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Water Technology

Water Technology

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Water Technology

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  1. Water quality parameters: physical, chemical & biological. Types of water - Boiler troubles: Scale and sludge, Boiler corrosion, Caustic embrittlement, Priming & foaming. Treatment of boiler feed water: Internal treatment (phosphate, colloidal, sodium aluminate and calgon conditioning) and External treatment–Demineralization and zeolite process. Specifications for drinking water BIS - WHO standards. Domestic water treatment- break-point chlorination. Desalination of brackish water: Reverse Osmosis. UNIT IWATER TECHNOLOGY 1

  2. A) Surface Waters • Rain Water - Pure but contaminated with gases • River Water - High dissolved salts moderate organics • Lake Water - Const. composition but high organics • Sea Water - High salinity, pathogens, organics • B) Underground Waters • Spring/Well Water - Crystal clear but high dissolved salts and high purity from organics Sources ofWater 2

  3. MAJOR IMPURITIES OFWATER Ionic anddissolved Cationic Calcium Magnesium Anionic Bicarbonate Carbonate Hydroxide Nonionic and undissolved Turbidity, silt, mud, dirtand other suspendedmatter Color,Plankton Organicmatter, Microorganisms, Bacteria Gases CO2H2S NH3 CH4 O2 Alkalinity Sodium Potassium Ammonium Iron Manganese Sulfate Nitrate Phosphate ChlorideColloidal silica 3

  4. Does not produce leather with soap solution • Presence of various (CaHCO3, MgHCO3, CaCO3, CaSO4, MgSO4, CaCl2, MgCO3). • Ions come fromdissolved rock the water has passed through. • Affects properties oftap water HARDWATER 4 1

  5. DETECTION OFWATER • Minerals in hardwater interact withsoap. • Interferes with soap’s ability tolather. 5

  6. CHEMICAL REACTION (C17H35COO)2Ca +2NaCl Calciumstearate (Insoluble) 2C17H35COONa +CaCl2 Sodiumstearate (sodiumsoap) Hardness Hardness + Soap white precipitate 6

  7. METAL ION INDICATOR 7

  8. HARDNESS OFWATER • Hardness in Water is characteristic that prevents the‘lathering of soap’ thus water which does not produce lather with soap solution readily, but forms a white curd is called hardwater. • Type ofHardness • Temporary or CarbonateHardness • Permanent Hardness or non-carbonateHardness. 8

  9. Alkaline hardness • Carbonate hardness • Presence of bicarbonate • It can be removed by process such as boiling and softening TemporaryHardness 9

  10. Non Carbonate Hardness is due to the presence of chlorides, sulfatesof • calcium, Magnesium, iron and other heavymetals • Non-alkaline PERMANENTHARDNESS (C17H35COO)2Ca +2NaCl Calciumstearate (Insoluble) 2C17H35COONa +CaCl2 Sodiumstearate (sodiumsoap) Hardness (C17H35COO)2Mg+2Na2SO4 Magnesiumstearate (Insoluble) 2C17H35COONa +MgSO4 Sodiumstearate (sodiumsoap) Hardness 10

  11. Removal method 1. Lime soda process 2. Zeolite process Na2 Al2 Si2 O3 = Sodium alumino silicate 11

  12. DomesticUse IndustrialUse Draw backs (or) Disadvantages of HardWater TextileIndustry SugarIndustry DyeingIndustry PaperIndustry PharmaceuticalIndustry In Steam generation inBoilers Washing Bathing Drinking Cooking 12

  13. Water with very low concentrations ofminerals. • Soap lathers easily and is sometimes difficult to rinse off. • Absence of Ca2+, Mg2+ SOFTWATER 13

  14. Expression of hardness Units of Hardness: Parts Per Million: PPM Parts of CaCO3 equivalent hardness per 106 parts of water. 1 ppm = 1 part of CaCO3 equivalent hardness in 106 parts of water. 15

  15. Milligrams per litre: - mg / L Number of milli grams of Calcium Carbonate equivalent hardness present in 1 litre of water. 1 mg / L = 1 mg of CaCO3 equivalent hardness present in 1 liter of water. 16

  16. Clarke’s degree : (°cl) Equivalent hardness per 70,000 parts of water. 1°Clarke = 1 part of CaCO3 of hardness per 70,000 parts of water. 17

  17. Degree French (°Fr) : It is in the parts of CaCO3 eq hardness per 105 parts of water. 1°Fr = 1 part of CaCO3 in 105 parts of water. Milliequivalents per litre (meq/L) No of milli equivalents of hardness present per litre. 1 meq/L = 50 mg /L of CaCO3 eq = 50 ppm 1 meq/L = 50 ppm 18

  18. Relation between various units o hardness: • 1 ppm = 1 mg /L = 0.1 ° Fr = 0.07°CL = 0.02 meg / L • 1 mg /L = 1ppm = 0.1 °Fr = 0.07°CL = 0.02 meg /L • 1 ° cL = 1.433 °Fr = 14.3 ppm = 14.3 mg /L = 0.286 meq /L • 1°Fr = 10 ppm = 10 mg /L = 0.7 °cL =0.2 meq /L • 1 meq /L = 50 mg /L = 50ppm = 5°Fr = 0.35 °Cl. 19

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