Watertechnology1ppts

Engineering/Applied Chemistry CODE NO: 07A1BS07/07A1BS04 I B.TECH MECHNICAL/CIVIL ENGINEERING Unit No: I Nos. of slides: 150

Engineering/Applied Chemistry Water Technology-I Term: 2008-09 Unit-1 Power Point Presentations Text Books: • A text book of Engineering Chemistry by Jain & Jain, • Chemistry of Engineering Materials by C.P. Murthy, C.V. Agarwal and A. Naidu

INDEX UNIT-I PPTS Srl. Module as per Lecture PPT No. Session Planner No. Slide No. ------------------------------------------------------------------------------------------------4. Introduction L-1 L1-1 to L1-19 2. Hardness & Units L-2 L2-1 to L2-28 3. Estimation of Hardness by EDTA L-3 L3-1 to L3-18 4. Problems on Hardness L-4 L4-1 to L4-18 5. Analysis of water L-5 L5-1 to L5-19 6. Methods for treatment of water (Domestic) L-6,7 L6,7-1 to L6,7-33

INTRODUCTION • Water is the most abundant and useful compound essential for survival of all plants and animals. • It is used for various industrial purposes also

L1-1

SOURCES • • • • •

L1-2

Rain water River water Lake water Sea water Underground water-spring and well water

Effect of water on rocks and minerals: • Dissolution: some mineral constituents of rocks such as NaCl and CuSO4.2H2O readily dissolve in water. • Hydration: Some minerals are easily hydrated with the consequent increase in volume leading to disintegration of the rocks in which these minerals are present. Eg. • CaSO4 Hydration  CaSO4.2H2O (Gypsum) • Mg2SiO4 Mg2SiO4.XH2O (Serpentine) L1-3

Effect of water on rocks and minerals: • Effect of dissolved oxygen: This leads to oxidation and Hydration • Fe3O4 Fe2O3 3 Fe2O3.2H2O(Limonite) • 2FeS2 + 7O2 + 2H2O 2FeSO4 + 2H2SO4 • Effect of dissolved CO2: Water containing dissolved CO2 converts the insoluble carbonate of calcium, Magnesium and iron into their relatively soluble bicarbonates. • CaCO3 + H2O + CO2  Ca(HCO3)2 L1-4

Types of Impurities present in water: The natural water is usually contaminated by different types of impurities.

L1-5

They are mainly 3 types..

1. Physical impurities 2. Chemical impurities 3. Biological impurities

L1-6

PHYSICAL IMPURITIES:

1. Colour 2. Turbidity 3. Taste 4. odour L1-7

Colour

Colour in water is caused by metallic substances like salts. L1-8

Turbidity is due to the colloidal, extremely fine suspensions such as insoluble substances like clay, slit, and micro-organisms.

L1-9

Taste presence of dissolved minerals in water produces taste. Bitter taste can be due to the presence of Fe, Al, Mn, Sulphates, lime. Soap taste can be due to the presence of large amount of sodium bicarbonate. L1-10

Odour

In water is undesirable for domestic as well as industrial purpose.

L1-11

CHEMICAL IMPURITIES: In-organic chemicals Organic chemicals

L1-12

Inorganic chemicals: 1. Cations: Ca+2, Mg+2, Na+, K+, Fe+2, Al+3, Zn+2, Cu+2 2. Anions:Cl-, SO-2, NO3-, HCO3-, F-, NO2L1-13

Organic chemicals dyes, paints, petroleum products, pesticides, detergents, drugs textile materials, other organic related materials.

L1-14

Acidity surface and ground water attains acidity from industrial wastes like acid, mine, drainage, pickling liquors, usually acidity caused by the presence of free CO2, mineral acids, and weakly dissociated acids. L1-15

Gases All natural waters contains dissolved atmosphere CO2, O2, NH3 gases, pollutant and sewage water contains nitrogen in the form of nitrogenous organic compounds and urea , which are partially converted into NH3. L1-16

Mineral matters have organs from rocks and industrial effluents. These include mineral acids, Ca+2, Mg+2, Na+, K+, Fe+2, Al+3, Zn+2, Cu+2, Mn+2, Cl-, SO-2, NO3-, HCO3-, F-, NO3-, SiO2 L1-17

BIOLOGICAL IMPURITES • Biological impurities are Algae, pathogenic bacteria, fungi, viruses, pathogens, parasite-worms, etc. the source of these contamination is discharge of domestic and sewage wastes, excreta, etc.

L1-18

A. Micro-organisms: algae, fungi, viruses, etc. B. Water bodies in water include 1. Bacteria, organisms’ inhabitating the bottom sludge, and 2. Organisms and planktons developed at the water surfaces. These are inhibitated by different groups of worms like flat worms, hair worms, tiny roundworms, oligochetes, etc.

L1-19

HARDNESS OF WATER L2-1

Definition • Hardness of water defined as which prevent the lathering of soap.

• This is due to presence of in water of certain salts of Ca+2, Mg+2 and other Heavy metals dissolved in it. L2-2

Soap • Soaps are Sodium or Potassium salts of Higher fatty acids like Oleic acid or Palmitic acid or stearic acids.

L2-3

SODIUM STEAREATE formulae is….

C17H35COONa L2-4

POTASSIM STEARATE formulae is….

C17H35COOK L2-5

SOAP with HARDWATER REACTIONS

2C17H35COONa+CaCl2

(C17H35COO)2Ca+2NaCl L2-6

2C17H35COONa+MgSO4

(C17H35COO)2Mg+Na2SO4 L2-7

Types of Hardness Hardness of water is mainly TWO types

1.Temporary Hardness 2. Permanent Hardness L2-8

Temporary Hardness Temporary Hardness is mainly caused by Bicarbonates of Calcium, Magnesium and other heavy metals.

L2-9

Temporary Hardness mainly caused by the presence of dissolved Bicarbonates of Calcium, Magnesium and other Heavy metals and the carbonate of Iron. L2-10

Temporary hardness of water mainly responsible salts are…

Calcium bicarbonate

Ca(HCO3)2 Magnesium bicarbonate

Mg(HCO3)2 L2-11

Temporary Hardness can be largely removed by mere Boiling of water.

L2-12

Equations

Ca(HCO3)2 Calcium bicarbonate

Heat CaCO3+H2O+CO2 L2-13

When bicarbonates are decomposed a, yielding insoluble carbonates or hydroxides, which are deposited as a crust at the bottom of vessel. L2-14

Mg(HCO3)2 Magnesium bicarbonate

Heat

L2-15

Mg(OH)2+2CO2

Temporary Hardness is also known as….

3.Carbonate Hardness 4. Alkaline Hardness L2-16

Permanent hardness It is due to the presence of dissolved Chlorides and sulphates of Calcium, Magnesium , Iron and other metals. L2-17

L2-18

Permanent hardness responsible salts are…. • CaCl2 • MgCl2 • CaSO4 • MgSO4 • FeSO4 • Al2(SO4)3 L2-20

Permanent Hardness cannot removed by boiling but it can be removed by the use of chemical agents. L2-21

Permanent Hardness also known as….

• Non-Carbonate Hardness • Non-Alkaline Hardness

L2-22

Differences between

Temporary and

Permanent Hardness L2-23

Differences between Temporary and Permanent Hardness • Temporary Hardness is mainly caused by Bicarbonates of Calcium, Magnesium and other heavy metals.

L2-24

• Permanent hardness is due to the presence of dissolved Chlorides and sulphates of Calcium, Magnesium , Iron and other metals

Differences between Temporary and Permanent Hardness • • Calcium bicarbonate • Ca(HCO3)2 • • Magnesium bicarbonate • • Mg(HCO3)2 • L2-25

CaCl2 MgCl2 CaSO4 MgSO4 FeSO4 Al2(SO4)3

Differences between Temporary and Permanent Hardness 1. Carbonate Hardness • Non-Carbonate Hardness 2. Alkaline Hardness • Non-Alkaline Hardness

L2-26

Differences between Temporary and Permanent Hardness

• Temporary Hardness can be largely removed by mere Boiling of water.

L2-27

• Permanent Hardness cannot removed by boiling but it can be removed by the use of chemical agents.

Assignment work • Distinguish between Temporary and Permanent Hardness of water. • Explain the terms Temporary and permanent hardness of water. • Write a short notes on hardness of water.

L2-28

L2-29

ES TIMA TION OF HAR DNES S BY

EDT A M ET HOD

L3-1

T he har dne ss of water is not a pol lu tion par am eter but indi ca te the ea ter quali ty i n ter ms of Cal ci um and ma gnesi um expr essed in t er ms o f CaC O3.

L3-2

T he a nal ysis is d one by c omplex me tric tit r atio n usin g st andar d ED TA and EBT a s an in dic ator. L3-3

PRI NC IP LE: In t his comp le x metric Tit r ation, the wate r sa mple i s titr ated w ith st andar d so lutio n o f Di so diu m sa lt o f EDT A usin g EBT

L3-4

PR OC EDUR E: To 50 mL of sampl e ad d 10-15 mL buf fer sol uti ons adj ust the pH about 10 and then war m the sol uti on. A dd 2 dr ops o Eri oc hrome Bl oc k- T and it is tit r ated with 0. 01 M EDT A sol uti on changes its colour f rom r ed- wine to L3-5

REA CT IO NS INV OL VED : EBT indica tor w hen ad ded to har d w ater a t pH = 10, for ms w eak comple xes with calci um a nd ma gnesi um pr esent in har d w ater . It r esults in the f or ma tion of Ca-EBT or M g-E BT comple xes w hic h is wine-r ed, these ar e u nsta ble.

L3-6

Duri ng ti tr ati on with ED TA, cal ci um fir st reacts to f or m rel ati vel y sta bl e foll owed by ma gnesiu m to gi ve Mg- EDTA +2 compl ex rel easi ng the fr ee indi ca tor (bl ue) the col our changes fr om win e-r ed to bl ue at the

L3-7

EDTA COMPLEX STRUCTURE

L3-8

Pr epar ation of st andar d har d w ate r :

L3-9

di ssol ve 1g of pur e, dr y CaCO 3 i n mini mum quan ti ty of di l.H Cl and then eva por ate the sol ution to dr yness on a water ba th. Di ssol ve the r esi due in di sti ll ed wate r to mak e 1 Liter soluti on. Eac h mL of thi s soluti on thus contain s 1mg of CaCO 3 equal ent har dness.

1 mL h ar d w ater so lu tio n = 1mg o f Ca CO 3 e quale nt har dness.

L3-10

Pr epar ation o f EDT A so lu tion: Disso lv e 4 g of pur e EDT A cr ystals + 0.lg MgCl2 i n 1 Liter o f dist il le d wate r. L3-11

Pr epar ation o f EDT A so lu tion:

Di ssol ve 4 g of pur e EDTA cr ystal s + 0.l g MgCl 2 i n 1 Li ter of di stilled water . L3-12

EDTA STRUCTURE

L3-13

EDTA STRUCTURE

L3-14

EDTA STRUCTURE

L3-15

Pr epar ation o f Indic ator (EB T):

Di ss ol ve 0. 5 g of Erioc hr om e B lac kT in 100mL alcohol . L3-16

FORMULAE OF EBT

L3-17

Pr epar ation o f Bu f fer so lu tion: add 6 7.5g o f NH 4Cl to 5 70 mL of Con. Am mo nia so lu tio n and th en d ilu te wit h d ist il le d w ate r to 1 Lit r e. L3-18

Standar diza tion of EDT A so lu tion: Rinse and fill the burette with EDTA solution. Pipette out 50 mL of standard hard water in a conical flask. Add 10-15 mL of buffer solution and 4 to 5 drops indicator. Titrate with EDTA solution till wine-red colour changes to clear blue. Let volume used by V1 mL. L3-19

Titr atio n o f Un known Ha r d w ate r : Ti tr ate 50 mL of wate r sampl e just in Step- 5. let vol ume used be V2 mL. L3-20

Titr atio n o f Per manent h ar dness: take 250 mL o f t he w ater sample in a l ar ge b eaker. B oil i t. T ill the v olume is r educed to a bout 50 mL, filte r it , w ash th e precip ita te wi th d ist il le d water, colle cting f iltr ate and washin g i n a 2 50 mL measuring fla sk. F inall y ma keup the volu me to 2 50 m L wi th d istil le d wate r. T hen, titrate 50 mL o f boil ed water s ample ju st a s i n Ste p (5 ). L et volume u sed b y V3 mL. L3-21

Calculations

L3-22

V1 mL of EDTA is consumed by 50 mL standard hard water. 50 mL of standard hard water = V1 mL of EDTA 50 * 1 mg of CaCO3 = V1 mL of EDTA 1 mL of EDTA

L3-23

=50/V1 mg of CaCO3 eq.

Now 50 mL of given/ unknown hard water =V2 mL of EDTA V2*50/V1 mg of CaCO3

eq. 1 Lt or 1000 mL of given hard water = 1000*V2/V1 mg of CaCO3 eq. Total hardness of water = 1000*V2/V1 mg/L 1 mg/L = 1ppm so, L3-24

= 1000*V2/V1 ppm

Now 50 mL of boiled water = V3 mL of EDTA = V3*50/V1 mg of

CaCO3 eq. 1000 mL of boiled water = 1000*V3/V1 mg of CaCO3 eq. = 1000*V3/V1 ppm

So, Permanent hardness L3-25 1000*V3/V1 ppm

=

Temporary hardness = [Total hardness – Permanent hardness] = 1000*V2/V1 – 1000*V3/V1 = 1000[(V2-V3)/ V1] ppm

L3-26

• Total har dness of water = 1000* V2/ V1 pp m • Per manent har dness = 1000*V 3/ V1 ppm • Temp or ar y har dness = 1000[( V2- V3)/ V1] ppm

L3-26

Ad va nta ges of EDT A me thod: L3-27

T hi s method is def initel y pr ep ar abl e to the other me thods, b ec ause of the 1. Gr ea ter accur ac y 2. C on ven ience and 3. Mor e r api d p rocedur e L3-28

Assignment • Discuss the EDTA method for the determination of permanent hardness of water.

L3-29

EDTA & HARDNESS problems

L4-1

EDTA problems • 1 g of CaCO3 was dissolved in dil.HCl and diluted to 1000 mL. 50 mL of this solution required 48 mL of EDTA solution for titration. 50 mL of hard water sample required 15 mL of EDTA solution for titration. 50 mL of same water sample on boiling, filtering etc, required 10 mL of EDTA solution. Calculate the different kinds of hardness in ppm. L4-2

50 mL of standard hard water = V1 mL of EDTA V1= 48 50 mL of sample hard water = V2 mL of EDTA V2= 15 50 mL of after boiling sample hard water = V3 mL of EDTA V3= 10

L4-3

• Total hardness of water = 1000XV2/V1 = 1000 X 15/48 =312.5 ppm • Permanent hardness = 1000XV3/V1 = 1000 X 10/48 = 208.3 ppm • Temporary hardness = 1000[(V2-V3)/ V1] = 1000 X (15-10)/48) = 104.16 ppm L4-4

EDTA problems • 50 ml of a sample of hardwater was titrated against 0.01 M EDTA, required 48 ml of EDTA for titration. 50 ml of the same hardwater after boiling and filtering etc. required 35ml of the EDTA for titration. Calculate the total and temporary hardness of water in degree clark.

L4-5

50 mL of standard hard water = V1 mL of EDTA V1= 50 50 mL of sample hard water = V2 mL of EDTA V2= 48 50 mL of after boiling sample hard water = V3 mL of EDTA V3= 35

L4-6

• Total hardness of water = 1000XV2/V1 = 1000 X 48/50 = 960 ppm • Permanent hardness = 1000XV3/V1 = 1000 X 35/50 = 700 ppm • Temporary hardness = 1000[(V2-V3)/ V1] = 1000 X (15-10)/48) = 260 ppm L4-7

• Total hardness of water = 960 ppm In Clarks 960 X 0.07 =67.2 0Cl • Permanent hardness = 700 pp In Clarks 700 X 0.07 = 49 0Cl • Temporary hardness = 260 ppm In Clarks 260 X 0.07 =18.2 0Cl

L4-8

Hardness problems • Calculate the temporary, permanent and total hardness in mg/litre of a sample of water containing the following salts :Mg(HCO3)2 = 14.6mg/litre; Ca(HCO3)2 =16.2 mg/litre MgCl2 = 9.5mg/litre; CaSO4 = 13.6 mg/litre ( Assume the atomic mass of Ca to be 40 and that of Mg to be 24).

L4-9

S.No. Constituent

Amount mg/lit

MF

CaCO3 equivale nt

1

CaSO4

13.6

100/136

10

2

MgCl2

9.5

100/95

10

3

Ca(HCO3)2

16.2

100/162

10

4

Mg(HCO3)2

14.6

100/146

10

L4-10

• Temporary Hardness = Ca(HCO3)2 + Mg(HCO3)2 = 10 + 10 = 20 ppm = 20 ppm = 20 mg/L • Permanent Hardness = CaSO4 + MgCl2 = 10 +10 = 20 ppm = 20 ppm = 20 mg/L Total hardness = Temporary + permanent hardness = 20 + 20 = 40 ppm = 40 ppm = 40 mg/L L4-11

Hardness problems • 2 litres of water obtained from a borewell in Patancheru near Hyderabad gave the following analysis for salts :- FeSO4 = 30.4 mg; CaSO4 = 13.6mg; MgCl2 = 38 mg;Ca(HCO3)2 = 32.4 mg; Mg(HCO3)2 = 14.6mg; NaCl = 11.7 mg. Find out the total hardness of water in ppm units, giving temporary and permanent hardness assuming the atomic mass of Fe to be 56, that of Ca to be 40, that of Mg to be 24 and that of Na to be 23.

L4-12

S.No. Constituent

Amount mg/lit

MF

1

FeSO4

30.4

100/152

CaCO3 equivale nt 20

2

CaSO4

13.6

100/136

10

3

MgCl2

38

100/95

40

4

Ca(HCO3)2

32.4

100/162

20

5

Mg(HCO3)2

14.6

100/146

10

6

NaCl

11.7

L4-13

---

--

• Temporary Hardness = Ca(HCO3)2 + Mg(HCO3)2 = 20 + 10 = 30 ppm • Permanent Hardness = FeSO4 + CaSO4 + MgCl2 = 20 +10 +40 = 70 ppm Total hardness = Temporary + permanent hardness = 30 + 70 = 100 ppm L4-14

Hardness problems •

L4-15

Determine the total hardness of a sample of water in 0Fr and 0Clarke which showed the following analysis :Suspended matter = 100 mg; Ca(NO3)2 = 16.4 mg/litre; MgSO4 = 24 mg/litre; MgCl2 = 19 mg/litre; NaOH = 40 mg/litre; KOH = 56 mg/litre ( At. Masses of Na = 23, Mg = 24, K = 39 and Ca = 40.)

S.No. Constituent

1

Ca(NO3)2

2

Amount mg/lit

MF

CaCO3 equivalent

16.4

100/164

10

MgSO4

24

100/120

20

3

MgCl2

19

100/95

20

4

NaOH

40

-

-

5

KOH

56

-

-

L4-16

• Temporary Hardness = Ca(NO3)2 = 10 ppm • Permanent Hardness = MgSO4 + MgCl2 = 20 +20 = 40 ppm Total hardness = Temporary + permanent hardness = 10 + 40 = 50 ppm

L4-17

• Total hardness of water = 50 ppm In Clarks 50 X 0.07 = 3.5 0Cl In French 50 X 0.1 = 5 0Fr • Permanent hardness = 40 pp In Clarks 40 X 0.07 = 2.8 0Cl In French 40 X 0.1 = 4 0Fr • Temporary hardness = 10 ppm In Clarks 10 X 0.07 = 0.7 0Cl In French 10 X 0.1 = 1 0Fr L4-18

ANALYSIS OF WATER  Alkalinity  Estimation

of Chlorides  Estimation of Dissolved Oxygen

L5-1

1. ALKALINITY: 







Alkalinity of water is due to the presence of carbonates and bicarbonates of calcium and magnesium. So the carbonate and bicarbonates in water can be determined by titrating the solution with standard acid like N/50 H2SO4 using first Phenolphthalein and later methyl orange as an indicator conversion of carbonates into bicarbonates occur. In the II step to the solution using methyl orange as indicator (pH 4.2 to 5.4) bicarbonates present in the water and bicarbonates formed from carbonates gets neutralized. The alkalinity of water is attributed to the presence of the 100 Caustic alkalinity (due to hydroxide & carbonate ions) Temporary hardness (due to Bicarbonate ions) L5-2

Analysis of water  PROCEDURE:

Pipette out 100 mL of the water sample in a clean titration flask. Add to it 2 to 3 drops of a phenolphthalein indicator. Run in N/50 H2SO4 from burette. Till the pink colour is just dis-charged. Then to the same pink colour reappears.

L5-3

Analysis of water CALCULATION:  let volume of acid used to phenolphthalein endpoint = V1 mL And  Extra volume of acid used to Methyl orange endpoint = V2 mL  Phenolphthalein alkalinity P= V1 x 50 x 1000000 = 10 V1 ppm  50 x 100 x 1000  Methyl orange alkalinity M = (V1+V2) x 50 x 1000000 = 10(V1+V2) ppm  50 x 100 x 1000 

L5-4

ESTIMATION OF CHLORIDES IN WATER L5-5

 The

chloride ions are present in water in the form of one or more of the compounds like CaCl2, MgCl2, NaCl etc.  The estimation of chloride ions is generally made by titrating the water sample against a standard solution of silver nitrate using potassiumchromate as indicator.  The added silvernitrate precipitate chloride ions as white precipitate of silverchoride, L5-6

as per the following chemical reactions.. NaCl + AgNO3  AgCl + NaNO3 Cl- + Ag+  AgCl

L5-7

 When

all the chlorine ions are removed as AgCl ppt, the excess drop of silvernitrate reacts with potassiumchromate forming silverchromate, which is red in colour.

K2CrO4 + 2AgNO3  Ag2CrO4 + 2KNO3 CrO4-2 + 2Ag+  Ag2CrO4 L5-8

 The

L5-9

endpoint is therefore the change in colour from bright yellow due to brick red colour due to formation of Ag2CrO4 indicator. Then titrate it against N/50 AgNO3 solution till the colour changes from yellow to permanent brick-red.

Calculation: Let volume of N/50 AgNO3 used = V mL  50

x Normality of free chlorine = V x [N/50]

Strength of free chlorine = V x 35.5 x 1000000 25 x 100 x 1000 L5-10 = 14.2 x V 

DISSOLVED OXYGEN L5-11

The oxidation of KI by dissolved Oxygen determines the amount of dissolved Oxygen in water. Iodine is titrated with standard sodiumthiosulphate solution using starch as final indicator. The dissolved molecular Oxygen in water does not react with KI. So, an oxygen carrier is used to bring about the reaction between Oxygen and KI. Hence manganese hydroxide is produced as a result L5-12

 2KOH

+ MnSO4  Mn(OH)2 + K2SO4  2Mn(OH)2 + O2  2 MnO(OH)2  MnO(OH)2 + H2SO4  MnSO4 + 2H2O + [O]  2KI + H2SO4 + [O]  K2SO4 + H2O + I2  2Na2S2O3 + I2  2NaI + Na2S4O6 L5-13

 The

presence of sulphates, nitrates, etc. in rain water gives wrong results in determination of dissolved Oxygen in water because these ions also liberates Iodine from KI. In order to prevent the liberation of Iodine solution from nitrate, If present in water, sodium-azide (NaN3) is added to the alkaline Iodide solution.

L5-14

This reacts with nitrates and gives..  2NaN3

+ H2SO4  2HN3 + Na2SO4

 HNO2

H2O

L5-15

+ HN3  N20 + N2 +

PROCEDURE: 250 mL of water is taken in bottle preventing it to contact with air. 2 mL of Manganoussulphate solution is added by means of pipette, and also 2 mL of alkalineiodine solution is also added. Shake the bottle thoroughly, and repeat the process thrice. The ppt is allowed to settle half-way and mix again.

L5-16

PROCEDURE: Then add to it Conc. H2SO4 insert the stopper and shake the bottle again allow the yellow solution to stand for 5 min. Withdraw 100 mL of solution and titrate it against N/100 solution (Na2S2O3 solution) using freshly prepared starch as indicator. The endpoint will be the disappearance of blue- colour. L5-17

When

the volume of Thiosulphate is V mL then dissolved oxygen content in water is 8 V ppm.

L5-18

Assignment Write a short notes on b. Estimation of Chlorides c. Estimation of Dissolved Oxygen 

L5-19

Methods of Treatment of Water for Domestic Purposes L6,7-1

Methods of Treatment of Water for Domestic Purposes is mainly 3 types, they are..  SEDIMENTATION  Chemical

coagulation  FILTRATION

L6,7-2

SEDIMENTATION 

Sedimentation is the process of removing large suspended particles at the bottom of the reservoir. Which are collected due to gravity.



Sedimentation is the process of allowing to stand undisturbed in big tanks, about 5M deep, when most of the suspended particles settle down at the bottom, due to the force of gravity. The clear supernatant water is then drawn from tank with the help of pumps. The retention period in a sedimentation tank ranges from 2-6 hours.

L6,7-3

SEDIMENTATION 

In order to carry out the sedimentation process successfully, some of the chemicals are added to water before sedimentation. These chemicals are called Coagulants.



The coagulants when added to water, form an insoluble gelatinous, flocculent precipitates, which descends through the water and mixes-up into very fine suspended impurities forming bigger impurities forming bigger impurities or flocs, which easily settles down.



Some of the coagulants used for sedimentation process are alum, ferrous sulphate, sodium aluminate etc.

L6,7-4

SEDIMENTATION TANK

L6,7-5

SEDIMENTATION TANK

L6,7-6

SEDIMENTATION TANK

L6,7-7

SEDIMENTATION TANK

L6,7-8

Chemical coagulants: Alum: K2SO4. Al2(SO4)3. 24H2O NaAlO2 FeSO4

L6,7-9

•Alum is the most widely used in water treatment plants. Al2(SO4)3 + Ca(HCO3)2  2Al(OH)3 + 3CaSO4 + 6CO2 NaAlO2 + 2H2O  Al(OH)3 + NaOH FeSO4 + Mg(HCO3)2  Fe(OH)2 + MgCO3 + CO2

L6,7-10

FILTRATION It

is the process of removing colloidal matter and most of the bacteria, micro-organisms etc, by passing water through a bed of fine sand and other proper-sized granular materials. Filtration is carried out by using Sand-filter.

L6,7-11

Sand filter Sand

filter is used for removing suspended particles, micro-organisms, bacteria etc., by passing through a finely graded sand bed. All the colloidal matter and sediments gets accumulated and water is purified.

L6,7-12

SAND FILTER

L6,7-13

L6,7-14

L6,7-15

L6,7-16

Sand filter

L6,7-17

Sand filter

L6,7-18

SAND FILTER

L6,7-19

ULTRA SAND FILTER

L6,7-20

SAND FILTER

L6,7-21

 In

OPERATION OF SAND FILTER

sand filtration process, water is passed through different sized sand beds like fine, coarse and gravel beds.  The suspended particles are unable to pass through the gaps in the sand bed, because of their size.  Water is first passed through a fine sand-bed.  The suspended particles are first collected and gets clogged in fine sand and then passed through sand-bed.  Here colloidal matter gets collected and the water is free from sediments. L6,7-22

OPERATION OF SAND FILTER  After

passing of water from coarse sand bed it enters through a bed of medium sized stones called gravel or gravel-bed.  This is the last bed where water is filtered completely and the filtered wate is collected at the filter outlet.  During the filtration process the fine pores of sand bed gets seized and clogged.  In order to continue the filtration process, 2-3 cm of fine sand at the top is scrapped, replaced and leveled with L6,7-23

Sand filter operation

L6,7-24

Sand filter operation

L6,7-25

Disinfection:  The

process of destroying/killing the disease producing Bacteria, microorganisms, etc. from the water and making it safe for use, is called disinfection.  Disinfectants: the chemicals or substances which are added to eater for killing the Bacteria. L6,7-26

The disinfection of water can be carried out by following methods:  A)

BOILING  B) BLEACHING POWER  C) CHLORINE  D) OZONISATION

L6,7-27

The disinfection of water can be carried out by following methods: 

A). BOILING: Water for 10-15 minutes, all the diseaseproducing bacteria are killed and water becomes safe for use.

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B) BLEACHING POWER: It is used to purify the drinking water from micro organisms. The purification process is achieved by dissolving 1 kg of bleaching powder in 1000 kiloliters of water. This dissolved water solution is left undisturbed for many hours. When bleaching powder is mixed with water, the result of chemical reaction produces a powerful Germicide called Hypochlorous acid. The presence of chlorine in the bleaching powder produces disinfecting action, kills germs and purifies the drinking water effectively.

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CaOCl2 + H2O  Ca(OH)2 + Cl2  H2O + Cl2  HCl + HOCl L6,7-28  HOCl + Germs  Germs are killed  water purified. 

C) CHLORINE: 

 

Chlorination is the process of purifying the drinking water by producing a powerful Germicide like Hypochlorous acid. When this Chlorine is mixed with water it produces Hypochlorous acid which kills the germs present in water. H2O + Cl2  HOCl + HCl Chlorine is basic (means pH value is more than 7) disinfectant and is much effective over the germs. Hence Chlorine is widely used all over the world as a powerful disinfectant. Chlorinator is an apparatus, which is used to purify the water by chlorination process.

L6,7-29

BREAK-POINT CHLORINATION:  

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

 

Break-point chlorination is a controlled process. In this process suitable amount of Chlorine is added to water. In order to kill all the bacteria present in water, to oxidize the entire organic matter and to react with free ammonia the chlorine required should be appropriate. Break-point determines whether chlorine is further added or not. By Chlorination, organic matter and disease producing Bacteria are completely eliminated which are responsible for bad taste and bad odour in water. When certain amount of Chlorine is added to the water, it leads to the formation of Chloro-organic compounds and chloramines. Addition of some more chlorine leads to destruction of chloro-organic compounds and chloramines. The point at which free residual chlorine begins to appear is terms as “Break-point”.

L6,7-30

BREAK-POINT CHLORINATION:

L6,7-31

OZONISATION:  Ozone

is powerful disinfectant and is readily dissolved in water.  Ozone being unstable decomposes by giving Nascent Oxygen which is capable of destroying the Bacteria.  This ascent Oxygen removes the colour and taste of water and oxidizes the organic matter present in water. O3  O2 + [O] L6,7-32

Assignment  1.

Write short notes on the following :(a) Break-point chlorination (b) Slow sand filtration (c) Impurities in water and their effects. [5+5+6]  2. What is meant by disinfection? What is its significance? Explain the different methods used for disinfection of surface water. L6,7-33