Cvl_100_lecture1_water_finalver_31jan2019.pdf

Water Engineering Gazala Habib

Definition • Potable water: A water that can be consumed at any desired amount without concern for adverse health effects is termed as potable water (Davis and Cornwell, 1998). Potable does not necessarily mean that the water tastes good. • Potable and palatable (which is pleasing to drink) • Water should be both potable and palatable • Physical characteristics of water: • Mass density: Mass per unit volume (1000 kg/m3) • Specific weight: Weight (force) per unit volume 𝛾 = 𝜌 × 𝑔 = 1000 ∗ 9.81 = 9.81 𝐾𝑁/𝑚2 • Specific gravity: 𝜌 𝑆= 𝜌0 Where 𝜌 0 = 1000 𝑘𝑔/𝑚3 • Dynamic viscosity: (Pa.S=Ns/m2=kg/m.s) 𝜇=

𝑚 𝑙×𝑡

• Kinematic viscosity: (m2/s) 𝜐=

𝜇 𝜌

Water quality • Physical characteristics • Turbidity: The presence of suspended material such as clay, silt, finely divided organic material, plankton, and other particulate material in water is know as turbidity. • Color: Dissolved organic material from decaying vegetation and some inorganic matter cause color in water. • Taste and odor: taste in water can be caused by organic compounds, inorganic salts and dissolved gases. These material may come from domestic agricultural and natural sources. • Temperature

Water pollutants and their removal mechanisms • The impurities can be found in three states • Suspended • Colloidal • Dissolved • A dissolved substance is homogeneously dispersed in the liquid. Dissolved substances can be simple atoms or complex molecular compounds.

• The substance cannot be removed from the liquid without accomplishing a phase change such as distillation, precipitation, adsorption, extraction, or passage through "ionic" pore-sized membranes. • In distillation either the liquid or the substance itself is changed from a liquid phase to a gas phase in order to achieve separation. • In precipitation the substance in the liquid phase combines with another chemical to form a solid phase, thus achieving separation from the water. • Adsorption also involves a phase change, wherein the dissolved substance reacts with a solid particle to form a solid particle-substance complex. • Liquid extraction can separate a substance from water by extracting it into another liquid, hence a phase change from water to a different Iiquid.

• A membrane with pore sizes in the ionic-size range can separate dissolved substances from the solution by a high-pressure filtering process.

Suspended solids • Suspended solids are large enough to settle out of solution or be removed by filtration. In this case there are two phases present, the liquid water phase and the suspended-particle solid phase. • The lower size range of this class is 0.1 to 1.0 mm, about the size of bacteria. • In environmental engineering, suspended solids are defined as those solids that can be filtered by a glass fiber filter disc and are properly called filterabIe solids. • Suspended solids can be removed from water by physical method such as sedimentation, filtration, and centrifugation.

Colloidal impurities • Colloidal particles are between dissolved substances and suspended particles. They are in a solid state and can be removed from the liquid by physical means such as very high-force centrifugation or filtration through membranes with very small pore spaces. However, the particles are too small to be removed by sedimentation or by normal filtration processes. • Colloidal particles exhibit the Tyndall effects that is when light passes through a liquid containing colloidal particles, the light is reflected by the particles. The degree to which a colloidal suspension reflects light at a 90° angle to the entrance beam is measured by turbidity. • Turbidity is a relative measure. The most common standard unit is a nephelometric turbidity unit (NTU) or turbidity unit (TU). • For a given particle size, the higher the turbidity, the higher the concentration of colloidal particles.

Color • Another useful term in environmental engineering that is used to describe a solution state is color. • Color is not separate from the above three categories, but rather is a combination of dissolved and colloidal materials. • Color is widely used in environmental engineering because it, in itself, can be measured. • However, it is very difficult to distinguish "dissolved color" from "colloidal color”. • Some color is caused by colloidal iron or manganese complexes, although the most common cause of color is from complex organic compounds that originate from the decomposition of organic matter. • One common source of color is the degradation of soil humus, which produces humic acids.

• Humic acids impart reddish-brown color to the water. Humic acids have molecular weight between 800 and 50,000 the lower being dissolved and the greater, colloidal • Most color seems to between 3.5 and 10 mm which is colloidal. • Color is measured by the ability of the solution to absorb light.

• Color particles can be removed by the methods for dissolved and colloidal particles depending upon the state of the color.

Microbiological characteristics • Drinking water should be free from disease causing bacteria (pathogens). These organisms include viruses, bacteria, protozoa, worms. • Fecal discharge of infected individuals • Fecal discharge of animals

• Test: Number count of coliform group including Escherichia coli and Aerobacter aerogenes • E. coli are common inhabitants of intestinal tracts of human and other mammals. • Aerobacter are common in soil, on leaves, and on grain. • Total coliform test has been selected for detection of microbial contamination of water. •

Reasons for total coliform test • Detection of microorganism is complex and time consuming • Presence of e. coli indicates fecal contamination of water. • In acutely ill individual the number of coliform organisms excreted in the feces out number the disease producing organisms. Therefore, they are easier to culture than disease producing organisms. • Coliform group of organisms survive in natural water for long time but does not reproduce effectively. Therefore, their presence in water body indicates the fecal contamination rather than growth of organism because of favourable environmental condition. • The absence of coliform is indicator of safe water. • The organisms of coliform group are easy to culture in laboratory.

Size range of impurities in water

Water quality: Chemical characteristics • Chloride: Cl can come from leaching of marine sedimentary deposits or by pollution from sea water, brine, or industrial or domestic wastes. Chloride concentrations in excess of about 250 mg/L produces noticeable taste in drinking water. Domestic water should contain ;less than 100 mg/l of chloride. • Fluoride: In some areas, water sources contain natural fluoride. The F concentrations at optimum levels is beneficial for dental health. Excessive fluoride in drinking water supplies may produce fluorosis (mottling) of teeth, which increases as the optimum fluoride level is exceeded. Acceptable levels are generally between 0.8 and 1.3 mg/L fluoride. • Iron: Small amounts of iron frequently are present in water because of the large" amount of iron in the geologic materials. The presence of iron in water is considered objectionable because it imparts a brownish colour and affects the taste of beverages such as tea and coffee. • Lead: Exposure of the body to lead even in low amount can be seriously damaging to health. Prolonged exposure to relatively small quantities may result in serious illness or death. • Manganese: Manganese imparts a brownish colour to water and to cloth that is washed in it. It flavours coffee and tea with a medicinal taste.

• Sodium: The presence of sodium in water can affect persons suffering from heart, kidney, or circulatory ailments. When a strict sodium-free diet is recommended. Home water softeners may be of particular concern because they add large quantities of sodium to the water. • Sulfate: Waters containing high concentrations of sulfate, caused by the leaching of natural deposits of magnesium sulfate (Epsom salts) or sodium sulfate (Glauber's salt), may be undesirable because of their laxative effects.

• Zinc: Zinc is found in some natural waters, particularly in areas where zinc ore dep()sitspayebeelllIlined. Zinels not considered detrimental to health, but it will impart an undesirable taste to drinking water. • Arsenic: Arsenic occurs naturally in the environment and it is also used in timber treatment, agricultural chemicals (pesticides) and manufacturing of gallium arsenide wafers, glass and alloy. Arsenic in drinking water is associated with lungs and urinary cancer.

Water quality: Chemical characteristics • Toxic inorganic substances: (NO3, CN, Heavy metals).

• Methemoglobinemia (Blue baby syndrome) has occurred in infants who have been given water or fed formula prepared with water having high concentrations of nitrate. • CN ties up the haemoglobin sites that bind oxygen to red blood cells. This results in oxygen deprivation. A characteristic symptom is that the patient has a blue skin colour. This condition is called cyanosis. CN causes chronic effects on the thyroid and central nervous system.

• The toxic heavy metals include arsenic (As), barium (Ba), cadmium (Cd), chromium(Cr), lead (Pb), mercury (Hg), selenium (Se), and silver (Ag). The heavy metals may be acute poisons (As and Cr6+ for example),or they may produce chronic disease (Pb, Cd, and Hg for example). • Toxic organic substances (pesticides, insecticides, and solvents): The effects may be chronic or acute.

Water chemistry: common terms • Molarity is the number of moles in a liter of solution. • A 1 molar (I M) solution has 1 mole of substance per liter of solution. • Molarity is related to mg/L by

mg/L = molarity X molecular weight X 103 = (moles/L) X (g/mole) X ( 103 mg/g)

• A second unit, equivalent weight (EW), is frequently used in softening and redox reactions. • The equivalent weight is the molecular weight divided by the number (n) of electrons transferred in redox reactions or the number of protons transferred in acid/base reactions. • Normality (N)is the number of equivalent weights per liter of solution and is related to molarity(M) by N=M.n

Alkalinity of water • Alkalinity: alkalinity is defined as the sum of all titrable bases down to about pH 4.5. It can be found by experimentally determining how much acid it takes to lower the pH of water to 4.5. • In most waters the only significant contribution to alkalinity are the carbonate species and any free H+ and OH- ions. The total H+ that can be taken up by water containing primarily carbonate species is • Alkalinity=[HCO3-] +2[CO32-]+[OH-]-[H+] • 𝐻2 𝐶𝑂3 ⇌ 𝐻+ + 𝐻𝐶𝑂3 −

•

− 3 𝐻𝐶𝑂

⇌ 𝐻+ + 𝐶𝑂3 2− 𝐸𝑊𝐶𝑎𝐶𝑂3 𝑚𝑔 𝑚𝑔 𝑎𝑠 𝐶𝑎𝐶𝑂3 = ( 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠) × 𝐿 𝐿 𝐸𝑊𝑠𝑝𝑒𝑐𝑖𝑒𝑠

Alkalinity calculation • Example 1: A water contains 100.0 mg/l of 𝐶𝑂3 2− and 75.0 mg/l of 𝐻𝐶𝑂3 − at a pH of 10. Calculate the alkalinity exactly at 25 C. Approximate the alkalinity by ignoring [OH-] and [H+] ions.

Equivalent weights (EW) 𝐶𝑂3 2− Molecular Weight=12+3*16=60; n=2; EW=30 𝐻𝐶𝑂3 − MW=1+12+3*16=61; n=1; EW=61

𝐾𝑤 =

𝐻+

MW=1; n=1; EW=1

𝑂𝐻 −

MW=16+1=17; n=1; EW=17

𝐶𝑎𝐶𝑂3 MW=40+12+3*16=100; n=2; EW=50

10−14

pH=10 𝑝𝐻 = − log10 [𝐻+ ] [𝐻+ ] = 10−10 𝑚𝑜𝑙𝑒𝑠/𝐿 𝑂𝐻− =

𝐾𝑤 [𝐻− ]

Alkalinity calculation •

𝑂𝐻 −

=

𝐾𝑤 [𝐻 − ]

=

10−14 10−10

= 10−4 𝑚𝑜𝑙𝑒𝑠/𝐿

• Concentration in mg/l • • • •

𝑚𝑜𝑙𝑒𝑠 −10 = 10 ( ) 𝐿 + −10 𝑚𝑜𝑙𝑒𝑠 [𝐻 ] = 10 𝐿 𝑚𝑜𝑙𝑒𝑠 − −4 𝑂𝐻 = 10 𝐿

[𝐻 + ]

𝑚𝑔 𝑎𝑠 𝐿

• 𝐶𝑂3

𝐶𝑎𝐶𝑂3 =

2−

𝑚𝑔 3 × 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 × 10 𝑔 𝑔 3 𝑚𝑔 −7 𝑚𝑔 ×1 × 10 = 10 𝑚𝑜𝑙𝑒 𝑔 𝐿 𝑔 𝑚𝑔 𝑚𝑔 3 × 17 × 10 = 1.7 𝑚𝑜𝑙𝑒 𝑔 𝐿

𝑚𝑔 ( 𝑜𝑓 𝐿

𝑎𝑠 𝐶𝑎𝐶𝑂3 (𝑖𝑛

𝑔 𝑚𝑜𝑙𝑒

𝑠𝑝𝑒𝑐𝑖𝑒𝑠) ×

𝑚𝑔 ) 𝐿

𝐸𝑊𝐶𝑎𝐶𝑂3 𝐸𝑊𝑠𝑝𝑒𝑐𝑖𝑒𝑠

= 100.0 ×

50 30

= 167

Alkalinity calculation −

• 𝐻𝐶𝑂3 𝑎𝑠 𝐶𝑎𝐶𝑂3 (𝑖𝑛 • •

𝑚𝑔 ) 𝐿

= 75.0 ×

𝑚𝑔 50 −7 𝐻 𝑎𝑠 𝐶𝑎𝐶𝑂3 (𝑖𝑛 ) = 10 × 𝐿 1 𝑚𝑔 50 − 𝑂𝐻 𝑎𝑠 𝐶𝑎𝐶𝑂3 (𝑖𝑛 ) = 1.7 × 𝐿 17 +

50 61

= 61.5

= 5 × 10−7 = 5.0

• Total alkalinity as 𝑚𝑔 𝑚𝑔 −7 𝐶𝑎𝐶𝑂3 𝑖𝑛 = 167 + 61.5 + 5 × 10 + 5.0 = 233.5 𝐿 𝐿

Problem • Using following equations derive two equations which allow calculation of the bicarbonate and carbonate alkalinities in mg/L as CaCO3 from measurements of the total alkalinity (A) and the pH. • 𝐾𝑤 = 𝑂𝐻 − 𝐻+ • 𝐾𝑤 = 𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 10−14 𝑎𝑡 25 ℃ • 𝑝𝐾𝑤 = 14 𝑎𝑡25 ℃ • 𝐴𝑙𝑘𝑎𝑙𝑖𝑛𝑖𝑡𝑦 = 𝐻𝐶𝑂3− + 2 𝐶𝑂32− + 𝑂𝐻 − − 𝐻 + • 𝑟𝑒𝑓𝑒𝑟𝑠 𝑡𝑜 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝑜𝑙𝑒𝑠/𝐿 • 𝐻2 𝐶𝑂3 ⇌ 𝐻 + + 𝐻𝐶𝑂3− 𝑝𝐾𝑎1 = 6.35 𝑎𝑡 25℃ • 𝐻𝐶𝑂3− ⇌ 𝐻 + + 𝐶𝑂32− 𝑝𝐾𝑎2 = 10.33 𝑎𝑡 25℃

Solution

• 𝐴𝑙𝑘𝑎𝑙𝑖𝑛𝑖𝑡𝑦 = 𝐻𝐶𝑂3− + 2 𝐶𝑂32− + 𝑂𝐻− − 𝐻+ • 𝐴 = 𝐻𝐶𝑂3− + 2 𝐶𝑂32− + 𝑂𝐻 − − 𝐻+ • 𝐴 𝑖𝑛

𝑚𝑔 𝑎𝑠 𝑙

𝐶𝑎𝐶𝑂3 = 𝐻𝐶𝑂3− 𝑖𝑛

𝑚𝑔 𝑎𝑠 𝑙

• 𝐻𝐶𝑂3− 𝑖𝑛

𝑚𝑔 𝑎𝑠 𝑙

𝐶𝑎𝐶𝑂3 =

• 𝐻𝐶𝑂3− 𝑖𝑛

𝑚𝑔 𝑎𝑠 𝑙

𝐶𝑎𝐶𝑂3 = [𝐻𝐶𝑂3− ](

• 𝐻𝐶𝑂3− 𝑖𝑛

𝐶𝑎𝐶𝑂3 = [𝐻𝐶𝑂3− ](

•

𝑚𝑔 𝑎𝑠 𝑙 𝑚𝑔 𝐻𝐶𝑂3− 𝑖𝑛 𝑙 𝑎𝑠

𝐶𝑎𝐶𝑂3 = 50000 [𝐻𝐶𝑂3− ]

• 𝐶𝑂32− 𝑖𝑛

𝑚𝑔 𝑎𝑠 𝑙

𝑚𝑔 𝑎𝑠 𝑙 𝑚𝑔 𝐶𝑂32− 𝑖𝑛 𝑙 𝑎𝑠

𝑚𝑔 − 𝐻𝐶𝑂 3 𝑙

𝐶𝑎𝐶𝑂3 + 𝐶𝑂32− 𝑖𝑛

𝐶𝑎𝐶𝑂3 = 𝐶𝑂32− 𝑖𝑛

×

𝑚𝑔 𝑎𝑠 𝑙

𝐶𝑎𝐶𝑂3 + 𝐻 +

𝑀𝑊 𝑜𝑓𝐻𝐶𝑂3− (𝑚𝑜𝑙𝑒) × 1000 (

𝑚𝑜𝑙𝑒 )× 𝑙

61(𝑚𝑜𝑙𝑒) × 1000 (

𝑚𝑜𝑙𝑒 𝑙

𝑎𝑠 𝐶𝑎𝐶𝑂3 − 𝐻 +

𝐸𝑊 𝑜𝑓𝐶𝑎𝐶𝑂3 𝐸𝑊 𝑜𝑓𝐻𝐶𝑂3−

𝑚𝑜𝑙𝑒 )× 𝑙

𝑚𝑔 𝑙

𝑚𝑔 𝑙

×

𝑔

𝑔

𝑚𝑔 50 ) × 𝑔 61

𝐸𝑊 𝑜𝑓𝐶𝑎𝐶𝑂3 𝐸𝑊 𝑜𝑓𝐶𝑂32−

• 𝐶𝑂32− 𝑖𝑛

𝐶𝑎𝐶𝑂3 = 𝐶𝑂32−

•

𝐶𝑎𝐶𝑂3 = 2 𝐶𝑂32− × 50000

𝑔

𝑚𝑔 50 ) × 𝑔 30

× 60(𝑚𝑜𝑙𝑒) × 1000(

𝑚𝑔 50 ) × 𝑔 61

𝑚𝑔 𝑙

𝑎𝑠 𝐶𝑎𝐶𝑂3

𝑚𝑔 + × 50000 𝑎𝑠 𝐶𝑎𝐶𝑂 = 𝐻 3 𝑙 𝑚𝑔 𝑂𝐻 − 𝑙 𝑎𝑠 𝐶𝑎𝐶𝑂3 = 𝑂𝐻− × 50000

• 𝐻+ •

• 𝐴 𝑖𝑛 •

𝐴 𝑖𝑛

𝑚𝑔 𝑎𝑠 𝑙

𝐶𝑎𝐶𝑂3 = 50000 [𝐻𝐶𝑂3− ] + 2 𝐶𝑂32− × 50000 + 𝑂𝐻− × 50000 − 𝐻+ × 50000

𝑚𝑔 𝑎𝑠 𝐶𝑎𝐶𝑂3 𝑙

50000

= [𝐻𝐶𝑂3− ] + 2 𝐶𝑂32− + 𝑂𝐻− − 𝐻+ 𝑚𝑔

•

𝐻𝐶𝑂3−

• 𝐾𝑎 =

=

𝐴 𝑖𝑛 𝑙 𝑎𝑠 𝐶𝑎𝐶𝑂3 50000

[𝐻 + ][𝑊 − ] 𝐻𝑊

• 𝑝𝐾𝑎 = − log 𝐾𝑎 • 𝐾𝑎2 = •

[𝐻 + ][𝐶𝑂32− ] 𝐻𝐶𝑂3−

[𝐶𝑂32− ]

=

𝐾𝑎2 𝐻𝐶𝑂3− [𝐻 + ]

• 𝐾𝑤 = 𝐻+ 𝑂𝐻− •

𝑂𝐻− =

𝐾𝑤 𝐻+

− 2 𝐶𝑂32− − 𝑂𝐻− + 𝐻+

•

𝐻𝐶𝑂3−

=

𝑚𝑔 𝑎𝑠 𝐶𝑎𝐶𝑂3 𝑙

−

𝐾𝑎 𝐻𝐶𝑂3− 2 [𝐻 + ]

𝐴 𝑖𝑛 𝑙 𝑎𝑠 𝐶𝑎𝐶𝑂3 50000

−

𝐾𝑎2 𝐻𝐶𝑂3− 2 [𝐻 + ]

𝐾𝑎2 𝐻𝐶𝑂3− 2 [𝐻 + ]

𝐴 𝑖𝑛 𝑙 𝑎𝑠 𝐶𝑎𝐶𝑂3 50000

𝐴 𝑖𝑛

50000 𝑚𝑔

•

𝐻𝐶𝑂3−

=

• 𝐻𝐶𝑂3− + • 𝐻𝐶𝑂3−

1+

𝐾 2 𝑎2 [𝐻 + ]

• 𝐴𝑙𝑘𝑎𝑙𝑖𝑛𝑖𝑡𝑦 𝑑𝑢𝑒 𝑡𝑜 𝐴 𝑖𝑛

• 𝐻𝐶𝑂3− =

=

𝑚𝑔 𝑎𝑠 𝐶𝑎𝐶𝑂3 𝑙 50000

𝐾𝑤 𝐻+

−

+ 𝐻+

[𝐻 ]

• 𝐴𝑙𝑘𝑎𝑙𝑖𝑛𝑖𝑡𝑦 𝑑𝑢𝑒 𝑡𝑜 𝐻𝐶𝑂3− • 𝐻𝐶𝑂3− =

𝐾𝑤 𝐻+

+ 𝐻+

−

𝐾𝑤 𝐻+

+ 𝐻+

𝑚𝑔

𝐴 𝑖𝑛 𝑙 𝑎𝑠 𝐶𝑎𝐶𝑂3 = 50000 𝐻𝐶𝑂3−

𝑚𝑔 𝑎𝑠 𝐶𝑎𝐶𝑂3 𝑙 50000

𝐾 + − 𝑤 + + 𝐻

𝐾 1+2 𝑎2 +

𝐻

[𝐻 ]

+ 𝐻+

−

𝑚𝑔

𝐾 1+2 𝑎2 +

𝐴 𝑖𝑛

𝐾𝑤 𝐻+

−

−

𝐾𝑤 𝐻+

+ 𝐻+

• Alkalinity (A) is in mg/l as CaCo3 •

𝐻𝐶𝑂3− =

•

𝐻𝐶𝑂3− =

• 𝐻𝐶𝑂3− 𝑖𝑛

•

𝐴 50000

𝐾 + − 𝑤 + + 𝐻

𝐻 𝐾 1+2 𝑎2 [𝐻+ ] 𝑚𝑔 𝐻𝐶𝑂3− (𝑖𝑛 𝑎𝑠 𝐶𝑎𝐶𝑂3 ) 𝑙

𝑚𝑔 𝑎𝑠 𝑙

50000

𝐶𝑎𝐶𝑂3 = 50000 [𝐻𝐶𝑂3− ]

𝒎𝒈 𝑯𝑪𝑶− 𝒊𝒏 𝒂𝒔 𝑪𝒂𝑪𝑶𝟑 𝟑 𝒍

= 𝟓𝟎𝟎𝟎𝟎

𝑨 𝑲𝒘 − + 𝟓𝟎𝟎𝟎𝟎 𝑯+ 𝑲 𝟏+𝟐 𝒂𝟐 [𝑯+ ]

• Alkalinity due to [𝐶𝑂32− ] •

• •

•

2𝐾𝑎2 𝐻𝐶𝑂3− = [𝐻 + ] 𝑚𝑔 𝐶𝑂32− 𝑖𝑛 𝑎𝑠 𝐶𝑎𝐶𝑂3 𝑙

[𝐶𝑂32− ]

𝑚𝑔 𝐶𝑂32− 𝑖𝑛 𝑙 𝑎𝑠 𝐶𝑎𝐶𝑂3

50000

=

= 2 𝐶𝑂32− × 5000

2𝐾𝑎2

𝒎𝒈 𝑪𝑶𝟐− 𝒊𝒏 𝒂𝒔 𝑪𝒂𝑪𝑶𝟑 𝟑 𝒍

𝑚𝑔 𝑎𝑠 𝐶𝑎𝐶𝑂3) 𝑙 50000 [𝐻 + ]

𝐻𝐶𝑂− 3 (𝑖𝑛

𝒎𝒈

=

𝟐𝑲𝒂𝟐 (𝑯𝑪𝑶− 𝟑 𝒊𝒏 𝒍 𝒂𝒔 𝑪𝒂𝑪𝑶𝟑 ) [𝑯+ ]

𝑯+

Problems • What is the exact alkalinity of a water that contains 0.6580 mg/L of bicarbonate, as the ion at a pH of 5.66? No carbonate is present. • Calculate the approximate alkalinity (in mg/L as CaCO3) of a water containing 120 mg/L of bicarbonate ion and 15 mg/L of carbonate ion. • If the water has a carbonate alkalinity of 120.00 mg/L as the ion and pH of 10.30. what is the bicarbonate alkalinity in mg/L as the ion. • What is the pH of water that contains 120.00 mg/L of bicarbonate ion and 15.00 mg/L of carbonate ion?

Surprise quiz solution • What is the pH of water that contains 120.00 mg/L of bicarbonate ion and 15.00 mg/L of carbonate ion? 𝐸𝑊𝐶𝑎𝐶𝑂3 𝑚𝑔 𝑚𝑔 𝑎𝑠 𝐶𝑎𝐶𝑂3 = ( 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠) × 𝐿 𝐿 𝐸𝑊𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝑚𝑔 50 𝐴𝑙𝑘𝑎𝑙𝑖𝑛𝑖𝑡𝑦 𝑑𝑢𝑒 𝑡𝑜 𝑏𝑖𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 𝑖𝑜𝑛 𝑎𝑠 𝐶𝑎𝐶𝑂3 = 120 × = 98.36 𝐿 61 𝑚𝑔 50 𝐴𝑙𝑘𝑎𝑙𝑖𝑛𝑖𝑡𝑦 𝑑𝑢𝑒 𝑡𝑜 𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 𝑖𝑜𝑛 𝑎𝑠 𝐶𝑎𝐶𝑂3 = 15 × = 25 𝐿 30 𝒎𝒈 𝟐𝑲𝒂𝟐 (𝑯𝑪𝑶− 𝒊𝒏 𝒂𝒔 𝑪𝒂𝑪𝑶𝟑 ) 𝒎𝒈 𝟑 𝒍 𝟐− 𝑪𝑶𝟑 𝒊𝒏 𝒂𝒔 𝑪𝒂𝑪𝑶𝟑 = 𝒍 [𝑯+ ]

𝐻𝐶𝑂3− ⇌ 𝐻 + + 𝐶𝑂32−

𝑝𝐾𝑎2 = 10.33 𝑎𝑡 25℃

𝑝𝐾𝑎2 = 10.33 = − log 𝐾𝑎2

𝐾𝑎2 = 10−10.33 = 4.68 × 10−11

𝟐𝟓 =

𝒎𝒈 𝒂𝒔 𝑪𝒂𝑪𝑶𝟑 ) 𝒍 [𝑯+ ]

𝟐𝑲𝒂𝟐 (𝑯𝑪𝑶− 𝟑 𝒊𝒏

2 × 4.68 × 10−11 × 98.36 25 = [𝐻+ ] 𝐻 + = 3.68 × 10−10 𝑝𝐻 = − log(3.68 × 10−10 ) = 9.4

Hardness • The hardness is used to characterized the water that does not lather well, causes a scum in bath tubs, leaves hard white crusty deposits (scale) on coffee pot tea kettles and hot water heater. • This characteristics results from reaction of calcium and magnesium with soap. • 𝐶𝑎2+ + (𝑠𝑜𝑎𝑝)− ⇌ 𝐶𝑎 𝑆𝑜𝑎𝑝 2 (𝑠) • Calcium soap complex form an undesirable precipitate and limit the reaction of soap with dirt on cloth. • Hardness is defined as sum of polyvalent cations (in consistent units). • Common unit of expression are mg/L as CaCO3 or meq/L.

Hardness

• Undesirable to use water containing hardness >150 mg/L. • Water suppliers provide water with hardness 75-120 mg/L as CaCO3. • Predominate compounds imparting hardness are calcium bicarbonate [Ca(HCO3)2] and magnesium bicarbonate [Mg(HCO3)2]. Bicarbonates are soluble in water. • MgCO3 and CaCO3 are insolub;le in water. • Gypsum(CaSO4) and MgSO4 may also go into solution to contribute to hard ness. Hardness range (mg/L as CaCO3)

Description

0-75

soft

75-100

Moderately soft

100-300

Hard

>300

Very hard

Hardness • Total hardness=𝐶𝑎2+ + 𝑀𝑔2+ Ions mg/L as ion EW CaCO3/EW ion • Total hardness = carbonate 2+ 103 50/20=2.5 𝐶𝑎 hardness (associated with HCO3 anion, abbreviated as CH) +non 5.5 50/12=4.16 𝑀𝑔2+ carbonate hardness (associated 16 50/23=2.17 𝑁𝑎+ with other anion, abbreviated as NCH) 255 50/61=0.82 𝐻𝐶𝑂3− • Carbonate Hardness=alkalinity 49 50/48=1.04 𝑆𝑂42− • Carbonate hardness is called 37 50/35=1.42 𝐶𝑙 − temporary hardness as it can be removed when water is heated. 𝐸𝑊𝐶𝑎𝐶𝑂3 𝑚𝑔 𝑚𝑔 𝑎𝑠 𝐶𝑎𝐶𝑂3 = ( 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠) × • Non carbonate harness is in excess 𝐿 𝐿 𝐸𝑊𝑠𝑝𝑒𝑐𝑖𝑒𝑠 of carbonate hardness and cannot 𝑚𝑔 be removed by heating the water. 𝐶𝑎2+ 𝑖𝑛 𝑎𝑠 𝐶𝑎𝐶𝑂3 = (103) ×2.5=258 𝐿

Total hardness= 258+23= 281 mg/L as 𝐶𝑎𝐶𝑂3

Mg/L as CaCO3 258

23 35 209 51 52

Problems of Hardness • A water has an alkalinity of 200 mg/L as CaCO3. The Ca2+ concentration is 160 mg/L as the ion and mg2+ concentration is 40 mg/L as the ions. The pH is 8.1. Find the total carbonate, and noncarbonated hardness.

• The following mineral analysis was reported for water sample taken from well. Determine the total, carbonate and non carbonate hardness in mg/L as CaCO3 using the predominant polyvalent cation definition. • Calculate the total, carbonate and noncarbonate hardness using all of the polyvalent cations. What is the percent error in using only the predominant cations? mg/L as the ion

mg/L as the ion

Iron

0.2

Silica (SiO2)

20.0

Manganese

0.0

Fluoride

0.35

Ammonium

0.5

Boron

0.1

Sodium

4.7

Nitrate

0.0

Potassium

0.9

Chloride

4.5

Calcium

67.2

Sulfate

29.0

Magnesium

40.0

Alkalinity

284.0 as CaCO3

Barium

0.5

pH as recorded

7.6units

Solution Table Ions

mg/L as the ion

Molecular weight

valence

In mg/L as Ions CaCo3

mg/L as the ion

Molecular weight

valence

In mg/L as CaCo3

Iron

0.2

56

2

0.2*(50/2 8)=0.36

Silica (SiO2)

20.0

28

4

142.8

Manganes e

0.0

0.0

Fluoride

0.35

19

1

0.92

Ammoniu m

0.5

18

1

0.5*(50/1 8)=1.39

Boron

0.1

11

3

1.36

Sodium

4.7

23

1

10.22

Nitrate

0.0

Potassium

0.9

39

1

1.15

Chloride

4.5

35

1

6.43

Calcium

67.2

40

2

167.5

Sulfate

29.0

96

2

30.21

Magnesiu m

40.0

24

2

166.7

Alkalinity

284.0 as CaCO3

Barium

0.5

137

2

0.5*50/68 .65=0.36

pH as recorded

7.6units

Total hardness 167.5+166.7=334.2 mg/L as CaCO3 Carbonate hardness= 284.0 mg/L as CaCO3 Noncarbonate hardness= 334.2-284.0=50.2

0.0

CPCB criteria for water quality Designated-Best-Use

Class of water

Criteria

Drinking Water Source without conventional treatment but after disinfection

A

•Total Coliforms Organism MPN/100ml shall be 50 or less •pH between 6.5 and 8.5 •Dissolved Oxygen 6mg/l or more •Biochemical Oxygen Demand 5 days 20C 2mg/l or less

Outdoor bathing (Organised)

B

•Total Coliforms Organism MPN/100ml shall be 500 or less pH between 6.5 and 8.5 Dissolved Oxygen 5mg/l or more •Biochemical Oxygen Demand 5 days 20C 3mg/l or less

Drinking water source after conventional treatment and disinfection

C

•Total Coliforms Organism MPN/100ml shall be 5000 or less pH between 6 to 9 Dissolved Oxygen 4mg/l or more •Biochemical Oxygen Demand 5 days 20C 3mg/l or less

Propagation of Wild life and Fisheries

D

•pH between 6.5 to 8.5 Dissolved Oxygen 4mg/l or more •Free Ammonia (as N) 1.2 mg/l or less

Irrigation, Industrial Cooling, Controlled Waste disposal

E

•pH betwwn 6.0 to 8.5 •Electrical Conductivity at 25C micro mhos/cm Max.2250 •Sodium absorption Ratio Max. 26 •Boron Max. 2mg/l

Below-E

Not Meeting A, B, C, D & E Criteria

Water quality parameters

• PH

• Dissolved Oxygen: • Anything that can be oxidized in the receiving water with the consumption of dissolved molecular oxygen is termed oxygen-demanding material. • This material is usually biodegradable organic matter but also includes certain inorganic compounds. • The consumption of dissolved Oxygen, DO (pronounced "dee oh"), poses a threat to fish and other higher forms of aquatic life that must have oxygen to live. • The critical level of DO varies greatly among species. For example, brook trout may require about 7.5 mg/L of DO, while carp may survive at 3 mg/L.

• As a rule, the most desirable commercial and game fish require high levels of dissolved oxygen. • Oxygen-demanding materials in domestic sewage come primarily from human waste and food residue. • Particularly noteworthy among the many industries that produce oxygen-demanding wastes are the food processing and paper industries.

• Almost any naturally occurring organic matter, such as animal droppings, crop residues, or leaves, point sources, contribute to the depletion of DO.

• ThOD: The amount of oxygen required to oxidize a substance to carbon dioxide and water may be calculated by stoichiometry if the chemical composition of the substance is known. This amount of oxygen is known as the theoretical oxygen demand (ThOD). • BOD: The oxidation of an organic compound is carried out by microorganisms using the organic matter as food source, the oxygen consumed is known as biochemical oxygen demand (BOD).

Theoritical Oxygen demand Problem 1: Compute the ThOD of 108.75 mg/L of glucose (C6H120 6). 𝐶6 𝐻12 𝑂6 + 6𝑂2 ⇌ 6𝐶𝑂2 + 6𝐻2 𝑂 Gram molecular weight Glucose: 6*12+12*1+6*16=72+12+96=180 g Oxygen: 6*32=192 g

• 180 𝑔 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒 192 𝑔 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛 • ∴ 108.75 𝑚𝑔 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 𝑤𝑖𝑙𝑙 𝑟𝑒𝑞𝑢𝑖𝑟𝑒 108.75 ×

192 180

=

𝑚𝑔 116 𝑂2 𝑙

BOD

𝑑𝐿𝑡 = −𝑟𝐴 𝑑𝑡 𝐿𝑡 = 𝑂𝑥𝑦𝑔𝑒𝑛 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑜𝑟𝑔𝑎𝑛𝑖𝑐𝑠 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡, 𝑚𝑔/𝐿 −𝑟𝐴 = −𝑘𝐿𝑡 𝑘 = 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑑 −1 𝑑𝐿𝑡 = −𝑘𝐿𝑡 𝑑𝑡 𝐿

𝑑𝐿𝑡 = −𝑘 𝐿 𝐿0 𝑡

• Actual BOD is less than ThOD due to incorporation of some of the carbon into new bacterial cell. • The test is bioassay means to measure by biological means.

ln

𝑡

𝑑𝑡 0

𝐿𝑡 = −𝑘𝑡 𝐿0

𝐿𝑡 = 𝐿0

𝑒 −𝑘𝑡

𝐵𝑂𝐷 = 𝐿0 − 𝐿𝑡 = 𝐿0 − 𝐿0 𝑒 −𝑘𝑡

𝐵𝑂𝐷 = 𝐿0 (1 − 𝑒 −𝑘𝑡 ) 𝐵𝑂𝐷 = 𝐿0 (1 − 10−𝐾𝑡 )

𝑘 = 2.303(𝐾) Ultimate BOD (L0) is defined as maximum BOD excreted by the waste, which will be equal to concentration of organic matter at t=0.

BOD and COD

• A water sample is inoculated by bacteria which consume organic matter to obtain energy for their life process. • The organisms utilize the oxygen in process of consuming the waste, the process is called aerobic decomposition. • The BOD test is indirect measurement of organic matter, because we only measure the change in concentration of dissolved oxygen caused by microorganisms as they degrade organic matter. Although not all organic matter is biodegradable. • Chemical oxygen demand (COD) is a measure of the capacity of water to consume oxygen during the decomposition of organic matter and the oxidation of inorganic chemicals such as Ammonia and nitrite. • Strong chemical oxidizing agent (chromic acid, H2CrO4) is mixed with water sample and then boiled. The difference between the amount of oxidizing agent at the beginning of the test and that remaining at the end of the test is used to calculate the COD.

BOD Rate constant • The value of rate constant (k) depends upon following properties. • The nature of waste Easily degradable: sugar, sucrose, starch etc. Slowly degradable: cellulose (toilet paper) Undegradable: hair fingernails • The ability of organisms to utilize the waste Not all microorganisms degrade organic matter. Difficulty come specially in case of industrial waste. The bacterial seed need to be used for inoculation. • The temperature The biological process speedup as temperature increases and slow down as the temperature drops. Sample

k (base e, at 20 °C) day-1

K (base 10, at 20 °C) day-1

Raw sewage

0.35-0.70

0.15-0.30

Well treated sewage

0.12-0.23

0.05-0.10

Polluted river water

0.12-0.23

0.05-0.10

BOD

• Five days BOD is selected as standard value for most waste water analysis for the regulatory purpose. • Ultimate BOD is better indicator of total strength of waste. • Any type of waste having a define BOD the ratio of BOD5 to ultimate BOD is constant so that BOD5 indicates a relative waste strength. • Some of the compounds such as protein which contain N also oxidize with consumption of molecular oxygen. • The mechanism and rates of N oxidation is distinctly different from carbon oxidation. • The oxygen consumption due to oxidation of carbon is called carbonaceous BOD (CBOD) and while that due to nitrogen oxidation is called nitrogenous BOD (NBOD). • The organisms that oxidize carbon in organic compounds for energy cannot oxidize the nitrogen in these compounds. Instead the nitrogen is release into surrounding water as ammonia (NH3). • At normal pH the ammonia is in the form of ammonium cation (NH4+). • Ammonia is oxidized to nitrate by nitrifying bacteria (special group) as their source of energy in a process called nitrification. 𝑁𝐻4+

+ 2𝑂2

𝑚𝑖𝑐𝑟𝑜𝑜𝑟𝑔𝑎𝑛𝑖𝑠𝑚𝑠

⇌

𝑁𝑂3− + 𝐻2 𝑂 + 2𝐻 +

Problem Problem 1. If the BOD3 of a waste is 75 mg/L and K is 0.150 d-1, what is the ultimate BOD?

• 75 = 𝐿0 1 − 100.150×3 • 𝐿0 = 116 𝑚𝑔/𝐿 Problem 2. A waste is being discharged into a river that has a temperature of 10 °C. What fraction of the maximum oxygen consumption has occurred in four days if BOD rate constant determined in the laboratory under standard conditions is 0.115 d-1. 𝑘 𝑇 = 𝑘20 (𝜃)𝑇−20 [𝜃 𝑖𝑠 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 = 1.135 𝑓𝑜𝑟 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 4 − 20℃ 𝑎𝑛𝑑 1.056 𝑓𝑜𝑟 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 20 − 30 ℃] Problem 3 (a) Compute the theoretical NBOD of a waste water containing 30 mg/L of ammonia as nitrogen (we often use NH3.N expression for ammonia as nitrogen). (b) If waste water analysis was reported as 30 mg/L of ammonia (NH3), what would be theoretical NBOD be?

DO sag curvee • The water quality management in rivers require to assess the dissolved oxygen concentration profile down stream from the waste discharge. • This profile is called DO sag curve because the DO concentration dips as oxygen demanding materials are oxidized and then rises again further downstream as the oxygen is replenished from the atmosphere. • Mass balance approach is used to determine the DO

Oxygen Sag curve

DO sag curve

Mass of DO in wastewater

Mass of DO in River after mixing

Mass of DO in river

Conservative mass balance

Mass of DO in waste water= 𝑄𝑤 𝐷𝑂𝑤 Mass of Do in river= 𝑄𝑟 𝐷𝑂𝑟

𝑚3 𝑄𝑤 𝑎𝑛𝑑 𝑄𝑟 = 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒𝑠 𝑜𝑓 𝑤𝑎𝑠𝑡𝑒 𝑤𝑎𝑡𝑒𝑟 𝑎𝑛𝑑 𝑟𝑖𝑣𝑒𝑟 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 ( ) 𝑠 𝑔 𝐷𝑂𝑤 𝑎𝑛𝑑 𝐷𝑂𝑟 = 𝐷𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 𝑜𝑥𝑦𝑔𝑒𝑛 𝑖𝑛 𝑤𝑎𝑠𝑡𝑒 𝑤𝑎𝑡𝑒𝑟 𝑎𝑛𝑑 𝑟𝑖𝑣𝑒𝑟 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 ( 3 ) 𝑚 The mass of DO in river after mixing equals the sum of mass flows: Mass of DO after mixing= 𝑄𝑤 𝐷𝑂𝑤 + 𝑄𝑟 𝐷𝑂𝑟 In similar way the ultimate BOD: 𝑄𝑤 𝐿𝑤 + 𝑄𝑟 𝐿𝑟

𝐿𝑤 𝑎𝑛𝑑 𝐿𝑟 = 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝐵𝑂𝐷 𝑜𝑓 𝑤𝑎𝑠𝑡𝑒 𝑤𝑎𝑡𝑒𝑟 𝑎𝑛𝑑 𝑟𝑖𝑣𝑒𝑟 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 (

𝑚𝑔 ) 𝑙

DO sag curve The concentrations of DO and BOD after mixing are the respective masses per unit time divided by the flow rate

•

•

𝑄𝑤 𝐷𝑂𝑤 +𝑄𝑟 𝐷𝑂𝑟 𝐷𝑂 = 𝑄𝑤 +𝑄𝑟 𝑄𝑤 𝐿𝑤 +𝑄𝑟 𝐿𝑟 𝐵𝑂𝐷 = 𝑄𝑤 +𝑄𝑟

Problems • The town discharges 17360 m3/d of treated wastewater into the river. The treated wastewater has a BOD5 of 12 mg/L and a k of 0.12 d-1 at 20 C. The river has flow rate of 0.43 m3/s and an ultimate BOD of 5.0 mg/L. The DO of the river is 6.5 mg/L and the DO of the wastewater is 1.0 mg/L. Compute the DO and initial ultimate BOD after mixing.

Treatment systems • Surface water

Treatment systems • Ground water

Sedimentation Horizontal flow clarifier

Up flow clarifier

• Suspended particle that will settle within a reasonable period of time can be removed in sedimentation tank. • Rectangular and circular. • Inlet, settling, outlet, and sludge storage. •

Important criteria for sedimentation tank • Purpose of inlet to evenly distribute flow and suspended particles across the cross section of the settling zone. • Inlet water velocity: At inlet is decided in such a way that the due to turbulence the suspended particle can be evenly distributed but the velocity slowdown to the design velocity of the settling tank. • Length of the settling tank: Inlet and settling zone are designed separately and their length added together. • Depth of the sludge storage zone: depends upon method of cleaning, the frequency of cleaning and quantity of sludge estimated to be produced. • With a well-flocculated solid and good inlet design, over 75 percent of the solids may settle in the first fifth of the tank. For coagulant floc, Hudson recommends a sludge storage depth of about 03 m near the outlet and 2 m or more near the inlet (Hudson, 1981). • Mechanically-cleaned basins may be equipped with a bottom scraper, such as shown in Figure 4-33. The sludge is continuously scraped to a hopper where it is pumped out. For mechanically cleaned. basins, a one percent slope toward the sludge withdrawal point is used. A sludge hopper is designed with sides sloping with a vertical to horizontal ratio of 1.2: 1 to 2: 1. • Outlet Zone: The outlet zone is designed so as to remove the settled water from the basin without carrying away any of the floc particles. 𝑣=

𝑄 𝐴𝑐

𝑚 𝑣 = 𝑤𝑎𝑡𝑒𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑠 𝑚3 𝑄 = 𝑤𝑎𝑡𝑒𝑟 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑠 𝐴𝑐 = 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑚2

Sedimentation tank outlet • Scouring: Within the sedimentation tank, the flow is going through a very large area (basin depth times width); consequently, the velocity is slow. To remove the water from the basin quickly, it is desirable to direct the water into a pipe or small channel for easy transport, which will produce a significantly higher velocity. If a pipe" were to be-placed at the end of the sedimentation basin, all the water would "rush" to the pipe. This rushing water would create high velocity profiles in the basin, which would tend to raise the settled floc from the basin and into the effluent water. This phenomenon of washing out the floc is called scouring. • Rather than put a pipe at the end of the sedimentation basin, it is desirable to first put a series of troughs, called weirs, which provide a large area for the water to flow through and minimize the velocity in the sedimentation tank near the outlet zone. The weirs then feed into a central channel or pipe for transport of the settled water. • The length of weir required is a function of the type of solids. The heavier the. solids, the harder it is to scour them, and the higher. the allowable outlet velocity. Therefore, heavier particles require a. shorter length of weir than do light particles.

Sedimentation Concepts • Settling velocity 𝑣𝑠 • The velocity at which the tank is designed to operate, called the over flow rate 𝑣0 . • The particle settling velocity must be greater than the liquid-rise velocity (𝑣𝑠 > 𝑣0 ). • If 𝑣𝑠 > 𝑣0 100 percent particle removal • If 𝑣𝑠 < 𝑣0 is less than 0 % particle removal • In design the procedure is to determine the particle settling velocity and set the over flow rate at some lower value. • Over flow rate 50-70% of settling velocity for upflow clarifier. • It is sometimes3referred to as the surface loading rate because it 𝑚 has units of . 𝑑×𝑚2 • This can be thought of as the amount of water that-goes through each m2 of tank surface area per day.

Sedimentation Concepts • 𝑣0 =

𝑣𝑜𝑙𝑢𝑚𝑒/𝑡𝑖𝑚𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎

• 𝑣0 =

𝐷𝑒𝑝𝑡ℎ×𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎/𝑡𝑖𝑚𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎

=

𝑑𝑒𝑝𝑡ℎ 𝑡𝑖𝑚𝑒

=

ℎ 𝜃

• However the particle removal does not depend upon depth of the tank as long as 𝑣𝑠 > 𝑣0 the particle will settle downward and removed from the bottom. • An ideal horizontal sedimentation tank is based upon three assumptions (Hazen, 1904, and Camp, 1946). • 1. Particles and velocity vectors are evenly distributed across the tank cross section. This is the function of the inlet zone. • 2. The liquid moves as an ideal slug down the length of the tank. • 3. Any particle hitting the bottom of the tank is removed.

Type of sedimentation • Type I Sedimentation. Type I sedimentation is characterized by particles that settle discretely at a constant settling velocity. They settle as individual particles and do not flocculate or stick to other particles during settling. Examples of these particles are sand and grit material. Application of Type I settling is during pre-sedimentation for sand removal prior to coagulation in a potable water plant, in settling of sand particles during cleaning of rapid sand filters, and in grit chambers. • Type II Sedimentation. Type II sedimentation is characterized by particles that flocculate during sedimentation. Because they flocculate, their size is constantly changing; therefore, the settling velocity is changing. Generally the settling velocity is Increasing. These types of particles occur in alum or iron coagulation, in primary sedimentation, and in settling tanks in trickling filtration. • Type III or Zone Sedimentation. In zone sedimentation the particles are at a high concentration (greater than 1,000 mg/L) such that the particles tend to settle as a mass, and a distinct clear zone and sludge zone are present. Zone settling occurs in limesoftening sedimentation, activated-sludge sedimentation, and sludge thickeners.

Settling velocity of a particle 𝐹𝐷

𝐹𝐷

𝑣2 𝐹𝐷 = 𝐷𝑟𝑎𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝐶𝐷 𝐴𝑝 𝜌 2 𝐹𝐵 = 𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝜌𝑔𝑉𝑝 𝐹𝐺 = 𝐺𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 = 𝜌𝑠 𝑔𝑉𝑝

𝐹𝐵

𝐶𝐷 = 𝐷𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

𝐴𝑝 = 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒, 𝑚2 𝑚 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒, 𝑠 𝑘𝑔 𝜌 = 𝐹𝑙𝑢𝑖𝑑 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 3 𝑚 𝑉𝑝 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒, 𝑚2 𝑘𝑔 𝜌𝑠 = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒, 3 𝑚 𝑚 𝑔 = 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑔𝑟𝑎𝑣𝑖𝑡𝑦, 2 𝑠 Driving force for acceleration of the particle is the difference between the gravitational and buoyancy force: 𝐹𝐺

𝐹𝐺 − 𝐹𝐵 = 𝜌𝑠 − 𝜌 𝑔𝑉𝑝

𝐹𝐺 − 𝐹𝐵 = 𝐹𝐷

When the drag force is equal to the driving force, the particle velocity reaches to a constant value called the terminal velocity or settling velocity 𝑣𝑠 𝑣𝑠2 𝜌𝑠 − 𝜌 𝑔𝑉𝑝 = 𝐶𝐷 𝐴𝑝 𝜌 2

For laminar flow in settling zone 24 24𝜇 𝐶𝐷 = = 𝑅𝑒 𝜌𝑑𝑝 𝑣𝑠

For spherical particles of diameter 𝑑𝑝 𝑑𝑝3 4 ×𝜋× 8 𝑉𝑝 2 3 = = 𝑑𝑝 𝜋 2 𝐴𝑝 3 4 𝑑𝑝 𝟒 𝝆𝒔 − 𝝆 𝒈𝒅𝒑 𝒗𝒔 = 𝟑𝑪𝑫 𝝆

4 𝜌𝑠 − 𝜌 𝑔𝑑𝑝 × ρ𝑑𝑝 𝑣𝑠 𝑣𝑠 = 3 × 24 × 𝜇 × 𝜌 𝟏/𝟐

𝑣𝑠2

=

𝒗𝒔 =

𝜌𝑠 − 𝜌 𝑔𝑑𝑝 × 𝑑𝑝 𝑣𝑠 18 × 𝜇 𝝆𝒔 − 𝝆 𝒈𝒅𝟐𝒑 𝟏𝟖 × 𝝁

1/2

Summary of Settling velocity/terminal velocity in sedimentation tank Sir George Gabriel Stokes showed that, for spherical particles

𝟒 𝝆𝒔 − 𝝆 𝒈𝒅𝒑 𝒗𝒔 = 𝟑𝑪𝑫 𝝆 𝜌𝑑𝑣𝑠 • 𝑅𝑒 = 𝜇

𝟏/𝟐

• 𝑅𝑒 < 0.5, 𝐶𝐷 = • 𝑅𝑒 = 0.5 𝑡𝑜 • • • •

falling under laminar (quiescent) conditions, Equation reduces to the following (Stokes, 1845):

𝒗𝒔 =

24 𝑅𝑒

104 , 𝐶𝐷

=

24 𝑅𝑒

+

3 1/2 𝑅𝑒

+ 0.34

Where, 𝑅𝑒 Reynolds number . d diameter of spherical particle, 𝑣𝑠 = velocity of spherical particle, m/s

• 𝜇 Dynamic viscosity, Pa . S 𝜌 𝜇

• 𝜐 = Kinematic viscosity

𝑚2 𝑠

𝝆𝒔 − 𝝆 𝒈𝒅𝟐𝒑 𝟏𝟖 × 𝝁

Coagulation • Colloidal stability: Colloidal particles are too small to settle in a reasonable time period and too small to be trapped in the pores of filter. • The colloidal particles are stable because they carry negative charge that repels other particle before they collide with one another. • They involve in Brownian motion. • Colloid destabilization • Schulze and Hardy found that 1 mole of trivalent ion can reduce the charges of colloidal particles as much as 30-50 mole of divalent ion and as much as 1500 to 2500 moles of a monovalent ion (Schulze and Hardy rule).

Coagulation • The purpose of coagulation is to alter the colloids so that they can adhere to each other. • Properties of Coagulants: • Trivalent cations • Non toxic • Insoluble in natural pH range • Two most common coagulants are aluminium and ferric ions • Aluminium: dry or liquid alum 𝐴𝑙2 𝑆𝑂4 3 . 14𝐻2 𝑂 , molecular weight 594. • Liquid alum is 48.8% alum and 51.2% water. This solution has a crystallization point of -15.6 ◦C. • A 50.7% percent alum solution will crystallize at +18.3℃. • 𝐴𝑙2 𝑆𝑂4 3 . 14𝐻2 𝑂 + 6𝐻𝐶𝑂3− ⇌ 2𝐴𝑙 𝑂𝐻 3 𝑠 + 6𝐶𝑂2 + 14𝐻2 𝑂 + 3𝑆𝑂42− • Each mole of alum added uses six moles of alkalinity and produces six moles of carbon dioxide. The reaction shifts the carbonate equilibrium and decrease the pH. • As long as sufficient alkalinity is not present to neutralize the sulfuric acid production, the pH may be greatly reduced: • 𝐴𝑙2 𝑆𝑂4 3 . 14𝐻2 𝑂 ⇌ 2𝐴𝑙 𝑂𝐻 3 𝑠 + 8𝐻2 𝑂 + 3𝐻2 𝑆𝑂4 • Optimal pH range for alum is 5.5 to 6.5.

Coagulation • Iron • Sulphate salt 𝐹𝑒2 𝑆𝑂4 3 . 𝑥𝐻2 𝑂or chloride salt 𝐹𝑒𝐶𝑙3 . 𝑥𝐻2 𝑂 . • 𝐹𝑒𝐶𝑙3 + 3𝐻𝐶𝑂3− ⇌ 𝐹𝑒 𝑂𝐻 3 𝑠 + 3𝐶𝑂2 + 3𝐶𝑙 − • Without alkalinity • 𝐹𝑒𝐶𝑙3 + 3𝐻2 𝑂 ⇌ 𝐹𝑒 𝑂𝐻 3 𝑠 + 3𝐻𝐶𝑙 • Forms HCl which lowers the pH. • pH range 4-9.

Coagulant aids • The four basic types of coagulant aids are pH adjusters, activated silica, clay, and polymers. Acids and alkalies are both used to adjust the pH of the water into the optimal range for coagulation. The acid most commonly used for lowering the pH is sulfuric acid. Either lime [Ca(OH)2] or soda ash (Na2CO3) are used to raise the pH. • When activated silica is added to water, it produces a stable solution that has a negative surface charge. The activated silica can unite with the positively charged alumimim or with iron flocs, resulting in a larger, denser floc that settles faster and enhances enmeshment. The addition of activated silica is especially useful for treating highly colored, low-turbidity waters because it adds weight to the floc. However, activation of silica does require proper equipment and close operational control, and many plants are hesitant to use it. • Clays can act much like activated silica in that they have a slight negative charge and can add weight to the floes. Clays are also most useful for colored, low-turbidity waters; but are rarely used. • Polymers can have a negative charge (anionic), positive charge (cationic), positive and negative charge(polyamphotype) or no charge (nonionic). • Polymers are long chain carbon compounds of high molecular weight that have many active sites. The active sites adhere to flocs, joining them together and producing a larger, tougher floc that settles better. This process is called inter particle bridging. The type of polymer, dose and point of addition must be determined for each water, and requirements may change within a plant on a seasonal, or even daily, basis.

Mixing • Through rapid mixing the chemicals are quickly and homogenously disperse in water. The process is used in softening and coagulation. • The degree of mixing is measured by velocity gradient. • Velocity gradient G is a function of power input into per unit volume of water.

• 𝐺= • • • •

𝑃 𝜇𝑉

𝐺 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑠 −1 𝑃 = 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡, 𝑊 𝑉 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑖𝑛 𝑚𝑖𝑥𝑖𝑛𝑔 𝑡𝑎𝑛𝑘, 𝑚3 𝜇 = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑐𝑖𝑡𝑦, 𝑃𝑎. 𝑠

• The chemical reaction in coagulation is completed in less than 0.1 s.

Rapid mixing

• Therefore it is necessary that mixing should be as instantaneous and complete as possible. • Vertical shaft mixed is used for rapid mixing • Coagulation occurs predominantly by two mechanisms: (a) adsorption of the soluble hydrolysis species on the colloid and destabilization (b) Sweep coagulation where the colloid is trapped in the hydroxide precipitate. • The reactions in adsorption-destabilization occurs very fast within 1 s. • Sweep coagulation is slow and occurs in the range of 1 to 7 s.

• If charge reversal is apparent from the dose-turbidity curve then adsorption-destabilization is the predominant mechanism. • If dose-turbidity curve does not show charge reversal (i.e. curve is relatively flat at higher doses) then the predominant mechanism is sweep coagulation. • G values in the range of 3000 to 5000 s-1 and detention time on the order of 0.5 s are recommended for adsorption-desorption reactions. • For sweep coagulation, detention time of 1-10 s and G values in the range of 600-1000 s-1 are recommended.

Flocculation • Flocculation is important process that affect particle removal efficiency. • The aim of flocculation is to bring the particles into contact so that they will collide, stick together and grow to a size that will readily settle. • Enough mixing is required to bring the floc into contact and to keep the floc from settling in the flocculation basin. • To much mixing will shear the floc particles. • Therefore the velocity gradient should be maintained within a narrow range. • Heavier the floc and the higher the suspended solid concentration, the more mixing required to keep the floc in suspension. • Softening flocs are heavier than coagulation floc therefore requires a higher G value. • Increase in floc concentration also increases the G value. • At 20 C the detention time in flocculation basin is 20 min, at 15 °C the detention time is increased by 7% and at 10 °C it is increased by 15% and then at 5 °C it is increased by 25%

Disinfection

• ·Disinfection is used in water treatment to reduce pathogens (disease-producing microorganisms) to an acceptable level. Disinfection is not the same as sterilization. Sterilization implies the destruction of all living organisms. • Three categories of human enteric pathogens are normally of consequence: bacteria, viruses, and amebic cysts. Purposeful disinfection must be capable of destroying all three. • The water disinfectants must possess the following properties: • They must destroy the kinds and numbers of pathogens that may be introduced into water within a practicable period of time over an expected range in water temperature. • They must meet possible fluctuations in composition, concentration, and condition of the waters or wastewaters to be treated. • They must be neither toxic to humans and domestic animals nor unpalatable or otherwise objectionable in required concentrations. • They must be dispensable at reasonable cost and safe and easy to store, transport, handle, and apply. • Their strength or concentration in the treated water must be deterrnined easily, quickly, and (preferably) automatically. • They must persist within disinfected water in a sufficient concentration to provide reasonable residual protection against its possible recontamination before use, or-because this is not a normally attainable property-the disappearance of residuals must be a warning that recontamination may have taken place.

Chick’s law of disinfaction • Under ideal conditions, the rate of die-off follows Chick's law, which states that the number of organisms destroyed in a unit time is proportional to the number of organisms remaining (Chick, 1908): •

𝑑𝑁 − 𝑑𝑡

= 𝑘𝑁

• This is a first-order reaction. Under real conditions the rate of kill may depart significantly from Chick's law. • Increased rates of kill may occur because of a time lag in the disinfectant reaching vital centres in the cell. • Decreased rates of kill may occur because of declining concentration of disinfectant in solution or poor distribution of organisms and disinfectant.

Chlorine as disinfectant • Chlorination • Chlorine may be used as the element (Cl2), or as sodium hypocrlorite(NaOCl), or as calcium hypochlorite [Ca(OCl)2].

• 𝑪𝒍𝟐 𝒈 + 𝑯𝟐 𝑶 ⇌ 𝑯𝑶𝑪𝒍 + 𝑯+ + 𝑪𝒍− • Reaction pH • Dilute solution and pH above 1.0 equilibrium displaced to right and very little 𝐶𝑙2 𝑔 exists in solution. • Hypochlorous acid is week acid and dissociate poorly at levels of pH about 6 • Between 6.0 and 8.5 the HOCl completely dissociate to H+ and OCl- ions. • 𝑯𝑶𝑪𝒍 ⇌ 𝑯+ + 𝑶𝑪𝒍− , 𝒑𝑲 = 𝟕. 𝟓𝟑𝟕 𝒂𝒕 𝟐𝟓℃ • Cl exists as HOCl at pH levels between 4.0 to 6.0. • Below pH 1.0 the chlorine revert back to Cl2. • Chlorine in the form of HOCl and OCl- is defined as free chlorine. • Chlorine reacts with protein and amino acid of bacterial cell and alter the same and ultimately destroy the cell protoplasm.

CT Concept • Empirical expression for CT for defining the nature of biological inactivation. • 𝐶𝑇 = 0.9847 𝐶 0.1758 𝑝𝐻 2.7519 𝑇𝑒𝑚𝑝−0.1467 • C= disinfectant concentration • T= Contact time between the microorganisms and disinfectant • pH=-log[H+] • Temp= temperature in ℃

Combined available chlorine/Reaction with Ammonia • 𝑁𝐻3 + 𝐻𝑂𝐶𝑙 ⇌ 𝑁𝐻2 𝐶𝑙 + 𝐻2 𝑂 monochloramine

• 𝑁𝐻2 𝐶𝑙 + 𝐻𝑂𝐶𝑙 ⇌ 𝑁𝐻𝐶𝑙2 + 𝐻2 𝑂 dichloramine

• 𝑁𝐻𝐶𝑙2 + 𝐻𝑂𝐶𝑙 ⇌ 𝑁𝐶𝑙3 + 𝐻2 𝑂 Trichloramine or nitrogen trichloride

• In general high 𝐶𝑙2 : 𝑁𝐻3 ratios, low temperatures and low pH levels favour dichloramine formation. • Cl2 also reacts with organic nitrogenous material such as protein and amino acid to form organic chloramine complexes. • The chloramines are known as combined available chlorine. • Disinfecting ability of chloramines is much lower than that of free chlorine

Break point chlorination • With molar 𝐶𝑙2 : 𝑁𝐻3 ratio 1:1 (5:1 on mass basis) chloramine and dichloramine will be formed and relative amounts of each depend on pH and other factors. • Further increase in 𝐶𝑙2 : 𝑁𝐻3 ratio results in oxidation of ammonia and reduction of chlorine • At molar ratio 2:1 and if sufficient time is provided for reactions to complete the break point occurs. At this point the oxidation/reduction reactions are complete and further addition of chlorine produces free residual chlorine.

Other disinfectants • Chlorine dioxide (ClO2): form on-site by combining chlorine and sodium chlorite. • Ozonation • Ultraviolet radiation