Water Technology

Introduction 

Like air, water is one of the few basic materials which is of prime importance for the preservation of life on this earth. All are aware of the uses of water for drinking, cooking, bathing and farming etc., but few know the importance of water as an engineering material. As an engineering material water is used for producing steam in boilers to generate hydro-electric power, furnishing steam for engines, for refrigeration and air conditioning, for construction of concrete structures, for manufacturing purposes and as a solvent in chemical processes. Hence an engineer should possess knowledge regarding the composition, troubles that arise due to composition of water and prevention of these troubles.

Occurrence

Water is widely distributed in nature. It has been estimated that about 75% matter on the earth’s surface consists of water. Besides visible water on earth, there is a large amount of water under earth to an average depth of over three kilometers. The air contains 12 to 15% of volume of water vapour. Water is found in all living things. The body of human beings consists of about 60% of water. Plants, fruits and vegetables contain 90 to 95% of water.

Sources of Water 

The main sources of natural water are the following :
1) Rain water
2) River water
3) Spring or well water
4) Sea water
1) Rain water : Rain water is the most important source of water and the purest form of natural water. It is naturally distilled water as the water from sea, rivers etc., first evaporates in the great heat of sun to form clouds which afterwards condense and water comes down in the form of rain. As the rain water falls on the surface of the earth and flows as streams or seep through earth’s crust, it takes up dust particles, oxygen, carbondioxide, nitrogen and other dissolved and suspended solids present in the atmosphere and soil.
2) River water : Water in the form of rivers comes from rain water which is not absorbed into the soil and melting of snow on the mountains. While flowing, on the surface of earth, the water dissolves various organic, inorganic and suspended impurities.
3) Spring and well water :  About one third of rain water falls on the surface of earth seeps inside the soil and goes down through it till it is prevented by impervious rock etc. After sometime it then moves up and appears in the form of deep well water or spring water. This water dissolves minerals and because of filtering action of soil contains more of              dissolved solids like Ca(HCO3)2, CaSO4, MgSO4, MgCl2 etc. But it is free from suspended impurities hence more suitable for drinking purposes.
4) Sea water : Sea is the largest source of water. The river water which carries many dissolved and suspended impurities finally collects into sea. As water continuously evaporates from sea leaving behind the dissolved salts, the concentration of dissolved impurities in the form of carbonates, sulphates, chlorides, bromides and iodides of sodium, potassium, calcium, magnesium and a number of various other components is more. The percentage of impurities in sea water is around 3.6 in which about 2.7% is the common salt.

impurities of water

Chemically pure water is composed of two parts of hydrogen and one part of oxygen by volume and dissolves many substances. So natural water is not pure. It dissolves various impurities.
The various impurities present in natural waters may be broadly categorised as follows :
  1)   Suspended matter : It is of two types :
    a) Inorganic suspended matter, like clay and sand.
b) Organic suspended matter like animal and vegetable matter.
River water mainly contains suspended matter, which is responsible for turbidity of water.
2)   Dissolved substances : The dissolved substances are classified as :
  Inorganic salts like bicarbonates, chlorides, sulphates of calcium and magnesium in the dissolved form causes hardness to water. The bicarbonates of calcium and magnesium dissolve in water when it passes through the rocks of dolomite or limestone etc., in presence of air.
CaCO3 + H2O + CO2  Ca(HCO3)2
MgCO3 + H2O + CO2  Mg(HCO3)2
The dissolved impurities like  the salts of sodium, potassium cause alkalinity to water.
The other dissolved salts include iron, manganese, silica and alumina. Sometimes trivalent cations like Fe3+ and Al3+ may also be present in water. Different waters have different degrees of hardness.
Dissolved gases like CO2, N2, O2, H2S are responsible for the odour and acidity in water. The presence of O2 causes corrosion in boilers.
The total dissolved salts may range from a few parts per million in snow water to several  thousand parts per million in water obtained from mineral springs.
3) Colloidal impurities : The matter which exists in very finely divided state such that their small quantities are not visible to naked eye is called colloidal matter, which comprise of very finely divided silica, clay, aluminium hydroxide and micro-organisms. The micro-organisms include protozoa and some forms of vegetable life. These cause colour, odour and some of these are responsible for causing diseases in human beings.
4) Organic impurities : The organic impurities include alage, fungi and plant matter etc., which are vegetative by origin and micro organisms, bacteria are animal by their origin.

Sources of Impurities

How are different types of impurities introduced into water ?
a) When rain water floods to nearby streams it carries floating impurities along with it.
b) Gases like CO2, O2 etc. are picked up from atmosphere.
c) When water percolates through the layers of soil it dissolves the salts of calcium, magnesium, potassium, sodium etc. which remain as dissolved salts in the composition of water.
d) The organic impurities are introduced into water due to the decomposition of plant and animal remains in water and due to the contamination of water with dead biological matter, sewage and industrial waste producing organic impurities.
1.4.2 Effect of Impurities Present in Natural Water
Water containing impurities has a number of bad effects when used for domestic and industrial purposes. Various bad effects of impurities of water are mentioned below.
1)  Bad effect for domestic purposes :
The dissolved salts of calcium and magnesium in water does not give lather with soap. Soap is a mixture of sodium and potassium salts of higher fatty acids like stearic acid, oleic and palmetic acids. When worked with hard water lather is not produced due to the precipitation of soap as calcium or magnesium salts. Hence large amount of soap is consumed. In cooking also impurities dissolved in water increase the boiling point of water and scale formation on the inner sides of heating vessels and utensils leading to the wastage of fuel. The suspended impurities in water are harmful for drinking purposes, as it contains harmful pathogenic bacteria.
2)  Disadvantages for industrial purposes :
Water finds a great use in number of industries such as paper, dyeing, textile, sugar etc. The disadvantages of impurities present in water when used in some common industries are given below.
a) Paper industry : The dissolved impurities react with the chemicals used to give paper a smooth glossy surface effecting the colour of paper.
b) Dyeing industry : The dissolved salts will react with dyes to form undesirable precipitates which give improper shade and often cause spots on the fabric.
c) Textile industry : When soap is used for washing the yarns, undesirable precipitates produced will adhere to the fabric while dyeing and does not give exact colour. Iron and manganese salts may cause spots on fabrics.
d) Sugar industry : Water for sugar refineries must be free from sulphates, alkali carbonates, nitrates and bacteria, because in presence of these impurities sugar may not crystallise well and it may be deliquescent and decompose during storage.
e) Bakeries : The presence of organic matter is harmful and it may affect the action of yeast.
f) Boilers : The dissolved salts in water produce scale, sludges, caustic embrittlement etc., during the steam generation in boilers.
Similarly the impurities present in water have their own effects, when water is used for various other numerous industries.

Uses of Water 

Water is essential for the sustenance and continuation of life of both animal and vegetable kingdom. The various other uses of water are as follows.
1) Domestic purposes : Large quantities  of water is used for household purposes such as drinking, bathing, cooking, washing etc.  The water safe to drink is called potable water and it should be free from floating impurities turbidity and biological impurities.
2) Industrial purposes : Water finds lots of uses in number of industries. Vast amount of running water is required for various industries such as paper, textile, sugar, soap making, starch, bleaching and dyeing, chemical works, distilleries, refining of metals, extraction of sulphur and in manufacturing of fuel gases, inks etc. It is also used as a solvent, as a cooler, for cleansing, for carrying industrial wastes and for generation of hydroelectric power. Moreover various industries require water for boilers to generate steam. Water used for industrial purpose must be free from turbidity and dissolved salts.
3) Municipal purposes : Besides the above uses large quantity of water is required for public use such as flushing out sewers and underground drains, sprinkling public lawns and parks, swimming pools and fire fighting etc. The water used for this purpose must be free from floating impurities and turbidity.
4) Agricultural purposes : Quite a large amount of water is used for agricultural purposes and this water must be free from biological impurities and dissolved salts.

HARD  WATER 

The water which does not produce lather or produce very little lather with soap is called Hard water. On the other hand, soft water readily produces a lot of lather when mixed with little soap. The hardness of water is caused by the presence of dissolved salts  such as bicarbonates, sulphates, chlorides and nitrates of bivalent metal ions like calcium and magnesium. Soap is sodium or potassium salt of higher fatty acids like stearic, oleic and palmetic acids. When soap is mixed with soft water lather is produced due to sodium stearate.
Na-stearate + H2O  NaOH + Stearic acid
Stearic acid + Na-stearate  formation of lather
When soap comes in contact with hard water, sodium stearate will react with dissolved calcium and magnesium salts and produce calcium stearate or magnesium stearate which is white precipitate called ‘scum’.
2Na-stearate +      Ca++          Ca-stearate +     2Na+
                          (soluble)        (soluble)                (insoluble)       (soluble)
    (scum)
Different types of water have different degrees of hardness. The different types of water are commercially classified on the basis of degree of hardness as follows :                    Table - 1.1
   Hardness Name of water
  0-70 mg/litre soft water
 70-150 mg/litre moderate hard water
 150-300 mg/litre hard water
  300 mg/litre and above very hard water

Degree of Hardness 

The total hardness of water is caused by eight different dissolved salts of calcium and magnesium. Ca(HCO3)2, MgH(CO3)2, CaCl2, MgCl2, CaSO4, MgSO4, Ca(NO3)2 and Mg(NO3)2. Hence the hardness of water is expressed in terms of calcium carbonate equivalents. The weights of different salts causing hardness are converted to weights equivalent to that of calcium carbonate. If a sample of water contains two or more than two salts, their quantities are converted equivalent to CaCO3 as mentioned above and then the sum will give the total hardness in CaCO3 equivalents. CaCO3 is selected for expression of the degree of hardness because the molecular weight of CaCO3 is 100, which is easy for calculation and CaCO3 is an insoluble salt. All the dissolved salts of calcium are precipitated as CaCO3. Thus 120 parts by weight (1gm molecules weight) of MgSO4 would react with the same amount of soap as 100 parts by weight of CaCO3 (1 gm.mole) hardness. Hence weight in terms of CaCO3 would be equal to weight of MgSO4 in water multiplied by 100/120. Therefore 100 parts by weight of CaCO3 hardness must be equivalent to
1. 162 parts by weight of Ca(HCO3)2 hardness
2. 146 parts by weight of Mg(HCO3)2 hardness
3. 136 parts by weight of CaSO4 hardness
4. 111 parts by weight of CaCl2 hardness
5. 164 parts by weight of Ca(NO3)2 hardness
6.   120 parts by weight of MgSO4 hardness
7. 95 parts by weight of MgCl2 hardness and
8. 148 parts by weight of Mg(NO3)2 hardness
The method of calculating degree of hardness will be clear from the following formula.
Hardness of the hardness causing salt in terms of CaCO3   =
Problem 1.1 :
Calculate the CaCO3 equivalents of the following dissolved salts present in 1000 litre of a sample of water containing in ppm. Ca(HCO3)2 = 16.2, CaCl2 = 11.1, MgSO4 = 60,                              MgCl2 = 19.
Sol.  
Salt Quantity of   Molecular Equivalent of CaCO3 Equivalent salt present   weight CaCO3 (mg/l)     of H.C.S in 1000 l of water            
Ca(HCO3)2 16.2 162 = 10
CaCl2 11.1 111 = 10
MgSO4 60 120 = 50  
MgCl2 19 95 = 20    

Units of Hardness

There are four different units in which the hardness of water is expressed as given below.
1) Parts per million (ppm) : The number of parts by weight of CaCO3 equivalents of hardness causing salt present in one million parts of water. For example, 50 ppm of Ca(HCO3)2 is the expression of 50 parts by weight of Ca(HCO3)2 in one million parts that is 106 parts of water. One million is equal to 10 lakhs.
2) Milligrams per litre (mg/l) : The amount of the CaCO3 equivalents hardness causing salt in milligrams present in one litre of water. Taking the density of water as unity, one kilogram of water at 4 0C is equal to one litre. One kilogram of water is equal to 1000 × 1000 mgs. That is again 106 parts. It becomes number of parts of hardness causing salt in one million (106) parts of water which is equal to parts per million.
3) Degree Clark (0Cl) : In the English system hardness of water is expressed in terms of degree clark (0Cl), that is number of grains (1/7000 lb) of CaCO3 equivalents of hardness causing salt present in one gallon (10 lb) of water, which is otherwise expressed as number of parts of CaCO3 equivalents of hardness causing salt in 70,000 parts of water.
4) Degree French (0Fr) : It is a French unit which is expressed as number of parts of CaCO3 equivalents hardness causing substance in 105 parts (1,00,000) of water.
The hardness of water can be converted into all the four units by making use of the following interconversion formula.
1 ppm = 1 mg / litre = 0.07 0Cl = 0.1 0Fr
1 0Cl = 1.43 0Fr = 14.3 ppm = 14.3 mg/l
Problem 1.2 :
A sample of hard water contains 120 mg/l of hardness. Express the hardness of water in ppm, degree french and degree clark.
Sol. Hardness of water sample = 120 mg/l
The conversion of hardness is = 1 ppm = 1 mg/l = 0.07  0Cl = 0.1 0Fr
Hardness of water in degree Clark (0Cl) = 120  0.07 = 8.4 0Cl    
Hardness of water in degree French (0Fr) = 1200.1 = 12 0Fr    
Hardness of water in ppm = 120 ppm.
Problem 1.3 :
A water sample found to contain hardness of 80 0Cl. Express the hardness of water in milligrams per litre, parts per million and degree french.
Sol.     Hardness of water sample = 80 0Cl
Conversion of hardness is 1 0Cl = 1.43 0Fr = 14.3 ppm = 14.3 mg/l.
Hardness of water in ppm = 80  14.3 = 1144 ppm
Hardness of water in mg/l = 80  14.3 = 1144 mg/l
Hardness of water in 0Fr = 80  1.43 = 114.4 0Fr
1.6.3  Types of Hardness
The hardness of water is of two types :
1) Temporary hardness or carbonate hardness
2) Permanent hardness or non-carbonate hardness
1)  Temporary hardness : Temporary hardness is caused by two dissolved bicarbonate salts Ca(HCO3)2 and Mg(HCO3)2. The hardness is called temporary because it can be removed easily by means of  boiling.
Ca(HCO3)2  CaCO3  + H2O + CO2
Mg(HCO3)2   Mg(OH2)  + 2CO2
2)  Permanent hardness : Permanent hardness of water is due to the dissolved chlorides, sulphates and nitrates of calcium and magnesium. These salts are CaCl2, CaSO4, Ca(NO3)2, MgCl2, MgSO4, Mg(NO3)2. This hardness cannot be removed easily by boiling. Hence it is called permanent hardness. Only chemical treatment can remove this hardness.
Total hardness of water = Temporary hardness + Permanent hardness
Problem 1.4 :
A sample of hard water contains the following dissolved salts per litre. CaCl2 = 111 mgs, CaSO4 = 1.36 mgs, Ca(HCO3)2 = 16.2 mgs, Mg(HCO3)2 = 14.6 mgs, Silica = 40 mgs, Turbidity = 10 mgs. Calculate the temporary, permanent and total hardness of water in ppm, degree Clark and degree French.                                               (JNTU MID Exam)
Sol :
Hardness causing Quantity Mol. wt. of Equivalent of salt (H.C.S) of (H.C.S) H.C.S.      CaCO3
CaCl2 111 mg/l 111  = 100 mg/l
CaSO4 1.36 mg/l 136 = 1 mg/l
Ca(HCO3)2 16.2 mg/l 162  = 10 mg/l
Mg(HCO3)2 14.6 mg/l 146  = 10 mg/l
Silica and turbidity must not be considered because they do not cause hardness to water.
Total hardness of water = 100 + 1 + 10 + 10 = 121 mg/l
Temporary hardness of water = hardness of Ca(HCO3)2 and Mg(HCO3)2 in  terms                                                          of CaCO3 equivalents
= 10 + 10 = 20 mg/l
      Permanent hardness = hardness of CaCl2 + CaSO4 in terms of CaCO3 equivalents  = 100 + 1 = 101 mg/l
Conversion of hardness :
1 ppm = 1 mg/l = 0.07 0Cl = 0.1 0Fr
Total hardness of the sample of water = 121 ppm = 121 mg/l
121 × 0.07 = 8.47 0Cl and 121 × 0.1 = 12.1 0Fr
    Permanent hardness = 101 mg/l, 101 ppm, 7.07 0Cl, 10.1 0Fr
Temporary hardness = 20 mg/l, 20 ppm, 1.4 0Cl and 2 0Fr
Problem 1.5 :
Calculate the temporary and permanent hardness in degree french of a water sample collected in Anantapur district. The analysis of water is as follows. Ca(HCO3)2 = 16.2 ppm, Mg(HCO3)2 = 14.6 ppm, CaSO4 = 13.6 ppm, MgSO4 = 12 ppm, MgCl2 = 9.5 ppm.                                                      (JNTUA 2012)
Sol :
Hardness causing Amount Mol. wt. of CaCO3 equivalent salt (HCS) of H.C.S. in ppm H.C.S of H.C.S
Ca(HCO3)2 16.2 162     = 10 ppm
Mg(HCO3)2 14.6 146 = 10 ppm
CaSO4 13.6 136 = 10 ppm
MgSO4 12 120 = 10 ppm
MgCl2 9.5 95 = 10 ppm
Temporary hardness of water = Hardness of water due to Ca(HCO3)2 and Mg(HCO3)2 in CaCO3 equivalents
= 10 + 10 = 20 ppm
Permanent hardness of water = Hardness of water due to CaSO4 + MgSO4 + MgCl2 in CaCO3 equivalents
= 10 + 10 + 10 = 30 ppm
Problem 1.6 :
Calculate the temporary and permanent hardness of 100 litres of water containing the following impurities per litre. MgCl2 = 19 mgs,  MgSO4 = 60 mgs, NaCl = 36.5 mgs, CaCl2 = 11.1 mgs, Ca(HCO3)2 = 32.4 mgs, Mg(HCO3)2= 7.3 mgs.      (JNTUH, Dec 2012)
Sol :
Hardness Causing     Amount     Mol. wt. of CaCO3 equivalents          Salt (HCS)       of HCS in mg/l H.C.S H.C.S
1. MgCl2 19 95     = 20
2. MgSO4 60 120 = 50
3. CaCl2 11.1 111 = 10
4. Ca(HCO3)2 32.4 162 = 20
5. Mg(HCO3)2 7.3 146 =  5
Temporary hardness of water = Hardness of  Ca(HCO3)2 + Mg(HCO3)2 in CaCO3 equivalents
= 20 + 5 = 25 mg/l
Temporary hardness of 100 litres of water = 25  100 = 2500 mg / 100 l
Permanent hardness of water = hardness of MgCl2 + MgSO4 + CaCl2 in terms of  CaCO3 equivalents
= 20 + 50 + 10 = 80 mg/l
Problem 1.7 :
100 ml of a water sample has a hardness equivalent to 12.5 ml of 0.08 N MgSO4. What is the hardness of water in ppm?                                               (JNTU MID Exam)
Sol :
Assuming 1 N of 1 ml MgSO4 = 1N of 1 ml of CaCO3
Normality of water in 100 ml of
V1 = Volume of MgSO4 = 12.5
N1 = Normality of MgSO4 = 0.08
V2 = Volume of water = 100
N2 = Normality of water =  = 0.01
Amt of MgSO4 in water = Normalityequivalent = 0.0150 = 0.5 gms/litre
Hardness of water = 0.51000 = 500 mg/l
Hardness of water sample = 500 ppm (Hint : 1 ppm = 1 mg/l)
Problems for Practice :
1) A sample of water contains 100 ppm of total hardness and 25 ppm of temporary hardness. Calculate the permanent hardness of water in degree Clark and degree French.                              [Ans : 5.25 °cl, 7.5 °Fr]
2) One litre of water from an underground reservoir in Nalgonda town in Andhra Pradesh, found to have the following dissolved salts. Mg(HCO3)2 = 0.0146 gms, Ca(HCO3)2 = 0.0081 gms, MgSO4 = 0.0012 gms, CaSO4 = 0.0136 gms, NaCl = 0.585 gms, organic impurities 100 mgs. Calculate the temporary, permanent and total hardness of water in degree french.                       (JNTUH, Dec-2012 Set 3)                                              [Ans : Total Hardness = 2.98  oFr,                     Temporary = 1.50  oFr, Permanent = 1.48 oFr]
3) A water sample was analysed for its dissolved salt CaSO4. 50ml of the water sample was found to contain CaSO4 hardness equivalent to 20 ml of 0.01N. Calculate the hardness of water in ppm, degree clark and degree french.
   [Ans : 200 ppm,20 °Fr, 14 °cl]
4) A sample of water contains the following dissolved salts : Mg(HCO3)2 = 22 mg/l, MgCl2 = 30 mg/l, CaCl2 = 85 mg/l, CaSO4 = 28 mg/l. Calculate the temporary and permanent hardness of water.                         (Mangalore Sept - 1994)            [Ans : 15.07 ppm, 128.75 ppm]
5) Calculate the hardness of water containing the following per litre: CaSO4 = 16.2 mg, Mg(HCO3)2 = 1.4 mg, MgCl2 = 9.5 mg            (Ravishanker July 1994)                                       [Ans : 22.87 ppm]
6) A sample of water on analysis has been found to contain the following dissolved salts in ppm. Ca(HCO3)2 = 4.86, Mg(HCO3)2 = 5.84, CaSO4 = 6.8, MgSO4 = 8.4. Calculate the temporary and permanent  hardness of the water. (At wt of Ca = 40, Mg = 24, C = 12, S = 32, O = 16, H = 1)                 (Anna - Jan 2001)                    [Ans : Temporary hardness = 7 ppm, permanent hardness = 12 ppm]
7) A water sample contains Ca(HCO3)2 = 32.4mg/l, Mg(HCO3)2 = 29.2 mg/l, MgCl2= 50mg/l and CaSO4 = 13.5 mg/l. Calculate the temporary and total hardness.
(JNTUK - 2014)
[Ans : T.H = 40 mg/l, total hardness = 102.6 mg/l)

 Estimation of Hardness by EDTA Method  

The sample of water is analysed quantitatively for the amount of dissolved ca2+ and mg2+ salts present in it. Three methods are employed for the estimation of hardness. They are
1) Soap titration method
2) O.Hehner’s alkalimetric method
3) EDTA method
EDTA method :
This is a complexometric method where Ethylenediaminetetra acetic acid (EDTA) is the reagent. EDTA forms complexes with different metal ions at different pH. Calcium and magnesium ions form complexes with EDTA at pH 9-10. To maintain the pH 9-10 NH4Cl, NH4OH buffer solution is used. An alcoholic solution of Eriochrome black-T is used as an indicator. The disodium salt of EDTA under the trade name Triplex-III is used for complexation.

             EDTA                       Disodium salt of EDTA






M – EDTA Complex (Stable, colourless)
Basic Principle :
When hard water comes in contact with EDTA, at pH 9-10, the Ca++ and Mg++ forms stable, colourless complex with EDTA.
To the hard water sample the blue coloured indicator Eriochrome Black-T (EBT) is added along with the NH4Cl, NH4OH buffer solution. EBT forms an unstable, wine red complex with Ca++ and Mg++.
 Ca2+     +  Mg2+   +      EBT   complex
(from hard water)                   (unstable, wine red coloured)
The wine red coloured [Ca-EBT, Mg-EBT] complex is titrated with EDTA, where EDTA replaces EBT from  complex and forms stable, colourless  [Ca - EDTA] [Mg - EDTA] complex releasing the blue coloured indicator EBT into H2O. Hence the colour change at the end point is wine red to blue colour.
   
(unstable winered               (stable, colourless     (Blue colour)
complex) complex)
The titration is carried out in the following steps.                                            
Preparation of standard hard water : Dissolve 1 gm of pure, dry CaCO3 in minimum quantity of dilute HCl and evaporate the solution to dryness on a water bath. Dissolve the residue in distilled water to make 1 litre in a standard flask and shake well.
Molarity of standard hard water solution = =  = 0.01M
Preparation of EDTA solution : Dissolve 4 gms of pure EDTA crystals along with  0.1 gm of MgCl2 in one litre of distilled water.
Preparation of indicator : Dissolve 0.5 gms of Eriochrome Black-T in 100 ml of alcohol.
Preparation of buffer solution : Add 67.5 gm of NH4Cl to 570 ml of concentrated ammonia solution and dilute with distilled water to one litre.
Standardisation of EDTA solution  : Pipette out 20 ml of standard hard water solution into a conical flask. Add 2-3 ml of buffer solution and 2-3 drops of EBT indicator. Titrate the wine red coloured complex with EDTA taken in a burette after rinsing it with EDTA solution till the wine red colour changes to clear blue. Note the burette reading and let the volume be x ml. Repeat the titration to get concurrent values.
Standardisation of hard water sample : Pipette out 20 ml of the water sample collected from the tap into a 250 ml conical flask, add 2-3 ml of buffer solution and 2-3 drops of EBT indicator. Titrate the winered coloured solution with EDTA taken in the burette till a clear blue coloured end point is obtained. Let the volume of EDTA be y ml. Repeat the titration to get concurrent values.
Standardisation for permanent hardness : Pipette out 100 ml of hard water sample in a beaker and boil till the volume reduces to 20 ml. [All the bicarbonates of Ca++ and Mg++ decomposes to CaCO3 and Mg(OH)2]. Cool the solution and filter the water into a conical flask, wash the beaker and precipitate with distilled water and add the washings to conical flask. Add 2-3 ml of buffer solution and 2-3 drops of EBT indicator and titrate with EDTA solution taken in the burette till a clear blue colour end point is obtained. Note the burette reading. Let the volume be z ml.
Calculations :
      Molarity of standard hard water solution = 0.01M (calculated in the preparation of standard hard water)
Molarity of EDTA solution (M2) :
    =
Where,
n1 & n2 are no. of moles of Ca++ and EDTA = 1 each
 V1 = Volume of standard hard water, M1 = Molarity of standard hard water
V2 = Volume of EDTA [titre value (x ml)], M2 = Molarity of EDTA
M2 =  =
Molarity of hard water sample (M3) :
=
 n2 = n3 = 1
V2 = Volume of EDTA ( ml);  M2= Molarity of EDTA (calculated in the previous step)
V3 = Volume of hard water sample, M3 = Molarity of hard water sample
M3 =  =
Total hardness of water = M3×100 gms/1 litre = M3 × 100 × 1000 mg/l or ppm
Permanent hardness of water :
 =
n2 = n4 = 1
V2 = Volume of EDTA (titre value z ml) ;  M2 = molarity of EDTA
V4 = Volume of water sample containing permanent hardness (100 ml)
M4 = Molarity of water sample containing permanent hardness.
M4 = =
Permanent hardness of the water sample = M4 × 100 × 1000 ppm
Temporary hardness of the water sample
= (Total hardness – Permanent hardness)
                                   =  [(M3 × 100 × 1000) – (M4 × 100 × 1000)] ppm
Problem 1.8 :
1 gm of CaCO3 was dissolved in HCl and the solution was made upto one litre with distilled water. 50 ml of the above solution required 30 ml of EDTA solution on titration. 50 ml of hard water sample required 40 ml of the same solution of EDTA for titration. 50 ml of the hard water after boiling, filtering etc., required 30 ml of the same EDTA solution for titration. Calculate the temporary hardness of the water. (JNTU 1985)
Sol : Molarity of CaCO3 solution (M1) =  = 0.01M
Molarity of EDTA solution (M2) =
V1 = Vol. of CaCO3 solution = 50 ml
M1 = Molarity of CaCO3 solution = 0.01M
V2 = Vol. of EDTA = 30 ml
   =  = 0.016M
Molarity of hard water solution (M3) =
V2 = Vol. of EDTA = 40 ml
M2 = Molarity of EDTA = 0.016M
V3 = Volume of hard water = 50 ml
M3=  = 0.0128M
Total hardness of water = 0.0128 × 100 × 1000 = 1280 ppm
Permanent hardness of water :
=        
M4 =
n4 = n2 = 1
V2 = Vol. of EDTA = 30 ml
M2 = Molarity of EDTA = 0.016
V4 = Vol. of permanent hardness containing water = 50
M4 = = 0.0096 M
Permanent hardness of water = 0.0096 × 100 × 1000
= 960 ppm
Temporary hardness = Total hardness – Permanent hardness
  = 1280 – 960 = 320 ppm
Problem 1.9 :
Calculate the temporary, permanent and total hardness of water in ppm, 0Cl and 0Fr from the following determination. 20 ml of 0.05 solution of standard hard water required 40 ml of EDTA for titration. 20 ml of a sample of hard water consumed 30 ml of the same EDTA and 20 ml of hard water after boiling, filtering etc. required 20 ml of EDTA for titration.                                    [JNTU 1987]
Sol : Molarity of EDTA (M2) =
V1 = Vol. of standard hard water (SHW) = 20 ml
M1 = Molarity of SHW = 0.05M
V2 = Vol. of EDTA = 40 ml
M2 =  = 0.025M
Molarity of hard water sample (M3) =
V2 = Vol. of EDTA = 30 ml
M2 = Molarity of EDTA
V3 = Vol. of hard water = 20 ml
M3 =  = 0.0375M
Total hardness of water = 0.0375 × 100 × 1000 = 3750 ppm
                                                            = 375 0Fr
= 3750 × 0.07 = 262.5 0Cl
Molarity of permanent hardness containing water (M4) =
V2 = Vol. of EDTA = 20 ml
M2 = Molarity of EDTA = 0.0025
V4 = Vol. of permanent hardness containing water = 20 ml
M4 = = 0.0025M
Permanent hardness of water = 0.0025 × 100 × 1000 = 250 ppm
250 ppm = 25 0Fr = 250 × 0.07 = 17.50 0Cl
Temporary hardness of the water sample
                           = (Total hardness – Permanent hardness)
            = 375 – 250 = 125 ppm
125 ppm = 12.5 0Fr = 12.5 × 0.07 = 8.75 0Cl
Problems for Practice :
1. A standard hardwater contains 15 g. of CaCO3 per litre. 20 ml of this required 25 ml of EDTA solution. 100 ml of sample of water required 18 ml of EDTA solution. The same sample after boiling required 12 ml of EDTA solution. Calculate the temporary hardness of water in terms of ppm.                      (Poona I BE 1992)  [Ans : 720 ppm]
2. 50 ml of a standard hardwater containing 1 mg of pure CaCO3 per ml consumed 20 ml of EDTA. 50 ml of a water sample consumed 25 ml of same EDTA solution using Erio-T indicator. Calculate the total hardness of water sample in ppm.           (Vikram II BE 1987) [Ans : 1250 ppm]
3. Calculate the hardness of a water sample whose 10 ml required 20 ml of EDTA 20 ml of CaCl2 solution, whose strength is equivalent to 1.5 gm of calcium carbonate per litre, required 30 ml of EDTA solution.                      (Kurukshetra 1987) [Ans : 200 ppm]

Potable Water  

Access to good quality drinking water is essential to human health. The World Health Organisation estimates that nearly a quarter of the world’s six billion people currently lack access to good quality water for drinking, personal hygiene, and domestic use or sanitation. Water free from contaminants or water that is safe for human consumption is called potable water.
The common specifications or standards recommended for drinking water are as follows :
1) It should be colorless, odourless and clear.
2) It should be of good taste.
3) Its turbidity should be less than 10 ppm.
4) Its pH should be in the range of 7.0-8.5.
5) Its total hardness should be less than 125 ppm.
6) It should be free from pathogenic bacteria.
7) It should be free from objectionable dissolved gases like H2S.
8) It should be free from metals such as lead, arsenic, chromium and manganese.
9) Its chloride, fluoride and sulphate contents should be less than 250 ppm, 1.5 ppm and 250 ppm respectively.

Treatment of Water for Municipal Supply

The treatment of water for drinking purposes mainly includes the removal of  suspended impurities, colloidal impurities and harmful pathogenic bacteria. The following is the flow diagram of the water treatment for domestic purposes (Fig. 1.1) and various stages involved in purification are given as :

Fig. 1.1  Flow diagram of treatment of water for municipal supply
1) Screening : The water is passed through screens having large number of holes in it, to remove floating impurities like wood pieces, leaves etc.
2) Aeration : The water is then subjected to aeration which
   i)  helps in exchange of gases between water and air
ii)  increases the oxygen content of water
iii)   removes the impurites like Fe and Mn by precipitating as their hydroxides.
3) Sedimentation with coagulation : The suspended and colloidal impurities are allowed to settle under gravitation. The basic principle of this treatment is to allow water to flow at a very slow velocity, so that the heavier particles settle under gravitation. For setting of fine particles, coagulants like alums, sodium aluminate and salts of iron are added, which produces gelatinous precipitates called floc. Floc attracts and helps accumulation of the colloidal particles resulting in setting of the colloidal particles. The following is the sedimentation tank used for the removal of colloidal impurities.

Fig. 1.2  Sedimentation tank for purification of colloidal impurities
  4)   Filtration : Filtration helps in removal of the colloidal and suspended impurities not removed by sedimentation. Usually sand filters are employed. There are two types of sand filters, slow sand filters and rapid sand filters or pressure filters. In slow sand filter, the filter bed consists of three layers of sand of different particle size. A fine sand  layer on the top is supported by coarse sand layer, which is supported by gravel. The colloidal impurities are retained by the fine sand layer resulting in the very slow filtration of water. Periodically the top  layer of the fine sand layer is scraped  off, washed, dried and introduced into the filter bed for reuse (Fig. 1.3). Rapid sand filters make use of compressed air for fast filtration             (Fig. 1.3(b)).
(a) Slow sand filter                (b) Rapid sand filter
Fig. 1.3

Sterilisation or Disinfectation 

Destruction of harmful pathogenic bacteria from drinking water is carried out by sterilisation or disinfectation. The following are the methods adopted for sterilisation and disinfectation.
a) Boiling : By boiling water for 15-20 minutes, harmful bacteria are killed. This is not possible for the municipal supply of water. This method of sterilisation is adopted for domestic purpose.
b) Passing ozone : Ozone when passed into water acts as disinfectant. Ozone is an unstable isotope of oxygen, produces nascent oxygen which is a powerful disinfectant.
O3O2 + [O]
                          nascent oxygen
This treatment is costly and ozone is unstable and cannot be stored for a long time.
c) Chlorination : The process of utilising chlorine as a powerful disinfectant is called chlorination. There are three types of chlorinating reagents.
i) By passing chloramines : Chlorine is mixed with ammonia in the ratio 2 : 1 by volume to form a stable chloramine which generates hypochlorous acid, a powerful disinfectant, that kill bacteria as shown below.
Cl2 + NH3  ClNH2 + HCl
ClNH2 + H2O  HOCl + NH3
Hypochlorous acid (HOCl) inactivates the enzymes of bacteria and kills bacteria. Chloramine is useful for disinfecting swimming pools.
ii)  By bleaching powder : Bleaching powder contains 80% chlorine. When bleaching powder is used as disinfectant, this method is also called hypo-chlorination because the disinfectation is due to hypochlorous acid.
CaOCl2 + H2O  Ca(OH)2 + Cl2
                               (bleaching powder)
                           Cl2 + H2O  HOCl  + HCl
                                                                             (disinfectant)
iii)  Chlorination : The process of applying calculated amount of chlorine to water inorder to kill the pathogenic bacteria is called chlorination. Chlorine also reacts with water and generates hypochlorous acid, which kills bacteria.
Cl2 + H2O  HOCl + HCl
                 (hypochlorous acid)
Chlorine is a powerful disinfectant than chloramine and bleaching powder. Calculated amount of chlorine must be added to water because chlorine after reacting with bacteria and organic  impurities or ammonia, remains in water as residual chlorine which  gives  bad taste, odour and toxic to human beings. The amount of chlorine required to kill bacteria and to remove organic matter is called break point chlorination.
The water sample is treated with chlorine and estimated for the residual chlorine in water and plotted a graph as shown below which gives the break point chlorination.

From graph it is clear that
‘a’ gms of chlorine added oxidises reducing impurities of water.
‘b’ gms of chlorine added forms chloramines and other chloro compounds.
‘c’ gms of chlorine added causes destruction of bacteria.
‘d’ gms of chlorine is residual chlorine.
‘c’ gms is the break point for the addition of chlorine to water. This is called break point chlorination.
Advantages of break-point chlorination :
1. It removes bad taste, colour, oxidises completely organic compounds, ammonia and other reducing impurities.
2. It destroys completely (100%) all disease producing bacteria.
3. It prevents growth of any weeds in water.
Dechlorination :
Overchlorination after ‘break point’ produces unpleasant taste, odour, toxicity to water. The overchlorination is  removed by passing water through a bed of granular carbon and also by the addition of SO2 and sodium thiosulphate.
  SO2 + Cl2 + 2H2O H2SO4 + 2HCl
Na2S2O3 + Cl2 + H2O  Na2SO4 + 2HCl

Boiler Feed Water and boiler troubles

A boiler is a closed vessel in which water under pressure is transformed into steam by the application of heat. The steam so generated is used in industries and for generation of power. The water  used for steam generation in boilers  should be free from impurities which lead to operational troubles in boilers.
The commonly associated problems of boiler feed water are :
1) Hardness causing constituents like dissolved salts of calcium and magnesium. Produce sludges and scales which would result in wastage of fuel, lowering of boiler safety, decrease in its efficiency and danger of explosion of the boiler. Hence the hardness of boiler feed water should be below 0.2 ppm.
2) Presence of caustic alkalies lead to caustic embrittlement. Hence the alkalinity (due to OH–) of boiler feed water should lie between 0.15 ppm to 0.45 ppm.
3) Boiler feed water should be free from dissolved gases like O2, CO2, etc. in order to prevent boiler corrosion and
4) Turbidity, oil etc. should not be present in boiler feed water as they result in priming and foaming in the boilers.
Hence water must be softened before being used in boilers.
The following are the boiler troubles that arise due to the presence of  dissolved salts.
1) Priming and foaming
2) Caustic embrittlement
3) Boiler corrosion
4) Scale and sludge formation

Priming and Foaming 

Priming is the phenomenon of carrying out droplets of water with steam in boilers.  Because of very rapid and violent boiling of water inside the boiler, the water particles mix up with steam and pass out from the boiler. This process of wet steam generating is caused by several reasons.
Reasons for priming :
Priming in boilers arise due to the following reasons.
1) Due to the presence of large quantity of suspended organic matter, oily matter and alkalies.
2) High steam velocities.
3) Sudden boiling.
4) Due to defective design of the boiler and wrong operation. If the steam gets           condensed in the outer tubes, the pressure outside the boiler in these tubes may be decreased with the result that water inside the boiler jumps with ebulition and passes out with steam into the boiler tubes.
5) Sudden increase in steam demand also causes priming.
6) When the level of water rises due to foaming, priming arises. So priming is generally associated with foaming.
Prevention of priming : Priming can be avoided by the following methods :
1) Fixing mechanical purifiers.
2) Avoiding rapid change in steam rate.
3) Maintaining low water levels in boilers.
4) Effective softening and filtration of boiler feed water.
5) Blow down of the boiler (replacing the water concentrated with impurities of the boiler with fresh water).
Foaming : Foaming is the phenomenon of formation of foam or bubbles on the surface of water inside the boiler with the result of carrying of foam along with steam.
Causes for the presence of foaming : It has been observed that there is very little foaming in the case of pure water whereas the water containing dissolved impurities and suspended matter has a greater tendency to produce foam. The presence of large quantity of suspended impurities and oils lowers the surface tension of boiler water, producing foam.
Prevention of foaming : Foaming can be avoided by
1) Adding anti-foaming chemicals like castor oil. The amount of castor oil to be added varies with the boiler. Excess of castor oil can cause foaming.
Besides castor oil other substances like gallic and tannic acids, corn oil, cotton seed oil, sperm oil, beeswax, carnana wax, pyrogallol, tartaric and citric acids are also used as anti foaming agents.
2) Blow down of the boiler can prevent foaming.
Disadvantages of Priming and Foaming : Priming and foaming may cause the following boiler troubles.
1) The actual height of the water in boiler cannot be judged.
2) Wastage of heat with the result that it becomes difficult to keep up steam pressure and efficiency of the boiler is lowered.
3) Water concentrated with dissolved salts may deposit on the parts of the machinery, which cause corrosion due to the formation of concentration cell in that part resulting in the decay of metal on that spot.
1.8.2  Sludge and Scale Formation
Boilers are employed for the generation of steam in power plants, where water is continuously heated to produce steam.  As more and more H2O is removed from water in the form of steam, the boiler water gets concentrated with dissolved salts progressively reaches the saturation point. At this point the dissolved salts are precipitated out and slowly settle on the inner walls of the boiler plate. The precipitation takes place in two ways.
1) The precipitation in the form of soft, loose and slimmy deposits formed comparatively in the colder portions of boiler is called sludge.
2) The precipitation in the form of hard deposits that stick very firmly on the inner walls of the boiler is called scale.
Sludge formation : Sludges are soft, loose, slimmy and non-sticky precipitates                   produced due to the higher concentration of dissolved salts (Fig. 1.4 (a)).
Reasons for the formation of sludges : The dissolved salts whose solubility is more  in hot water and less in cold water produce sludges. For eg. MgCO3, MgCl2,  CaCl2 and MgSO4. The sludges were formed at comparatively colder portions of the boiler and get collected where rate of flow of water is low.
Disadvantages of sludges :
1) Sludges are bad conductors of heat and results in the wastage of heat and fuel.
2) Sometimes sludges were entrapped in the scale and gets deposited as scale, which causes more loss of efficiency of the boiler.
3) Excessive sludge formation leads to the settling of sludge in slow circulation areas such as pipe connections, plug openings, gauge-glass connections leading to the choking of the pipes.
Prevention of sludge formation :
1) By using soft water which is free from dissolved salts like MgCO3, MgCl2, CaCl2 and MgSO4 can prevent sludge formation.
2) By blowdown operation carried out frequently can prevent sludge formation.
Scales : Scales are hard, sticky deposits formed on the inner walls of the boiler. The main source of boiler troubles are scales which are very  difficult to remove once they are deposited on the surface of the boiler (Fig 1.4 (b)).
                                                               

Reasons for the formation of scales :

1)  Decomposition of Ca(HCO3)2 : Due to the  high temperature and pressure present in the boilers, the Ca(HCO3)2  salt decomposes to CaCO3, an insoluble salt, forms scale in low pressure boilers. CaCO3 is soluble in high pressure boilers.
Ca(HCO3)2  CaCO3 + H2O + CO2
                      (scale)                                    
2)  Decomposition of CaSO4 : CaSO4 is more soluble  in cold water, and less soluble in hot water hence its solubility decreases as the temperature of the boiler increases and precipitates out to produce hard scale on the surface of the boiler. The solubility of CaSO4 is              3200 ppm at 15 0C, reduces to 27 ppm at 320 0C and completely insoluble in super heated water in high pressure boilers. This is the main reason for the formation of scales in high pressure boilers. CaSO4 scale is very hard, highly adherent and difficult to remove.
3)  Hydrolysis of Magnesium salts :  Dissolved magnesium salts undergo hydrolysis at high temperature prevailing inside the boiler forming magnesium hydroxide precipitate, which form salt type of scale.
                   MgCl2+ 2H2O  Mg (OH)2 + 2HCl
4)  Presence of silica : SiO2 present even  in small quantities, deposits as calcium silicate (CaSiO3) or magnesium  silicate (MgSiO3). The deposits form

Disadvantages of scale formation :

 hard scale and are very difficult to remove.
The following are the disadvantages of scale formation.
1)  Wastage of fuel : Scales are bad conductors  of heat due to which the flow of heat from boiler to inside water is decreased, hence excessive heating is required which increases the fuel consumption causing wastage of fuel. The wastage of fuel increases with increase in the thickness of the scale as shown below.
Thickness of scale (mm)     0.325 0.625 1.25 2.5
Wastage of fuel                   10% 15% 50% 80%
2)  Lowering the boiler safety : Due to scale formation overheating of boiler is done to maintain the constant supply of steam. Due to overheating the boiler material becomes softer and weaker, which causes distortion of boiler. Thus the boiler safety is lowered.
3)  Decrease in efficiency : Scales deposited in the valves and condensers of the boiler cause choking which results in decrease in efficiency of the boiler.
4)  Danger of explosion : Because of the formation of the scales, the boiler plate faces higher temperature outside and lesser temperature inside due to uneven heat transfer and the coefficient of thermal expansion of boiler scale is less than boiler plate resulting in the development of cracks in the layer of scales. When water passes through the  crack and comes in contact with boiler plate having high temperature, results in the formation of large amount of steam suddenly developing sudden high pressure. This causes the explosion of the boiler.
Removal of scales :
1) If the scale formed is soft, it can be removed by a scrapper, wire brush etc.
2) By giving  thermal shocks, done by heating the boiler to high temperature and suddenly cooling with cold water. If the scale is brittle in nature, produces cracks on the surface of scale that can be removed by scrubbing with wire brush.
3) If the scale is very adherent and hard, chemical treatment  must be given, for example, CaCO3 scale is removed by washing with 5-10% HCl and CaSO4 scale is removed by washing the boiler plate with EDTA solution.
4) Frequent Blowdown operation can remove the scales which are loosely adhering.
Prevention of scale formation :
Scale formation can be prevented by softening water which is discussed separately in treatment of water.
1.8.3   Boiler Corrosion
The decay of boiler material by chemical or electrochemical attack of its environment is called boiler corrosion. The prime reasons for boiler corrosion are listed below.
a) Dissolved oxygen
b) Dissolved carbon dioxide
c) Acids from dissolved salts
a) Dissolved oxygen :
Among the dissolved gases oxygen is the most corroding impurity. At room temperature water contains 8 mgs of oxygen per litre. Hence the source of oxygen introduced into water is natural. Water, when saturated with air, contains 8 mg/l of oxygen.
Disadvantages of dissolved oxygen: In presence of prevailing high temperature, oxygen attacks the boiler plate creating serious corrosion problem.
2Fe + 2H2O + O2 2Fe(OH)2
   2Fe(OH)2 + O2  Fe2O3 . 3H2O
                        rust
Removal of dissolved oxygen :
1) Addition of sodium sulphite or sodium sulphide removes O2 by  converting O2 to  sodium sulphate.
2Na2SO3 + O2 2Na2SO4
Na2S + 2O2 Na2SO4
A very less amount of concentration of Na2SO3 / Na2S (5-10 ppm) should be added to water because the residual sodium sulphite may decompose to SO2 under prevailing high pressure, which finally appear as sulphurous acid (H2SO3) in the steam condensate.
2) Addition of hydrazine  (NH2NH2) is an ideal reagent added to boiler to remove dissolved oxygen as H2O.
NH2 NH2 + O2   N2 + 2H2O
N2 is harmless to boilers. Hence hydrazine removes dissolved oxygen without increasing the concentration of dissolved solids / salts.
3) A polyvalent organic compound under the trade name A2 aming 8001-RD is employed to degasify water nowadays.
4)  Mechanical deaeration is another method of degasification (Fig. 1.5). Water is sprayed through a perforated plate, fitted in the degasification tower, heated from sides and connected to vacuum pump as shown in the figure. High temperature, low  pressure and large exposed surface reduces dissolved oxygen and other gases in water.
b) Dissolved carbon dioxide :
Dissolved carbon dioxide has a slow corrosive effect on the materials of boiler plate. Source of carbon dioxide into water is the boiler feed water which contains bicarbonates. Under the high temperature and pressure, maintained in the boiler the bicarbonates decompose to produce CO2.
Ca(HCO3)2  CaCO3 + CO2 + H2O
Mg(HCO3)2  Mg(OH)2 + 2CO2
The disadvantage of the carbon dioxide is slow corrosive effect on boiler plates by producing carbonic acid.
CO2 + H2O  H2CO3
Removal of CO2  is carried out by two ways :
i) by the addition of calculated quantity of ammonia
NH4OH +  CO2  (NH4)2CO3 + H2O ii) by mechanical deaeration process (Fig. 1.5), as explained in section 1.7.3 (4)
c) Acids :
Presence of acids in boiler water is another main reason for boiler corrosion.
Sources of acid production are dissolved magnesium salts which undergo hydrolysis to produce acids for e.g.,
MgCl2 + 2H2O   Mg (OH)2 + HCl
Disadvantages of the acid  production is that the acids react with iron of the boiler plate in a  chain reaction to produce rust(decay) of the metal.
Fe + 2HCl  FeCl2 + H2
FeCl2 + H2O  Fe(OH)2 + 2HCl
                             Fe(OH)2 + H2O + O2  Fe2O3 . 3H2O
                                                                              (rust)  
Consequently even a small amount of MgCl2 can cause corrosion to a large extent. Prevention of acid corrosion  is carried by
i)  Softening of the boiler water to remove MgCl2 from the water.
ii) By frequent blowdown operation i.e., removal of water, concentrated with dissolved salts and feeding the boiler with fresh soft water.
iii) Addition of inhibitors, which are chemicals that form a thin film on the surface of the boiler protecting the metal from attack. Sodium silicates, sodium phosphate and sodium chromate act as good corrosion inhibitors.
1.8.4  Caustic Embrittlement
Caustic embrittlement is a term used for the appearance of cracks inside the boiler particularly at those places which are under stress such as rivetted joints due to the high concentration of alkali leading to the failure of the boiler. The cracks have appearance of brittle fracture hence the failure is called caustic embrittlement.
Reasons for the formation of caustic embrittlement :
The boiler feed water containing carbonates and bicarbonates of alkali metals,             sodium hydroxide and a small quantity of silica or sodium silicate is purified by lime-soda process. During the softening process by lime soda, free Na2CO3 is usually present in small portion in the soft water which decomposes to give sodium hydroxide and carbon dioxide at high pressure of the boilers.
            Na2CO3 + H2O  2NaOH + CO2
The precipitation of NaOH makes the boiler water ‘caustic’. The NaOH containing water flows into the small pits and minute hair-cracks present on the inner walls of the boiler. As the water evaporates, the concentration of caustic soda increases progressively, creating a concentration cell as given below thus dissolving the iron of the boiler as sodium ferrate.
          (--)        Iron at bends Concentrated Dilute Iron at           (+)             rivets and joints       NaOH solution NaOH solution plane surfaces                           anode                                                                                                                                                   cathode
The cracking of the boiler occurs particularly at stressed parts like bends,  joints, rivets etc., causing the failure of the boiler. The iron at plane surfaces surrounded by dilute NaOH becomes cathodic while the iron at bends, rivets and joints surrounded by highly concentrated NaOH becomes anodic which consequently decayed or corroded. Thus the cracks present at such places are intercrystalline, very fine and irregular running from one rivet to another without joining each other. These cracks have the appearance of brittle fracture hence known as caustic embrittlement.
Prevention of caustic embrittlement :
1) By using sodium phosphate as softening reagent instead of sodium carbonate. Disodium hydrogen phosphate (Na2HPO4) is the best softening reagent because it not only forms complex with Ca2+ and Mg2+ resulting in the softening of water but also maintains pH of water 9-10. The other phosphates used are trisodium phosphate, (Na3PO4) sodium dihydrogen phosphate (NaH2PO4) etc.
2) By adding tannin or lignin to boiler water which blocks the hair cracks and pits that are present on the surface of the boiler plate thus preventing the infiltration of caustic soda solution.
3) By adding sodium sulphate to boiler water, which also blocks the hair cracks and pits present on the surface of the boiler plate, preventing the infiltration of the caustic soda solution. The amount of sodium sulphate added to boiler water should be in the ratio [Na2SO4 concentration] : [NaOH concentration] kept as  1 : 1, 2 : 1 and 3 : 1 in boilers working as pressures upto 10, 20 and above 30 atmospheres respectively.
Disadvantages of caustic embrittlement :
The cracking or weakening of the boiler metal causes failure of the boiler.
The boiler troubles are summarised as follows. (Fig. 1.6)

1.9  TURBINE DEPOSITS

A turbine is a rotary mechanical device that extracts energy from a fluid flow and converts it into useful work i.e. electricity. A turbine is a turbo machine with one moving part called rotor assembly, a shaft or drum with blades attached. As the moving fluids or steam acts on the blades, the blades move and impart rotational energy to rotor (Fig. 1.7). A steam turbine is a device that extracts thermal energy from pressurised steam and uses it to do mechanical work on a rotating output shaft. Because the turbine generates rotary motion, it is particularly suited to be used to drive an electric generator to generate electricity. About 90% of all electricity is produced by using steam turbines
When steam is carried over the turbine plates (blades) due to the boiler trouble carry over some droplets of water etc. are deposited on the plates which are called turbine deposits.
1.9.1  Disadvantages of Turbine Deposits
The following are the disadvantages of turbine deposits
a) Turbine deposits cause corrosion of blades, rotors, and discs by electrochemical corrosion
b) Turbine deposits reduce the turbine efficiency.
c) The megawatt (M.W) generating capacity of the turbine is reduced by turbine deposits.
d) Turbine deposits cause malfunctioning of valves and seals.
1.9.2  Estimation of the Turbine Deposits
It is very difficult to evaluate the turbine deposits. The following are some of the met- hods of evaluation of turbine deposits.
a) Deposit collector/simulator : It is a device which simulates the turbine conditions governing deposition from superheated steam and it can be installed anywhere in piping or in turbine. The deposits are collected from superheated steam flowing through collector and the concentration of the impurities in steam can be determined.
b) Deposit analysis : The deposit from the collector is chemically analysed for its composition by elemental analysis, optical and SEM microscopy, x-ray diffraction and wet chemistry methods.
c) Optional flow totalizer : Measures both the average flow rate of total flow over the length of the test and concentration of impurities in the steam.
1.9.3  Composition of Turbine Deposits
The turbine deposits when analysed has the following composition. Most deposits are silicates and amorphous silica (SiO2). They are sodium silicate (Na2SiO3), sodium metasilicate hydrates (Na2SiO3xH2O, x = 5, 9), sodium aluminum silicates (NaAlSiO4) and their hydroxides, sodium iron silicate (NaFeSi2O6), potassium aluminum silicate (KAlSi3O8), magnesium silicates, calcium silicates noselite (SiO3), sodium aluminium molybdenum oxide silicate (Na8Al6Si6MoO4) etc.

 Causes of Turbine Deposition 

The following are some of the causes for turbine deposition
a) Entertainment is the cause of turbine deposition due to carry over where minute drops of boiler water in the steam along with some solid produce and dissolved salt deposit on the turbine. Priming and foaming are responsible for high level deposits leading to superheater tube failures.
b) Attemperating water impurity : Controlled cooling in a steam boiler is called attemperating. Turbine deposits are also formed due to impurities of water used for steam  attemperation and leakage in closed heat exchangers used for attemperation. The attemperating water should be investigated and it should be of same purity as steam.
c) Vaporisation of boiler water salts : The vaporisation of salts present in water at operating pressures below 2400psi cause deposition problem. This is because the solubility silica in steam increases with increasing temperature, hence more silica is present in hot steam. As the steam is cooled by expansion through the turbine silica deposits are formed causing turbine problems. Hence less than 0.02 ppm silica level must be maintained into steam. Uneven deposition on turbines cause vibrational problems.
d) Localised silica saturation : Turbine deposits are formed when localised silica saturation increases and silica condenses from steam in those areas of the turbines causing turbine deposits.
e) Turbine velocity : The velocity in the turbine is another reason for turbine deposits. The steam flows from inlet to the outlet of the turbine within a fraction of a second. As a result, the deposit is shifted downstream from saturation point by steam velocities.

Prevention of Turbine Deposits 

The following are some of the measures to prevent turbine deposits.
a) To maintain low silica concentration
b) To monitor the condensate
c) To make use of external treatment equipment like steam purifier
d) To increase the number of blow down operations.
1.9.6  Removal of Turbine Deposits
The removal of turbine deposits are carried out by
1) Water washing under the supervision and recommendation of the turbine vendor should be carried out
2) To remove water insoluble deposits blasting of turbines with aluminium oxide or other soft grit materials is carried out.
1.10  Softening of Boiler Feed Water
Water used for industrial purposes especially for generation of steam should be  sufficiently pure. The treatment of water includes the removal of hardness causing salts either by precipitation or by complex formation. Hence two types of treatments are given as shown below.

 External Treatment/Softening of Water

The treatment given to water for the removal of hardness causing salts before it is taken into the boiler is called external treatment or softening of water. Softening of boiler feed water includes the following methods.
1.10.2  Lime-Soda Process
In this process lime [Ca(OH)2] and soda [Na2CO3] are the reagents used to precipitate the dissolved salts of Ca2+ and Mg2+ as CaCO3 and Mg(OH)2. The precipitated CaCO3 and Mg(OH)2 are filtered off. Lime reacts with temporary hardness causing salts, magnesium permanent hardness, CO2, acids, bicarbonates and alums. Lime cannot remove calcium permanent hardness, which should be removed by soda.
Precautions to be followed during the process :
1) Only calculated amounts of lime and soda must be added. Excess of lime soda causes boiler troubles like corrosion and caustic embrittlement.
2) The chemical reactions of lime and soda are slow and  the precipitates CaCO3 and Mg(OH)2 are very fine, which cause after deposition troubles in the pipes and boiler tubes producing sludges. To avoid these troubles coagulants or flocculants like alums, accelerators like charcoal were added to water during filtration to enhance the formation of coarse particles and fast settling of the fine precipitates.
3) Proper time must be given for the reactions of  the softening process to complete, because all the reactions of the process are slow.
Calculation of lime and soda required for the process
The amount of lime and soda required for the process is calculated based on the chemical reactions involved as given below (Table-1.2).
Table - 1.2
Dissolved salt               Chemical reaction Treatment
  Required lime (L) or Soda (S)

1. Ca(HCO3)2         Ca(HCO3)2+ Ca(OH)2 2CaCO3+2H2O Lime (L)
(temp. hardness)    
2. Mg(HCO3)2    Mg(HCO3)2+2Ca(OH)2
(temp. hardness)                   Mg(OH)2+2CaCO3+2H2O 2  L
(One mole of Mg (HCO3)2   reacts with 2 moles of lime. Hence twice the amount of lime required.)

Permanent hardness causing calcium salts cannot be treated by lime as they do not react with lime.
Only soda treatment is required.
CaCl2 No lime treatment
CaSO4      – No –
Ca(NO3)2     – No –
Soda (Na2CO3) reacts with only permanent hardness causing calcium salts.
CaCl2 + Na2CO3  CaCO3+ NaCl
CaSO4 + Na2CO3 CaCO3 + Na2SO4
Ca(NO3)2 + Na2CO3  CaCO3 + 2NaNO3
Substances like NaCl, KCl, Na2SO4, SiO2, Fe(SO4)3  etc., do not impart any hardness hence do not consume any lime or soda. These salts should not be taken for calculation of lime and soda.
Dissolved                   Chemical                Treatment
Salt                    Reaction               required lime
       (L) or soda (S)
As the permanent magnesium hardness produces, permanent calcium hardness causing salts into the water, during lime treatment these salts require both lime and soda treatment.
MgCl2   MgCl2 +Ca(OH)2  Mg(OH)2+ CaCl2            L+S
MgSO4   MgSO4+Ca(OH)2Mg(OH)2+CaSO4            L+S
Mg(NO3)2   Mg(NO3)2+Ca(OH)2Mg(OH)2+Ca(NO3)            L+S
   Apart from hardness causing salts,
    lime reacts with the following salts
CO2   CO2+Ca(OH)2CaCO3+H2O            L (dissolved gas) HCl and H2SO4  2HCl+Ca(OH)2CaCl2+H2O    
(free acids)     [One mole of lime reacts with mole of HCl.]
H2SO4+Ca(OH)2CaSO4+2H2O L+S NaHCO3,KHCO3 2NaHCO3+Ca(OH)2CaCO3+Na2CO3+2H2O (bicarbonates) 2KHCO3+Ca(OH)2CaCO3+K2CO3+2H2O 1) One mole of lime reacts with 1/2 mole   of NaHCO3. Hence 1/2 the amount of   lime is required.
2) One mole of K2CO3 and Na2CO3 are
  produced, which should be subtracted from   Na2CO3  treatment.    FeSO4 FeSO4+Ca(OH)2Fe(OH)2+CaSO4 L + S (Coagulant) 2Fe(OH)2+H2O+O22Fe(OH)3 Al2(SO4)3 Al2(SO4)3+3Ca(OH)22Al(OH)3+3CaSO4+H2O 3L+ 3S (Coagulant)       1 mole of Al2(SO4)3 reacts with 3 moles of         lime and produces 3 moles of CaSO4.   NaAlO2 NaAlO2+2H2OAl(OH)3+NaOH   (Coagulant) 2NaOH + CaCl2Ca(OH)2+2NaCl
 2NaOH = Ca(OH)2
NaOH produced reacts with hardness causing salts and produces lime.                    Hence its hardness must be subtracted from lime requirement -
Amount of lime required for softening :
100 parts by weight of CaCO3
    = 74 parts by weight of Ca(OH)2 (lime) and 106 parts by weight of Na2CO3
Amount of lime required for softening =

Amount of soda required for softening :
   
1.   Cold lime soda process :
In this method the calculated quantities of lime are mixed with water at room temperature, and precipitates formed are finely divided.

Fig. 1.8  Continuous cold lime-soda softener
They cannot settle down easily. Filtration cannot be done easily. Hence small amounts of alums were added. The addition of sodium aluminate (NaAlO2) can act as a coagulant and also helps in the removal of silica and oil.
Raw water along with calculated amounts of chemicals (lime+soda+coagulant) are fed from the top into the inner chamber, fitted with a vertical rotating shaft carrying a number of paddles. As the water and chemicals flow down the chamber vigorous mixing and softening of water takes place. As the water comes out into the outer chamber and rises up, the settling of the sludge takes place. The water passes through the wood-fibre filter and flows out        continuously from the outlet at the top. The sludge settling at the bottom of the outer chamber is drawn off occasionally. The water sample contains residual hardness 50-60 ppm.
2.  Hot lime-soda process :
In this process the raw water is treated with softening chemicals at 80–150 0C. Because the hot lime soda process is carried at the temperature of the boiling point. (1) The reaction proceeds faster, (2) The precipitates formed settle down rapidly, (3) No coagulant is        required, (4) Filtration is fast and (5) Dissolved gases like CO2 are driven out. The residual hardness of water will be 15 to 30 ppm.
  Hot lime-soda process contains essentially 3 parts.
1) A reaction tank in which water + chemical + steam are thoroughly mixed.
2) A conical sedimentation vessel in which sludge settle down.
3) A sand filter, which ensures complete removal of sludge from the softened water.
Advantages of lime–soda process :
1) The process decreases the pH value of treated water thereby corrosion of  the distribution pipes is reduced.
2) Due to alkaline nature of water the pathogenic bacteria in water is reduced.
3) Besides removal of hardness causing salts, minerals, iron and manganese present in water are removed.
4) This process is economical.
Disadvantages of lime-soda process :
1) Skilled supervision is required.
2) Disposal of large amounts of sludge is a problem.
3) The water softened by this process contains 15 ppm of residual hardness which is not good for boilers.
Problems on lime-soda process
Points to be followed while solving the problems in lime-soda process.
1) The first step in all the problems of lime-soda process is to convert the hardness of the  hardness causing salts in terms of CaCO3 equivalents.
2) If CaCO3 is given as a hardness causing salt, it should be considered as  temporary hardness causing Ca(HCO3)2, which is expressed in terms of CaCO3.
3) If MgCO3 is given as hardness causing salt, it should be considered as temporary hardness causing Mg(HCO3)2 and it should be converted to CaCO3 equivalents.
4) Some salts like NaCl, KCl, Silica, K2SO4, Fe2O3 neither cause hardness nor react with lime and soda. Hence they must not be considered for the calculation of lime and soda required.
5) The hardness of Al2(SO4)3 must be multiplied by 3 for both lime and soda, HCl hardness must be multiplied by 0.5, Mg(HCO3)2 hardness must be multiplied by  2, NaHCO3, KHCO3, NaAlO2 hardness must be multiplied by 0.5 and  should be subtracted for the calculation of the amount of soda and lime required respectively.
6) Soda (Na2CO3) reacts only with permanent hardness causing Ca+ salts.
Problem 1.10 :
Calculate the amount of lime and soda required for the treatment of 10,000 litres of raw water containing the following dissolved salts per litre. CaCO3= 50 mgs, CaCl2=11.1 mgs, MgSO4=12 mgs, NaHCO3 = 7.25 mgs, Silica = 10 mgs.  Calculate the total cost of lime and soda, if the cost of lime is Rs 40/- per kg and soda is Rs 90 per kg.   (JNTU MID Exam)          
Sol : Conversion of the hardness in terms of CaCO3 equivalents :
   Sl.  Hardness Amount in  Mol.  CaCO3 equivalent
   No     causing salt      Mg/l                wt.
1. CaCO3 50 100 [Ca(HCO3)2 hardness already  converted CaCO3] = 50
2. CaCl2 11.1 111
3. MgSO4 12 120
4. NaHCO3 7.25 84
5. Silica 10 Does not require treatment. Amount of lime required                        = (CaCO3 + MgSO4 +  NaHCO3 hardness in CaCO3 equivalents)
                   
             
Amount of lime required for 10,000 litres of water
    = 47.589 × 10,000 = 4,75,890 mgs/10,000 litre
    = 475.89 gms / 10,000 litre
    = 0.4759 kg / 10,000 litre
Cost of lime per kg = Rs 40/-
Total cost of lime = 400.4759 = 19.036/-  = 19.04/-
Amount of soda required =(CaCl2 + MgSO4 – NaHCO3 hardness converted to CaCO3 equivalents)
                                                                       

                       15.69 = 16.6314 mg/litre
Soda required for 10,000 litres of water = 16.63 × 10,000
    = 166314 mg/10,000 litres
    = 0.1663 kg/10,000 litres
Cost of soda = Rs 90/- per kg
Total cost of soda = 0.166390 = Rs.14.97/-
Total expenditure on Lime & Soda = 19.04 + 14.97 = Rs. 34.01/-
Problem 1.11 :
A sample of water contains the following dissolved salts in mgs/litre.  CaSO4 = 6.8, MgCO3 = 8.4, Al2(SO4)3 = 34.2, CO2= 4.4, HCl = 3.65. Calculate the amount of lime and soda required for the treatment of 5000 litres of water. The purity of lime 80% and soda is 90%. 10% of excess chemicals were added.
Sol : Conversion of the amounts of hardness causing salts into CaCO3 equivalents.
Hardness  Amount of  Mol. wt. of CaCO3 equivalents
causing  H.C.S. mg/l     H.C.S
salt (H.C.S)
1.  CaSO4 6.8 136      
2.  MgCO3 8.4 84      
3.  Al2(SO4)3 34.2 342      
4.  CO2 4.4 44      
5.  HCl 3.65 36.5      
Amount of lime required = [(2×MgCO3) +(3 × Al2(SO4)3)+CO2+HCl hardness    in terms of CaCO3 equivalents]


Lime required for 5,000 litres of water = 48.1 × 5,000
                                                          = 240500 mg/5000 lit. = 0.2405 kg/5000 lit.
Purity of lime = 80%.
Out of 100 kg of raw lime used only 80 kg is pure lime
? raw lime should be used  if 0.2405 kg of pure lime required
The amount of raw lime required  = 0.3006 kg / 5000 lit.
10% excess lime is added.
Total amount of lime along with 10% excess chemicals =
                                                                            = 0.33066 kg / 5000 litres
Amount of soda required
                =  (CaSO4+(3×Al2 (SO4)3)+HCl hardness in CaCO3 equivalents)
          =  [5+(3×10)+5]
          =  = 42.4 mg/l.
Lime required for 5000 litres of water = 42.4 × 5000
                                                         = 21200 mg/1000 lit = 0.212 kg/5000 lit
Purity of soda = 90%
Amount of raw soda to be required =  = 0.2356 kg / 5000 litres
10% excess soda is added.
Total amount of soda along with 10% excess chemicals
  =  = 0.2591 kg / 5000 litres
Problem 1.12 :
Calculate the amount of lime and soda required for softening 10,000 liters of water containing the following salts per litre.
Ca(HCO3)2 = 162 mgs, CaSO4 = 136 mgs, MgCl2 = 95 mgs and NaCl = 56.1 mgs Purity of lime is 93% and that of soda is 99%.        (BIT Mesra IBE 1993)
Sol :     S.No. Hardness  Amt in  Mol. wt. CaCO3 equivalents
causing   mgs/l  
salt (H.C.S)
1.  Ca(HCO3) 162 162        mgs/l
2.  CaSO4 136 136        mgs/l
3.  MgCl2 95 95        mgs/l
4.  NaCl does not require lime soda treatment
Amount of lime required = [Temporary hardness of Ca(HCO3)2) + Permanent hardness of MgCl2]
         
Amount of lime required
for the treatment 10,000 litres  = 148 × 10000 = 14,80,000 mgs/10000 l
                                             =  = 1480 gms/10000 l
               =  = 1.48 kg/10000 l
Purity of lime is 93%.
Total amount of lime required  = 1.59 kg / 10,000 lit.
Amount of soda required
for the 10000 litres  =  = 212 mgs/l
                                = 2.12 kg / 10,000
Purity of soda is 99%
Total amount of soda required =  = 2.14 kg / 10,000
Problem 1.13 :
A water works has to supply 1000 liters of water, that contains Mg2+ = 36 ppm, HCO3– = 18.3 ppm and H+ = 1.5 ppm. Calculate the amount of lime and soda required.
Sol :
S.No. Hardness  Amt in  Mol. wt. CaCO3 equivalents
Causing    ppm  
Salt (H.C.S)
1. Mg2+ 36 24        ppm
2. HCO3– 18.3 122        ppm
3. H+ 1.5   2        ppm
Amt. of lime required =  [Mg2+ + HCO3– + H+]
     =  [150+15+75] =  [240] = 177.6 mg/l
Lime required for 1000 litres of water =
Soda required for the treatment =  [Mg2+ + H+– HCO3–]
                 =  [150 + 75 – 15]
                 = = 222.6 mg/l
Amount of soda required for the treatment of 1000 litres of water
                 = = 0.2226 kg.
Problem 1.14 :
Analysis of water gave the following results : H2SO4 = 196 mg/l, MgSO4 = 24 mg/l, CaSO4 = 272 mg/l and NaCl = 25 mg/l. Water is to be supplied to the town of the population one lakh only. The daily consumption of water is 50 litres per head. Calculate the cost of lime and soda required for softening of hardwater for town for the month April 2012, if the cost of lime is Rs 5.00 per kg and cost of soda is Rs 8.00 per kg.                 (MREC 2002)
Sol :
  S.No. Hardness  Amt in  Mol. wt. CaCO3 equivalents
Causing  mg/l  
Salt (H.C.S)
1.   H2SO4     196       98      
2.   MgSO4      24     120      
3.   CaSO4       272     136      
4.   NaCl                       no treatment with lime & soda required
Amount of lime required = [Hardness of H2SO4 + MgSO4 in terms of CaCO3                    equivalents]
       
Daily consumption of water = 50 l per head
Population of the town = 1,00,000
 Lime required for the treatment for the population  =  = 814 kg
For April 2002 (30 days) total amount of lime required
= 81430 = 24420 kg
Given cost of lime = 5/- per kg
Total cost of lime in Rs. = 244205 = 1,22,100/-
Amount of soda required = [Hardness of H2SO4 + MgSO4 + CaSO4 in terms of CaCO3 equivalents]
                = (420) = 445.2 mg/l
Soda required for the treatment for the population =  = 2226 kg
For the April 2012 (30 days) the amount of soda required = 222630 = 66780 kg
Cost of soda = Rs 8/- per kg
Total cost of soda = 667808 = Rs. 5,34,240/-
Problem 1.15 :
Calculate the amount of lime and soda required for softening 10,000 litres of water which was analysed as follows :
Analysis of water
Ca2+ = 380 ppm Dissolved CO2 = 120 ppm
Mg2+ = 144 ppm FeSO4.7H2O = 278 ppm
HCO3– = 1500 ppm
Analysis of treated water
OH– = 34 ppm, CO32– = 32 ppm      (MREC 2000)
Sol :
 S.No. Hardness  Amt in  At. wt CaCO3 equivalents
Causing   ppm  Mol. wt.
Salt (H.C.S)
Raw water
1.   Ca2+ 380 40      
2.   Mg2+ 144 24      
3.   HCO3– 1500 122      
S.No. Hardness  Amt in  At. wt CaCO3 equivalents
Causing   ppm  Mol. wt.
Salt (H.C.S)
4.   Dissolved CO2   120 44      
5.   FeSO4.7H2O   278 278      
Treated Water
6.   OH–   34 34      
7.   CO32–   32 60      
Amount of lime required = [Hardness of Mg2+ + HCO3– + CO2 + FeSO4.7H2O           Converted in terms of CaCO3 equivalents]

Amount of lime required
for the treatment 10,000 liters of raw water =  = 16.296 kg/10000 l
Amount of soda required for raw water [Hardness of Ca2+ + Mg2+  + FeSO4.7H2O              – HCO3– in terms of CaCO3 equivalents]
                      =
=  = 445.73 mg/l
Amount of soda required
10,000 litres of raw water =  = 4.4573 kg/10,000 l
Amount of lime required for excess  in treated water =
Amount of lime required for excess  in 10,000l of treated water                               =
Total Amount of lime required for the treatment of raw + treated 10,000 litre of water = 16.296 + 7.4 = 23.696 kg/10,000 l
Amount of soda required for excess of  and HCO3– in treated water =
Amount of soda required for excess  and HCO3– in 10,000l treated water                                =
Total Amount of soda required for the treatment of raw + treated 10,000 litre of water              = 4.4573 + 1.625 = 6.0823 kg/10,000 l
Problems for Practise :
1.    A water sample gave the following constituents on analysis in mgs/litre. Mg(HCO3)2  = 73, CaSO4 = 68, MgCl2 = 95, MgSO4 = 12, Ca(HCO3)2 = 81, NaCl = 4.8. Calculate the cost of the chemicals required for softening 20,000 litres of water. The purity of lime and soda are 95% and 90%. The costs per 100 kg. each of lime and soda are Rs. 75 and Rs. 2480 respectively.       (Nagpur IBE 1997) (Ans : Rs. 3.04 + 93.50 = 96.54/-)
2. Calculate the amount of lime and soda required to soften 1 million litres of a water sample containing the following dissolved salts (calcium permanent hardness) Ca2+ = 20 ppm, Mg2+ = 24 ppm, HCO3– = 150 ppm, and CO2 = 30 ppm. The purity of lime is 87% and soda is 91%. 10% of excess chemicals were added.
(Ans : lime = 272.46 kg, soda = 34.6k)
3. A sample of raw water contains the following dissolved salts. Ca(HCO3)2 = 8.1 ppm, Mg = 14.6 ppm, HCO–3 = 91.5 ppm, CaCl2 = 33.3 ppm, MgCl2 = 38 ppm,               Al2 (SO4)3 = 17.1 ppm. Calculate amounts of lime (90% pure) and soda (98% pure) for the treatment of 10,00,000 litres of water.              (JNTU 1988)
(Ans :  Lime = 214.05 kg, Soda = 10.82 kg)
4. Calculate the quantity of lime and soda required for softening 60,000 litres of water containing CO2 = 20 ppm, CaCO3 = 45.45 ppm, Mg (HCO3)2 = 25 ppm,  HCl = 8.4 ppm, Al2 (SO4)3 = 40 ppm and MgCl2 = 12 ppm. 10% of excess chemicals were added.                                   (Ans : Lime = 14.07 kg, Soda = 11.39 kg)
5. A water works has to supply 1 m3/s of water. The raw water contain Mg(HCO3)2 = 219 ppm, Mg2+ = 36 ppm, HCO3– = 18.3 ppm and H+ = 1.5 ppm. Calculate the cost of treating water per day, if lime (90% pure) and soda (95% pure) has the cost  Rs. 500/- per tonne and Rs. 7000/- per tonne respectively.
(Hint : Amount of water purified per day = 1 m3/s  60  60  24 s/day = 8.64 104 m3 / day = 8.64 107 litres / day. 1 tonne = 103 kg = (109 mgs)).
         (Ans :  Rs.1,60,894/-)
6. Calculate the amount of lime and  soda required for softening 10,000 litres of hard water containing Ca(HCO3)2 = 81 mgs, CaSO4 = 13.6 mgs, MgCl2 = 95 mgs, and NaCl = 56.1 mgs. Purity of lime is 93% and soda is 99%.                                    (JNTU 1993)
      (Ans : Lime = 1.19 kg/10,000l, Soda = 1.178 kg/10,000l)
7. Calculate the amount of lime and soda required for the softening of 10,000 litres of water, containing the salts CaCO3 = 10 mg/l, MgCO3 = 8.4 mg/l, CaCl2 = 11.1mg/l, MgSO4 = 6 mg/l, SiO2 = 1.2 mg/l. The purity of lime is 90%, and soda is 95%.
  (JNTUK 2014)
      (Ans : Lime = 0.1924 kg/10,000l, Soda = 0.1674 kg/10,000l)
1.10.3  Zeolite Process
The term zeolite stands for ‘boiling stones’ (zeo-boiling, olite-stone). Zeolites are         porous, when water passes through it, it gives the appearance as though water is boiling hence the name ‘zeolite’ which was  discovered and named by cronstedt, a swedish geologist.
The chemical composition of zeolites is hydrated sodium aluminium silicate, represented as Na2O Al2O3  x SiO2 y H2O where x = 2 – 10 and y = 2 – 6. Zeolites are capable of exchanging reversibly its sodium ion for hardness causing Ca2+ and Mg2+ in water, hence zeolites are cation exchangers.
Zeolites are of two types.
1) Natural zeolites which are natural and non-porous green sands.
For eg : Natrolite Na2O. Al2O3. 4SiO2.2H2O
2) Synthetic zeolites are porous, possess gel structure and are prepared from china clay, felspar and soda ash. The synthetic zeolites possess higher exchange capacity.
Process :
The  hard water is passed through a zeolite bed fixed in a cylinder (Fig. 1.10) at a specific rate.

Fig. 1.10  Permutit process
The hardness causing ions, Ca2+, Mg2+ etc. are retained by the zeolite as CaZe and MgZe respectively. An equivalent amount of sodium salts were introduced into water. The reaction taking place during the softening process are
                           Na2Ze + Ca (HCO3)2   CaZe + 2NaHCO3
                           Na2Ze + Mg (HCO3)2   MgZe + 2NaHCO3
Na2Ze + CaCl2 or CaSO4 or Ca(NO3)2   CaZe + 2NaCl or Na2SO4 or 2NaNO3
Na2Ze + MgCl2 or MgSO4 or Mg (NO3)2  MgZe + 2NaCl or Na2SO4 or 2NaNO3
Regeneration :  After sometime the zeolite bed is completely converted to calcium and magnesium zeolites and no purification of raw water takes place. i.e., the zeolite  bed gets exhausted.  At this stage the purification of hard water is stopped and the zeolite bed is regenerated by treating the bed with 10% brine (NaCl) solution.
                           CaZe + 2NaCl  Na2Ze + CaCl2
                           MgZe + 2NaCl  Na2Ze + MgCl2
The washings containing CaCl2 and MgCl2 are discarded.
Limitations of Zeolite Process :
1) Raw water should not contain turbidity. Turbidity will clog the pores of  zeolite bed and makes it inactive. Turbidity of water must be removed by coagulation, filtration etc.
2)   Raw water must not contain any coloured ions like Mn2+ and Fe2+, because they form manganese zeolite and ferrous zeolite which cannot be regenerated.
3) Mineral acids if present in water will destroy the zeolite bed permanently. Water must be neutralized with soda before it is fed into the zeolite bed.
Advantages of Zeolite Process :
1) It removes hardness almost completely and the treated water contains hardness upto 10 ppm.
2) No precipitation of the products takes place. Hence no disposal of sludge is required.
3) The equipment is compact and requires less skilled assistance.
4) The process adjusts itself for variation of hardness of incoming water.
Disadvantages of  Zeolite Process :
1) The treated water contains more sodium salts than in lime–soda process.
2) The method replaces only Ca2+ and Mg2+  ions by Na+ ions, leaves acidic ions like HCO3– and CO2–3  as NaHCO3 and Na2CO3.
3) These NaHCO3 salt decompose and liberate CO2  which causes corrosion.
4) The Na2CO3 salt produced in water is decomposed due to the high temperature maintained in the boiler to NaOH, which causes caustic embrittlement.
Problems on zeolite process :
Problem 1.16 :
The hardness of 50,000 litres of a sample of water was removed by passing it through a zeolite softner. The softner required 200 litres of sodium chloride solution containing 150 gms/litre of NaCl for regeneration. Calculate the hardness of the sample of water in ppm and OCl. (JNTU 1980)
Sol : Amount of NaCl required for purification = 150 x 200 = 30,000 gms
1 gm equivalent of NaCl = 1 gm equivalent of CaCO3
58.5 gms of NaCl   = 50 gms of CaCO3
30,000 gms of NaCl       = ? of CaCO3
Amount of NaCl in CaCO3 equivalents
                   =  = 25,641 gms equivalents of CaCO3
CaCO3 equivalent of 25,641 gms  = 25,641  1000 mgs = 2.5641107 mgs
2.5641107 mgs of hardness is possessed by 50,000 litres of water
The hardness of 1 litre of water = = 512.8 mg / l.
Hardness of water sample  = 512.8 ppm = 512.8  0.07 = 35.896 OCl
Problem 1.17 :
An exhausted zeolite softner was regenerated by passing 100 litres of NaCl solution having a strength of 200 gm/litre of NaCl. If the hardness of water is 500 ppm, calculate the total volume of water softened by the softner.                         (JNTU 1979)
Sol : Amount of NaCl present in the solution = 100  200 = 20,000 gms
1 gm equivalent of NaCl = 1 gm equivalent of CaCO3
58.5 gm of NaCl = 50 gms of CaCO3
20,000 gms of NaCl == 17094 gms of CaCO3 equivalents
CaCO3 equivalent of 17094 gms = 170941000 mgs = 1.7094107 mgs
hardness of the water sample = 500 ppm
Total volume of water softened = = 34188 litres
Problem 1.18 :
A zeolite softner was 90% exhausted when 10,000 litres of hard water was passed through it. The softner require 200 litres of NaCl of strength 20 gm/litre. Calculate the hardness of water.                                                       (JNTU MID)
Sol : Amount of NaCl required = 20200 = 4000 gm
CaCO3  equivalent of NaCl =  = 3418.8 gms
3418.8 gms of CaCO3  equivalent of NaCl
    = 3418.8 1000 mgs of CaCO3 equivalent of NaCl                 = 3418800 mgs
Volume of water purified = 10,000 litres.
Hardness of water =  = 341.88 ppm
Problems for Practise :
1) A zeolite softner was completely exhausted and regenerated by passing 200 litres containing 50 gm/l of NaCl solution. Calculate the volume of hard water containing 400 ppm of hardness can be softened by the softner.                  (Ans : 21367.5 litres)
2) A zeolite  softner was regenerated by passing 100 litres of sodium chloride solution containing 120 gm/l of NaCl, after softening 20,512 litres of hard water. Calculate the hardness of the sample of water.                                      (Ans : 500 ppm)
3) The total hardness of 1000 litres of water was completely removed by a zeolite softner. A zeolite softner required 30 litres of sodium chloride solution containing 15 gms/litre of NaCl for regeneration. Calculate the hardness of water.
 (JNTU, July 1994), (Ans : 384.6 ppm)
1.10.4  Ion-exchange Process
Ion-exchange process includes the exchange of the cations and anions of the dissolved salts with H+  and OH– ions respectively. For this two types of ion-exchangers are used, which are insoluble, cross-linked long chain organic polymers with microporous structure.
1) Cation exchangers  are capable of exchanging their H+ ions with cations of the dissolved salts, which comes in their contact. The cation exchangers are represented by general formula RH, are mainly styrene-divinyl benzene copolymers containing the functional groups – COOH or – SO3H. R is the general structure of resin and H is exchangeable with cation.
2) Anion exchangers are phenol-formaldehyde or amine formaldehyde copolymer          resins which  exchange  their OH– ion with any anion present in the dissolved salts. The anion exchangers are represented by the formula R|OH. R| is the representation of the general structure of the resin and OH is exchangeable with anion.
Thus all the cations and anions present in the dissolved salts are exchanged with cation exchanger and anion exchanger.
Process :  Raw water is first  passed through  cation exchanger  and the removal of cations takes place in the  following way.
2RH + Ca(HCO3)2   R2Ca + 2H2CO3
2RH + Mg (HCO3)2   R2Mg + 2H2CO3
2RH + CaCl2 R2Ca + 2HCl
2RH + CaSO4 R2Ca + H2SO4
2RH + Ca(NO3)2 R2Ca + 2HNO3
         2RH + MgCl2   R2Mg + 2HCl
         2RH + MgSO4   R2Mg + H2SO4
         2RH + Mg(NO3)2 R2Mg + 2HNO3
The Ca2+ and Mg2+ are retained by the cation exchanges as CaR and MgR releasing H+ into water. Thus the water coming out of the resin is highly acidic because the H+  released by the exchange combine with anion  of the dissolved salt to produce the corresponding acids.
Then the water is passed through anion exchanger  where the anions of the acids present in water are removed by the exchanger releasing OH– into water. The H+ and OH– released from exchangers get combined and produce H2O.
                      R|OH + HCl         R|Cl + H2O
2R|OH + H2SO4    R|2SO4 + 2H2O
R|OH + H2CO3     R|HCO3 + H2O
R|OH + HNO3      R|NO3 + H2O
Thus the water coming out from exchanger is free from all ions known as deionised or demineralised water.
When water sample is completely deionised, it has the tendency to absorb gases like CO2, O2 etc, from atmosphere which cause boiler corrosion. Hence deionisation must be followed by degasification.
Regeneration : After the deionisation of certain amount of raw water the cation and anion exchangers will be exhausted. Regeneration of cation exchanger is carried out by passing dil.HCl or H2SO4 solution into the cation exchanging bed. The H+ ions of the acid are exchanged with the cations (Ca2+ and Mg2+) present in the cation exchanger regenerating it the following way.
R2Ca + 2HCl   2RH + CaCl2            (Washings)
R2Mg + H2SO4  2RH + MgSO4                               (Washings)
The washings containing CaCl2 and MgSO4 etc., were passed to sink or drain.
Similarly the exhausted anion exchanger is treated with dil.NaOH solution. The regeneration can be represented as below
R|2SO4 + NaOH  2R|OH + Na2SO4
 (Washings)
R|Cl + NaOH  R|OH + NaCl
       (Washings)
                  R|HCO3+NaOH  R|OH + NaHCO3             (Washings)
R|NO3+ NaOH   R|OH + NaNO3             (Washings)
The washings are discarded into sink.
The regenerated ion-exchanger is used for softening.
Thus deionisation and regeneration are the alternate processes.
Advantages :
1)  Highly acidic or alkaline water samples can be purified by this process.
2)  The hardness possessed by the deionised water is 2 ppm.
3)  The deionised water is most suitable for high pressure boilers.
Disadvantages :
1)  The ion exchanging resins are expensive, hence the cost of purification is high.
2) Raw water should contain turbidity below 10 ppm. Otherwise pores in the resin will be blocked and output of the process is reduced.
Mixed–bed deionisor :
Mixed-bed deionisor consists of a single cylinder containing an intimate mixture of cation exchanger and anion exchanger. When raw water is passed through the bed it comes in contact for a number of times with two kinds of ion exchangers alternately. The net result is equivalent to passing the raw water through several series of cation and anion exchangers. The hardness of this deionised water is less than 1 ppm.
Regeneration of mixed-bed ioniser :
The mixed-bed is back washed (by forcing water in the upward direction). The anion exchanger which is lighter (low specific gravity) gets displaced to form upper layer and heavier cation exchanger remains as lower layer. The cation exchanger is separated and treated with dil.H2SO4 and the anion exchanges with dil.NaOH solutions and then washed with water. The two beds were mixed again by forcing compressed air (Fig. 1.12).
The advantages of mixed bed deioniser is more convenient to use and efficient purification takes place.
The disadvantages of mixed bed deioniser is that it is expensive.
           a) Purification of water                       b) Separation of mixed bed          c) Regeneration of mixed bed
Fig. 1.12  Regeneration of mixed-bed ion exchanger
Table - 1.3 Comparison of Lime-soda process, Permutit process and Ion-exchange process

Internal treatment or conditioning of water :

The softening of water carried out  inside the boiler is called conditioning / internal treatment of water. In this process the hardness causing dissolved salts were prohibited.
1) By complexing (also called sequestring) the hardness causing soluble salt by  adding appropriate reagents.
2) By precipitating the scale forming impurities in the form of sludges which can be removed by blowdown  operation.
3) By converting the scale  forming salts into compounds which stay in ‘dissolved form’ and donot cause any trouble to the boilers.
All internal treatment methods must be followed by blowdown operation  so that accumulated sludges are removed.
Important internal conditioning methods are :
1) Colloidal conditioning : The scale formation in low pressure boilers is prevented by the addition of kerosene, tarnnin, agar agar  etc. which get coated over the scale forming precipitates preventing the coagulation of the particles. These form loose, non-sticky deposits that can be removed by blowdown. This type of treatment is colloidal conditioning.
2) Phosphate conditioning : The scale formation due to permanent hardness causing calcium salts is avoided by complexation with sodium phosphate in high pressure boilers. The complex formed is soft, non-adherent and easily removable.
                   3CaCl2 + 2Na3PO4   Ca3(PO4)2 + 6NaCl
                   3MgCl2 + 2Na3PO4  Mg3(PO4)2 + 6NaCl
The calcium phosphate and magnesium phosphate complexes were removed by blow down operation.
The three phosphates employed for conditions are :
1) NaH2PO4 – Sodium dihydrogen phosphate (acidic)
2) Na2HPO4 – Disodium hydrogen phosphate (weakly alkaline)
3) Na3PO4 – Trisodium phosphate (alkaline)
Trisodium phosphate is the most preferred reagent because it not only  forms complex  with Ca2+ and Mg2+  ions, but also maintains the pH of water between 9-10, where the calcium and magnesium ions undergo complexation.
If the alkalinity of the boiler water  is insufficient then disodium phosphate is selected for conditioning. If the alkalinity of the boiler water  is very high then sodium dihydrogen phosphate  is selected for the treatment to reduce the alkalinity of the water.
3) Carbonate conditioning : The hard and strong adherant scales  formed due to CaSO4 are avoided by the addition of sodium carbonate to boiler water and this is called carbonate conditioning.
                       Na2CO3+CaSO4 CaCO3+ Na2SO4
The CaSO4  is converted to CaCO3  which is loose sludge and it can be removed by blow down. Carbonate conditioning is used only for low pressure boilers. In high pressure boilers excess Na2CO3 is hydrolysed to NaOH, which causes caustic embrittlement.
Na2CO3 + H2O  NaOH + CO2 + H2O
4) Calgon conditioning : Sodium hexameta phosphate Na2[Na4(PO3)6] or (NaPO3)6 is called  calgon. This forms soluble complex compounds with CaSO4 which causes no boiler troubles.
                      Na2 [Na4(PO3)6]  2Na++ [Na4P6O18]2–
                                (calgon)
                2CaSO4+[Na4P6O18]2–  [Ca2P6O18]2–  + 2Na2SO4
                                  (soluble complex ion)
The treatment of boiler water with calgon is called calgon conditioning.
5) Treatment with sodium aluminate (NaAlO2) : Sodium aluminate is hydrolysed at the high temperature of boiler to Al(OH)3 and NaOH. The sodium hydroxide so formed precipitates some of the magnesium salts as Mg(OH)2
              NaAlO2 + H2O     Al(OH)3  + NaOH
              MgCl2 + 2NaOH  Mg(OH)2  + NaCl
The flocculent precipitates of Mg(OH)2 plus Al(OH)3 produced inside the boiler entraps the finely suspended and colloidal impurities including oil drops and silica which can be removed by blow-down operation. This type of conditioning is called sodium aluminate conditioning, majorly removes MgCl2, which otherwise will produce HCl due to reaction with water at high temperatures causing corrosion to boiler plate MgCl2 + 2H2O  Mg(OH)2 + 2HCl (corrosion)
6) Electrical conditioning : Sealed glass bulbs containing mercury connected to a battery are set rotating in the boiler. When water boils mercury bulbs emit electrical discharges which prevent scale forming particles to adhere/stick  together to form scale. This type of conditioning is called electrical conditioning.
7)  Radioactive conditioning : In this type of conditioning radioactive salts in the form of tablets  are placed inside the boiler water at a few points. The radiation emitted by the tablets of radioactive material prevents scale formation.
The internal conditioning of water can be summarised as follows.

DESALINATION OF BRACKISH WATER

Water containing high concentrations of dissolved solids with a peculiar salty or brackish taste is called brackish water. Sea water is an example of brackish water  containing about 3.5% of dissolved salts. This water cannot be used for domestic and industrial applications unless the dissolved salts are removed by desalination.
Commonly used methods are summarised below.
Reverse Osmosis                                                         Distillation

                                     Desalination of brackish water

      Electrodialysis  Freezing
1.11.1  Electrodialysis
Electrodialysis is based on the principle that the ions present in saline water migrate               towards their respective electrodes through ion selective membranes under the influence of applied e.m.f. The process of electrodialysis is explained with help of the following (Fig. 1.13).
The unit consists of a chamber with two electrodes, the cathode and anode. The chamber is divided into three compartments with the help of thin, rigid, ion-selective membranes which are permeable to either cation or anion. The anode is placed near anion selective membrane while the cathode is placed near cation selective membrane. The anion selective membrane is containing positively charged functional groups such as R4 N+ and is permeable to anions only. The cation selective membrane consists of negatively charged functional groups. Such as RSO–3 and is permeable to cations only. Under the influence of applied e.m.f. across the electrodes the cations move towards cathode through the membrane and the anions move towards anode through the membrane. The net result is depletion of ions in the central compartment while it  increases in the cathodic and anodic compartments. Desalinated water is periodically drawn from the central compartment while concentrated brackish water is replaced with fresh sample.
Advantages of electrodialysis :
1) The unit is compact.
2) The process is economical as for as capital cost and operational expenses are concerned.

Reverse Osmosis

When two solutions of unequal concentration  are separated by a semipermeable membrane which does not permit the passage of dissolved solute particles, (i.e molecules and ions) flow of solvent takes place from the dilute solution to concentrated solution this is called Osmosis. If a hydrostatic pressure  in excess of osmotic pressure is applied on the concentrated side of the solvent, the solvent is forced to move from higher concentration to lower concentrated side across. Thus the solvent flow is reversed hence this method  is called reverse osmosis. Thus in reverse osmosis pure water is separated from the contaminated water. This membrane filtration is also called super filtration or hyper-filtration.

Fig. 1.14 Reverse Osmosis Cell
Method of Purification : The reverse osmosis cell as shown above (Fig. 1.14) consists of a chamber fitted with a semipermeable membrane, above which sea water/impure water is taken and a pressure of 15 to 40 kg/cm2 is applied on the sea water/impure water. The pure water is forced through the semipermeable membrane which is made of very thin  films of cellulose acelate. However superior membrane made of polymethacrylate and polyamide polymers have come to use.
Advantages :
1) Both ionic and non-ionic, colloidal and high molecule weight organic matter is re-   moved from the water sample.
2) Cost of purification of water is less and maintenance cost is less.
3) This water can be used for high pressure boilers.
EXERCISE
I. Short Answer Questions :
1. If the hardness  of water sample is 14 degree clarke, what is the hardness in terms of ppm?
2. What is hardness of water due to ?
3. What is degree of hardness ?
4. How temporary hardness differs from permanent hardness ?
5. What is sedimentation ?
6. Write any two harmful affects of silica present in water ?
7. Is water obtained from zeolite process free from all impurities? Justify.
8. What is reverse osmosis ?
9. Why magnesium bicarbonate requires double amount of lime for softening ?
10. Explain why water containing dissolved calcium and magnesium salt cause hardness.
11. Why is NH4Cl and NH4OH buffer added during the determination of hardness ?
12. What is the indicator used in EDTA method?
13. During sterilization why chloramine is preferred to bleaching powder ?
14. What is meant by softening of water ?
15. What is break point chlorination ?
16. Name any two coagulants.
17. Name the gases dissolved in water which causes corrosion.
18. How is temporary hardness removed ?
19. What are the advantages of ion-exchange process ?
20. Define priming and foaming [JNTU - 2014]
21. What is meant by wet steam ?
22. What is caustic embrittlement ? [JNTU - 2014]
23. A sample of hard water contains 14.6 gm of Magnesium bicarbonate and 9.5 gm of  MgCl2 and 13.6 gms of CaSO4. What is the permanent, temporary and total hardness of the water sample ?
24. How is dissolved oxygen removed from boiler feed water ?
25. What is meant by dechlorination ?
26. Explain the role of chemical coagulants.
27. Explain the role of anion exchange.
28. Why hot lime-soda process is advantageous over cold lime-soda process ?
29. How does boiler corrosion occur ?
30. How is caustic embrittlement avoided ?
31. How do you minimise silica turbine deposits?
32. What is a turbine?
33. Explain why CaCO3 is selected for expression of hardness of water?
34. How is mixed bed deioniser regenerated?
35. What is the composition of zeolites?
ii. ESSAY ANSWER QUESTIONS :
1. a) Mention the different sources of water and the probable impurities present in these  samples.
    b) How are scales formed in boilers and what are the affects of scales on boiler plates?                                                   (JNTU 1989)
2.   a) Describe the zeolite and ion exchange methods of treatment of water.
        b) Calculate the quantity of lime and soda required for softening of 1 million litres water having the following composition.
      Free CO2 = 20 ppm, Ca(HCO3)2 = 200 ppm, Mg(HCO3)2 = 5 ppm,  Ca+2 = 30 ppm, Mg2+ = 10 ppm.     (Ans : Lime = 160.9 kg, Soda = 123.67 kg) (JNTU 1989)
3. a) What is the composition of scale and sludge in boilers? How are these formed ?
b) In determination of hardness of a sample of water by EDTA method,
i) 20 ml of standard hard water containing 0.1 gm of CaCO3  per 100 ml required                 15 ml of EDTA solution.
ii) 100 ml of hard water required 12 ml of EDTA solution.
iii) After boiling the hard water and filtering the filtrate required 6 ml of EDTA.
    Calculate : a) temporary and permanent hardness.
                        b) the amounts of  lime and soda required for softening of water.
           (Ans : Temporary hardness : 82 ppm, permanent hardness : 78 ppm. Amt. of lime : 60.7m/l, Soda : 82.68 ppm)
4. a) How is hard water estimated by EDTA method ?
b) How is hard water softened by zeolite method ?
c) Calculate the amounts of lime and soda required for softening 15,000 litres of       water containing the following ?
  Ca (HCO3)2 = 121.5 ppm, Mg (HCO3)2  = 116.8 ppm, MgCl2 = 79.2 ppm,   CaSO4 = 102 ppm.               (Ans : Lime : 3.534 kg, Soda = 2.519 kg)  (JNTU 1981)
5. a) What are the requirements of boiler feed water ?
b) How are boiler scales formed ?
c) What is foaming ?
d) Explain the process of a phosphate conditioning of boiler feed water. (JNTU 1981)
6. a) Discuss the various boiler troubles, their causes and prevention.
b) The total hardness of a sample of water was determined by titrating the sample with         standard EDTA solution. From the following results calculate the total hardness of water sample and express the hardness of water in ppm. 50 ml of water required 14 ml of 0.01M EDTA. The indicator used is EBT (1000 ml of 1M EDTA = 100 gms of   CaCO3)                                         [Ans : Total hardness : 280 ppm]  (JNTU 1985)
7. Discuss the lime soda and zeolite process of softening of water giving merits and demerits of the process.   (JNTU 1985)
8. a) Describe the various affects caused by different impurities present in boiler feed     water.
b) 20,513 litres of hard water was softened by zeolite process. The zeolite was           regenerated by 100 litres of sodium chloride solution containing 120 gms of NaCl. Calculate the hardness of water in mg/l.(atomic weights of Na = 23, Cl = 35.5)                                                                [Ans : 500mg/l] (JNTU 1982)
9. a) Compare lime soda and zeolite process.
b) Write a brief account on break point chlorination. (JNTU 1982)
10. a) Explain the formation and composition of scales in boilers.  How are they different from sludges?  What are their bad affects ?
b) A sample of water contains the following impurities per litre Mg(HCO3)2  = 73 mg, CaCl2  = 222 mg, MgSO4 = 73 mg, Ca(NO3)2 = 164 mg, NaCl = 58.5 mg. The sample of lime available is 74% pure and  soda is 90% pure. 20% excess chemicals have to be added. Calculate the amount of lime and soda required for the treatment of 10,000 litres of water.     [Ans : Lime : 1.93 kg, Soda : 4.2 kg] (JNTU 1983)
11. a) Explain how boiler scales and caustic embrittlement are caused in boilers and how are they minimized.
b) What is hardness of water ? How do you express hardness ? Express the hardness    of a sample of water containing 156 parts of Mg(HCO3)2.
   [Ans : 106.9], (JNTU 1990, Supp)
12. Write short notes on          
a) Break point chlorination  [JNTUK - 2014]
b) Internal treatment of boiler feed water
c) Calculate the lime and soda required for softening 20,000 litres of water, if the water contains free acidity 1.5 ppm bicarbonate ions = 396.5 ppm, Ca ions =  90 ppm, ferrous sulphate = 14 ppm.
                                          (Ans : Lime : 7.684 kg, Soda = 0.106 kg)  (JNTU 1987)
13. With the help of a neat diagram describe the reverse osmosis method for the             desalination of brackish water.
14. a) What are requisites of drinking water ?
b) How is natural water sterilized by chlorine, bleaching powder, chloramine and ozone?
15. What are scales ? How are they formed?  What are the disadvantages and prevention methods for scales ? [JNTUK - 2014]
16. Write short notes on
a)  Calgon treatment b) Phosphate conditioning
c)  Ion-exchange process d) Colloidal conditioning
17. What are zeolites ? How do they function in removing the hardness of water ?
[JNTUK - 2014]
18. Discuss the basic principles involved in the estimation of hardness of water by EDTA method.
19. Distinguish between
a)  purification and softening of water
b)  boiler scales and sludges
c)  internal and external treatment of boiler feed water.
20. What is desalination ? Explain the various methods available for desalination ?
21. a) Give the chemical reactions involved in the determination of hardness of water ?
b) Explain why NH4Cl + NH4OH buffer is added in the determination of hardness in       water by EDTA?
22. Write a brief account on
a)  caustic embrittlement [JNTU - 2014]
b)  boiler corrosion
23. Give a brief account on reverse osmosis and electrodialysis.
24. a) Explain the hot lime-soda process. What are its advantages ?
b) Write an account on mixed-bed ion exchange.
25. Differentiate permutit, lime-soda and ion exchange process.
26. What are the basic reactions involved in the calculation of lime and soda for purification of water ?
27. What are the natural and synthetic zeolites ? Explain the zeolite process for the external treatment of boiler feed water and what are its limitations and advantages?
28. Give an account on ion exchanging process for the external treatment of boiler feed water with special reference to mixed bed ionizer.
29. Write short notes on the following
i) reverse osmosis ii) carry over
30. a) What is hardness of water due to ? What are its units ? Explain the different types of  hardness of water.
b) Briefly describe disinfectation of municipal water.
31. a) Differentiate between lime-soda and zeolite processes for softening of water giving merits and demerits of the two processes.
b) How is the hardness of water expressed ? What are the various units employed ?               Explain their interconversion.                          [JNTU-2010, Set-1, Q.No.7]
32. a) What are the factors that lead to caustic embrittlement in boilers ? How can   this be prevented ?                              [JNTU-2010, Set-2, Q.No.7]
b) Distinguish between Zeolite process and Ion-Exchange process.
33. Write a note on complexometric titrations used for estimation of hardness of water by EDTA.                                         [JNTU-2010, Set-3, Q.No.6]
34. a) Why is hard water harmful to boilers ?                   [JNTU-2010, Set-4, Q.No.4]
b) Describe the causes and harmful effects of scale formation.
35. One litre of water from Khammam Dist. in Andhra Pradesh showed the following analysis :
 Nacl = 0.0167 gms, CaSO4 = 0.0065 gms, and  MgSO4 = 0.0054 gms. Calculate the lime & soda required  for softening of 10,000 litres of water.                    [JNTU-2010, Set-4, Q.No.4]
36. a) What are turbine deposits?  [Ans : Lime : 3.64 kg,  Soda : 0.984 kg]
b) How are they formed?
c) How do you prevent the turbine deposits?
37. Write a brief account on
a) Disadvantages of turbine deposits
b) Causes of turbine deposits
c) Prevention and removal of turbine deposits
38. a) Write notes on priming & forming and caustic embrittlement.
b) Explain the process of treatment of water for domestic use.             (JNTUK 2014)
iII. multiple choice Questions
1. Temporary hardness in water is removed by
a) filtration b) sedimentation c) boiling d) coagulation
2. Blow-down operation causes the removal of
a)  scales b)  sludges c) acidity d)  sodium chloride
3. Solubility of calcium sulphate in water
a) increases with rise of temperature
b) decreases with rise of temperature
c) remains unaltered with rise of temperature
d) does not adopt any definite pattern with rise of temperature
4. Permanent hardness of water cannot be removed by
a) treatment with lime soda b)  by permutit process
c) boiling d)  ion-exchange process
5. Hard water is unfit for use in boilers for generating steam because
a) its boiling point is higher
b) hardwater does not produce lather inside the boiler
c) water decomposes into O2 and H2  
d) it produces scales inside the boiler
6. Estimation of hardness water by EDTA method is used to determine
a) alkaline hardness b)  temporary hardness only
c) permanent hardness only d)  all the above
7. Hardwater can be softened by passing it through
a) lime stone b)  sodium hexameta phosphate
c) ion-exchange resin d)  sodium silicate
8. Calgon is a trade name given to
a) sodium silicate b) sodium hexametaphosphate
c) sodium meta phosphate d) calcium phosphate
9. Brackish water mostly contains dissolved
a) calcium salts b)  magnesium salts
c) turbidity d)  sodium chloride
10. The method by which the ions are pulled out of salt water by direct current, and employing thin, rigid membrane pair is called
a) electro dialysis b)  reverse osmosis
c) zeolite   d)  ion exchange
11. The purification of brackish water by reverse osmosis is also called as
a) super-filtration b)  supra-filtration
c) hypo-filtration d)  filtration
12. One part of CaCO3 equivalent hardness per 105 parts of water is called
a) degree Clark b)  ppm
c) degree French d)  mg/l
13. Boiler corrosion caused by using highly alkaline water in a boiler is called
a) corrosion b) boiler corrosion
c) caustic embrittlement d) erosion
14. Caustic embrittlement can be avoided by using
a) sodium phosphate b)  hydrogen
c) ammonium hydroxide d)  sodium sulphate
15. Caustic embrittlement is a type of
a)  boiler corrosion b)  conditioning
c)  scale formation d)  sludge formation
16. The soft, loose and slimy precipitate formed within the boiler is called
a) scale b)  sludge
c) embrittlement d)  coagulation
17. Sodium meta aluminate used in internal treatment of boiler water produces flocculant precipitates of
a) Mg(OH)2 and Al(OH)3   b)  NaOH and Al(OH)3
c) Ca(OH)2 and Al(OH)3 d)  Mg(OH)2 and Ca(OH)2
18. One of the following chemical acts on both coagulant and softening agent
a) lime b)  soda
c) alum d)  sodium aluminate
19. The process of allowing water to stand undisturbed in big tanks for settling of the suspended particles due to force of gravity
a) coagulation b)  conditioning
c) sedimentation d)  screening
20. The composition of alum is
a) K2SO4. Al2 (SO4)3. 24 H2O b)  K2(SO4)3. Al2. (SO4)3. 24 H2O
c) K2SO4 . Al2 (SO4)3 . 20 H2O   d)  K2SO4. Al2SO4. 24 H2O
21. Ferrous sulphate is commonly used in the treatment of municipal water for
a) filtration b) flocculation
c) sedimentation d) disinfection
22. Liquid chlorine is a most effective
a) disinfectant b) coagulant
c) flocculant d)  sterilizing agent
23. Disinfection by ozone is due to liberation of
a) oxygen b) nascent oxygen
c) molecular oxygen d)  oxide
24. The formula of chloramine is :
a) ClNH2 b)  NHCl2 c) NCl3 d)  NH2Cl2
25. Phosphate conditioning of boiler feed is carried out by
a) Na3PO4 b)  Ca3(PO4)2 c) Mg3(PO4)2 d)  H3PO4
26. Ultraviolet rays are used in the treatment of water for
a) Filtration b)  Sedimentation
c) Screening d)  Sterilisation
27. The external treatment of boiler water feed done by
a) lime-soda process b)  sodium sulphate treatment
c) calgon process d)  sodium aluminate treatment
28. The process of wet-steam formation is called
a) foaming b)  priming
c) corrosion d)  caustic embrittlement
29. Mechanical steam purifiers avoid
a) corrosion b)  priming c) scale formation d)  sludge formation
30. Castor oil is a
a) antiskinning agent b)  antifoaming agent
c) anti-ageing agent d)  anti-corrosive agent
31. The largest source of water is
a) lake water b)  river water
c) rain water d)  sea water
32. One of the following impurity is introduced into rain water due to the decomposition of plant and animal remains in water.
a) Organic impurities b)  Dissolved impurities
c) Floating impurities d)  Colloidal impurities
33. A water sample found to possess 16.2 mg/l of Ca(HCO3)2. Its hardness in terms of CaCO3 equivalents is
a) 100 b)  10 c) 16.2 d)  1000
34. Hardness of water is caused due to the presence of
a) Undissolved salts of Ca2+ and Mg2+
b) Dissolved sulphates of potassium
c) Dissolved salts of Ca2+ and Mg2+
d) Undissolved CaCO3
35. Which of the following methods separates both ionic and non-ionic impurities from water?
a) Electrodialysis b)  Deionisation
c) Reverse osmosis d)  Zeolite process
36. Potable water treatment does not involve
a) demineralisation b)  sedimentation
c) filtration d)  disinfectation
37. The cofficient of thermal expansion of boiler plant is
a) more than boiler scale b)  less than boiler scale
c) equal to boiler scale d)  no comparison between two
38. Which of the following salts cause least hardness to water when converted to CaCO3 equivalents?
a) 10 mgs of CaCO3 b)  19 mgs of CaSO4
c) 10 mgs of MgCl2 d)  10 mgs of CaCl2
39. Ultraviolet rays are used in the treatment of water for
a) Filtration b) Sedimentation
c) Screening d) Sterilisation
40. One of the following chemical acts as both coagulant and softening agent
a) lime b) soda
c) alum d) sodium aluminate
41. The ion-exchange resins used for softening of water are a) Cross-linked polymers with microporous structure
b) Branched polymers with porous structure
c) Cross-linked polymers with non-porous structure
d) Branched polymers with non-porous structure
42. One of the following is an example of cation exchanging resin.
a) Copolymer of phenol formaldehyde-amine formaldehyde
b) Copolymer of styrene-divinyl benzene
c) Copolymer of phenol formaldehyde and styrene
d) Copolymer of amine formaldehyde and divinyl benzene.
43. The membrane filtration adopted in reverse osmosis is also called
a) Super filtration b) Suprafiltration        
c) Ultrafiltration d) Hypofiltration
44. Uneven deposition on turbines causes
a) vibrational problems b) heat problems
c) power problems d) super heating
45. Attemperating is
a) uncontrolled heating b) controlled heating
c) controlled cooling d) uncontrolled cooling
IV. FILL IN THE BLANKS :
 1. On addition of chlorine to water __________ acid is produced which is powerful           germicide.
 2. Hardness of water is due to the dissolved salts of  __________ and ___________.
 3. The chemical which removes dissolved oxygen of water without adding hardness
is __________
 4. Hardness of water is expressed in equivalents of __________.
 5. In lime-soda process of softening, calcium and magnesium ions are precipitated as __________  and __________.
 6. Sodium aluminate is used as __________  during purification of water.
 7. Anion exchange resins are regenerated by using __________
 8. Best method of removing hardness of water is __________ process.
 9. Among chloramine, bleaching powder and chlorine __________  is a powerful disinfectant.
10. The hardness of water in CaCO3 equivalents containing MgSO4  (mol. wt. 120) with concentration 12 mg/l is __________.
11. Presence of residual __________  in boiler water causes caustic embrittlement.
12. __________   causes the flow of solvent from lower  concentration to higher concentration, which is separated by a semipermeable membrane.
13. In lime-soda process the addition of lime cannot remove __________  hardness of water.
14. The suspended and colloidal impurities are allowed to settle under gravitation in ___________ process.
15. Priming and foaming in boilers produce __________  steam.
16. Cation exchange resin contains __________ mobile ions.
17. The hardness of a sample of water is 10 ppm,  which can be expressed as _________  degree clarke.
18. A sample of water contains 11.1 mg/l of CaCl2. Its hardness of CaCO3 equivalent is __________.
19. __________  is used as an indicator in the determination of hardness by EDTA method.
20. To maintain the pH between 9-10 during complexometric titration, estimation of hardness of water is __________.
21. Temporary hardness of water can be removed by __________.
22. Calgon treatment is used for the removal of dissolved __________.
23. The chemical structure of zeolite is __________.
24. Natrolite is a _________ zeolite.
25. Ion free water is known as __________.
26. The exhausted zeolite is regenerated by __________.
27. __________   is a process of allowing water to stand undisturbed in big tanks.
28. Al2(SO4)3  alum produce __________ as flocculant precipitates during softening of water.
29. __________ membranes are selected for efficient separation of ions.
30. The presence of even small amounts of MgCl2 will cause __________ of boiler plate to a large extent.
31. __________ is considered to be the naturally distilled water.
32. __________  water contain the largest concentration of dissolved impurities.
33. Rain water picks up CO2, O2 etc., from atomsphere which introduces __________  impurities into water.
34. The pure water is pushed out through a semipermeable membrane during   _________.
35. Extra-pure water can be obtained by using ___________.
36. A semipermeable membrane allows the flow of ___________.
37. Chlorine, when treated with water produce ___________ acid, which acts as a powerful germicide.
38. The exhausted anion exchanging resin is regenerated by ___________.
39. Rain water picks up gases like CO2, O2 etc. from ___________.
40. The suspended and colloidal impurities are allowed to settle under gravitation in _____. process.
41. Among the dissolved gases ________ is the most corroding impurity.
42. Zeolites are _____.
43. A copolymer of __________ or _____ is used as anion exchanging resin.
44. Deionisation must be followed by ____________.
45. Steam turbines convert ______ energy to rotary motion.
46. A turbine is a rotary mechanical device that extracts energy from a ________.
47. CaCO3 is an ________ salt present in water.
48. Disinfectation by bleaching powder is also called ________.
49. CaSO4 is _________ soluble in cold water.
50. The precipitates of CaCO3 and Mg(OH)2 are very fine and causes ________ problems.
51. During carbonate conditioning the excess Na2CO3 remaining causes _________ in high pressure boilers.
V. Indicate True or False FOR the following :
1. Rain water is the purest form of natural water. [ T / F ]
2. Suspended impurities are present in natural water. [ T / F ]
3. Dissolved oxygen do not cause corrosion to boiler plate. [ T / F ]
4. Calcium and magnesium salts in water are removed by lime-soda process. [ T / F ]
5. It is possible to remove permanent hardness of water by boiling. [ T / F ]
6. Chloramine is better than chlorine for sterilisation. [ T / F ]
7. The common unit to express hardness of water is ppm. [ T / F ]
8. Sludge is a soft, loose, slimy precipitate formed inside the boiler. [ T / F ]
9. NH4Cl and NH4OH buffer is used to maintain pH 12. [ T / F ]
10. Water softened by zeolite process causes caustic embrittlement. [ T / F ]
11. Hardness of water is expressed in calcium chloride equivalents. [ T / F ]
12. Hot lime-soda process is better than cold process. [ T / F ]
13. The presence of CO2 in water produces carbonic acid. [ T / F ]
14. The process of removing hardness producing salts is called softening. [ T / F ]
15. Hydrated sodiumaluminium silicate is called zeolite. [ T / F ]
16. The process of removing common salt from water is called salination. [ T / F ]
17. Addition of calgon to boiler water prevents calcium sulphate scale formation. [ T / F ]
18. Hard water precipitates, soap as calcium soap. [ T / F ]
19. Reverse osmosis removes all ionic, non-ionic and colloidal impurities
from water. [ T / F ]
20. Dissolved calcium bicarbonate in water causes permanent hardness. [ T / F ]
21. pH of acidic water is more than 7. [ T / F ]
22. Hard water softened by EDTA method. [ T / F ]
23. Break point chlorination gives the exact amount of chlorine to be added to raw water. [ T / F ]
24. The exhausted permutit is regenerated by ZnCl2. [ T / F ]
25. Extra pure water is obtained by electrodialysis. [ T / F ]
26. Coagulant helps in setting of colloidal particles only. [ T / F ]
27. A permeable membrane allows the flow of solvent molecules. [ T / F ]
28. Calgon is a trade name of sodium hexa meta phosphate. [ T / F ]
29. Calgon conditioning is better than phosphate conditioning. [ T / F ]
30. Water is softened before using in boilers by external treatment. [ T / F ]
ANSWERS
iII. multiple choice Questions :
1) c 2) b 3) b 4) c 5) d 6) d 7) c 8) b 9) d 10) a
11) a 12) c 13) c 14) d 15) a 16) b 17) a 18) d 19) c 20) a
21) b 22) a 23) b 24) a 25) a 26) d 27) a 28) b 29) b 30) b
31) d 32) a 33) b 34) c 35) c 36) a 37) a 38) b 39) d 40) d
41) a 42) b 43) a 44) a 45) c
IV. FILL IN THE BLANKS :
1)   hypochlorous 2) calcium and magnesium 3) hydrazine
4) calcium carbonate 5) CaCO3 and Mg(OH)2 6) coagulant
7) sodium hydroxide 8) ion-exchange process  9) chlorine
10) 10 mgl 11) NaOH or caustic soda 12)   Osmosis
13) calcium permanent 14) Sedimenation 15) wet
16) H+ 17) 0.7 0Cl 18) 10 mgl
19) Eriochrome black - T 20) NH4Cl, NH4OH 21) boiling
22) CaSO4 23) Na2O. Al2O3 , nSiO2. y H2O 24) natural
25) deionised or demineralised water 26) NaCl
27) Sedimentation 28) Al(OH)3
29) Ion-selective membrane 30) corrosion 31) Rain water 32) Sea water 33) dissolved gas 34) reverse osmosis 35) electrodialysis 36) solvent molecules 37) hypochlorous 38) NaOH 39) atmosphere 40) sedimentation 41) Oxygen 42) Cation exchangers
43) Phenol formaldehyde and amine formaldehyde
44) degasification 45) thermal/heat 46) fluid flow
47) insoluble 48) hypo - chlorination 49) more
50) after deposition 51) Caustic embrittlement
V. TRUE OR FALSE QUESTIONS :
1) T 2) T 3) F 4) T 5)  F 6) F 7) T 8) T 9) F 10) T
11) F 12) T 13) T 14) T 15) T 16) F 17) T 18) T 19) T 20) F
21) F 22)  F 23) T 24) F 25) T 26) F 27) T 28) T 29) T 30) T