Domestic Water Treatment and Supply

Domestic Water Treatment and Supply Module 1: Municipal Water Supply: Sources and Quality Lecture 1: Raw Water Source and Quality Raw Water Source The various sources of water can be classified into two categories: 1. Surfac Surface e sourc sources, es, such such as a. Po Pond nds s and and lake lakes; s; b. Stream Streams s and and rivers rivers;; c. Stor Storag age e reserv reservoi oirs rs;; and d. Oceans, Oceans, generally generally not used for water supplie supplies, s, at present. 2. Sub-surface Sub-surface sources sources or underg underground round sources, sources, such as as a. Spri Sprin ngs; b. Infiltration wells ; and

c. Well Wells s and and Tube Tube-w -wel ells. ls. Water Quality The raw or treated water is analyzed by testing their physical, chemical and bacteriological characteristics: Physical Characteristics: Characteristics: Turbidity If a large amount of suspended solids are present in water, it will appear turbid in appearance. The turbidity depends upon fineness and concentration of particles present in water. Originally turbidity was determined by measuring the depth of  column of liquid required to cause the image of a candle flame at the bottom to diffuse into a uniform glow. This was measured by Jackson Jacks on candle turbidity meter. meter . The calibration was done based on suspensions of silica from Fuller's earth. The depth of sample in the tube was read against the part per million (ppm) silica scale with one ppm of suspended silica called one Jackson one  Jackson Turbidity (JTU). Turbidity unit (JTU). Because standards were prepared from materials found in nature such as Fuller's earth, consistency in standard formulation was difficult to achieve.

These days turbidity is measured by applying Nephelometry, a technique to measure level of light scattered by the particles at right angles to the incident light beam. The scattered light level is proportional to the particle concentration in the sample. The unit of  expression is Nephelometric Nephelometric Turbidity Unit (NTU). Unit (NTU). The IS values for drinking water is 10 to 25 NTU . Colour Dissolved organic matter from decaying vegetation or some inorganic materials may impart colour to the water. It can be measured by comparing the colour of o f water sample with other standard glass tubes containing solutions of different standard colour intensities. The standard unit of colour is that which is produced by one milligram of platinum cobalt dissolved in one litre of distilled water. The IS value for treated water is 5 to 25 cobalt  units. Taste and Odour Odour depends on the contact of a stimulating substance with the appropriate human receptor cell. Most organic o rganic and some inorganic chemicals, originating from municipal or industrial wastes, contribute taste and odour to the water. Taste and odour can be expres expressed sed in in terms terms of odou odourr inten intensit sity y or thresh threshold old values values.. A new method to estimate taste of water sample has been developed based on flavour known as 'Flavour Profile Analysis' (FPA). The character and intensity of taste and odour discloses the nature of pollution or the presence of microorganisms. Temperature The The inc incre reas ase e in in temp temper erat atur ure e decr decrea eases ses pala palata tabi bili lity ty , beca becaus use e at at elevated temperatures carbon dioxide and some other volatile . gases are expelled .The ideal temperature of water for drinking purposes is 5 to 12 °C °C  - above 25 °C, water is not recommended for drinking. Chemical Characteristics: Characteristics: pH pH value denotes the acidic or alkaline condition of water. It is expressed on a scale ranging from 0 to 14, which is the common logarithm of the reciprocal of the hydrogen ion concentration. The recommended pH range for treated drinking waters is 6.5 to 8.5 .

Acidity The acidity of water is a measure of its capacity to neutralize bases. Acidity of water may be caused by the presence of uncombined carbon dioxide, mineral acids and salts of strong acids and weak bases. It is expressed as mg/L in terms of calcium carbonate. Acidity is nothing but representation of carbon dioxide or carbonic acids. Carbon dioxide causes corrosion in public water supply systems. Alkalinity The alkalinity of water is a measure of its capacity to neutralize acids. It is expressed as mg/L in terms of calcium carbonate. The various forms of alkalinity are (a) hydroxide alkalinity, (b) carbonate alkalinity, (c) hydroxide plus carbonate alkalinity, (d) carbonate plus bicarbonate alkalinity, and (e) bicarbonate alkalinity, which is useful mainly in water softening and boiler feed water processes. Alkalinity is an important parameter in evaluating the optimum coagulant dosage. Hardness If water consumes excessive soap to produce lather, it is said to be hard. Hardness is caused by divalent metallic cations. The principal hardness causing cations are calcium, magnesium, strontium, ferrous and manganese ions. The major anions associated with these cations are sulphates, carbonates, bicarbonates, chlorides and nitrates. The total hardness of water is defined as the sum of calcium and magnesium concentrations, both expressed as calcium carbonate, in mg/L. Hardness are of two types, temporary or carbonate hardness and permanent or non-carbonate hardness. Temporary hardness is one in which bicarbonate and carbonate ion can be precipitated by prolonged boiling. Non-carbonate ions cannot be precipitated or removed by boiling, hence the term permanent hardness. IS value for drinking water is 300 mg/L as CaCO3. Chlorides Chloride ion may be present in combination with one or more of the cations of calcium, magnesium, iron and sodium. Chlorides of these minerals are present in water because of their high solubility in water. Each human being consumes about six to eight grams of  sodium chloride per day, a part of which is discharged through urine and night soil. Thus, excessive presence of chloride in water

indicates sewage pollution. IS value for drinking water is 250 to 1000 mg/L.

Sulphates Sulphates occur in water due to leaching from sulphate mineral and oxidation of sulphides. Sulphates are associated generally with calcium, magnesium and sodium ions. Sulphate in drinking water causes a laxative effect and leads to scale formation in boilers. It also causes odour and corrosion problems under aerobic conditions. Sulphate should be less than 50 mg/L, for some industries. Desirable limit for drinking water is 150 mg/L. May be extended upto 400 mg/L.

Iron Iron is found on earth mainly as insoluble ferric oxide. When it comes in contact with water, it dissolves to form ferrous bicarbonate under favourable conditions. This ferrous bicarbonate is oxidized into ferric hydroxide, which is a precipitate. Under anaerobic conditions, ferric ion is reduced to soluble ferrous ion. Iron can impart bad taste to the water, causes discolouration in clothes and incrustations in water mains. IS value for drinking water is 0.3 to 1.0 mg/L. Solids The sum total of foreign matter present in water is termed as 'total solids'. Total solids is the matter that remains as residue after evaporation of the sample and its subsequent drying at a defined temperature (103 to 105 °C). Total solids consist of volatile (organic) and non-volatile (inorganic or fixed) solids. Further, solids are divided into suspended and dissolved solids. Solids that can settle by gravity are settleable solids. The others are non-settleable solids. IS acceptable limit for total solids is 500 mg/L and tolerable limit is 3000 mg/L of  dissolved limits. Nitrates Nitrates in surface waters occur by the leaching of fertilizers from soil during surface run-off and also nitrification of organic matter. Presence of high concentration of nitrates is an indication of  pollution. Concentration of nitrates above 45 mg/L cause a disease methemoglobinemia. IS value is 45 mg/L.

Bacteriological Characteristics: Bacterial examination of water is very important, since it indicates the degree of pollution. Water polluted by sewage contain one or more species of disease producing pathogenic bacteria. Pathogenic organisms cause water borne diseases, and many non-pathogenic bacteria such as E.Coli , a member of coliform group, also live in the intestinal tract of human beings. Coliform itself is not a harmful group but it has more resistance to adverse condition than any other group. So, if it is ensured to minimize the number of  coliforms, the harmful species will be very less. So, coliform group serves as indicator of contamination of water with sewage and presence of pathogens. The methods to estimate the bacterial quality of water are: Standard Plate Count Test In this test, the bacteria are made to grow as colonies, by innoculating a known volume of sample into a solidifiable nutrient medium (Nutrient Agar), which is poured in a petridish. After incubating (35°C) for a specified period (24 hours), the colonies of  bacteria (as spots) are counted. The bacterial density is expressed as number of colonies per 100 ml of sample. Most Probable Number Most probable number is a number which represents the bacterial density which is most likely to be present. E.Coli is used as indicator of pollution. E.Coli ferment lactose with gas formation with 48 hours incubation at 35°C. Based on this E.Coli density in a sample is estimated by multiple tube fermentation procedure, which consists of identification of E.Coli in different dilution combination. MPN value is calculated as follows: Five 10 ml (five dilution combination) tubes of a sample is tested for E.Coli. If out of five only one gives positive test for E.Coli and all others negative. From the tables, MPN value for one positive and four negative results is read which is 2.2 in present case. The MPN value is expressed as 2.2 per 100 ml. These numbers are given by Maccardy based on the laws of statistics. Membrane Filter Technique

In this test a known volume of water sample is filtered through a membrane with opening less than 0.5 microns. The bacteria present in the sample will be retained upon the filter paper. The filter paper is put in contact of a suitable nutrient medium and kept in an incubator for 24 hours at 35°C. The bacteria will grow upon the nutrient medium and visible colonies are counted. Each colony represents one bacterium of the original sample. The bacterial count is expressed as number of colonies per 100 ml of sample.

Module 2: Water Quantity and Intake Details Lecture 2: Water Quantity Estimation Water Quantity Estimation The quantity of water required for municipal uses for which the water supply scheme has to be designed requires following data: 1. Water consumption rate (Per Capita Demand in litres per day   per head) 2. Population to be served. Quantity= Per capita demand x Population Water Consumption Rate It is very difficult to precisely assess the quantity of water demanded by the public, since there are many variable factors affecting water consumption. The various types of water demands, which a city may have, may be broken into following classes: Water Consumption for Various Purposes: Types of Consumption

Normal Range (lit/capita/day)

Average

%

1 Domestic Consumption

65-300

160

35

2 Industrial and Commercial Demand

45-450

135

30

3 Public Uses including Fire Demand

20-90

45

10

45-150

62

25

4 Losses and Waste Fire Fighting Demand:

The per capita fire demand is very less on an average basis but the rate at which the water is required is very large. The rate of fire

demand is sometimes treated as a function of population and is worked out from following empirical formulae:

Authority American 1 Insurance Association

Formulae (P in thousand)

Q for 1 lakh Population)

Q (L/min)=4637 √P (10.01 √P)

41760

2

Kuchling's Formula

Q (L/min)=3182 √P

31800

3

Freeman's Formula

Q (L/min)= 1136.5(P/5+10)

35050

Ministry of  Q (kilo liters/d)=100 √P for Urban P>50000 4 Development Manual Formula

31623

Factors affecting per capita demand: a. Size of the city: Per capita demand for big cities is generally large as compared to that for smaller towns as big cities have sewerage houses. b. Presence of industries. c. Climatic conditions. d. Habits of people and their economic status. e. Quality of water: If water is aesthetically $ medically safe, the consumption will increase as people will not resort to private wells, etc. f. Pressure in the distribution system. g. Efficiency of water works administration: Leaks in water mains and services; and unauthorized use of water can be kept to a minimum by surveys. h. Cost of water.

i. Policy of metering and charging method: Water tax is charged in two different ways: on the basis of meter reading and on the basis of certain fixed monthly rate. Fluctuations in Rate of Demand Average Daily Per Capita Demand = Quantity Required in 12 Months/ (365 x Population) If this average demand is supplied at all the times, it will not be sufficient to meet the fluctuations. •

•

•

Seasonal variation: The demand peaks during summer. Firebreak outs are generally more in summer, increasing demand. So, there is seasonal variation . Daily variation depends on the activity. People draw out more water on Sundays and Festival days, thus increasing demand on these days. Hourly variations are very important as they have a wide range. During active household working hours i.e. from six to ten in the morning and four to eight in the evening, the bulk of the daily requirement is taken. During other hours the requirement is negligible. Moreover, if a fire breaks out, a huge quantity of water is required to be supplied during short duration, necessitating the need for a maximum rate of hourly supply.

So, an adequate quantity of water must be available to meet the peak demand. To meet all the fluctuations, the supply pipes, service reservoirs and distribution pipes must be properly proportioned. The water is supplied by pumping directly and the pumps and distribution system must be designed to meet the peak demand. The effect of monthly variation influences the design of storage reservoirs and the hourly variations influences the design of pumps and service reservoirs. As the population decreases, the fluctuation rate increases. Maximum daily demand = 1.8 x average daily demand Maximum hourly demand of maximum day i.e. Peak demand  = 1.5 x average hourly demand = 1.5 x Maximum daily demand/24 = 1.5 x (1.8 x average daily demand)/24 = 2.7 x average daily demand/24 = 2.7 x annual average hourly demand Design Periods & Population Forecast

This quantity should be worked out with due provision for the estimated requirements of the future . The future period for which a provision is made in the water supply scheme is known as the design period. Design period is estimated based on the following: • •

Useful life of the component, considering obsolescence wear, tear, etc. Expandability aspect.

,

•

Anticipated rate of growth of population, including industrial, commercial developments & migration-immigration.

•

Available resources.

•

Performance of the system during initial period.

•

Population Forecasting Methods The various methods adopted for estimating future populations are given below. The particular method to be adopted for a particular case or for a particular city depends largely on the factors discussed in the methods, and the selection is left to the discretion and intelligence of the designer. 1. Arithmetic Increase Method This method is based on the assumption that the population increases at a constant rate; i.e. dP/dt=constant=k; P t= P0+kt. This method is most applicable to large and established cities.

2. Geometric Increase Method This method is based on the assumption that percentage growth rate is constant i.e. dP/dt=kP; lnP= lnP 0+kt. This method must be used with caution, for when applied it may produce too large results for rapidly grown cities in comparatively short time. This would apply to cities with unlimited scope of expansion. As cities grow large, there is a tendency to decrease in the rate of growth. 3. Incremental Increase Method Growth rate is assumed to be progressively increasing or decreasing, depending upon whether the average of the incremental

increases in the past is positive or negative. The population for a future decade is worked out by adding the mean arithmetic increase to the last known population as in the arithmetic increase method, and to this is added the average of incremental increases, once for first decade, twice for second and so on. 4. Decreasing Rate of Growth Method In this method, the average decrease in the percentage increase is worked out, and is then subtracted from the latest percentage increase to get the percentage increase of next decade. 5. Simple Graphical Method In this method, a graph is plotted from the available data, between time and population. The curve is then smoothly extended up to the desired year. This method gives very approximate results and should be used along with other forecasting methods. 6. Comparative Graphical Method In this method, the cities having conditions and characteristics similar to the city whose future population is to be estimated are selected. It is then assumed that the city under consideration will develop, as the selected similar cities have developed in the past. 7. Ratio Method In this method, the local population and the country's population for the last four to five decades is obtained from the census records. The ratios of the local population to national population are then worked out for these decades. A graph is then plotted between time and these ratios, and extended up to the design period to extrapolate the ratio corresponding to future design year. This ratio is then multiplied by the expected national population at the end of  the design period, so as to obtain the required city's future population. Drawbacks: 1. Depends on accuracy of national population estimate. 2. Does not consider the abnormal or special conditions which can lead to population shifts from one city to another. 8. Logistic Curve Method

The three factors responsible for changes in population are : (i) Births, (ii) Deaths and (iii) Migrations. Logistic curve method is based on the hypothesis that when these varying influences do not produce extraordinary changes, the population would probably follow the growth curve characteristics of  living things within limited space and with limited economic opportunity. The curve is S-shaped and is known as logistic curve.

Lecture 3: Intake, Pumping and Conveyance Intake Structure The basic function of the intake structure is to help in safely withdrawing water from the source over predetermined pool levels and then to discharge this water into the withdrawal conduit (normally called intake conduit), through which it flows up to water treatment plant. Factors Governing Location of Intake 1. As far as possible, the site should be near the treatment plant so that the cost of conveying water to the city is less. 2. The intake must be located in the purer zone of the source to draw best quality water from the source, thereby reducing load on the treatment plant. 3. The intake must never be located at the downstream or in the vicinity of the point of disposal of wastewater. 4. The site should be such as to permit greater withdrawal of water, if required at a future date. 5. The intake must be located at a place from where it can draw water even during the driest period of the year. 6. The intake site should remain easily accessible during floods and should noy get flooded. Moreover,

the flood waters should not be concentrated in the vicinity of the intake. Design Considerations 1. sufficient factor of safety against external forces such as heavy currents, floating materials, submerged bodies, ice pressure, etc. 2. should have sufficient self-weight so that it does not float by up thrust of water. Types of Intake Depending on the source of water, the intake works are classified as follows:

Pumping Module 3: Unit Processes in Municipal Water

Treatment Lecture 4: Water Treatment Philosophy The available raw waters must be treated and purified before they can be supplied to the public for their domestic, industrial or any other uses. The extent of treatment required to be given to the particular water depends upon the characteristics and quality of the available water, and also upon the quality requirements for the intended use. . The layout of conventional water treatment plant is as follows:

Depending upon the magnitude of treatment required, proper unit operations are selected and arranged in the proper sequential order for the purpose of modifying the quality of raw water to meet the desired standards. Indian Standards for drinking water are given in the table below. Indian Standards for drinking water

Parameter

Desirable-Tolerable

 If no alternative source available, limit extended  upto

Turbidity (NTU unit)

< 10

25

Colour (Hazen scale)

< 10

50

Taste and Odour

Un-objectionable

Un-objectionable

 pH

7.0-8.5

6.5-9.2

Total Dissolved Solids mg/l

500-1500

3000

Total Hardness mg/l (as CaCO 3)

200-300

600

Chlorides mg/l (as Cl)

200-250

1000

Sulphates mg/l (as SO 4)

150-200

400

Fluorides mg/l (as F )

0.6-1.2

1.5

 Nitrates mg/l (as NO3)

45

45

Calcium mg/l (as Ca)

75

200

Iron mg/l (as Fe )

0.1-0.3

1.0

Physical

Chemical

The typical functions of each unit operations are given in the following table:

Functions of Water Treatment Units Unit treatment

Function (removal)

Aeration, chemicals use Colour, Odour, Taste Screening

Floating matter  

Chemical methods

Iron, Manganese, etc.

Softening

Hardness

Sedimentation

Suspended matter  

Coagulation

Suspended matter, a part of colloidal matter and bacteria

Filtration

Remaining colloidal dissolved matter, bacteria

Disinfection

Pathogenic bacteria, Organic matter and Reducing substances

The types of treatment required for different sources are given in the following table: Source

Treatment required

1. Ground water and spring water fairly free from  No treatment or Chlorination contamination 2. Ground water with chemicals, minerals and gases

Aeration, coagulation (if  necessary), filtration and disinfection

3. Lakes, surface water reservoirs with less amount of pollution

Disinfection

4. Other surface waters such as rivers, canals and Complete treatment impounded reservoirs with a considerable amount of pollution

Lecture 5: Preliminary Treatment: Silt Excluder Design Aeration • •

•

Aeration removes odour and tastes due to volatile gases like hydrogen sulphide and due to algae and related organisms. Aeration also oxidise iron and manganese, increases dissolved oxygen content in water, removes CO2 and reduces corrosion and removes methane and other flammable gases. Principle of treatment underlines on the fact that volatile gases in water escape into atmosphere from the air-water interface and atmospheric oxygen takes their place in water, provided the water body can expose itself over a vast surface to the atmosphere. This process continues until an equilibrium is reached depending on the partial pressure of each specific gas in the atmosphere.

Types of Aerators 1. Gravity aerators 2. Fountain aerators 3. Diffused aerators 4. Mechanical aerators. Gravity Aerators (Cascades): In gravity aerators, water is allowed to fall by gravity such that a large area of water is exposed to atmosphere, sometimes aided by turbulence. Fountain Aerators : These are also known as spray aerators with special nozzles to produce a fine spray. Each nozzle is 2.5 to 4 cm diameter discharging about 18 to 36 l/h. Nozzle spacing should be such that each m3 of water has aerator area of 0.03 to 0.09 m2 for one hour. Injection or Diffused Aerators : It consists of a tank with perforated pipes, tubes or diffuser plates, fixed at the bottom to release fine air bubbles from compressor unit. The tank depth is kept as 3 to 4 m and tank width is within 1.5 times its depth. If  depth is more, the diffusers must be placed at 3 to 4 m depth below

water surface. Time of aeration is 10 to 30 min and 0.2 to 0.4 litres of air is required for 1 litre of water. Mechanical Aerators : Mixing paddles as in flocculation are used. Paddles may be either submerged or at the surface.

Lecture 6: Sedimentation Tank Design

. Settling Solid liquid separation process in which a suspension is separated into two phases – • •

Clarified supernatant leaving the top of the sedimentation tank (overflow). Concentrated sludge leaving the bottom of the sedimentation tank (underflow).

Purpose of Settling • •

To remove coarse dispersed phase. To remove coagulated and flocculated impurities.

•

To remove precipitated impurities after chemical treatment.

•

To settle the sludge (biomass) after activated sludge process / tricking filters.

Principle of Settling •

• •

Suspended solids present in water having specific gravity greater than that of water tend to settle down by gravity as soon as the turbulence is retarded by offering storage. Basin in which the flow is retarded is called settling tank . Theoretical average time for which the water is detained in the settling tank is called the detention period .

Types of Settling Type I: Discrete particle settling - Particles settle individually without interaction with neighboring particles. Type II: Flocculent Particles – Flocculation causes the particles to increase in mass and settle at a faster rate. Type III: Hindered or Zone settling –The mass of particles tends to settle as a unit with individual particles remaining in fixed positions with respect to each other.

Type IV: Compression – The concentration of particles is so high that sedimentation can only occur through compaction of the structure. Type I Settling • •

Size, shape and specific gravity of the particles do not change with time. Settling velocity remains constant.

If a particle is suspended in water, it initially has two forces acting upon it: (1) force of gravity : Fg=ρpgVp (2) the buoyant force quantified by Archimedes as: F b=ρgVp If the density of the particle differs from that of the water, a net force is exerted and the particle is accelaratd in the direction of the force: Fnet=(ρp-ρ)gVp This net force becomes the driving force. Once the motion has been initiated, a third force is created due to viscous friction. This force, called the drag force, is quantified by: Fd=CDApρv2 /2 CD= drag coefficient. Ap = projected area of the particle. Because the drag force acts in the opposite direction to the driving force and increases as the square of the velocity, accelaration occurs at a decreasing rate until a steady velocity is reached at a point where the drag force equals the driving force: (ρp-ρ)gVp = CDApρv2 /2 For spherical particles, Vp=πd3 /6 and Ap=πd2 /4 Thus, v2= 4g(ρp-ρ)d 3 CDρ Expressions for CD change with characteristics of different flow regimes. For laminar, transition, and turbulent flow, the values of  CD are: CD = 24 (laminar) Re CD= 24 + 3 +0.34 (transition) 1/2 Re Re CD= 0.4 (turbulent) where Re is the Reynolds number: Re=ρvd

µ Reynolds number less than 1.0 indicate laminar flow, while values

greater than 10 indicate turbulent flow. Intermediate values indicate transitional flow. Stokes Flow  For laminar flow, terminal settling velocity equation becomes: v= (ρp-ρ)gd2 18µ which is known as the stokes equation.

Transition Flow  Need to solve non-linear equations: v2= 4g(ρp-ρ)d 3 CDρ CD= 24 + 3 +0.34 1/2 Re Re Re=ρvd

µ • •

Calculate velocity using Stokes law or turbulent expression. Calculate and check Reynolds number.

•

Calculate CD.

•

Use general formula.

•

Repeat from step 2 until convergence.

Types of Settling Tanks •

•

Sedimentation tanks may function either intermittently or continuously. The intermittent tanks also called quiescent type tanks are those which store water for a certain period and keep it in complete rest. In a continuous flow type tank, the flow velocity is only reduced and the water is not brought to complete rest as is done in an intermittent type. Settling basins may be either long rectangular or circular in plan. Long narrow rectangular tanks with horizontal flow are generally preferred to the circular tanks with radial or spiral flow.

Long Rectangular Settling Basin •

Long rectangular basins are hydraulically more stable, and flow control for large volumes is easier with this configuration.

•

A typical long rectangular tank have length ranging from 2 to 4 times their width. The bottom is slightly sloped to facilitate sludge scraping. A slow moving mechanical sludge scraper continuously pulls the settled material into a sludge hopper from where it is pumped out periodically.

A long rectangular settling tank can be divided into four different functional zones:  Inlet zone: Region in which the flow is uniformly distributed over the cross section such that the flow through settling zone follows horizontal path. Settling zone: Settling occurs under quiescent conditions. Outlet zone: Clarified effluent is collected and discharge through outlet weir. Sludge zone: For collection of sludge below settling zone.  Inlet and Outlet Arrangement   Inlet devices: Inlets shall be designed to distribute the water equally and at uniform velocities. A baffle should be constructed across the basin close to the inlet and should project several feet below the water surface to dissipate inlet velocities and provide uniform flow; Outlet Devices: Outlet weirs or submerged orifices shall be designed to maintain velocities suitable for settling in the basin and to minimize short-circuiting. Weirs shall be adjustable, and at least equivalent in length to the perimeter of the tank. However, peripheral weirs are not acceptable as they tend to cause excessive short-circuiting. Weir Overflow Rates Large weir overflow rates result in excessive velocities at the outlet. These velocities extend backward into the settling zone, causing particles and flocs to be drawn into the outlet. Weir loadings are generally used up to 300 m 3 /d/m. It may be necessary to provide special inboard weir designs as shown to lower the weir overflow rates.

 Inboard Weir Arrangement to Increase Weir Length Circular Basins •

Circular settling basins have the same functional zones as the long rectangular basin, but the flow regime is different. When the flow enters at the center and is baffled to flow

•

radially towards the perimeter, the horizontal velocity of the water is continuously decreasing as the distance from the center increases. Thus, the particle path in a circular basin is a parabola as opposed to the straight line path in the long rectangular tank. Sludge removal mechanisms in circular tanks are simpler and require less maintenance.

Settling Operations •

Particles falling through the settling basin have two components of velocity: 1) Vertical component: vt=(ρp-ρ)gd2 18µ 2) Horizontal component: v h=Q/A

•

•

The path of the particle is given by the vector sum of  horizontal velocity vh and vertical settling velocity v t. Assume that a settling column is suspended in the flow of the settling zone and that the column travels with the flow across the settling zone. Consider the particle in the batch analysis for type-1 settling which was initially at the surface and settled through the depth of the column Z 0, in the time t 0. If  t0 also corresponds to the time required for the column to be carried horizontally across the settling zone, then the particle will fall into the sludge zone and be removed from the suspension at the point at which the column reaches the end of the settling zone. All particles with v t>v0 will be removed from suspension at some point along the settling zone. Now consider the particle with settling velocity < v 0. If the initial depth of this particle was such that Z p /vt=t0, this particle will also be removed. Therefore, the removal of  suspended particles passing through the settling zone will be in proportion to the ratio of the individual settling velocities to the settling velocity v 0. The time t 0 corresponds to the retention time in the settling zone. t= V = LZ 0W Q Q

Also, t0= Z0 v0 Therefore, Z0 = LZ0W and v0= Q v0 Q LW or v0= Q

AS Thus, the depth of the basin is not a factor in determining the size particle that can be removed completely in the settling zone. The determining factor is the quantity Q/A s, which has the units of  velocity and is referred to as the overflow rate q 0. This overflow rate is the design factor for settling basins and corresponds to the terminal setting velocity of the particle that is 100% removed. Design Details 1. Detention Detention period: period: for plain plain sedimentat sedimentation: ion: 3 to 4 h, and for coagulated sedimentation: 2 to 2.5 h. 2. Velocity Velocity of flow: Not greate greaterr than 30 cm/min cm/min (horizonta (horizontall flow). 3. Tank dimensi dimensions: ons: L:B = 3 to 5:1. Genera Generally lly L= 30 m (common) maximum 100 m. Breadth= 6 m to 10 m. Circular: Diameter not greater than 60 m. generally 20 to 40 m. 4. Depth Depth 2.5 2.5 to to 5.0 5.0 m (3 (3 m). m). 5. Surface Surface Overflow Overflow Rate: Rate: For plain sediment sedimentation ation 12000 12000 to 18000 L/d/m2 tank area; for thoroughly flocculated water 24000 to 30000 L/d/m2 tank area. 6. Slopes: Slopes: Rectangular Rectangular 1% towards towards inlet inlet and circula circularr 8%.

Lecture 7: Coagulation - Flocculation Theory General Properties of Colloids 1. Colloidal Colloidal particle particles s are so small that their their surface  surface area in relation to mass is very large. 2. Electrical properties: properties : All colloidal particles are electrically charged. If electrodes from a D.C. source are placed in a colloidal dispersion, the particles migrate towards the pole of  opposite charge. 3. Colloidal Colloidal particles particles are in constan constantt motion because because of  bombardment by molecules of dispersion medium. This motion is called Brownian motion (named after Robert Brown who first noticed it). 4. Tyndall effect: Colloidal particles have dimensioThese are reversible upon heating. e.g. organics in water. 5.  Adsorption: Colloids have high surface area and hence have a lot of active surface for adsorption to occur. The stability of 

colloids is mainly due to preferential adsorption of ions. There are two types of colloids: i.

Lyoph Lyophobi obic c colloi colloids: ds: that that are solven solventt hating hating.. These These are irreversible upon heating. e.g. inorganic colloids, metal halides.

ii.

Lyophi Lyophilic lic colloi colloids: ds: that that are solvent solvent loving loving.. These These are reversible upon heating. e.g. organics in water.

Coagulation and Flocculation •

• •

Colloidal particles are difficult to separate from water because they do not settle by gravity and are so small that they pass through the pores of filtration media. To be removed, the individual colloids must aggregate and grow in size. The aggregation of colloidal particles can be considered as involving two separate and distinct steps: 1. Particle Particle transport transport to effect effect interpart interparticle icle collision. collision. 2. Particle Particle destabiliz destabilization ation to permit permit attachm attachment ent when when contact occurs.

Transport step is known as flocculation whereas coagulation coagulationis is the overall process involving destabilization and transport. Electrical Double Layer  Although individual hydrophobic colloids have an electrical charge, a colloidal dispersion does not have a net electrical charge. The diffuse layer in a colloidal dispersion contains a quantity of counter ions sufficient to balance the electrical charge on the particle. The charge distribution in the diffuse layer of a negatively charged colloid can be represented by the curve ABCD in the figure. The ions involved in this electroneutrality are arranged in such a way as to constitute what is called electrical double layer .

Net repulsion force, which may be considered as energy barrier must be overcome before aggregation occurs. The magnitude of energy barrier depends on (1) charge on the particle, and (2) ionic composition of water.

Destabilization of Colloidal Dispersion Particle destabilization can be achieved by four mechanisms:

• Change characteristics of medium-Compression of double layer . Double Layer Compression Colloidal systems could be destabilized by the addition of ions having a charge opposite to that of the colloid. The coagulating power of ions increased in the ratio of 1:10:1000 as the valence of  the ions increased in the ratio from 1 to 2 to 3. This is called Schulze-Hardy rule.

• Change characteristics of colloid particlesAdsorption and Charge Neutralization Some chemical species are capable of being adsorbed at the surface of colloidal particles. If the adsorbed species carry a charge opposite to that of the colloids, such adsorption causes a reduction of surface potential and a resulting destabilization of the colloidal particle. Reduction of surface charge by adsorption is a much different mechanism than reduction by double layer compression. 1. The sorbable species are capable of destabilizing colloids at much lower dosage than nonsorbable "double layer compressing" ions. 2. Destabilization by adsorption is stoichiometric. Thus, the required dosage of coagulant increases as the concentration of colloids increases. 3. It is possible to overdose a system with an adsorbable species and cause restabilization as a result of a reversal of charge on the colloidal particle. •

Provide bridges:-

1. Enmeshment in a Precipitate If certain metal salts are added to water or wastewater in sufficient amounts, rapid formation of precipitates will occur. Colloids may serve as condensation nuclei for these precipitates or may become enmeshed as the precipitates settle. Removal of colloids in this manner is frequently referred to as sweep-floc coagulation. Several characteristics that distinguish sweep-floc coagulation from double layer compression and adsorption have been reported.

i.

ii.

An inverse relationship exists between the optimum coagulant dosage and the concentration of colloids to be removed. At low colloid concentrations a large excess of coagulant is required to produce a large amount of precipitate that will enmesh the relatively few colloidal particles as it settles. At high colloid concentrations, coagulation will occur at a lower chemical dosage because the colloids serve as nuclei to enhance precipitate formation. Optimum coagulation conditions do not correspond to a minimum zeta potential but depends on pH depending on solubility-pH relationship for that coagulant.

2. Adsorption and Interparticle Bridging Many different natural compounds such as starch, cellulose, polysaccharide gums, and proteineous materials, as well as a wide variety of synthetic polymeric compounds are known to be effective coagulating agents. Research has revealed that both positive and negative polymers are capable of destabilizing negatively charged colloidal particles

Lecture 8: Rapid Mixing, Coagulation - Flocculation Flocculation Flocculation is stimulation by mechanical means to agglomerate destabilized particles into compact, fast settleable particles (or flocs). Flocculation or gentle agitation results from velocity differences or gradients in the coagulated water, which causes the fine moving, destabilized particles to come into contact and become large, readily settleable flocs. It is a common practice to provide an initial rapid (or) flash mix for the dispersal of the coagulant or other chemicals into the water. Slow mixing is then done, during which the growth of the floc takes place. Rapid or Flash mixing is the process by which a coagulant is rapidly and uniformly dispersed through the mass of water. This process usually occurs in a small basin immediately preceding or at the head of the coagulation basin. Generally, the detention period is

30 to 60 seconds and the head loss is 20 to 60 cms of water. Here colloids are destabilized and the nucleus for the floc is formed. Slow mixing brings the contacts between the finely divided destabilized matter formed during rapid mixing. Per kinetic and Ortho kinetic Flocculation The flocculation process can be broadly classified into two types, per kinetic and ortho kinetic. Perikinetic flocculation refers to flocculation (contact or collisions of  colloidal particles) due to Brownian motion of colloidal particles. The random motion of colloidal particles results from their rapid and random bombardment by the molecules of the fluid. Orthokinetic flocculation refers to contacts or collisions of colloidal particles resulting from bulk fluid motion, such as stirring. In systems of stirring, the velocity of the fluid varies both spatially (from point to point) and temporally (from time to time). The spatial changes in velocity are identified by a velocity gradient, G. G is estimated as G=(P/ ηV)1/2, where P=Power, V=channel volume, and η= Absolute viscosity. Mechanism of Flocculation Gravitational flocculation: Baffle type mixing basins are examples of  gravitational flocculation. Water flows by gravity and baffles are provided in the basins which induce the required velocity gradients for achieving floc formation. Mechanical flocculation: Mechanical flocculators consists of revolving paddles with horizontal or vertical shafts or paddles suspended from horizontal oscillating beams, moving up and down.

Lecture 9: Coagulation - Flocculation Coagulation in Water Treatment • •

Salts of Al(III) and Fe(III) are commonly used as coagulants in water and wastewater treatment. When a salt of Al(III) and Fe(III) is added to water, it dissociates to yield trivalent ions, which hydrate to form aquometal complexes Al(H2O)63+ and Fe(H2O)63+. These complexes then pass through a series of hydrolytic reactions

in which H2O molecules in the hydration shell are replaced by OH- ions to form a variety of soluble species such as Al(OH)2+ and Al(OH)2+. These products are quite effective as coagulants as they adsorb very strongly onto the surface of  most negative colloids. Destabilization using Al(III) and Fe(III) Salts •

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Al(III) and Fe(III) accomplish destabilization by two mechanisms: (1) Adsorption and charge neutralization. (2) Enmeshment in a sweep floc. Interrelations between pH, coagulant dosage, and colloid concentration determine mechanism responsible for coagulation.

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Charge on hydrolysis products and precipitation of metal hydroxides are both controlled by pH. The hydrolysis products possess a positive charge at pH values below iso-electric point of the metal hydroxide. Negatively charged species which predominate above iso-electric point, are ineffective for the destabilization of negatively charged colloids.

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Precipitation of amorphous metal hydroxide is necessary for sweep-floc coagulation.

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The solubility of Al(OH) 3(s) and Fe(OH)3(s) is minimal at a particular pH and increases as the pH increases or decreases from that value. Thus, pH must be controlled to establish optimum conditions for coagulation.

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Alum and Ferric Chloride reacts with natural alkalinity in water as follows:

Al2(SO4)3.14H2O + 6 HCO32 Al(OH)3(s) + 6CO 2 +14 H2O + 3 2SO4 FeCl3 + 3 HCO3Fe(OH)3(S) +3 CO2 + 3 ClJar Test The jar test is a common laboratory procedure used to determine the optimum operating conditions for water or wastewater treatment. This method allows adjustments in pH, variations in coagulant or polymer dose, alternating mixing speeds, or testing of  different coagulant or polymer types, on a small scale in order to predict the functioning of a large scale treatment operation. Jar Testing Apparatus

The jar testing apparatus consists of six paddles which stir the contents of six 1 liter containers. One container acts as a control while the operating conditions can be varied among the remaining five containers. A rpm gage at the top-center of the device allows for the uniform control of the mixing speed in all of the containers.

Jar Test Procedure

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• The jar test procedures involves the following steps: Fill the jar testing apparatus containers with sample water. One container will be used as a control while the other 5 containers can be adjusted depending on what conditions are being tested. For example, the pH of the jars can be adjusted or variations of coagulant dosages can be added to determine optimum operating conditions. Add the coagulant to each container and stir at approximately 100 rpm for 1 minute. The rapid mix stage helps to disperse the coagulant throughout each container.

Turn off the mixers and allow the containers to settle for 30 to 45 minutes. Then measure the final turbidity in each container. • Reduce the stirring speed to 25 to 35 rpm and continue mixing for 15 to 20 minutes. This slower mixing speed helps promote floc formation by enhancing particle collisions which lead to larger flocs.

Residual turbidity vs. coagulant dose is then plotted and optimal conditions are determined. The values that are obtained through the experiment are correlated and adjusted in order to account for the actual treatment system.

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Lecture 10: Filtration Theory Filtration The resultant water after sedimentation will not be pure, and may contain some very fine suspended particles and bacteria in it. To remove or to reduce the remaining impurities still further, the water is filtered through the beds of fine granular material, such as sand, etc. The process of passing the water through the beds of such granular materials is known as Filtration. How Filters Work: Filtration Mechanisms There are four basic filtration mechanisms: SEDIMENTATION : The mechanism of sedimentation is due to force of gravity and the associate settling velocity of the particle, which causes it to cross the streamlines and reach the collector.  INTERCEPTION : Interception of particles is common for large particles. If a large enough particle follows the streamline, that lies very close to the media surface it will hit the media grain and be captured. BROWNIAN DIFFUSION : Diffusion towards media granules occurs for very small particles, such as viruses. Particles move randomly about within the fluid, due to thermal gradients. This mechanism is only important for particles with diameters < 1 micron.  INERTIA : Attachment by inertia occurs when larger particles move fast enough to travel off their streamlines and bump into media grains.

Filter Materials Sand: Sand, either fine or coarse, is generally used as filter media. The size of the sand is measured and expressed by the term called effective size. The effective size, i.e. D10 may be defined as the size of the sieve in mm through which ten percent of the sample of sand by weight will pass. The uniformity in size or degree of variations in sizes of particles is measured and expressed by the term called uniformity coefficient . The uniformity coefficient, i.e. (D 60 /D10) may be defined as the ratio of the sieve size in mm through which 60 percent of the sample of sand will pass, to the effective size of  the sand.

Gravel: The layers of sand may be supported on gravel, which permits the filtered water to move freely to the under drains, and allows the wash water to move uniformly upwards. Other materials: Instead of using sand, sometimes, anthrafilt is used as filter media. Anthrafilt is made from anthracite, which is a type of coal-stone that burns without smoke or flames. It is cheaper and has been able to give a high rate of filtration. Types of Filter Slow sand filter: They consist of fine sand, supported by gravel. They capture particles near the surface of the bed and are usually cleaned by scraping away the top layer of sand that contains the particles. Rapid-sand filter: They consist of larger sand grains supported by gravel and capture particles throughout the bed. They are cleaned by backwashing water through the bed to 'lift out' the particles. Multimedia filters: They consist of two or more layers of different granular materials, with different densities. Usually, anthracite coal, sand, and gravel are used. The different layers combined may provide more versatile collection than a single sand layer. Because of the differences in densities, the layers stay neatly separated, even after backwashing. Principles of Slow Sand Filtration In a slow sand filter impurities in the water are removed by a combination of processes: sedimentation, straining, adsorption, and chemical and bacteriological action. • During the first few days, water is purified mainly by mechanical and physical-chemical processes. The resulting accumulation of sediment and organic matter forms a thin layer on the sand surface, which remains permeable and retains particles even smaller than the spaces between the sand grains.

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As this layer (referred to as “Schmutzdecke”) develops, it becomes living quarters of vast numbers of micro-organisms which break down organic material retained from the water, converting it into water, carbon dioxide and other oxides.

Most impurities, including bacteria and viruses, are removed from the raw water as it passes through the filter skin and the layer of filter bed sand just below. The purification mechanisms extend from the filter skin to approx. 0.3-0.4 m below the surface of the filter bed, gradually decreasing in activity at lower levels as the water becomes purified and

contains less organic material. When the micro-organisms become well established, the filter will work efficiently and produce high quality effluent which is virtually free of disease carrying organisms and biodegradable organic matter. They are suitable for treating waters with low colors, low turbidities and low bacterial contents.

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Sand Filters vs. Rapid Sand Filters Base material: In SSF it varies from 3 to 65 mm in size and 30 to 75 cm in depth while in RSF it varies from 3 to 40 mm in size and its depth is slightly more, i.e. about 60 to 90 cm. • Filter sand: In SSF the effective size ranges between 0.2 to 0.4 mm and uniformity coefficient between 1.8 to 2.5 or 3.0. In RSF the effective size ranges between 0.35 to 0.55 and uniformity coefficient between 1.2 to 1.8.

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Rate of filtration: In SSF it is small, such as 100 to 200 L/h/sq.m. of filter area while in RSF it is large, such as 3000 to 6000 L/h/sq.m. of filter area. Flexibility: SSF are not flexible for meeting variation in demand whereas RSF are quite flexible for meeting reasonable variations in demand.

Post treatment required: Almost pure water is obtained from SSF. However, water may be disinfected slightly to make it completely safe. Disinfection is a must after RSF.

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Method of cleaning: Scrapping and removing of the top 1.5 to 3 cm thick layer is done to clean SSF. To clean RSF, sand is agitated and backwashed with or without compressed air.

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Loss of head: In case of SSF approx. 10 cm is the initial loss, and 0.8 to 1.2m is the final limit when cleaning is required. For RSF 0.3m is the initial loss, and 2.5 to 3.5m is the final limit when cleaning is required.

Lecture 11: Rapid Sand Filtration Typical Rapid Gravity Filter Flow Operation Isometric view of Rapid Sand Filter

Clean Water Headloss

Several equations have been developed to describe the flow of clean water through a porous medium. Carman-Kozeny equation used to calculate head loss is as follows: h= f (1-α)Lvs2 φα3dg h= f p(1- α)Lvs2 φα3dgg f =150 (1- α) + 1.75 Ng Ng= φdvsρ

µ where, h = headloss, m f = friction factor α = porosity φ = particle shape factor (1.0 for spheres, 0.82 for rounded sand, 0.75 for average sand,0.73 for crushed coal and angular sand) L = depth of filter bed or layer, m d = grain size diameter, m vs = superficial (approach) filtration velocity, m/s g = accelaration due to gravity, 9.81 m/s 2 p = fraction of particles ( based on mass) within adjacent sieve sizes dg = geometric mean diameter between sieve sizes d1 and d2 Ng = Reynolds number µ = viscosity, N-s/m2 Backwashing of Rapid Sand Filter •

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For a filter to operate efficiently, it must be cleaned c leaned before the next filter run. If the water applied to a filter is of very good quality, the filter runs can be very long. Some filters can operate longer than one week before needing to be backwashed. However, this is not recommended as long filter runs can cause the filter media to pack down so that it is difficult to expand the bed during the backwash. Treated water from storage is used for the backwash cycle. This treated water is generally taken from elevated storage tanks or pumped in from the clear well. The filter backwash rate has to be great enough to expand and agitate the filter media and suspend the floc in the water for removal. However, if the filter backwash rate is too high, media will be washed from the filter into the troughs and out of the filter.

When is Backwashing Needed The filter should be backwashed when the following conditions have been met: • • •

The head loss is so high that the filter no longer produces water at the desired rate; and/or Floc starts to break through the filter and the turbidity in the filter effluent increases; and/or A filter run reaches a given hour of operation.

Operational Troubles in Rapid Gravity Filters  Air Binding : •

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When the filter is newly commissioned, the loss of head of  water percolating through the filter is generally very small. However, the loss of head goes on increasing as more and more impurities get trapped into it. A stage is finally reached when the frictional resistance offered by the filter media exceeds the static head of water above the and bed. Most of this resistance is offered by the top 10 to 15 cm sand layer. The bottom sand acts like a vacuum, and water is sucked through the filter media rather than getting filtered through it.

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The negative pressure so developed, tends to release the dissolved air and other gases present in water. The formation of bubbles takes place which stick to the sand grains. This phenomenon is known as Air Binding as the air binds the filter and stops its functioning.

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To avoid such troubles, the filters are cleaned as soon as the head loss exceeds the optimum allowable value.

Formation of Mud Balls : •

The mud from the atmosphere usually accumulates on the sand surface to form a dense mat. During inadequate washing this mud may sink down into the sand bed and stick to the sand grains and other arrested impurities, thereby forming mud balls.

Cracking of Filters :

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The fine sand contained in the top layers of the filter bed shrinks and causes the development of shrinkage cracks in the sand bed. With the use of filter, the loss of head and, therefore, pressure on the sand bed goes on increasing, which further goes on widening these cracks.

Remedial Measures to Prevent Cracking of Filters and Formation of Mud Balls

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Breaking the top fine mud layer with rakes and washing off  the particles. Washing the filter with a solution of caustic soda.

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Removing, cleaning and replacing the damaged filter sand.

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Standard design practice of Rapid Sand filter : Maximum length of lateral = not less than 60 times its diameter. Spacing of holes = 6 mm holes at 7.5 cm c/c or 13 at 15 c/c. C.S area of lateral = not less than 2 times area of   perforations. C.S area area of manifold = 2 times total area of laterals. laterals. Maximum loss of head = 2 to 5 m. Spacing of laterals = 15 to 30 3 0 cm c/c. Pressure of  wash water at perforations = not greater than 1.05 kg/cm 2. Velocity of flow in lateral = 2 m/s. Velocity of flow in manifold = 2.25 m/s. Velocity of flow in manifold for wash water= 1.8 to 2.5 m/s. Velocity of rising wash water= 0.5 to 1.0 m/min. Amount of wash water = 0.2 to 0.4% of  total filtered water. Time of backwashing = 10 to 15 min. Head of  water over the filter = 1.5 to 2.5 m. Free board = 60 cm. Bottom slope = 1 to 60 towards manifold. Q = (1.71 x b x h 3/2) where Q is in m 3 /s, b is in m, h is in m. L:B = 1.25 to 1.33:1 .

Lecture 12: Disinfection Disinfection

The filtered water may normally contain some harmful disease producing bacteria in it. These bacteria must be killed in order to make the water safe for drinking. The process of killing these bacteria is known as Disinfection or Sterilization. Disinfection Kinetics When a single unit of microorganisms is exposed to a single unit of  disinfectant, the reduction in microorganisms follows a first-order reaction. dN/dt=-kN N=N0e-kt This equation is known as Chick’s Law:-

N = number of microorganism (N 0 is initial number) k = disinfection constant t = contact time Methods of Disinfection 1. Boiling: The bacteria present in water can be destroyed by boiling it for a long time. However it is not practically possible to boil huge amounts of water. Moreover it cannot take care of future possible contaminations. 2. Treatment with Excess Lime: Lime is used in water treatment plant for softening. But if excess lime is added to the water, it can in addition, kill the bacteria also. Lime when added raises the pH value o water making it extremely alkaline. This extreme alkalinity has been found detrimental to the survival of bacteria. This method needs the removal of  excess lime from the water before it can be supplied to the general public. Treatment like recarbonation for lime removal should be used after disinfection. 3. Treament with Ozone: Ozone readily breaks down into normal oxygen, and releases nascent oxygen. The nascent oxygen is a powerful oxidising agent and removes the organic matter as well as the bacteria from the water. 4. Chlorination: The germicidal action of chlorine is explained by the recent theory of Enzymatic hypothesis, according to which the chlorine enters the cell walls of bacteria and kill the enzymes which are essential for the metabolic processes of  living organisms. Chlorine Chemistry

Chlorine is added to the water supply in two ways. It is most often added as a gas, Cl 2(g). However, it also can be added as a salt, such as sodium hypochlorite (NaOCl) or bleach. Chlorine gas dissolves in water following Henry's Law. Cl2(g) Cl2(aq) KH =6.2 x 10-2 Once dissolved, the following reaction occurs forming hypochlorous acid (HOCl): Cl2(aq)+H2O HOCl + H+ + ClHypochlorous acid is a weak acid that dissociates to form hypochlorite ion (OCl-). HOCl

OCl- + H+

Ka = 3.2 x 10 -8

All forms of chlorine are measured as mg/L of Cl 2 (MW = 2 x 35.45 = 70.9 g/mol) Hypochlorous acid and hypochlorite ion compose what is called the free chlorine residual. These free chlorine compounds can react with many organic and inorganic compounds to form chlorinated compounds. If the products of these reactions posses oxidizing potential, they are considered the combined chlorine residual. A common compound in drinking water systems that reacts with chlorine to form combined residual is ammonia. Reactions between ammonia and chlorine form chloramines, which is mainly monochloramine (NH2Cl), although some dichloramine (NHCl 2) and trichloramine (NCl3) also can form. Many drinking water utilities use monochloramine as a disinfectant. If excess free chlorine exits once all ammonia nitrogen has been converted to monochloramine, chloramine species are oxidized through what is termed the breakpoint reactions. The overall reactions of free chlorine and nitrogen can be represented by two simplified reactions as follows: Monochloramine Formation Reaction. This reaction occurs rapidly when ammonia nitrogen is combined with free chlorine up to a molar ratio of 1:1. HOCl +NH3

NH2Cl + HOCl

Breakpoint Reaction: When excess free chlorine is added beyond the 1:1 initial molar ratio, monochloramine is removed as follows: 2NH2Cl + HOCl

N2(g)+ 3H++ 3Cl-+ H2O

The formation of chloramines and the breakpoint reaction create a unique relationship between chlorine dose and the amount and form of chlorine as illustrated below.

Free Chlorine, Chloramine, and Ammonia Nitrogen Reactions Chlorine Demand Free chlorine and chloramines readily react with a variety compounds, including organic substances, and inorganic substances like iron and manganese. The stoichiometry of chlorine reactions with organics can be represented as shown below: HOCl:

1/10C5H7O2 N + HOCl Cl- + 1/10H2O

4/10CO2 + 1/10HCO3- + 1/10NH4++ H+ +

OCl-:

1/10C5H7O2 N + OCl1/10H2O

4/10CO2 + 1/10HCO3- + 1/10NH4++ Cl- +

NH2Cl:

1/10C5H7O2 N + NH2Cl + 9/10H2O 11/10NH4++ Cl-

4/10CO2 + 1/10HCO3- +

Chlorine demand can be increased by oxidation reactions with inorganics, such as reduced iron at corrosion sites at the pipe wall. Possible reactions with all forms of chlorine and iron are as follows:

Module 4: Municipal Water Treatment Plant Design Details Lecture 13: Treatment Plant Siting and Hydraulics Treatment Plant Layout and Siting Plant layout is the arrangement of designed treatment units on the selected site. Siting is the selection of site for treatment plant based on features as character, topography, and shoreline. Site development should take the advantage of the existing site topography. The following principles are important to consider: 1. A site on a side-hill can facilitate gravity flow that will reduce pumping requirements and locate normal sequence of units without excessive excavation or fill. 2. When landscaping is utilized it should reflect the character of  the surrounding area. Site development should alter existing naturally stabilized site contours and drainage as little as possible. 3. The developed site should be compatible with the existing land uses and the comprehensive development plan. Treatment Plant Hydraulics Hydraulic profile is the graphical representation of the hydraulic grade line through the treatment plant. The head loss computations are started in the direction of flow using water surface in the influent of first treatment unit as the reference level. The total  available head at the treatment plant is the difference in water surface elevations in the influent of first treatment unit and that in the effluent of last treatment unit. If the total available head is less than the head loss through the plant, flow by gravity cannot be achieved. In such cases pumping is needed to raise the head so that flow by gravity can occur.

There are many basic principles that must be considered when preparing the hydraulic profile through the plant. Some are listed below: 1. The hydraulic profiles are prepared at peak and average design flows and at minimum initial flow. 2. The hydraulic profile is generally prepared for all main paths of flow through the plant. 3. The head loss through the treatment plant is the sum of head losses in the treatment units and the connecting piping and appurtenances. 4. The head losses through the treatment unit include the following: a. Head losses at the influent structure. b. Head losses at the effluent structure. c. Head losses through the unit. d. Miscellaneous and free fall surface allowance. 5. The total loss through the connecting pipings, channels and appurtenances is the sum of following: a. Head loss due to entrance. b. Head loss due to exit. c. Head loss due to contraction and enlargement. d. Head loss due to friction. e. Head loss due to bends, fittings, gates, valves, and meters. f. Head required over weir and other hydraulic controls. g. Free-fall surface allowance.

Module 5: Water Storage Tanks and Distribution Network Lecture 14: Water Storage Tanks and Water Supply Network Water Distribution Systems

The purpose of distribution system is to deliver water to consumer with appropriate quality, quantity and pressure. Distribution system is used to describe collectively the facilities used to supply water from its source to the point of usage.

Requirements of Good Distribution System 1. Water quality should not get deteriorated in the distribution pipes. 2. It should be capable of supplying water at all the intended places with sufficient pressure head. 3. It should be capable of supplying the requisite amount of  water during firefighting. 4. The layout should be such that no consumer would be without water supply, during the repair of any section of the system. 5. All the distribution pipes should be preferably laid one metre away or above the sewer lines. 6. It should be fairly water-tight as to keep losses due to leakage to the minimum. Layouts of Distribution Network The distribution pipes are generally laid below the road pavements, and as such their layouts generally follow the layouts of roads. There are, in general, four different types of pipe networks; any one of which either singly or in combinations, can be used for a particular place. They are: Dead End System It is suitable for old towns and cities having no definite pattern of  roads.

Advantages: 1. Relatively cheap. 2. Determination of discharges and pressure easier due to less number of valves. Disadvantages 1. Due to many dead ends, stagnation of water occurs in pipes.

Grid Iron System: It is suitable for cities with rectangular layout, where the water mains and branches are laid in rectangles.

Advantages: 1. Water is kept in good circulation due to the absence of dead ends. 2. In the cases of a breakdown in some section, water is available from some other direction. Disadvantages 1. Exact calculation of sizes of pipes is not possible due to provision of valves on all branches.

Ring System: The supply main is laid all along the peripheral roads and sub mains branch out from the mains. Thus, this system also follows the grid iron system with the flow pattern similar in character to that of dead end system. So, determination of the size of pipes is easy.

Advantages: 1. Water can be supplied to any point from at least two directions.

Radial System: The area is divided into different zones. The water is pumped into the distribution reservoir kept in the middle of each zone and the supply pipes are laid radially ending towards the periphery.

Advantages: 1. It gives quick service.

2. Calculation of pipe sizes is easy.

Distribution Reservoirs Distribution reservoirs, also called service reservoirs, are the storage reservoirs, which store the treated water for supplying water during emergencies (such as during fires, repairs, etc.) and also to help in absorbing the hourly fluctuations in the normal water demand. Functions of Distribution Reservoirs: • •

to absorb the hourly variations in demand. to maintain constant pressure in the distribution mains.

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water stored can be supplied during emergencies.

Location and Height of Distribution Reservoirs: • •

should be located as close as possible to the center of  demand. water level in the reservoir must be at a sufficient elevation to permit gravity flow at an adequate pressure.

Types of Reservoirs 1. Underground reservoirs. 2. Small ground level reservoirs. 3. Large ground level reservoirs. 4. Overhead tanks. Storage Capacity of Distribution Reservoirs The total storage capacity of a distribution reservoir is the summation of: 1. Balancing Storage: The quantity of water required to be stored in the reservoir for equalizing or balancing fluctuating demand against constant supply is known as the balancing storage (or equalizing or operating storage). The balance storage can be worked out by mass curve method .

2. Breakdown Storage: The breakdown storage or often called emergency storage is the storage preserved in order to tide over the emergencies posed by the failure of pumps, electricity, or any other mechanism driving the pumps. A value of about 25% of the total storage capacity of reservoirs, or 1.5 to 2 times of the average hourly supply, may be considered as enough provision for accounting this storage. 3. Fire Storage: The third component of the total reservoir storage is the fire storage. This provision takes care of the requirements of water for extinguishing fires. A provision of 1 to 4 per person per day is sufficient to meet the requirement. The total reservoir storage can finally be worked out by adding all the three storages.

Lecture 15: Water Supply Network Design Pipe Network Analysis Analysis of water distribution system includes determining quantities of flow and head losses in the various pipe lines, and resulting residual pressures. In any pipe network, the following two conditions must be satisfied: 1. The algebraic sum of pressure drops around a closed loop must be zero, i.e. there can be no discontinuity in pressure. 2. The flow entering a junction must be equal to the flow leaving that junction; i.e. the law of continuity must be satisfied. Based on these two basic principles, the pipe networks are generally solved by the methods of successive approximation. The widely used method of pipe network analysis is the Hardy-Cross method. Hardy-Cross Method This method consists of assuming a distribution of flow in the network in such a way that the principle of continuity is satisfied at each junction. A correction to these assumed flows is then computed successively for each pipe loop in the network, until the correction is reduced to an acceptable magnitude. If Qa is the assumed flow and Q is the actual flow in the pipe, then the correction δ is given by

δ=Q-Qa; or Q=Qa+δ Now, expressing the head loss (H L) as

HL=K.Qx we have, the head loss in a pipe =K.(Qa+δ)x =K.[Qax + x.Qax-1δ + .........negligible terms] =K.[Qax + x.Qax-1δ] Now, around a closed loop, the summation of head losses must be zero.

ΣK.[Qax + x.Qax-1δ] = 0 or ΣK.Qax = -ΣKx Qax-1δ Since, δ is the same for all the pipes of the considered loop, it can be taken out of the summation.

∴ΣK.Qax = -δ . ΣKx Qax-1 x Or δ=-ΣK.Qa / Σx.KQax-1

Since δ is given the same sign (direction) in all pipes of the loop, the denominator of the above equation is taken as the absolute sum of the individual items in the summation. Hence, x Or δ =-ΣK.Qa / Σ l x.KQax-1 l

Or δ =-ΣHL / x.Σ lH L/Qal where HL is the head loss for assumed flow Q a. The numerator in the above equation is the algebraic sum of the head losses in the various pipes of the closed loop computed with assumed flow. Since the direction and magnitude of flow in these pipes is already assumed, their respective head losses with due regard to sign can be easily calculated after assuming their diameters. The absolute sum of respective KQ ax-1 or H L/Qa is then calculated. Finally the value of  δ is found out for each loop, and the assumed flows are corrected. Repeated adjustments are made until the desired accuracy is obtained. The value of x in Hardy- Cross method is assumed to be constant (i.e. 1.85 for Hazen-William's formula, and 2 for Darcy-Weisbach formula)

Module 6: Rural Water Supply Lecture 16: Water Treatment and Supply for Rural Areas

Quizzes, examinations and Tutorials Coagulation and Flocculation-I SOLUTION

Design a conventional vertical-shaft rapid mix tank unit for uniformly dispersing coagulant in 10 MLD of settled raw water as per design  parameters given below: Detention time (t): Ratio of tank height (H) to diameter (D): Ratio of impeller diameter (D I) to tank diameter (D): Velocity gradient (G): Gt: Tank diameter (D): Paddle tip speed (v p): Velocity of paddle relative to water (v): Paddle area (A p)/Tank section area (A T): Coefficient of drag on impeller blade (C D): Maximum length of each impeller blade (L): diameter  Maximum width of impeller blade (B): Impeller height from bottom (H B): Kinematic viscosity

20 – 60 s (1:1 to 1:3) (0.2:1 to 0.4:1) >300 /s 10000 – 20000 <3m 1.75 – 2.0 m/s 0.75 x paddle tip speed 10:100 – 20:100 1.8 0.25 x impeller  0.20 x impeller diameter  1.0 x impeller diameter  1.003 x 10 -6 m2/s

:

Dynamic viscosity of water

:

1.002 x 10 -3 N.s/m2

Determine tank dimensions (provide a freeboard of 0.5 m), impeller  diameter, paddle dimensions, number of paddles, clearance of the impeller  from tank bottom, paddle rotation speed and power input requirement. Solution:

Let the detention time (t) be 40 s. Therefore, volume of tank (V) = Let the tank diameter (D) be 2 m

Tank cross-sectional area (A cs) = 3.14 m2 Tank height (H) = 1.47 m, provide freeboard of 0.5 m

Total height (HT)= 1.97 m, say 2 m.

Tank height (H) to tank diameter (D) ratio = (within the range of 0.33–1.0, hence okay) -1 Let velocity gradient, G = 400 s Therefore, G.t = (400).(40) = 16000 (within the limit of 10000-20000, hence okay) Let the paddle tip speed (v p) be 1.8 m/s Hence, velocity of paddle relative to water (v) = 0.75.(1.8) = 1.35 m/s

,

or,

or, A p = 0.335 m2 Tank sectional area = (D).(H) = (1.47).(2) = 2.94 m 2 , which is around 0.15, hence okay. Let the impeller diameter be 0.8 m , i.e.,

, which is okay

Choose length of each impeller blade (L) as 0.20m , i.e., (okay) Choose breadth of each impeller blade (B)

as

0.15m ,

i.e., (okay) Area of each blade = (L.B) = (0.2).(0.15) = 0.03 m 2 Therefore, number of blades to be provided = Clearance of the paddles from the tank bottom = 0.8 m Paddle rotation speed (w, radians/s) = i.e.,

Power

, say 12

radians/s,

revolutions per minutes

requirement

is

given

by,

,

or, or, P = 742 Watts, i.e., provide 1 KW motor for driving the impeller at 45 rpm.

Coagulation and Flocculation-II SOLUTION

Design a conventional rectangular horizontal-shaft flocculation tank unit for  10 MLD of settled raw water after coagulant addition and rapid mixing as  per design parameters given below: Detention time (t): Velocity gradient (G): Gt: Tank Depth (D): Paddle tip speed (v p): Velocity of paddle relative to water (v): Paddle area (A p)/Tank section area (A T): Coefficient of drag on paddle blade (C D): Maximum length of each paddle (l): Maximum width of each paddle (b): Kinematic viscosity : Dynamic viscosity of water

:

Freeboard:

10 – 30 minutes 20 – 75 /s 2 x 104 – 6 x 104 <5m 0.25 – 0.75 m/s 0.75 x paddle tip speed 10:100 to 20:100 1.8 5.0 m 0.50 m 1.003 x 10 -6 m2/s 1.002 x 10 -3 N.s/m2

0.50 m

Draw a net sketch of the designed tank (top and front view) clearly showing tank dimensions, paddle shaft position, paddle blade dimensions, water  level, etc. Also mention paddle rotation speed and power requirement. Solution:

Let the detention time (t) be 25 minutes Let the velocity gradient (G) be 30 /s Therefore G.t = 25.(60).(50) = 45000, i.e., within the 20000 – 60000 limit for Gt. Volume of the tank (V) = , say 175 m3. Let the tank depth (D) = 5 m, freeboard = 0.50 m, Total tank depth = 5.5 m Tank cross sectional area (A cs) = Let the tank length (L) be 7 m

Therefore, tank width (B) = Let the paddle be placed length-wise. Let the diameter of the paddle (D p) be 4 m. Let the paddle tip speed (v p) be 0.40 m/s Then the velocity of paddle relative to water (v) = 0.75.(0.40) = 0.30 m/s , or,

or, A p = 6.50 m2 Tank sectional area = (L.D) = 5.(7) = 35 m 2 , which is between 0.10 and 0.20,

hence okay. Let five paddles be provided. Therefore, area of each paddle = Let the length of each paddle (l) be 5 m Therefore, breadth of each paddle (b) = Paddle rotation speed (w, radians/s) =

i.e.,

radians/s,

revolutions per minutes

Power requirement is given by,

, or,

or, P = 157 Watts, i.e., provide 0.2 KW motor for driving the impeller at 2 rpm. EXAMINATION-II SOLUTION The bonus question carries 5 marks, over and above the 30 marks for the regular  questions. Marks obtained by the student in the bonus question will be added to the total marks obtained in other questions.

1. Give brief answers to the following questions with appropriate reasoning: A. Why is recycling necessary in activated sludge  process (2) Solution:

In activated sludge process we try to increase the substrate utilization rate without increasing the specific substrate utilization rate, i.e., in the equation, ,

we

want

to

increase

by lowering

. However, q cannot be increased, because, , and thus increasing q will increase S, which is undesirable. Hence X has to be increased. This is achieved by collecting the biomass leaving the aeration tank and re-circulating it back to the aeration tank. B. What are the differences between microbial physiology in high rate and an extended aeration activated sludge  processes. (2) Solution:

Assuming Q, So and Va (hence ) to be the same in both cases, In high rate aeration, is low, hence µ and q are high. Since q is high, S is high. Since µ is high ∆X is high. X in the aeration tank is comparatively low. Oxygen requirement is low, as rate of biomass  production is high. Nutrient requirement is high. In extended aeration, is high, hence µ and q are low. Since q is low, S is low. Since µ is low ∆X is low. X in the aeration tank is comparatively high. Oxygen requirement is high, as rate of biomass  production is low. Nutrient requirement is low. C. Why a pure oxygen aeration system is often used for activated sludge treatment of high strength waste (2) D. Why is there an upper and lower limit for detention time in a secondary settling tank used in the activated sludge process. (2) E. Raw wastewater influent to the aeration tank of activated sludge  processes has inorganic suspended solids of small size (<20 µm), due

to incomplete particle removal in the primary settling tank. This material is completely inert, and is generally completely removed from wastewater during secondary settling following aeration, and ultimately expelled from the system during sludge wasting. However, sludge recycling results in maintenance of a certain steady-state inorganic suspended solid concentration (C) in the aeration tank, which is much higher than the influent inorganic suspended solid concentration (C o). Based on the above information, derive the expression for steady-state inorganic suspended solids concentration (C) in the aeration tank of an ASP, and hence calculate C, given Co = 20 mg/L, θ = 4 hours, and θc = 6 days. (2) Solution:

Mass Balance for the whole system: Mass Output,

Rate of mass Input = Rate of  i.e.,

Also, Mass Balance for the secondary sedimentation tank, Rate mass input =

= Rate mass output =

or, or,

,

or, 2. 1 MLD of 400mg/L Glucose solution (C 6H12O6) is to be treated by a completely-mixed activated sludge process (ASP). Hydraulic detention time in the aeration tank is 4 hours. The effluent from the process should have 20mg/L glucose. Biological solids retention time (BSRT) of the ASP is 6 days. Based on this information, calculate the oxygen requirement in the aeration tank in Kg/d. Assume all necessary nutrients that are required for bio-mass growth are present in excess. Also assume that BODu = 1.5 x BOD5. YT and K d values for microbial degradation of  glucose are 0.5 mg/mg and 0.05 /d respectively (calculated based on BOD5 values). (10) Solution:

Thus, 180 g glucose requires 192 g oxygen for complete oxidation.

Therefore,

BODu for

400

solution is

,

BOD5 = Also,

,

and

glucose corresponding

i.e., S o = 284.44 mg/L

BODu for 200 mg/L glucose solution is

and corresponding BOD5 = Given,

mg/L

,

,

, i.e., S = 14.22 mg/L

therefore,

,

and Also,

,

So,

Also, Therefore, Oxygen Requirement (Kg/d) =

3. The aeration tank volume in an activated sludge plant was calculated to  be 5184 m3, to be provided in four parallel units. The depth of the tanks will be 3 m. Aerators are available with power of 1, 2, 5, 10, 25, 50 KW and their areas of influence are 25 m 2 for 1, 2, and 5 KW, and 36 m 2 for  10, 25 and 50 KW aerators respectively. The expected maximum oxygen requirement for the process was calculated to be 13700 Kg/d. Manufacturers specify that the oxygen transfer capacity of these aerators is 2.0 Kg O 2/kW-h under standard conditions. Based on this information, design an adequate aerator arrangement for each aeration tank. The operating temperature of the aeration tank is expected to be 30oC. The steady state dissolved oxygen concentration in the aeration tank should be 1 mg/L. Saturation concentration of oxygen in water at 20oC is 9.1 mg/L and at 30 oC is 7.5 mg/L. (10) , ,

where, Solution:

Volume of each tank = Surface area of each tank = Provide 36 m length and 12 m width for each tank  If a 1 kW aerator is used under standard conditions: Standard Where V = 25m2 or 36m2 c/s area x 3 m depth

Under actual conditions,

or,

= 1.232 kg/hr  Oxygen Requirement per tank = Power Requirement per tank = Using 12 aerators per tank, 2 breadth-wise and 6 length-wise, power  requirement per aerator = . Hence provide 10 KW aerators @12 per tank to take care of oxygen requirement. Bonus Question:

4. An activated sludge plant is designed for treating a flow (Q) and influent BOD5 concentration (S o) such that the effluent BOD 5 concentration is S. Due to some unforeseen circumstances, the value of So increases to

2.So, while Q remains the same. Explain what changes must be made to the treatment plant such that the effluent quality does not degrade. Building new tanks is not a viable option. (5)

EXAMINATION-I

Time: 180 Minutes

Full Marks: 90 (+10 bonus)

Question 1

Sketch the top and front view of a rectangular primary sedimentation tank, clearly showing the inlet and outlet arrangements and sludge and scum collection mechanisms. Label various relevant features. Also draw expanded and/or sectional views of the inlet and outlet regions of the tank to show all details and label all relevant features. Solution:  No solution provided.

Question 2

a.

To predict the performance of an activated sludge unit, one must have data on microorganism properties. To obtain this data, a laboratory experiment is conducted using a continuous flow completely mixed reactor without recycle. Data obtained after  two runs are as follows: Run No. So, mg/L S, mg/L X, mg/L (influent BOD5) (Effluent BOD5) (Biomass) , days 1. 250 3 10 120 2. 250 1 45 110 Based on the above data calculate qm, K d, K s, and

YT.

(8)

 b. Also determine the minimum value of for this reactor, below which no substrate removal shall occur. Explain the physical significance of this minimum value. (2) Solution:

We know, Hence,

, and

We also know,

,

Hence,

,

also,

or, or,

Solving,

;

Also,

,

Hence,

Also, Hence,

Solving, When there is no substrate removal in the reactor, Hence, Also,

Hence,  Physical Significance:

If for this reactor is less than , then no substrate removal will occur because  biomass will flow out of the reactor at a faster rate than it can be produced. Under such conditions, the biomass concentration in the reactor at steady state will be zero. Question 3

Draw a neat sketch of a trickling filter (plan and sectional elevation) and label all the essential  parts. Write an essay on trickling filter operation, covering the following points, mode of   biomass growth and removal from the reactor, mode of substrate utilization and oxygen uptake and importance of OLR, HLR, and recycling. (3 + 7) Solution:

 No solution provided.

Question 4

Draw a neat sketch (sectional elevation) of an Up-flow Anaerobic Sludge Blanket (UASB) reactor and explain its various parts and their operation. Explain why and how this reactor  configuration is the more successful in treating domestic wastewater as compared to other  anaerobic reactor configurations. (5 + 5) Solution:

 No solution provided.

Question 5

1 MLD of wastewater with influent soluble BOD 5 (So) = 75 mg/L, TKN = 13 mg/L (as N) and 2 mg/L phosphorus (as P) is to be treated in an oxidation pond such that effluent soluble BOD5 (S) is 5 mg/L. Calculate the oxidation pond surface area, assuming the depth of the  pond to be 0.5 m. Calculate oxygen requirement for microbial respiration and oxygen supplied by algal growth, and check adequacy of the design by assuming that 50 percent of  the oxygen produced by algal growth is available for microbial respiration. Neglect oxygen input into the pond by mass transfer from atmosphere. Calculate the effluent algal, effluent nitrogen and phosphorus concentration from the pond, and also calculate the total effluent BODU from the pond. Data:

•

• • • • • •

K = 0.1 L/mg/d, where K is the first order microbial substrate utilization rate ( YT = 0.5 mg/mg; K d = 0.05 /d (based on BOD 5) Formula for microbial biomass: C60H87O23 N12P Average intensity of solar radiation: 150 calories/cm2/d Solar energy utilization efficiency for algae: 6 percent Energy content of algal bio-mass: 6000 calories/g algae Equation for algal photosynthesis:

Solution:

)

We know,

;

Therefore, Also, Also, Sludge Production (

,

or,

) = Q.X =

Oxygen Requirement = 1.5Q.(S o - S) – 1.42( ∆X) = 1.5.(70) – (1.42).(28) = 65.24 Kg/d

Volume of Oxidation Pond = Assuming depth to be 0.5 m,

Surface Area (A) =

Algae production = Total algal production = (15).(10000) = 150 kg/d Assuming 1.3 Kg oxygen production per Kg al gal production, Oxygen Production = (1.3).(150) = 195 Kg/d Oxygen available for microbial respiration = (0.5).(195) = 97.5 Kg/d Since oxygen available is more that oxygen requirement for microbial respiration, the design is adequate. Biomass production = 28 Kg/d Formula weight of biomass = 60.(12) + 87.(1) + 23.(16) + 12.(14) + 1.(31) = 1374 Hence, 1374 Kg of biomass contains 168 Kg of nitrogen and 31 Kg of phosphorus 28 Kg of biomass contains kg nitrogen and  phosphorus Algae production = 150 Kg/d Formula weight of algae = 106.(12) + 263.(1) + 110.(16) + 16.(14) + 31 = 3550 Hence, 3550 Kg of algae contains 224 Kg of nitrogen and 31 Kg of phosphorus

kg of 

150 Kg of algae contains kg nitrogen and  phosphorus Therefore, total nitrogen uptake = 3.42 + 9.46 = 12.88 Kg/d Total phosphorus uptake = 0.63 + 1.31 = 1.95 Kg/d

kg of 

Influent nitrogen loading = Influent phosphorus loading =

Kg/d Kg/d

Hence effluent dissolved nitrogen concentration = 13-12.88 = 0.12 mg/L (as N) Effluent dissolved phosphorus concentration = 2 – 1.95 = 0.05 mg/L (as P) Effluent biomass concentration = 28 mg/L, corresponding BOD u = 1.42.(28) = 39.76 mg/L Also, 3550 Kg of algae has BOD u of 4416 Kg Effluent algae concentration = 150 mg/L, corresponding BOD u = mg/L Effluent soluble BOD u = 1.5.(5) = 7.5 mg/L Therefore, total effluent BOD u = 39.76 + 186.59 + 7.5 = 233.85 mg/L Question 6

0.25 MLD of sludge is generated from the activated sludge process of a wastewater treatment  plant. The solids concentration in this sludge is 13,000 mg/L. This sludge is first processed in a sludge thickener, where the solids content is increased to 4 percent (weight basis). Next the sludge is treated in the anaerobic digester for reduction in sludge solids. The digested sludge is then applied to sludge-drying beds for reduction of water content to 55%. Determine the weight and density of dried sludge that will be produced per day. Density of Water = 1000 Kg/m 3; Dry Density of Sludge Solids = 2200 Kg/m 3 COD reduction efficiency of sludge digester = 60 %; ; For sludge digestion, Y T = 0.06 mg/mg, and K d = 0.03 /d on COD basis  Assumptions:

• • •

The sludge thickener is 100 percent efficient. Entire effluent from the anaerobic digester is applied to sludge drying beds. Dried sludge is completely saturated with water.

Solution:

Consider 1 m3 of sludge after thickening Let the sludge density be Weight of this sludge = Weight of solids in the sludge =

; Volume =

Weight of liquid in the sludge =

; Volume =

Therefore, Total Volume = [ Given: ρl = 1000 kg/m3;

+ ρs = 2200 kg/m 3

]=1

So, Therefore,

Weight of solids in 1 m3 of sludge = (0.04).(1).(1022.305) =40.89 kg. Solids Concentration (X o) = 40890 mg/L

Let the discharge of thickened sludge be Q sl Therefore, assuming 100 percent solids capture,

Qsl. (40890) = Q w. (13000)

or, Qsl =

Solids Loading ( ∆X) = Digester Design:

;

Digester Volume (Vd) = ;

Assuming 60% treatment efficiency, Also,

,

Also,

Therefore, Anaerobic

Sludge

Production

(

= Therefore,

Reduction in Solids =

Therefore,

Solids Loading to Sludge Drying Beds = (3250-1891) = 1359 kg/d

After Drying:

Therefore,

Dried Sludge Production Rate =

Density of Dried Sludge = Question 7

A food-processing unit produces 0.1 MLD of wastewater with the following characteristics. COD: 10,000 mg/L; TKN: 500 mg/L; Total-P: 100 mg/L; pH: 7.5; Alkalinity: 500 mg/L as CaCO 3. Discharge standards for wastewater are, COD: <50 mg/L; TKN: < 2 mg/L; Total-P: < 1 mg/L; pH: 6.5–8.5. Based on the above information, provide a  preliminary sketch of appropriate treatment process train, including those for residuals management. Explain the function of each unit in the process train, and also mention the  probable percentage of pollutant reduction in each unit. Use of a mixture of engineered and natural methods for treatment is encouraged. Solution: This is one of many possible treatment trains.

Question 8

Consider settled raw water of three types; • Turbidity 10 NTU, alkalinity: 100 mg/L (as CaCO 3) • Turbidity 50 NTU, alkalinity 100 mg/L (as CaCO 3) • Turbidity 150 NTU, alkalinity 10 mg/L (as CaCO 3) Based on you understanding of water treatment unit operations suggest the best treatment option (in terms of cost) for all three cases to reduce the turbidity to < 2.5 NTU. Explain clearly why the selected option in each case is the best vis-à-vis other available options. Solution: Case A:

Mixing Case B:

Mixing

Recommended treatment train is as follows.

Coagulant/Poly-electrolyte Rapid Sand Filtration

addition

Rapid

Recommended treatment train is as follows

Coagulant/Ploy-electrolyte Flocculation

addition

Rapid

Rapid Sand (Dual Media) Filtration Case C:

Mixing

Recommended treatment train is as follows.

Coagulant Flocculation

and

Soda

addition

Rapid

Secondary Sedimentation

Rapid Sand Filtration In all above cases, excellent effluent quality might be obtained by the following treatment train: Coagulant (High Dose) and Soda Addition

Rapid Mixing

Flocculation to Encourage sweep floc formation

Secondary Sedimentation

Rapid Sand Filtration However, such a treatment chain is not recommended because of the large chemical cost (coagulant and soda), and the cost of disposing the large quantity of sludge produced.

Case A.

Flocculation will be useless in this case due to the low number of particles  present, unless sweep floc formation is desired (which is not the case). Consequently, secondary sedimentation will also be redundant. Under  the circumstances, direct filtration is the most logical option.

Case B.

Flocculation will be moderately effective in this case, but flocs large enough to settle efficiently may not be formed, unless sweep-floc formation is desired (which is not the case). Consequently, secondary sedimentation will be redundant. Flocculation will however reduce particle loading on the filter, thus extending filter runs. Dual media filter is proposed since the influent  particle size is higher in this case as compared to Case A.

Case C.

Flocculation will be very effective in this case due to large influent particle concentration. Consequently, secondary sedimentation will also be effective. However on coagulant addition, pH is likely to decline in this case to due the low alkalinity of water. Hence the conventional treatment chain, with soda addition along with coagulant to maintain pH drop, is thought to be the most logical option in this case.

Question 9

10 MLD of water after secondary sedimentation (average turbidity: 10 NTU) is to be filtered through a battery of rapid sand filters to reduce water turbidity to < 2.5 NTU. Based on pilot  plant studies, it was determined that 60 cm filter beds of sand (0.5 mm average sand diameter) were suitable for this purpose. It was further determined that such beds could be operated for  7.5 hours at a filtration rate of 10 m 3/m2/hr before the terminal head-loss of 3 m was reached. Filter backwashing rate was 1 m3/m2/min and the backwash time was 5 minutes. A filter unit will be off-line for 30 minutes during each backwash operation. Based on this information, determine the numbers of filter units to be provided and dimensions of each unit. Determine how much filtered water is required for backwashing each day and hence determine the filtered water production per day. Solution:

 Nominal filtration rate: 10 m3/m2/h Filter is off-line for 1.5 hours every 24 hours Therefore, effective filtration rate =

m3/m2/h

Therefore, required filter cross-sectional area = m2 Let two filters be provided for this purpose Let the length of each filter be 5.5 m and width 4.1 m (length: width = 1.34) Therefore total filter cross-section area provided = 2.(5.5).(4.1) = 45.1 m 2 Corrected actual filtration rate = m3/m2/h Filtered water required for backwashing = 1.(5).(45.1) .3 = 676.5 m 3/d Hence total filtered water production = 10 – 0.676 = 9.324 MLD Question 10

Consider water from a polluted river having BOD 5 = 5 mg/L, TKN = 1 mg/l (as N), and MPN: 106 organisms / mL. This water will be treated in a conventional water treatment plant and supplied for potable purposes. Compute the chlorine dose (in mg/L as Cl 2) required per  liter of this water (consider both pre and post-chlorination) such that after treatment BOD 5, TKN, NH3-N are negligible and MPN < 1organism/mL.  Assumptions:

• • • • •

Assume that 1 mg/L (as Cl 2) chlorine is required to destroy 1 mg/L of BOD 5. Assume TKN is completely converted to NH 3-N during pre-chlorination. The average time between post-chlorination and water consumption by the end users is 1 hour  The product of disinfectant dose (C in mg/L) and the contact time (t in minutes) for 5 and 6 log-kills using free residual chlorine as disinfectant is 96 and 120 respectively. Assume 2 log-kill of microorganism during water treatment up to just before the postchlorination step.

Solution:

Chlorine dose required during pre-chlorination for destruction of BOD 5 = 5 mg/L as Cl 2 All TKN in water is converted to NH 3-N during this process.

Hence ammonia concentration in water before post-chlorination = 1 mg/L (as N) Breakpoint chlorination has to be performed to destroy ammonia in water. Relevant equation: Ammonia concentration in water = mmoles/L Chlorine required for destruction of ammonia = 1.5.(0.0714 ) = 0.1071 mmoles/L, Therefore, breakpoint chlorination dose = 71.(0.1071) = 7.6 mg/L Initial microorganism concentration = 10 6 /mL Removal during water treatment up to post-chlorination = 2 Log Hence microorganism concentration just before post chlorination = 10 4 /mL To get this concentration below 1 /mL, 5 log kills are required “C.t” for 5 log kills = 96 Contact time = 1 hour = 60 minutes Therefore required free chlorine residual dose = mg/L as Cl2 Therefore, total chlorine dose required = 5 + 7.6 + 1.6 = 14.2 mg/L as Cl 2

EXAMINATION-III SOLUTION

The total number of marks allotted for various questions add up to 35. However, the marks obtained will be reported on 30.

1. A. Describe the difference between the recycling in the activated sludge and trickling filter    processes. ( 2) Solution:

The settled sludge is recycled in ASP, while the treated effluent is recycled in TF. The objective of recycling in ASP is to maintain the desired biomass concentration in the aeration tank, while in TF, recycling allows the HLR to be varied independent of the OLR, thus allowing uniform wetting of the filter media. B. Explain why anaerobic reactors may fail if organic loading rate (OLR) is increased suddenly. What are the early warnings of reactor  failure and how can such failure be  prevented. (2) Solution:

If OLR in an anaerobic reactor is increased suddenly, the acid formers will become more active, and VFA concentration in the reactor will increase. If the buffering capacity in the reactor is insufficient, all alkalinity in the reactor will be exhausted due to the increased acid formation, and reactor pH will go down. At reactor pH<7, the methane formers will become inactive, and methane production will thus stop. This will stop COD reduction in the reactor. Reactor  failure in such circumstances may be avoided by adding a buffering agent like soda (Na 2CO3) to the reactor to stop decline in pH. C. Calculate the minimum effluent COD that can be expected from an anaerobic reactor at 45 oC and 20oC, assuming the microbial kinetics of anaerobic reactors may be described using the following relationships, YT = 0.040; K d = 0.015 /d; q = 6.67 x 10 -0.015.(35-T) /d; K s = 2224 x 100.046.(35-T) mg/L. ‘T’ is the temperature expressed in degree Celsius. (2) Solution:

qm (20oC) = K s(20oC) =

3.97 /d 10892 mg/L

Putting µ = 0,

or,

, S = 1136 mg/L

or, (0.375).(10892) = (3.97- 0.375).S

qm(45oC) =

9.42 /d

K s(45oC) =

771.14 mg/L

Putting µ = 0,

or,

, S = 32 mg/L

or, (0.375).(771) = (9.42- 0.375).S

D. Explain why a suspended growth anaerobic reactor gives very poor  COD removal when used for treatment of domestic wastewater. (2) Solution:

Anaerobic reactors must be maintained at very low specific growth rates for obtaining low effluent COD values. This means that very high biomass concentration must be maintained in the reactor to achieve reasonable rates of substrate utilization. In suspended growth systems, maintenance of high biomass concentration is only possible if the biomass escaping from the reactor is settled and recycled back  efficiently. However this is difficult in anaerobic reactors because anaerobic sludge has poor settling characteristics. Also, biomass escaping with the treated effluent degrades the effluent quality. E. In a single stage aeration tank designed for carbon oxidation and nitrification, the dissolved oxygen concentration has to be maintained at 2.5-3.0 mg/L, as opposed to value of 1.0 mg/L in cases where only carbon oxidation is required. Why? (2) Solution:

In single stage nitrification systems typically only 1-2 percent of the  biomass consists of nitrifying organisms. Often such organisms are enmeshed in clusters of heterotrophic organisms responsible for  carbon oxidation, and thus have to compete for aqueous oxygen. Under the circumstances, high DO concentration must be maintained in such reactors to ensure that aqueous oxygen diffuses into the inner parts for biomass clusters where many nitrifying microorganisms may be found. 2. Consider a wastewater flow of 10 MLD with a BOD 5 value of 600 mg/L. It is proposed to treat this waste using an activated sludge process (ASP). The size of the proposed aeration tank will such that the water is detained for 8 hours. Complete mixing is assumed in the aeration tank. The BSRT (

) of the system is 10 days. In addition to the BOD5,

this water also contains nitrogen and phosphorus. The nitrogen concentration expressed as Total Kjeldahl Nitrogen (TKN) is 40 mg/L (as N), and phosphorus concentration is not limiting. Assuming that nitrification occurs in this system, calculate the effluent TKN and concentration (as N), and the total oxygen requirement for both carbon oxidation and nitrification. Data:

K S = 40 mg/L; (K S) N = 2.0 mg/L;

qm = 4 /d; (qm) N = 2 /d

YT = 0.5; K d = 0.05 /d; (YT) N = 0.2 ; (K d) N = 0.05/d

Assumptions:

• Biomass may be represented as C 5H7O2 N •  Nitrogen incorporation in heterotrophic biomass must be accounted for. •  Nitrogen incorporation in autoprophic biomass may be neglected. (15)

Solution:

Carbon Oxidation Calculations:

;

; ;

Therefore,

S = 3.24 mg/L

Also,

Hence,

Also, Oxygen Requirement (in Kg/d) =

1.5.Q.(So – S) –1.42.(∆X) = 1.5.

Formula weight of biomass, C 5H7O2 N = 5(12)+7(1)+2(16)+1(14) = 113 Therefore, production of 1990 Kg/d biomass requires:

Influent Nitrogen concentration = 40 mg/L (as N), or  Therefore, Nitrogen available for nitrification = 400 – 246.42 = 153.57 Kg/d Or, [TKN]i = 15.36 mg/L (as N)

 Nitrification Calculations:

;

Also,

;

Therefore, Effluent TKN = Effluent Nitrate =

(15.36 - 1.2) =

1.2 mg/L (as N) 14.16 mg/L (as N)

Also, ;i.e.,

Additional Oxygen Requirement (Kg/d) =

4.57.Q.

{

}

i.e., = Total Oxygen Requirement = 6127 + 617 = 6744 Kg/d

3. 10 MLD of wastewater with influent COD (S o) of 800 mg/L is treated in an UASB reactor. In addition, the influent wastewater contains 300 mg/L of sulfate, and negligible amounts of sulfide. 50% conversion of this sulfate to sulfide, as  per the following equation, , was reported due to the action of sulfur reducing bacteria**. In addition, 70% COD removal was reported. Calculate theoretical methane production, i.e., volume of methane  produced per day at STP. Neglect COD conversion to anaerobic biomass. ** These are anaerobic bacteria using organic acids produced by acid producing bacteria under anaerobic conditions as food and energy

 source and sulfate as the terminal electron acceptor. Sulfate reducing  bacteria are present in anaerobic reactors if sulfate concentration is high. (10) Solution:

Influent COD = 800 mg/L, i.e., Influent COD (Liquid Phase) = Methane COD (Gas Phase) + Sulfide COD (Liquid Phase) + Effluent COD (Liquid Phase) + Biomass COD (Solid Phase) Biomass COD (Solid Phase) is neglected. Sulfide COD (Liquid Phase) + Effluent COD (Liquid Phase) = 0.30.(8000) = 2400 Kg/d Therefore Methane COD (Gas Phase) = 8000 – 2400 = 5600 Kg/d 64 Kg of Methane COD = 16 Kg of Methane Methane Production = Kg/d 3 16 Kg of methane at STP = 22.4 m

1400 Kg of methane at STP = Therefore Methane Production = 1960 m 3/d

EXAMINATION-IV The bonus question carries 5 marks, over and above the 30 marks for the regular questions. Marks obtained by the student in the bonus question will  be added to the total marks obtained in other questions.

1. Give brief answers to the following questions with appropriate reasoning:

A. Why

is

 process

recycling

necessary

in

activated

sludge

(2)

B. All other things remaining constant, if the BSRT in an activated sludge plant is increased, what will happen to the effluent substrate concentration (2) C. Physically explain why BSRT is the inverse of specific growth rate (µ) (2) D. What are the differences between microbial physiology in high rate

and an  processes.

extended

aeration

activated

sludge

(2)

E. Why a pure oxygen aeration system is often used for aerobic

treatment waste

of

high

strength (

2)

2. 1 MLD of 400mg/L Glucose solution (C 6H12O6) is to be treated by a

completely-mixed activated sludge process (ASP). Hydraulic detention time in the aeration tank is 4 hours. The effluent from the process should have 20mg/L glucose. Biological solids retention time (BSRT) of the ASP is 6 days. Based on this information, calculate the oxygen requirement in the aeration tank in Kg/d. Assume all necessary nutrients that are required for bio-mass growth are present in excess. Also assume that BODu = 1.5 x BOD5. YT and K d values for microbial degradation of  glucose are 0.5 mg/mg and 0.05 /d respectively (calculated based on BOD5 values). (10)

3. a) The volume of an aeration tank in an activated sludge plant is 4700

m3. The oxygen requirement for the process was calculated to be 5400 Kg/d. It was decided that 50 kW surface aerators should be used for  aeration purposes. Manufacturers of such aerators specify that the oxygen transfer capacity of these aerators is 2.0 Kg O 2/kW-h under  standard conditions. Based on this information, calculate the total power  requirement for satisfying oxygen requirements in the aeration tank, and

hence calculate the number of aerators required. The operating temperature of the aeration tank is expected to be 30 oC. The steady state dissolved oxygen concentration in the aeration tank should be 1 mg/L. Saturation concentration of oxygen in water at 20 oC is 9.1 mg/L and at 30oC is 7.5 mg/L. (7) If a 1 kW aerator is used in the above aeration tank under standard conditions: Hint:

b) If the steady state bio-mass concentration (X) in the aeration tank is

3000 mg/L, calculate the power requirement of keep the bio-mass in suspension, and hence the number of aerators required. Compare results with those of 3(a) and comment on aerator requirements for the aeration (3) tank.

Star Question:

4. Raw wastewater influent to the aeration tank of activated sludge  processes often has inorganic suspended solids. This material is completely inert, and is generally removed from wastewater during secondary settling following aeration, and ultimately expelled from the system during sludge wasting. However, sludge recycling results in maintenance of a certain steady-state inorganic suspended solid concentration (C) in the aeration tank, which is much higher than the influent inorganic suspended solid concentration (C o). Based on the above information, derive the expression for steady-state inorganic suspended solids concentration (C) in the aeration tank of an ASP, and hence calculate C, given C o = 200 mg/L, θ = 4 hours, and θc= 6 days. (5) EXAMINATION-V SOLUTION

Problem No.1 Consider a wastewater flow of 10 MLD with a BOD 5 value of 600 mg/L. It is proposed to treat this waste using an activated sludge process (ASP). The

size of the proposed aeration tank will such that the water is detained for 8 hours. Complete mixing is assumed in the aeration tank. The BSRT ( ) of the system is 10 days. In addition to the BOD 5 , this water also contains nitrogen and phosphorus. The nitrogen concentration expressed as Total Kjeldahl Nitrogen (TKN) is 40 mg/L as N, and phosphorus concentration is not limiting. Assuming that nitrification occurs in this system, calculate the effluent TKN concentration, and the additional oxygen requirement for  nitrification. Assumptions:

X r  = 10000 mg/L; =0.1 L/mg/d.

Y T = 0.5;

(K S ) N = 2.0 mg/L; (q m )  N = 2 /d

(Y T ) N = 0.2;

K d = 0.05 /d;

K 

(K d ) N = 0.05/d

• Biomass may be represented as C5 H 7 O 2 N • Nitrogen incorporation in heterotrophic biomass must be accounted for. • Nitrogen incorporation in autoprophic biomass may be neglected. (15) Solution:

;

;

Therefore, Also,

Hence,

.

Also,

Oxygen Requirement (in Kg/d)

= 1.5.Q.(S o - S e ) -1.42.( D X)

= 1.5. Formula weight of biomass, C 5 H7 O2 N = 5(12)+7(1)+2(16)+1(14) = 113 Therefore, production of 1990 Kg/d biomass requires:

Influent Nitrogen concentration = 40 mg/L (as N), or  Therefore, Nitrogen available for nitrification = 400 - 247 = 153 Kg/d Or, [TKN]o = 15.3 mg/L (as N)

 Nitrification Calculations:

;

Also,

;

Also, ;

i.e.,

Problem No. 2

0.25 MLD of sludge is generated from the activated sludge process of a wastewater treatment plant. The solids concentration in this sludge is 13,000 mg/L. This sludge is first processed in a sludge thickener, where the solids content is increased to 4 percent (weight basis). Next the sludge is treated in the anaerobic digester for reduction in sludge solids. The digested sludge is then applied to sludge-drying beds for reduction of water content to 55%. Determine the weight and density of dried sludge that will be produced per  day. Density of Water = 1000 Kg/m 3 ; Dry Density of Sludge Solids = 2200 Kg/m3 COD reduction efficiency of sludge digester = 60 %;

;

For sludge digestion, Y T = 0.06 mg/mg, and K d = 0.03 /d on COD basis Assumptions: • •

The sludge thickener is 100 percent efficient. Entire effluent from the anaerobic digester is applied to sludge drying  beds.

•

Dried sludge is completely saturated with water.

Answer:

Consider 1 m3 of sludge after thickening Let the sludge density be Weight of this sludge =

Therefore, Weight of solids in 1 m 3 of sludge = (0.04).(1).(1022.305) =40.89 kg. Solids Concentration (X o ) = 40890 mg/L Let the discharge of thickened sludge be Q sl Therefore, assuming 100 percent solids capture,

Q sl . (40890) = Q w . (13000) or, Q sl = Solids Loading ( D X) = Digester Design: ; Digester Volume (V d ) = ;

Assuming 60% treatment efficiency,

Also,

Therefore,

, Also,

Anaerobic Sludge Production ( = Therefore, Reduction in Solids =

Therefore, Solids Loading to Sludge Drying Beds = (3250-1891) = 1359 kg/d After Drying

Oxidation Pond SOLUTION

1. 1 MLD of wastewater with influent soluble BOD 5 (S o ) = 75 mg/L is to be treated in an oxidation pond such that effluent soluble BOD 5 (S) is 5 mg/L. Calculate the oxidation pond surface area, assuming the depth of the pond to be 0.5 m. Calculate oxygen requirement for  microbial respiration and oxygen supplied by algal growth, and check  adequacy of the design by assuming that 50 percent of the oxyg en  produced by algal growth is available for microbial respiration.  Neglect oxygen input into the pond by mass transfer from atmosphere Data:

•

K = 0.1 L/mg/d, where K is the first order microbial substrate utilization rate

( ) • Y T = 0.5 mg/mg; K  d = 0.05 /d • Average intensity of solar radiation: 150 calories/cm 2 /d • Solar energy utilization efficiency for algae: 6 percent • Energy content of algal bio-mass: 6000 calories/g algae • Equation for algal photosynthesis:

Solution:

We know,

;

Therefore,

Also,

Also,

, or,

Sludge Production (

) = Q.X =

Volume of Oxidation Pond =

Assuming depth to be 0.5 m, Surface Area (A) =

Algae production = Total algal production = (15).(10000) = 150 kg/d

Assuming 1.3 Kg oxygen production per Kg algal production, Oxygen Production = (1.3).(150) = 195 Kg/d

Since oxygen available is more that oxygen requirement for microbial respiration, the design is adequate.

Rapid Sand Filtration SOLUTION

The effect of increase in certain rapid sand filter and influent particle characteristics on time (T B) for reaching particle breakthrough (turbidity > 2.5 NTU) in the effluent, and time (T H) of reaching terminal head-loss (>3 m) across filter bed are given below. Explain the results based on you understanding of filter operation and particle removal mechanisms operative in rapid sand filters. Assume that when one characteristic is changed, all other characteristics remain fixed. Parameter

Time to Breakthrough TB

Time to Terminal Headloss TH

Filter Depth (L) Superficial Velocity (V s), (m3/m2/h) Influent Particle Concentration, (Co), mg/L)

Increase

Decrease

Decrease

Decrease

Decrease

Decrease

Floc Strength

Increase

Decrease

Collector diameter (d), m

Decrease

Increase

Porosity, α Particles not Destabilized by Addition of Coagulants

Decrease

Increase

Decrease

Increase

Solution:

Effect of Increase in Filter Depth (L): TB : TB will increase since particles in water will have the opportunity to  potentially interact with more collectors, and hence the chance for   particle attachment to a collector is more.

TH:

Since bed depth is more, water is encounter more resistance is  passing through the bed. Hence TH will decrease.

Effect of Increase in Superficial Velocity (V s): TB : Since pore velocity is more, the shear forces experienced by collected particles will be more. Hence particle detachment will be more, leading to a decrease in TB. TH:

Since particle loading rate on the filter and rate of particle collection in the filter will increase, the porosity of the filter bed will decrease more rapidly, leading to more rapid head-loss buildup and hence adecrease in TH. Effect of Increase in Influent Particle Concentration (C o): TB : Particle loading rate on the filter, rate of particle collection and hence porosity of the filter bed will decrease more rapidly. Thus pre velocity in the filter bed will increase more rapidly, leading to more rapid particle detachment. This will lead to decrease in TB. TH:

Since particle loading rate on the filter and rate of particle collection in the filter will increase, the porosity of the filter bed will decrease more rapidly, leading to more rapid head-loss buildup and hence adecrease in TH.

Effect of Increase in Floc Strength: TB : Particle detachment due to shear forces will become less prominent leading to increase in TB. TH:

Since particle retention in the filter will be more, the porosity of the filter bed will decline faster leading to faster buildup of head-loss and hence TH will decrease .

Effect of Increase in Collector Diameter (d): TB : Increase in collector diameter will mean the presence of a lesser  number of collectors in the filter media. This will result is lesser  number of potential interaction between a particle and a collector. Thus TB is expected to decrease . TH:

Since lesser number of particles will be collected as mentioned above, the rate of head-loss buildup will be lower, leading to increase in TH.

Effect of Increase in porosity ( α): It must be realized that porosity of the filter media does not depend on  particle size, but on grading of particles. Well-graded sand, i.e., with a

 broader particle size distribution will have less porosity, as compared to  poorly graded sand with a narrower particle size distribution. TB : In case poorly graded sand (with more porosity) is used as the filter  media, the influent particles will potentially encounter lesser number  of collectors and will not be collected efficiently. Hence TB will decrease. TH:

Increase in porosity of sand and poor particle collection as described above will result in lower rate of head-loss build-up and hence increase in T H.

Effect of Non-Addition of Coagulant: TB : If no coagulants are added, the particles will be stable and hence will not attach efficiently to the filter media. Hence T B will decrease . TH:

Since rate of particle collection in the filter media will be lower as described above, the rate of head-loss buildup will also be lower, leading to an increase in TH.

Quiz No. 1 Pollution Monitoring Techniques -I SOLUTION

1. What, in your own words, do you understand when it is stated that MPN of a certain water sample is 25/100 mL. Solution:

MPN of a certain water sample is 25/100 mL does not mean that the sample necessarily has 25 microorganisms. Rather is means that the probability that the number of microorganisms in the sample is 25 is the highest among  probabilities corresponding to different possible numbers of microorganisms in the sample. MPN test was performed on a wastewater sample, and the results given in the table below obtained.

1. 2. 3. 4. 5.

1:10-5 1:10-6 Dilution Dilution + + + + + + + + -

1:10-7 1:10-8 Dilution Dilution + + + -

Describe in detail how you would determine the MPN of the sample  based on the above results. Solution:

Let the MPN of the wastewater sample be . With 1:105 dilution,

Probability of a negative result is, Probability of a positive result is, With 1:106 dilution, Probability of a negative result is, Probability of a positive result is, With 1:107 dilution, Probability of a negative result is, Probability of a positive result is, With 1:108 dilution, Probability of a negative result is, Probability of a positive result is, Hence, probability of results shown in the table above is,

Hence, solution of the equation, the MPN value.

, gives the value of , i.e.,

2.  Nitrite standards for gaseous NO2 me meas asur urem emen entt are are prep prepar ared ed by dissolving 2.03 g NaNO2 in 1000 mL of distilled distilled water. water. This primary primary standard standard is then diluted 100 times to get the secondary standard. standard. Prove that 1 mL of this secondary standard is equivalent to 10 µL of NO2 (at 298oK, 1 atm.), atm.), given that 0.72 moles of NaNO2 produces same color as 1 mole of  NO2.

Solution:

Molecular Weight of NaNO 2 = 69 69 g of NaNO2 has 46 g of  2.03 g of NaNO2 has That is, That is, mg/L as

Strength of the primary standard = 1.35 g/L Strength of the secondary standard (1 to 100 dilution) is 13.5

 

That is,

1 mL of the secondary secondary standard standard has 13.5 µg

= 2.93 2.93 x

10-7 moles of  Volume of 1 mole of NO2 at 298oK and 1 atmosphere pressure =

Therefore, Therefore,

1 Liter of NO2 = 10 µL of NO2 =

Also,

1 mole of NO2 = 0.72 moles of 

moles of NO2

Therefore, moles of NO2 = 2.93 x 10-7 moles of  Thus, 1 mL of the the secondary secondary standard standard solution solution is equival equivalent ent to 10 µL of   NO2. 3. A synthetic sample of water is prepared by dissolving 200 mg glucose, 168 mg sodium bicarbonate, 120 mg magnesium sulphate and 111 mg calcium chloride in one liter distilled deionized water. Assuming that the complete dissociation dissociation of the salts occur leading to presence p resence of Na +, Mg+2, Ca+2, Cl-, SO4-2 and and HCO HCO3- spec specie iess in addit dditio ion n to H + and and OH- ions, compute, Total solids (TS), Total dissolved solids (TDS), Volatile dissolved solids (VDS) and Fixed dissolved solids (FDS). Solution:

Total Total solids solids (TS) (TS) = mg/L Total dissolved solids (TDS) = Volat Volatile ile dissol dissolved ved solids solids (VDS) (VDS) = mg/L Fixed dissolved solids (FDS) =

200 + 168 + 120 + 111 =

599

599 mg/L 200 599-200 =

399 mg/L

 pH of the synthetic sample from electroneutrality electroneutrality consideration consideration (show your computations for  electroneutrality).

Solution:

Concentration of moles Concentration of moles

Added

=

Formed

Concentration of moles

=

Formed

=

Concentration of Added moles Concentration of Mg 2+ Formed moles Concentration of moles

2 milli2 milli 2 milli=

1 milli-

=

1 milli-

Formed

=

Concentration of Added moles Concentration of Ca 2+ Formed moles Concentration of Cl - Formed milli-moles

=

1 milli1 milli-

=

1 milli-

=

2

Electroneutrality Electroneutrality condition states that the sums of positive and negative charges in the solution must be equal. Solution:

Sum of positive charges: =

2.(0.001) + 2(0.001) + 1.(0.002)

+ Sum of negative charges: = + Equating:

0.006 +

Also, Hence,

or, pH = 7

1.(0.002) + 2.(0.001) + 1.(0.002)

Total alkalinity (TA), Hydroxyl alkalinity (HA), Carbonate alkalinity (CA), and Bicarbonate alkalinity (BA). Solution:

Total Alkalinity (TA) Hydroxyl Alkalinity (HA) Carbonate Alkalinity (CA) Bicarbonate Alkalinity (BA)

= = = =

0 0 200 mg/L as CaCO 3

Total hardness (TH), Carbonate hardness (CH), Non-carbonate hardness (BH), Calcium hardness (CaH), and Magnesium hardness (MgH), Solution:

Total Hardness (TH) Carbonate Hardness (CH)  Non-carbonate Hardness (BH) = Calcium Hardness (CaH) Magnesium Hardness (MgH)

= = 0 = =

200 mg/L as CaCO3 200 mg/L as CaCO3 100 mg/L as CaCO3 100 mg/L as CaCO3

(Report alkalinity and hardness values as mg/L CaCO 3) 4. You are given an unknown sample for the determination of nitrite. You do the following: Take 10 mL of the unknown sample. Add the required reagents for the colorimetric determination of nitrite by the NEDA method. Make up the volume to 25 mL and measure absorbance. The resulting absorbance is 0.402. Take another 10 mL of the unknown sample. Add 10 mL of standard nitrite solution (10 mg/L concentration) to the unknown sample, add required reagents as before and make up to 25 mL, and measure absorbance. The resulting absorbance is 0.503. Based on these readings, find the nitrite concentration in the unknown sample. Solution:

Let the concentration of the unknown sample be ‘C’

When 10 mL of the sample is diluted to 25 mL, concentration in the diluted sample is Beer-Lambert law States: Absorbance (A) = Where, = molar absorptivity b = path length c = concentration Hence,

for

Similarly,

the (A)

first

measurement,

for the second measurement,

0.402

=

0.503 =

(B) From equation (A) and (B)

or,

C = 39.7 mg/L

5. Propionitrile has a general formula of oxidized by dichromate during the COD test.

. It is completely

Write a balanced equation for reaction of . The .

nitrogen

end-product

of

with the

reaction is

Solution:

Writing balanced equation: Oxidation: Reduction: Final Equation:

What is of propionitrile.

the

COD

of

a

50

mg/L

solution

Solution:

While getting oxidized, 1 molecule of

releases 14 electrons.

Reduction of oxygen takes place as follows, , accepting 4 electrons per molecule of oxygen reduced, Thus reaction stoichiometry for reaction between follows:

Molecular weight of That is,

55 g of

Therefore,

50 mg of

Hence COD of a 50 mg/L

and O 2 is as

is 55 consumes consumes solution is 102 mg/L.

Preliminary and Primary Treatment –II SOLUTION

1. A rectangular primary sedimentation basin has to be designed for a flow (Q). The influent water contains uniform sized sand particles of settling velocity (v). Show that theoretically, the surface overflow rate (SOR) of the  basin required to remove all influent sand particles in the water is numerically equal to the settling velocity (v) of the sand  particles. 2. In a primary sedimentation basin, what are effluent launders, and why are they required. Draw a net sketch of such launders. 3. Give three reasons of why even though, theoretically speaking, depth has no effect on the particle removal efficiency in a primary settling tank used in wastewater treatment, a depth of 3-5 m is provided in practice.

TUTORIAL-I The schematic of a typical water supply scheme is shown below. Raw water  is collected from a source, which is a river in this case, through an intake structure, and pumped to the treatment plant. The treated water is collected in a sump well and then pumped to an elevated distribution reservoir for  distribution by gravity to the distribution system. The objective of  successive tutorial classes will be to design various components of this system.

Water

From River

Water to To Distribution System

For the purpose of this tutorial, we will assume that the community in question is the IIT Kanpur campus. Water is to be supplied to the campus  by pumping raw water from the Ganga River. The intake structure for this  purpose will located a little distance upstream of Bithoor. Before designing the system shown above, information is required regarding the capacities of various components of the system. This information is obtained through the estimation of the water demand of the community. Following tasks need to be completed for obtaining information as described above:

Task #1:

Population Estimation for year 2000 and year 2020

First task is to estimate the population of IIT Kanpur Campus in year 2000 and the expected population in year 2020, given the following information: Table 1.

Types and Numbers of Residential Units of  Various Types and Number of Residents in Each Unit in Year 2000

Type of Residence

Number

Avg. No. of   Residents per Unit

Type 1 Type 2 Type III Type IV Type V Hostel Servant’s Quarters

192 220 192 200 40 6 300

6 5 4 4 4 500 5

In addition, remember that IIT Kanpur Campus is a restricted access community, where population growth is controlled through policies adapted  by the Institute authorities. Current Institute policies envisage an increase in the student population of the Institute from 2100 in 1994 to 4200 by year  2004. No further increase in student population is planned after this doubling of population. In spite of student population doubling, neither the faculty, nor the staff strength of the Institute is expected in increase in the next twenty years. Hence, though new hostels will have to be built, construction of additional residential quarters for faculty and staff are not envisaged in the next twenty years. Task #2:

Estimate the water Demand in Year 2000 and Year 2020

Average domestic water demand is 180 lpcd (liters/per capita/day) in year  2000. This is expected to increase to 235 lpcd in 2020. In addition, temporary population, i.e., people who do not live on campus but visit the Institute daily for various reasons, is expected to be around 3000. Water  demand for this population is expected to be 40 lpcd in year 2000 and 60 lpcd in year 2020. Water demand of the commercial establishments, i.e., academic area, health centre, restaurants, community centres, shopping centres, etc., is expected to be about 50 % of the total domestic water  demand. Fire demand is given by the formula, Q = 100 , where P is the  population in thousands and Q is the water demand in kiloliters. Water  demand for horticultural purposes 0.5 cm-acre/d, with the total area requiring this water being 100 acres. 

Estimate the average daily demand in year 2000 and 2020 (which includes residential, commercial and horticultural demand) based on the above information. Also calculate the fire demand in years 2000 and 2020.



The water demand in summer months is expected to be more than the value calculated. Assume maximum daily demand to be 1.8 times the average demand.



In addition to daily variations, hourly variations in water demand must also be taken into account. Assume maximum hourly demand of the maximum day to be 2.7 times the average daily demand.

The critical water demand is obtained by comparing the sum of maximum daily demand and fire demand (A) with the maximum hourly demand of the maximum day (B). The larger of these two values will be used to design the distribution system to serve the community. The smaller of the two values will be used to design the size of treatment plant and the pipe for conveying raw water to the treatment plant. The clear water tank, i.e., that tank(s) constructed to hold treated water before distribution should be large enough to store the difference between the above two values (A and B) for twelve hours. Based on the above information, determine the capacities of various components of a water supply, treatment and distribution system.

Tutorial 2

In this tutorial your first task will be to design the conveyance system for the water from the intake point in Bithooor to the IIT Kanpur campus. The level of the pumps in the water intake structure in Bithoor is 124 m. The intake point is situated within the river, at an elevation of 119 m. The distance between the pump-house and the raw water storage tanks in IIT K  campus is 24 Km. Ground level at the vicinity of the raw water storage tanks is 129 m. A pictorial depiction of the above description is shown in Figure 1.

Pipeline

Figure 1.

Schematic of the Conveyance system

Other relevant information required for designing the conveyance system are as follows:

•

Four pumps and pipelines of equal capacity will be employed for pumping the water. However the capacities of the pump should be such that three pumps

working together shall be able to pump the required volume of water. In other  words, one pump and pipeline will constantly be on standby in rotation.

•

The pressure head at the end of the pipeline should be 1 m. Water from the supply pipe will fall freely into raw water reservoirs, from which it will be  pumped further into the treatment plant.

•

The raw water reservoirs will consist of three open water tanks at ground level, with a capacity to hold seven days of water supply.

•

Design will require sizing of the pipes, pumps and the raw water reservoir.

The second task in this tutorial will be to assess the Ganga water  characteristics, as given in Table 1 for the wet and dry seasons. Based on this assessment, prepare a brief report on the water characteristics and the type and nature of treatment required to improve water quality such that it is suitable for supply as potable water to the campus community. Table 1.

Sl. No.

Average Water Quality Characteristics of River Ganga at the Water Intake Point in the Dry and Wet Seasons.

Parameter

Desired Maximum Average Average Concentration in Concentration in Concentration in Treated Water Raw Water Raw Water From Nov. to From June to May October

1

Turbidity

2.5 NTU

2

Colour 

5 Units (Pt-Co Scale) 20 Units (Pt-Co Scale)

3 Units (Pt-Co Scale)

3

PH

7.0 – 8.5

7.8

7.5

4

Alkalinity

 No Limit

40 mg/L as CaCO3 35 mg/L as CaCO3

5

TDS

500 mg/L

450 mg/L

6

Total Hardness

200 mg/L as CaCO 3 60 mg/L as CaCO3 40 mg/L as CaCO3

7

Calcium Hardness

200 mg/L as CaCO 3 30 mg/L as CaCO3 25 mg/L as CaCO3

8

Chloride

500 mg/L

100 mg/L

75 mg/L

9

Sulfate

500 mg/L

50 mg/L

40 mg/L

1.0 mg/L

0.7 mg/L

0.5 mg/L

10 Fluoride

50 NTU

150 NTU

350 mg/L

11  Nitrate

45 mg/L

12 mg/L

7 mg/L

12 Ammonia

0.1 mg/L

0.5 mg/L

0.2 mg/L

13 Iron (II)

0.1 mg/L

 ND

 ND

14 Manganese (II)

0.05 mg/L

 ND

 ND

Sl. No.

Parameter

Desired Maximum Average Average Concentration in Concentration in Concentration in Treated Water Raw Water Raw Water From Nov. to From June to May October

15 Cyanide

0.1 mg/L

ND

ND

16 Chromium (VI)

0.1 mg/L

 ND

 ND

17 MPN

<1 106 Org./100 mL 105 Org./100 mL Organisms/100 mL

18 a. DDT

10 µg/L

 ND

 ND

 b.

Aldrin

10 µg/L

 ND

 ND

c.

Heptachlor 

10 µg/L

 ND

 ND

19 a. Vinyl Chloride

10 µg/L

 ND

 ND

 b.

Tetrachloroethylene

10 µg/L

 ND

 ND

c.

Trichloroethylene

10 µg/L

 ND

 ND

< 1 mg/L

6 mg/L

3 mg/L

20 BOD5

Tutorials 3 and 4 Based on the discussions of tutorial number 2, following process train for the treatment of water was decided upon: • Assume 30 % reduction in turbidity due to storage in raw water storage tanks, i.e., due to the settling and consequent removal of a part of the settleable solids.

•

Also assume that each NTU of remaining turbidity is equivalent to 3 mg/L of  suspended solids.

Prechlorination

Chlorine Addition

Raw Water  from Storage Tanks

Prechlorination: Chlorine dose required for prechlorination is 1 mg/L per mg/L

BOD to be destroyed. Liquified Chlorine stored in tanks is directly applied using jet mixers. Caculate the chlorine required per day. A contact time of 45 – 90 seconds in an open channel is required for chlorine reaction to be complete. Design a suitable open channel for this purpose. Rapid Mix: Assume the required coagulant (Alum) dose to vary between 20 – 40

mg/L depending on the raw water turbidity. Design conventional vertical-shaft rapid mix unit. Appreciable decrease in pH due to addition of alum, if any, must be  prevented by adding soda (sodium carbonate) along with alum.  Design Parameters:

Detention time (t): Ratio of tank height to diameter: Ratio of impeller diameter to tank diameter: Velocity gradient (G): Gt: Tank diameter: Paddle tip speed: Velocity of paddle relative to water: Paddle area/Tank section area: Coefficient of drag on impeller blade: Maximum length of each impeller blade: Maximum width of impeller blade: Impeller height from bottom:

20 – 60 s (1:1 to 1:3) (0.2:1 to 0.4:1) >300 /s 1000 – 2000 <3m 1.75 – 2.0 m/s 0.75 x paddle tip speed 15:100 1.8 0.25 x impeller diameter  0.20 x impeller diameter  1.0 x impeller diameter 

Use the following equation to calculate the soda dose to be added (if reqd):

Coagulation – Flocculation: Design horizontal paddle flocculation unit using

conventional design parameters.

 Design Parameters:

Detention time: Velocity gradient (G): Gt: Basin depth: Paddle tip speed: Velocity of paddle relative to water: Paddle area/Tank section area: Coefficient of drag of paddle: Maximum length of each paddle: Maximum width: Thickness:

10 – 30 minutes 20 – 75 /s 2x104 – 6x104 <5m 0.25 – 0.75 m/s 0.75 x paddle tip speed 15:100 1.8 5m 50 cm 5 cm

Secondary Sedimentation: Assume suitable surface overflow rate to design circular 

settling tank. Calculate the amount of sludge generated, both on volume any dry weight basis. Assume specific gravity of sludge to be 1.1 and solids content to be 5  percent. Assume 90 percent removal of turbidity in secondary sedimentation. Also assume that 100 percent of alum added has precipitated as aluminium hydroxide.  Design Parameters:

Surface over flow rate: Depth: Diameter: Detention time:

30 – 40 m 3/m2/d 5–6m 20 – 80 m 30 – 45 minutes

Rapid Sand Filtration: Assume filter depth to be 60 cm. Filter media is 0.5 mm

sand particles. Calculate clean bed headloss using Karmen – Kozeny equation. Terminal headloss is 2.5 m. Average filter run length is 8 hours, and filtration rate is 8 m 3/m2/hr. Calculate the water required for backwash, and hence calculate the size of the backwash tank. Also calculate the effective filtration rate, i.e., after accounting for filter down-time due to backwashing and water requirement for backwashing. Assume the turbidity of water post-filtration to be 1 NTU. Based on this information calculate the solids concentration in backwash water.  Design Parameters:

Length to width ratio: 1.3 – 1.5 : 1 Length: <7m Water Depth on top of Filter: 3m Free Board: 0.5 m Underdrainage system depth: 1.5 m 3 Backwash rate: 1.0 m /m2/min Backwash time: 5 minutes Filter down time due to backwashing: 30 minutes Post-Chlorination: Chlorine is added using jet mixers in closed conduits in a manner 

similar to during prechlorination. Assume that the initial demand for chlorine has  been satisfied due to prechlorination. Calculate the breakpoint chlorination dose to

destroy ammonia. A free chlorine residual of 2 mg/L must be maintained in the treated water.

Use the following equation to calculate the breakpoint chlorination dose: Assume the CT value for 6 Log kill of coliform organism is 120, and that for 5 log kill is 96. Based on this information, calculate the minimum time of contact required to achieve complete disinfection. Other Remarks:

When designing water treatment plants, it must be remembered that it is advisable to put several small unit processes in parallel, rather than designing a single large unit. However construction of very large number of  small unit processes will result in an increase in operation and maintenance costs. Proper provisions must be made for units being under repairs, and hence out of commission.

Tutorial 5 This tutorial is designed to impart basic concepts of plant siting and hydraulics of  water treatment systems. A map of IIT Kanpur campus is given in Figure 1 and Figure 2. Based on these maps, it was decided that the water treatment plant would be sited in the empty land available in the western end of the campus. A contour map of the IIT Kanpur campus is shown in Figure 3.

The plant should be designed such that water through the plant flows without pumping. This may be impossible in a place with flat terrain, as is the IIT Kanpur campus, unless the ground elevations at the plant site are changed significantly through earthwork. Luckily, as per our design, significant amount of soil is available for this purpose from the excavation of the raw water tanks, as described in tutorial 2. A part of this soil will be used to change the ground elevation at the water treatment plant site. For determining the hydraulic grade line, or the water surface level at various parts of the treatment plant, the head-loss through various treatment units must be known (see schematic in Figure 4). Following information regarding water surface levels are given:

•  pre-chlorination tank:

Difference in elevation of inlet and outlet

water surface level to be calculated using slope of the channel, as determined by Manning’s equation.

• rapid max tanks:

0.3 m difference in water level between inlet

and outlet • flocculation tank: 0.6 m difference in water level between inlet and outlet 0.6 m difference in water level between inlet • secondary clarifier: and outlet • Rapid sand filters: 4.0 m difference in water level between inlet and outlet Water will be conveyed from one process unit to another through open channels of 1:100 slope. Water from the rapid sand filters will be conveyed to a covered clear-water  sump with a storage capacity of 6 hours.

• Covered underground clearwater sump: and outlet

1.0 m difference in water level between inlet

Water from the clear-water sump will be pumped to four overhead tanks located in various parts of the campus. Water from the overhead tanks will  be fed to the distribution system by gravity.

Figure 4.

Water Surface Level Schematic of the Treatment Plant

Post-chlorination will be performed by addition of chlorine through jet mixers in the four pipes at the exit from the clear-water sump. Retention of 

water in overhead tanks must provide sufficient contact time for postchlorination to be effective.

• Design the layout of the water treatment plant, based on the plant sizing • • • • •

done in tutorials 2, 3 and 4. Clearly mention the ground level and corresponding water level at various points of the treatment plant. Mention the quantity and degree of earthwork required for changing the ground level at the plant location. Design the required pumps and other water conveyance units such as open channels and pipes. Design the overhead water tanks such that the tank near the MT section supplies 40 percent of the water, while the other tanks supply 20 percent each. Design the pipes and pumps required to supply water from the treatment  plant to these overhead tanks.

Tutorial 6 The objective of this tutorial is to impart knowledge about design of water  distribution network for a community. We will only be designing the basic  backbone of the network. Minor details, i.e., pipe sizes and water flow-rate into individual households, pipe connections, etc. are beyond the scope of  this tutorial. Consider Figure 1 given with this tutorial. Figure 1 shows the map of IIT Kanpur campus with the proposed water supply network superposed. The  proposed network consists of 10 loops, 15 nodes and 24 pipes. Water will  be fed to the network through four nodes shown in the figure. These nodes are the input nodes. Water will be abstracted from the network at various nodes (except the input nodes) which will be known as output nodes. The amount of withdrawal from a node will depend upon the expected water  demand at the vicinity of the node. The reference pressure head for the network is 15 m at the water-input point near the MT section (0.4Q). The network will be laid at a depth of 0.5 m below the ground surface. The network should be able to deliver the required flow to each node with a  pressure head of at least 12 m. Calculations must be done as follows for designating the output nodes and the outflow from such nodes: • Designate 10 nodes in the network as output nodes. • Designate an area in the immediate vicinity of the output node as the area whose water demand will be met by that node.

• Calculate the water demand from that area. The water abstraction rate from the output node will be equal to the water demand. • Similarly calculate the water abstraction rate from each output node. • Remember, the total water abstraction rate must be exactly equal to the water inflow to the water distribution system. Based on the available information as described above, the water supply network must be designed. Design will include determining the diameter of  various pipes and checking to ensure that the pressure head available at various nodes is at least 12 m. A computer program called LOOPS will be used for designing the water  distribution network. This software has already been sent to you by electronic mail. I hope you have already installed it on some PC of your  choice. For your convenience, this software has also been installed in five PCs of the VARUN laboratory. To use this software in VARUN laboratory, go the START menu of the PC, scroll to ‘program’ and then scroll to ‘LOOPS’.

Instructions for Using LOOPS (Can be Downloaded from Website): Consider the water distribution network shown in Figure 2. • Show the water abstraction rates at various designated output nodes based the calculations described earlier. •  Name the 10 loops in the network as 1, 2, 3, … 10 in any order you choose •  Name the 15 nodes in the network as A, B, C,…. O in any order you choose. •  Name the 24 pipes in the network as 1, 2, 3, ….. 24 in any order you choose. • Determine the length of each pipe in meters using the scale given in Figure 1. • Determine the position of each node (x, y), w.r.t any arbitrary origin. • Determine the approximate height of each node by consulting the contour map of  the campus. Remember, the pipe network is 0.5 m below the ground level. • Show an assumed direction of flow in each pipe. • In each loop, determine the number of pipes, and the numbers designated to each of those pipes. This number is positive if the assumed flow in the pipe is in clockwise direction, otherwise the number is negative. • Determine the co-ordinates (x, y, and z) of each node. • Determine inflow/outflow from nodes. Remember, inflow is positive, outflow is negative. • Determine for each pipe the initial and final node, as per assume direction of  flow.

The data generated above should be input to the LOOPS program, and the network  analyzed. If the pressure head in any node is less that 12 m. appropriate modifications in pipe diameters feeding that node should be made to increase the pressure head. A graphical depiction of the analyzed depiction showing flows in various pipes and  pressure head at various nodes must be presented. Also satisfy yourself that the solution is correct, i.e., the mass balance and head-loss criteria are satisfied.

Figure 1. Water Supply Network Superposed on the IIT Kanpur Campus Map

Tutorial 7 The objective of this tutorial is to calculate the quantity and characteristics of the wastewater generated in the IIT campus. For the purposes of this calculation, it is assumed that the water used for domestic and commercial establishments only will constitute the wastewater. Since the IIT Kanpur campus is a fully sewered community, it is expected that 80 % of the water supply to the domestic and commercial establishments will become wastewater. To determine the characteristics of the wastewater use the following unit loading factors, which are valid for  domestic wastewater: Soluble BOD5 : 50 g/capita/d Total Solids: 200 g/capita/d Organic Suspended Solids: 100 g/capita/d Grit: 10 g/capita/d Total Nitrogen: 8 g/capita/d Free Ammonia: 0.6 x Total Nitrogen Total Phosphorus: 2 g/capita/d The contributions from commercial establishments can be taken as half the above values. Based on the information given above, calculate the wastewater flow and characteristics. Wastewater generated will be conveyed to the wastewater treatment plant by three trunk sewers shown in Figure 1. Several branch sewers and minor sewers will also be required. Based on the sewer network shown in Figure 1, calculate the flow at each junction of  the three trunk sewers. Also calculate the flow at each junction of the branch sewers. The wastewater flow will not be the same at all times of the day. The wastewater flow at various times of the day are given as fraction of the design flow in Table 1. Table 1.

Flow Fractions at various times of the day

Time Fraction

M-1 0.89

1-2 0.72

2-3 0.53

3-4 0.42

4-5 0.34

5-6 0.32

6-7 0.39

7-8 0.66

8-9 1.15

9-10 1.34

10-11 1.38

11-N 1.40

Time Fraction

N-1 1.38

1-2 1.32

2-3 1.25

3-4 1.14

4-5 1.06

5-6 1.06

6-7 1.07

7-8 1.19

8-9 1.30

9-10 1.30

10-11 1.24

11-M 1.12

Based on the information given above, calculate the dimensions of an equilization tank to be constructed for dampening the variations of flow, thus ensuring that the wastewater input to the wastewater treatment plant remains constant all through the day.

Figure 2.

Water Distribution Network to be Analyzed Using LOOPS

Tutorial 8 The objective of this assignment is to design the sewer system on IIT Kanpur  campus. A map of IIT Kanpur campus (Figure 1), proposed alignment of sewers in IITK campus (Figure 2), and contour map of IIT K campus (Figure 3) are provided for this purpose. Please go through the accompanying handout on sewer design before attempting this assignment. As specified in this handout, the required calculations must be done in a tabular form. A sample table for this purpose is attached (Table 1). A separate table must be created for each trunk or main sewer designed. Design of branch sewers is not required. Following points should be remembered during the design: 1. Minimum diameter of a sewer is 150 mm. Other available diameters are 200mm and higher at increments of 100 mm. Take Manning’s coefficient to be 0.013. 2. Check that the sewers are not more than 0.8 full at ultimate peak flow in year  2020. 3. The velocity in the sewer should be at least 0.6 m/s at the present peak flow. This is required to ensure that particles are not deposited in sewers on a permanent  basis. 4. Maximum velocity in the sewer at no time should be more than 3 m/s. 5. Manholes should be built at every change of alignment, gradient or sewer  diameter. 6. For straight sewers of uniform diameter and constant slope, cleaning manholes should be provided every 100m.

7. Main sewers should be connected to trunk sewers (or, branch sewers should connect main sewers), at angles ranging from 30 to 90 degrees. The center-line of  the smaller sewer should be at least 0.4 m above the center-line of the larger sewer  at the manhole. This will ensure dissipation of excess energy of water in the smaller sewer. 8. Manholes are connected over the center-line of the sewer. A minimum drop of  0.03m is provided between center-line of sewers entering and exiting manholes. 9. The diameter of a sewer exiting a manhole must always be greater than or equal to diameter of the sewer entering the manhole. 10. The top of a sewer entering a manhole must never be at a lower elevation than that of a sewer exiting the manhole. 11. Typical slopes in sewers vary from 1 per 1000 to 10 per 1000, with larger  diameter sewers having less slope. Remember to put the least slope possible, while ensuring that the minimum velocity criterion is satisfied, such that sewers do not go too far underground. 12. Design of branch sewers is not required. However, assume them to have a uniform slope of 8 in 1000. This is required to fix the levels of the main and trunk  sewers. 13. A sewer (branch, main or trunk) should be at least 1 m below the ground surface. This should be kept in mind when fixing the depth of the main or the trunk sewers.

Table 1. 1

4 5 6 7 8 9 10 11 12 13 14 15 16 Avera Peak Diame Slop Prese ge Flow ter, e Discharg Velocity, nt Self- Tot Invert Manh Leng Sewa , mm e, (2020) Peak Clensi al Elevation, ole th, ge (202 (2020) m/s Flow, ng Fall, m m Flow, 0) Litre/s Year Veloc m (2020 Litre 2020 ity ) /s Litre/ (2000) Litre/ s m/s s Fr To FullActu FullActu Upp Low o al al er er m

• • • • • • • • • • • • •

2

Tabular Calculations for Sewer Design 3

A separate table is required for each trunk and main sewer  Locate manholes on a particular sewer, giving each an identification number  (columns 1-2) Determine the length of each section (between two manholes) (column 3) Determine the average sewage flow in each section, i.e., at the starting manhole in year 2020 (column 4) Determine the peak flow in each section by multiplying by the peaking factor  (column 5) Determine the diameter and slope such that the sewer flows 0.8 full at ultimate  peak flow (columns 6-7) Remember that a minimum diameter of 150 mm must be provided Calculate full discharge for the pipe diameter selected and compare with actual full discharge (col. 8-9) Calculate corresponding velocities and check that they are not less than 0.6m/s, or  more than 3.0 m/s (columns 10-11) Calculate the present peak flow (Year 2000) in each sewer section (column 12) Calculate the velocity corresponding to present peak flow, and ensure that this is more than 0.6 m/s (column 13) Calculate total fall for each section (column 14) Calculate the invert level, i.e., level at the bottom of the sewer, at the upper end (starting manhole) and lower end (ending manhole) of the sewer section (columns 15-16)

H VII

Figure 1. Map of IIT Kanpur Campus

Figure 2.

Proposed Sewers in IIT Kanpur Campus

H VII

Figure 3.

Contour Map of IIT Kanpur Campus

Tutorials 9 and 10 Based on the discussions of tutorial numbers 7 and 8 following information is available regarding wastewater characteristics of IIT Kanpur campus: • Wastewater Quantity, i.e., average and minimum and maximum wastewater  flow-rate per hour. • Wastewater Quality, i.e., BOD 5, TSS, Grit, TKN, Phosphate etc. Objective of wastewater treatment is to obtain effluent with BOD 5 <10 mg/L, TKN< 1 mg/L, Nitrate< 1 mg/L, and Phosphorus < 0.5 mg/L. Based on the above objectives, the following wastewater treatment chain was decided upon:

A. Solid Waste from Bar-Rack  B. Sludge From Grit Chamber  C. Sludge from Primary Sedimentation Tank  D. Sludge from Secondary Sedimentation Tank 

The extended aeration process is designed for combined carbon oxidation and nitrification, and the tertiary treatment should result in removal of  nitrogen and phosphorus to the desired levels. Assume 100 % removal of Grit in the grit chamber, 80% removal of  inorganic suspended solids, and 80% removal of volatile suspended solids in

the primary sedimentation tank.. Total suspended solids concentration effluent from secondary treatment is 25 mg/L. 50 kW surface aerators should be used for aeration purposes. Manufacturers of such aerators specify that the oxygen transfer capacity of these aerators is 2.0 Kg O 2/kW-h under standard conditions. Based on this information, calculate the total power  requirement for satisfying oxygen requirements in the aeration tank. The operating temperature of the aeration tank is expected to be 30 oC. The steady state dissolved oxygen concentration in the aeration tank should be 3 mg/L. Saturation concentration of oxygen in water at 20 oC is 9.1 mg/L and at 30 oC is 7.5 mg/L. Calculate the power  requirement to keep the bio-mass in suspension in aeration tank. Based on the above two values, decide on the aerator requirements.

•

Design and size the wastewater treatment units based on the above information. Calculate chemical requirements.

Hydraulic Design: Based on maps of IIT Kanpur campus given in tutorial 5, it was decided that the wastewater treatment plant would be sited in the empty land available in the western end of the campus, adjacent to the water treatment plant sited in this area earlier.

The plant should be designed such that wastewater through the plant flows without pumping. This may be impossible in a place with flat terrain, as is the IIT Kanpur campus, unless the ground elevations at the plant site are changed significantly through earthwork. Luckily, as per our design, significant amount of soil is available for this purpose from the excavation of the raw water tanks, as described in tutorial 2. A part of this soil will be used to change the ground elevation at the wastewater treatment plant site. For determining the hydraulic grade line, or the water surface level at various parts of the treatment plant, the head-loss through various treatment units must be known (see schematic in Figure 1). Following information regarding water surface levels are given:

• Bar Rack: • Grit Chamber: • Primary Sedimentation Tank: inlet and outlet • Aeration Tank: inlet and outlet • Secondary Settling Tank: inlet and outlet • Oxidation Pond: inlet and outlet

Headloss through bars, 0.2 m As per open Channel specifications 0.4m water level difference between 0.6m water level difference between 0.6m water level difference between 0.8m water level difference between

• Fish Pond:

0.8m water level difference between

inlet and outlet Water will be conveyed from one process unit to another through open channels of 1:100 slope. Water from the tertiary settling tank should be available with a head of  approximately 5 m above the ground level, i.e., at an elevation of 135 m. This water will flow over a cascade of 3 m height for aeration and odour  removal, and will be conveyed through an open channel to a small ‘nalla’ in the extreme western end of the campus flowing southwards towards Pandu River flowing in the Southern end of Kanpur City.

Figure 1.

Treatment Plant Hydraulics

• Design the layout of the wastewater treatment plant, based on the plant sizing done earlier. • Clearly mention the ground level and corresponding water level at various points of the treatment plant. • Mention the quantity and degree of earthwork required for changing the ground level at the plant location. • Design the required pumps (for wastewater pumping and sludge recycling) and other water conveyance units such as open channels and  pipes.

PHILOSOPHY OF WATER AND WASTEWATER TREATMENT SOLUTION 1. Discuss the philosophy behind water and wastewater treatment with reference to water/wastewater quality characteristics, standards, unit processes and process train. (3) Answer:

Water from natural sources, i.e., surface or ground water will contain certain impurities, either in particulate or dissolved form. A listing of the concentration of these impurities is known as water quality characteristics. This water is to be used for beneficial purposes like domestic, agricultural or industrial use. Maximum acceptable quantities of  impurities in water for various beneficial uses are specified. A listing of  these values are known as water quality standards. A comparison of the water quality characteristics and standards enables one to determine the impurities in the water whose concentration is unacceptably high for a  particular beneficial use. The purpose of water treatment is to remove these impurities, such that the water conforms to the standards, and hence is suitable for the designated beneficial use. Unit processes are devices/machines/reactors designated to remove one or more of these impurities during water treatment. Several such unit processes put together  constitute a process train, which will remove the required impurities, and thus make water conform to the water quality standards. During beneficial use of water, impurities are added to it, and water is thus converted to wastewater. Wastewater may be disposed to land or water   bodies or reused or recycled. Wastewater disposal or reuse/recycling standards are available. In case the concentrations of impurities in wastewater, i.e., wastewater characteristics are incompatible with the  prevailing standards, wastewater treatment is required. As with water  treatment, suitable process train must be provided for this purpose to reduce the concentration of the objectionable impurities.

2. Write a short note to clearly bring out the difference between disposal, reuse and recycling of wastewater. (3)

 

Answer:

Due to the addition of impurities during beneficial use, water is transformed to wastewater. This wastewater may be discharged to aquatic or   terrestrial environment, provided the relevant discharge standards are attained through wastewater treatment. This is known as wastewater  disposal. Alternatively, the wastewater may be used for certain other beneficial  purpose, provided adequate treatment is given to the wastewater to conform to the water quality standards for that beneficial use. This procedure of  using water for multiple beneficial use is known as reuse. Alternatively, the wastewater may be used for the same beneficial purpose as before, provided adequate treatment is given to the wastewater such that it again conforms to the standards for the original beneficial use. This  procedure for using water for the same beneficial use is known as recycling. 3. Why is equilization necessary in a wastewater treatment plant. With appropriate diagrams describe the difference between on-line and offlineequilization. (1+ 3) Answer:

Rapid changes in wastewater quantity and quality make operation of a wastewater treatment plant difficult. This is because of the necessity to constantly adjust the process parameters in the treatment units to compensate for the unsteady conditions under such circumstances. Provision of an equilization tank dampens the rapid changes described above, and hence the treatment plant can operate under steady state conditions.

On-Line Equilization:

QS

Equilization Tank 

QU

Variable flow, Q U is influent to the equalization tank. However, a steady flow Q S is effluent from this tank. When Q U>QS, excess wastewater is stored in the tank. When Q U
Worked-out Examples: Population Forecast by Different Methods Sedimentation Tank Design Rapid Sand Filter Design Flow in Pipes of a Distribution Network by Hardy Cross Method Trickling Filter Design Population Forecast by Different Methods Problem: Predict the population for the years 1981, 1991, 1994, and 2001 from the

following census figures of a town by different methods.