludge

For 1 Kg mass of sludge, volume of sludge = 1/(0.10*1.56) =6.41 liters

Q4. What are three stages of anaerobic digestion process? Discuss their importance and BOD

requirements at different stages.

Hint: See Lecture Notes.

Q5. Some substances can be toxic to bacterial growth. How can you incorporate this effect in

determining specific biomass growth rate constant?

Solution:

Some substance can impart toxicity to bacteria and thus it might affect their growth. For incorporating

this effect, microbial decay constant due to toxic substance (i.e., kdecay,substance) need to be determined.

Then this aspect needs to be incorporated in the overall biomass rate equation.

Roverall=Rlag+Rgrowth+Rstationary+Rendogenous decay+Rsubstance decay

Q6. Why do we keep SRT greater than HRT and how does it affect plant’s performance?

Solution:

Keeping SRT> HRT results in long retention of biological solids (i.e., biomass) in aeration tank where

aerobic biological processes occur. This increased retention of biological solids results in higher removal

of organic matter compared to SRT=HRT situation (i.e., no recirculation case).

Q7.Comment on performance of chlorination unit placed before and after aeration tank.

Solution:

Placing chlorination unit before aeration tank can kill bacteria however; it would not reduce

microorganisms and pathogen in wastewater effluent which we might discharge to river. In this case, the

discharge stream might have pathogens unsuitable for discharging in river water.

39

Q8. Calculate BOD and alkalinity requirements for removing 14 mg NH4+-N during nitrification

process? Also explain the need for maintaining alkalinity in the nitrification reactor.

Solution:

See web solution

Q9. A completely mixed activated sludge plant is to treat 10000 m3/d industrial wastewater (BOD5

= 1200 mg/L; BOD effluent = 100 mg/L prior to discharge). For 5-day SRT, 5000 mg/L MLSS is

required (Y= 0.7 kg/kg and kd=0.03/day). Calculate following:

(i) The volume of the reactor

(ii) The mass and volume of solids wasted each day

(iii)The sludge recirculation ratio.

(iv) To reduce BOD effluent to 30 mg/L would increase SRT a good idea? Discuss.

Q10. Write overall oxidation and cell synthesis reactions during nitrification process and calculate

amount of BOD required?

Hint: See solution of Q8.

Q11. Calculate total carbonaceous and nitrogen oxygen demand of a water sample that contains 5

mg/L organic compounds having chemical formula C6H6N2O2. Assume that nitrogen is converted

to ammonia and then to nitrate.

Hint: See notes. This question has been discussed in class and asked in HW and Exam.

Q12. Discuss different stages of anaerobic digestion process in the order they occur. Hint: See notes

Q13. Name three disinfection kinetics models generally used to model disinfection process. Can CT

concept be applied to all three kinetic models? Why or why not?

Hint:

For part 1: See notes.

For part 2 Ct concept is applied only for first order kinetic model (i.e., for Chick’s Law when n=1 in the

Watson’s Law).

Q14. Comment on difference between domestic and hospital solid waste management (SWM)

treatment methodologies.

Solution:

domestic SWM hospital SWM

more treatment is required for inert

and biodegradable materials

more treatment is required for sharps, infectious

microorganisms and unused drugs, etc.

generally, segregration, screening,

biological processes are used and

then land fill disposal and methan

gas capturing is done.

generally segregation, followed by incineration is done to

kill pathogens; biological process and advanced oxidation

processes such as ozonation, UV radiation and hydrogen

peroxide usage for properly treating unused drug before

doing land filling and methane gas capturing if its feasible.

Q15. Look at the following information for a completely mixed biological reactor.

Influent water information:

Parameter Influent water Effluent water

40

Flow rate 50 m3/d 50 m

3/d

Biomass 0 mg VSS/L ??

Substrate 95 mg BOD5/L ??

Tank volume 200 m3

VSS: volatile suspended solids

Calculate following:

1. effluent BOD5

2. biodegradable organic matter removal efficiency of tank

3. biomass leaving the tank

Solution:

Here HRT =θ= 200m3/(50m3/d)= 4 days

X in aerobic reactor =?

S0=95 mg BOD5/L

Sfinal=?

1/θ =[(µmS)/(KS+S)]-kd

so: determine expression for S.

S=[ KS (1+ θ kd)]/[( θµm -1- θ kd)]

Given KS=60 mg BOD5/L (this is average value)

kd=0.06/day (this is average value)

µm =3/day (this is average value)

Solve for S now. It is coming out to be: Sfinal= 6.9 mg BOD5/L

S removal efficiency =(1-6.9/95)100% = 92.74%

For determining X coming out of the system: Do a mass balance for substrate first.

0=QS0-QSf+VrS here rS is rate of substrate utilization.

so rS = -Q(S0-S)/V

And we also know that rS = -kXS/[ KS +S]

so {-Q(S0-S)/V} = -kXS/[ KS +S]

X = [ (KS +S)(S0-S)(Q/V)]/(kS)

here Q/V = 1/HRT

solve for X. X=43 mg VSS/L

Q16. Discuss removal efficiencies of different units for different contaminants in wastewater

treatment plant with following schematic:

Influent water �� Primary settling tank �� Biological aeration �� Secondary settling-�� Effluent

water:

Parameter Influent water After settling (and

influent to aeration tank)

Effluent

water

BOD 200 mg/L 130 mg/L 30 mg/L

suspended solids 240 mg/L 120 mg/L 30 mg/L

phosphorous 7 mg/L 6 mg/L 5 mg/L

nitrogen 35 mg/L 30 mg/L 26 mg/L

Solution:

41

Removal efficiency table

Parameter After settling after aeration

tank

remarks

BOD =(200-

130)/200=35%

=(130-

30)/130=76.9%

more removal is observed in aeration

tank than settling basin

suspended solids =50% 75% more removal is observed in aeration


phosphorous 14.29% 16.67% more removal is observed in aeration


nitrogen 14.29% 13.33% comparable removal is observed in

settling tank and in aeration tank

Now for every contaminant, remaining concentration values need to be compared with receiving body

standard (say for river) to determine if further water treatment is required and which parameters need to

be removed and then one needs to decides about another unit process in the given treatment train.

department of civil engineering-i.i.t. delhi cvl212...

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