Comprehensive Study Notes on Sanitary Engineering, Sewage Treatment, and Environmental Management

Fundamentals of Sanitary Engineering and Terminology

Sanitary engineering involves the collection, treatment, and disposal of liquid and solid wastes from communities to safeguard public health and protect the environment.

Definitions of Common Terms

• Sewage: All kinds of liquid waste or wastewater produced from the community. It includes sullage, discharge from toilets, urinals, industrial wastewater, groundwater, surface water, and storm water admitted into sewers.
• Domestic Sewage: Sewage derived from residential buildings, commercial establishments, institutions, and public buildings (offices, schools, cinemas, hotels).
• Industrial Sewage: Wastewater obtained from industrial production and manufacturing processes.
• Sanitary Sewage: Sewage produced during dry weather that excludes storm water.
• Storm Water: Runoff rainwater originating from a locality following precipitation.
• Sullage: Wastewater originating from bathrooms, kitchen sinks, wash basins, etc., excluding human excreta and urine. It is generally less harmful and does not emit severe foul odors.
• Sewer: An underground conduit or drain designed to convey sewage. It is typically closed and operates normally under open-channel flow (not flowing full).
• Sewerage: The overall process and infrastructure system of collecting, conveying, treating, and disposing of sewage via a water carriage system through sewers.
• Refuse: Rejected or worthless solid or semi-solid waste material. Refuse is categorized into seven types:
  1. Garbage
  2. Rubbish
  3. Sullage
  4. Sewage
  5. Subsoil water
  6. Storm water
  7. Night soil
• Dry Refuse: The dry combination of garbage and rubbish.
• Garbage: Putrescible organic wastes obtained from residential, commercial, and institutional kitchens, including waste food, vegetable and fruit peelings (food waste), animal dung, grass, tree leaves, and bird excreta.
• Rubbish: Non-putrescible, inorganic, combustible and non-combustible solid wastes excluding food waste. Examples include paper, plastic, bottles, textiles, rubber, and wood.

Classification of Sewers

• Separate Sewer: A sewer that carries only domestic and industrial sewage while excluding storm water.
• Storm Water Sewer: A sewer designed exclusively to carry storm water runoff.
• Combined Sewer: A single sewer designed to carry domestic sewage, industrial sewage, and storm water together.
• Main Sewer (Trunk Sewer): A major sewer line that receives discharge from multiple branch or sub-main sewers, serving as the primary outlet for a large drainage area.
• Branch Sewer (Sub-Main Sewer): A sewer that receives sewage from several lateral sewers and delivers it into a main or trunk sewer.
• Lateral Sewer: A sewer that collects sewage directly from house sewers and discharges it into a branch or sub-main sewer. It represents the primary collection stage.
• House Sewer: A pipe conveying sewage from a building to a lateral sewer.
• Outfall Sewer: A sewer that receives sewage from the entire collecting network and conveys it to a wastewater treatment plant or final point of disposal. It spans between the lowest branch sewer connection and the disposal point.

Sanitation Systems and Sewerage Classification

Public administrative authorities (such as municipal offices) manage the collection and disposal of waste products using two main administrative and technological systems.

Conservancy System

An ancient system where waste types are collected and disposed of separately. It is obsolete in major cities but remains in use in small towns, villages, and undeveloped areas.

• Operation: Dry refuse (garbage/rubbish) is collected in roadside dustbins and transported by trucks or carts daily or twice daily. Non-combustible waste is dumped in low-lying pits; combustible waste (paper, leaves) is burned; vegetable peelings are dried, burned, and dumped. Night soil (human excreta) is manually collected by human agencies, transported in closed carts/trailers, and buried in underground trenches. Over approximately 2 years, buried night soil converts into agricultural manure. Sullage and storm water flow through open or closed street drains into natural streams, rivers, or seas, or are used for land farming.
• Merits:
  • Low initial capital cost due to inexpensive open drains and dry latrines.
  • Quantity of sewage reaching treatment plants is minimal.
  • Storm water travels in open drains; underground sewer sections remain small and run full for most of the year, reducing silting.
  • Pumping cost at outfalls during high river flood levels is minimal.
• Demerits:
  • Requires extensive land area for trenching night soil.
  • High risk of storm water entering sanitary drains, overloading plants.
  • Liquid refuse can seep into subsoil, polluting groundwater sources.
  • Latrines must be detached from main living quarters; indoor placement creates foul odors.
  • Manual collection and handling of night soil creates highly unhygienic conditions.
  • Aesthetic appearance of the city is poor.
  • Highly vulnerable to labor strikes (e.g., sweepers), leading to rapid waste decomposition and epidemic risks.

Water Carriage System

Wastewater and human excreta are collected and conveyed using water as the hydraulic medium through underground closed conduits without manual labor. Excreta from water closets mixes with water to form a semi-solid mixture (approx. 99.9% liquid and 0.1% solid) that flows to treatment plants.

• Merits:
  • Hygienic; eliminates manual human handling of excreta.
  • Underground closed conveyance eliminates foul odor nuisances and epidemic risks.
  • Requires a single sewer line, saving underground space compared to separate dry system channels.
  • Allows compact building design with attached water closets equipped with water seals and traps.
  • Large flow volumes allow self-cleansing velocities at flatter gradients.
  • Independent of manual labor strikes for routine operation.
  • Treated sewage effluent can be reused for agricultural irrigation.
• Demerits:
  • High initial capital construction cost.
  • High maintenance cost.
  • Large variations in wastewater volume during monsoon seasons burden treatment plants.

Comparison: Conservancy System vs. Water Carriage System

Sewerage System Types

1. Separate Sewerage System: Sanitary sewage and storm water are carried separately in two distinct sets of sewers. Sanitary sewage goes to the Wastewater Treatment Plant (WWTP); storm water discharges directly into natural waterways without treatment.
   • Advantages: Reduced volumetric load on WWTP; smaller, more economical sanitary sewers; economical pumping; natural water is not unnecessarily contaminated.
   • Disadvantages: Small sanitary sewer diameters are difficult to clean; self-cleansing velocity is harder to achieve during low flow; storm sewers lie idle during dry seasons and accumulate dry refuse; higher overall maintenance cost; sanitary sewers must be laid deeper than storm sewers.
2. Partially Separate Sewerage System: Sewage and roof/yard storm water from private property are combined into one set of sewers, while street/roadway runoff is conveyed in open surface drains.
   • Advantages: Combines benefits of both systems; prevents silting due to periodic flushing by household storm water; reasonable pipe sizing.
   • Disadvantages: Minor drawbacks of combined systems (slight increase in treatment volume during rains).
3. Combined Sewerage System: Domestic sewage, industrial wastewater, and all storm water are conveyed together in a single sewer line to the WWTP.
   • Advantages: Larger sewer diameters make cleaning easier; lower overall maintenance cost; sewage is diluted by storm water, reducing its strength; lower construction cost compared to laying two underground lines.
   • Disadvantages: Sewer overflow during heavy storms poses severe public health risks; heavy silting and foul odors occur during dry weather; high treatment plant load; unnecessary treatment of clean rainwater; uneconomical if pumping is required.

Comparison: Separate System vs. Combined System

Wastewater Quantity Estimation and Stormwater Hydrology

Sources of Wastewater

1. Public Water Supply: Municipal water supplied to residential, commercial, and industrial facilities.
2. Private Water Supply: Water drawn independently from wells, tube wells, rivers, or canals by private entities, institutions, malls, or hotels.
3. Groundwater Infiltration: Water entering sewer lines through loose joints, cracks, or defective pipe walls when the surrounding groundwater table head exceeds the hydraulic head inside the sewer.
4. Unauthorized Connections: Rainwater or illegal plumbing discharges tied directly into sanitary sewers.

Flow Components: DWF and WWF

• Dry Weather Flow (DWF): The quantity of sanitary sewage flowing during dry weather periods without storm water runoff. It comprises domestic and industrial sewage. DWF is the minimum flow available throughout the year. It depends on population, industrial output, rate of water supply, and baseline infiltration/ex-filtration.
• Wet Weather Flow (WWF): The combined volumetric flow of sanitary sewage, industrial wastewater, groundwater infiltration, and storm water runoff during rainy periods.

Factors Affecting Sanitary Sewage Quantity

1. Population Growth: Wastewater volume scales directly with population. Population forecasting uses methods such as:
   • Arithmetic Increase Method
   • Geometric Increase Method
   • Incremental Increase Method
   • Graphical Method

   Design Periods for Sewerage Components:

• Sewer Network: $30\text{ years}$
• Pumping Stations: $30\text{ years}$
• Sewage Treatment Plants: $30\text{ years}$
• Efficient Disposal and Utilization Works: $30\text{ years}$

1. Rate of Water Supply: Wastewater generation ranges from $70\%$ to $80\%$ of the total public water supply rate.
2. Groundwater Infiltration and Ex-filtration: Infiltration increases sewage volume, while ex-filtration (leakage of sewage out into the soil) decreases volume.
3. Unauthorized Connections: Direct input of unmetered storm water or illegal drains.
4. Type of Area Served: Zoning characteristics (residential density, commercial commercial hubs, heavy industrial zones).

Factors Affecting Storm Sewage Quantity

1. Catchment Area Size and Shape: Runoff is higher for large, fan-shaped catchments than for small, fern-shaped catchments.
2. Topography: Steep ground slopes yield higher runoff velocity and volume.
3. Surface Characteristics: Impervious, rocky, or clay soils produce higher runoff; loose sandy soils yield lower runoff.
4. Storm Direction: Storms moving in the direction of streamflow produce higher peak runoff.
5. Precipitation Type: Rainfall produces immediate runoff; snow melting produces delayed runoff.
6. Rainfall Distribution: Spatial coverage of the storm across the basin.
7. Rainfall Intensity and Duration: High intensity increases runoff rates; longer duration prolongs peak flows.
8. Soil Moisture Deficiency: Dry soil absorbs substantial initial rainfall (low runoff); saturated soil yields maximum runoff.
9. Climatic Conditions: High ambient temperatures increase evaporation losses, reducing runoff.
10. Geological Conditions: Fissures, subsoil cracks, and high permeability reduce surface runoff.
11. Vegetative Cover: Dense vegetative canopy and root systems increase interception and infiltration, reducing runoff.

Estimation of Peak Sanitary Sewage Flow

Sanitary sewage quantity $Q_d$ is calculated using the Return Factor ($R.F. = 0.70\text{ to }0.80$) and Peak Factor ($PF$):

$\text{Average Sewage Flow } (Q_{avg}) = R.F. \times \text{Water Supply Rate (lpcd)} \times \text{Population}$$Q_d = PF \times Q_{avg}$

Peak Factor ($PF$) Values by Sewer Type:

• Domestic Sewer: $6.0$
• Lateral Sewer: $4.0\text{ to }6.0$
• Branch Sewer: $3.0$
• Sub-Main Sewer: $3.0$
• Main Sewer: $2.5$
• Outfall Sewer: $2.0$

Peak Factor ($PF$) Values by Population:

• Up to $20,000$: $3.0$
• $20,000\text{ to }50,000$: $2.5$
• $50,000\text{ to }75,000$: $2.25$
• Greater than $750,000$: $2.0$

Estimation of Storm Water Quantity

1. Rational Method

Used for small catchments ($\le 400\text{ hectares}$).

$Q = \frac{C I A}{360}$

Where:

• $Q$ = Storm water runoff ($\text{m}^3/\text{sec}$)
• $C$ = Runoff coefficient (overall dimensionless composite value)
• $I$ = Rainfall intensity ($\text{mm/hr}$)
• $A$ = Catchment area ($\text{hectares}$, where $1\text{ ha} = 10,000\text{ m}^2$)

Alternative units formulation:

• If $Q$ is in $\text{Litres/sec}$: $Q = 2.8 C I A$
• If $Q$ is in $\text{m}^3/\text{hr}$: $Q = 10 C I A$

Composite Runoff Coefficient ($C$): $C = \frac{C_1 A_1 + C_2 A_2 + C_3 A_3 + \dots + C_n A_n}{A_1 + A_2 + A_3 + \dots + A_n}$

Runoff Coefficients ($C$) by Area Type:

• Commercial and Industrial Areas: $0.70\text{ to }0.90$ ($70\%\text{ to }90\%$ impervious)
• Residential Area (High Density): $0.60\text{ to }0.75$ ($60\%\text{ to }75\%$ impervious)
• Residential Area (Low Density): $0.35\text{ to }0.60$ ($35\%\text{ to }60\%$ impervious)
• Parks and Underdeveloped Areas: $0.10\text{ to }0.20$ ($10\%\text{ to }20\%$ impervious)

Rainfall Intensity ($I$) Equations:

• British Ministry of Health Formulas ($I$ in $\text{mm/hr}$, $t$ in $\text{minutes}$):
  • For storm durations $t = 5\text{ to }20\text{ minutes}$:     $I = \frac{760}{t + 10}$
  • For storm durations $t = 20\text{ to }100\text{ minutes}$:     $I = \frac{1020}{t + 20}$
• Kuichling's Formulas:
  • For 10-year storm frequency: $I = \frac{2667}{t + 20}$
  • For 15-year storm frequency: $I = \frac{3048}{t + 20}$

Time of Concentration ($t_c$): $\text{Time of Concentration } (t_c) = \text{Inlet Time } (t_{\text{inlet}}) + \text{Travel Time in Sewer } (t_{\text{flow}})$$t_{\text{flow}} = \frac{\text{Sewer Length}}{\text{Flow Velocity in Sewer}}$$\text{Weighted Inlet Time } (t_{\text{inlet}}) = \frac{\sum (A_i \times t_{\text{inlet}, i})}{\sum A_i}$

Assumptions and Limitations of the Rational Method:

1. Restricted to small catchment areas ($\le 400\text{ ha}$).
2. Rainfall intensity is uniform over the entire catchment area.
3. Storm duration equals the time of concentration ($t = t_c$).
4. Runoff is dominated by overland sheet flow.
5. Catchment surface slope is not explicitly parameterized.
6. Selection of runoff coefficients and intensities is tedious for heterogeneous large areas.
7. Basin channel storage effects are neglected.

2. Empirical Formulas

• Burkli-Ziegler Formula:   $Q = 0.0173 C I A \sqrt[4]{\frac{S}{A}}$   Where $Q$ is in $\text{L/s}$, $C = 50\text{ to }90$ (average $70$), $I$ in $\text{mm/hr}$, $A$ in $\text{ha}$, $S$ is slope in $\text{mm per }1000\text{ m}$.
• Mc-Math Formula:   $Q = 2.30 C A I \sqrt[5]{\frac{S}{A}}$   Where $C = 0.3\text{ to }0.9$ (developed for St. Louis, USA).
• Metcalf and Eddy Formula:   $Q = 215.67 (A)^{0.73}$
• Fanning Formula:   $Q = 3125 A^{5/8}$ ($Q$ in $\text{L/s}$, $A$ in $\text{km}^2$).
• Talbat Formula:   $Q = 87000 A^{1/4}$ ($Q$ in $\text{L/s}$, $A$ in $\text{km}^2$).
• Dickens Formula:   $Q = C A^{3/4}$ ($Q$ in $\text{m}^3/\text{sec}$, $A$ in $\text{km}^2$, maximum $C = 35$).
• Ryves Formula:   $Q = C A^{2/3}$ ($Q$ in $\text{m}^3/\text{sec}$, $A$ in $\text{km}^2$).
• Inglis Formula:   $Q = \frac{124 A}{\sqrt{A + 10.4}}$ ($Q$ in $\text{m}^3/\text{sec}$, $A$ in $\text{km}^2$).

Solved Mathematical Examples

Example 1

Determine design discharge for separate and combined systems for a town of $45,000$ population, area $20\text{ ha}$, water supply rate $150\text{ lpcd}$, runoff coefficient $0.45$, time of concentration $30\text{ min}$, sewer slope $1\text{ in }500$, Manning's $n = 0.013$. Also find combined sewer diameter.

Solution:

1. Dry Weather Flow ($Q_{\text{DWF}}$):    $Q_{\text{DWF}} = 45000 \times 150 \times 0.8 \times 2.75 \text{ (taking } PF=2.75) = 14,850,000\text{ L/day}$$Q_{\text{DWF}} = \frac{14850000}{3600 \times 24 \times 1000} = 0.171\text{ m}^3/\text{sec}$
2. Wet Weather Flow ($Q_{\text{WWF}}$):    For $t = 30\text{ min}$ ($>20\text{ min}$):    $I = \frac{1020}{t + 20} = \frac{1020}{30 + 20} = 20.4\text{ mm/hr}$$Q_{\text{WWF}} = \frac{C I A}{360} = \frac{0.45 \times 20.4 \times 20}{360} = 0.51\text{ m}^3/\text{sec}$
3. Combined System Discharge:    $Q_{\text{combined}} = Q_{\text{DWF}} + Q_{\text{WWF}} = 0.171 + 0.51 = 0.681\text{ m}^3/\text{sec}$
4. Diameter Determination ($D$):    $Q = \frac{1}{n} A R^{2/3} S^{1/2}$    For pipe flowing full: $A = \frac{\pi D^2}{4}$, $R = \frac{D}{4}$$0.681 = \frac{1}{0.013} \left(\frac{\pi D^2}{4}\right) \left(\frac{D}{4}\right)^{2/3} \left(\frac{1}{500}\right)^{1/2}$$0.6351 = D^{8/3} \implies D = 0.8434\text{ m}$

Example 2

Catchment area $45\text{ ha}$. Composition: Forest ($10\%$, $C=0.15$), Open ground ($10\%$, $C=0.20$), Parks ($15\%$, $C=0.15$), Gravel road ($20\%$, $C=0.25$), Asphalt ($20\%$, $C=0.85$), Roofs ($25\%$, $C=0.90$). Entry time $20\text{ min}$, flow time $10\text{ min}$. Calculate storm water flow.

Solution:

1. Composite Runoff Coefficient ($C$):    $C = (0.15 \times 0.10) + (0.20 \times 0.10) + (0.15 \times 0.15) + (0.25 \times 0.20) + (0.85 \times 0.20) + (0.90 \times 0.25)$$C = 0.015 + 0.020 + 0.0225 + 0.050 + 0.170 + 0.225 = 0.5025$
2. Time of Concentration ($t$): $t = 20 + 10 = 30\text{ min}$.
3. Rainfall Intensity ($I$): $I = \frac{1020}{30 + 20} = 20.4\text{ mm/hr}$.
4. Storm Water Quantity:    $Q_{\text{storm}} = \frac{0.5025 \times 20.4 \times 45}{360} = 1.281\text{ m}^3/\text{sec}$

Example 3

Population $30,000$, area $60\text{ ha}$, $C = 0.60$, entry time $25\text{ min}$, travel time $5\text{ min}$, water supply rate $45\text{ lpcd}$, peak factor $2.7$. Calculate design discharge for combined sewer.

Solution:

1. Sanitary Flow ($Q_{\text{DWF}}$ with PF):    $Q_{\text{DWF}} = \frac{2.7 \times 30000 \times 45 \times 0.8}{24 \times 3600 \times 1000} = 0.03375\text{ m}^3/\text{sec}$
2. Time of Concentration: $t = 25 + 5 = 30\text{ min}$.
3. Rainfall Intensity: $I = \frac{1020}{30 + 20} = 20.4\text{ mm/hr}$.
4. Storm Discharge:    $Q_{\text{storm}} = \frac{0.60 \times 20.4 \times 60}{360} = 2.04\text{ m}^3/\text{sec}$
5. Total Design Discharge: $Q_{\text{total}} = 2.04 + 0.03375 = 2.074\text{ m}^3/\text{sec}$.

Example 4

Residential district area $200 \times 10^4\text{ m}^2$ ($200\text{ ha}$). Surface breakout: $40\%$ roofs/pavements ($C=0.85$), $40\%$ lawns/gardens ($C=0.20$), $20\%$ ground surface ($C=0.10$). Intensity $I = 40\text{ mm/hr}$. Calculate $C$ and storm sewage flow.

Solution:

1. Overall Coefficient ($C$):    $C = (0.40 \times 0.85) + (0.40 \times 0.20) + (0.20 \times 0.10) = 0.34 + 0.08 + 0.02 = 0.44$
2. Storm Sewage Discharge:    $Q = \frac{0.44 \times 40 \times 200}{360} = 9.78\text{ m}^3/\text{sec}$

Example 5

Calculate combined sewage quantity for:

• Area = $5\text{ ha}$, Population density = $700\text{ persons/ha}$ (Total pop = $3500$)
• $60\%$ pavement ($C_1 = 0.8$), $40\%$ open ground ($C_2 = 0.3$)
• Water supply rate = $165\text{ lpcd}$, Time of concentration = $10\text{ min}$, Peak factor = $3$

Solution:

1. Composite Coefficient ($C$): $C = (0.6 \times 0.8) + (0.4 \times 0.3) = 0.48 + 0.12 = 0.60$
2. Intensity ($I$): For $t = 10\text{ min}$, $I = \frac{760}{10 + 10} = 38\text{ mm/hr}$
3. Storm Discharge ($Q_{\text{WWF}}$): $Q_{\text{WWF}} = \frac{0.6 \times 38 \times 5}{360} = 0.3167\text{ m}^3/\text{sec}$
4. Sanitary Sewage ($Q_{\text{DWF}}$):    $Q_{\text{DWF}} = \frac{3 \times 0.8 \times 3500 \times 165}{24 \times 3600 \times 1000} = 0.016041\text{ m}^3/\text{sec}$
5. Total Combined Discharge: $Q = 0.3167 + 0.016041 = 0.3327 \approx 0.33\text{ m}^3/\text{sec}$

Example 6

Town area = $150\text{ ha}$. Breakdown: $20\%$ hard pavements ($C=0.85$), $15\%$ roofs ($C=0.80$), $30\%$ barren land ($C=0.20$), $20\%$ garden/lawns ($C=0.20$), $15\%$ forest ($C=0.15$). Intensity $I = 45\text{ mm/hr}$, Density = $300\text{ persons/ha}$, Water supply = $150\text{ lpcd}$, Peak factor = $2.7$. Find $C$ and discharge for (i) Separate, (ii) Combined, and (iii) Partially Combined systems.

Solution:

1. Composite Coefficient ($C$):    $C = (0.85 \times 0.20) + (0.80 \times 0.15) + (0.20 \times 0.30) + (0.20 \times 0.20) + (0.15 \times 0.15)$$C = 0.170 + 0.120 + 0.060 + 0.040 + 0.0225 = 0.4125 \approx 0.4128$
2. Storm Runoff ($Q_{\text{WWF}}$):    $Q_{\text{WWF}} = \frac{0.4128 \times 45 \times 150}{360} = 7.74\text{ m}^3/\text{sec}$
3. Sanitary Sewage ($Q_{\text{DWF}}$):    Total Population = $300 \times 150 = 45,000$$Q_{\text{DWF}} = \frac{2.7 \times 45000 \times 150 \times 0.8}{24 \times 3600 \times 1000} = 0.169\text{ m}^3/\text{sec}$
4. System Discharges:
   • Separate System: Sanitary Sewer $Q = 0.169\text{ m}^3/\text{sec}$; Storm Drain $Q = 7.74\text{ m}^3/\text{sec}$
   • Combined System: $Q = 7.74 + 0.169 = 7.909 \approx 7.91\text{ m}^3/\text{sec}$
   • Partially Combined System: Closed sewer carries $Q_{\text{DWF}} = 0.169\text{ m}^3/\text{sec}$; Open drain carries storm water $Q = 7.74\text{ m}^3/\text{sec}$.

Characteristics, Decomposition, and Examination of Sewage

Wastewater Sampling

Sampling must yield a representative volume that reflects true pollutant concentrations.

• Grab Sample: A single discrete sample collected at a specific spot and specific time. Reflects conditions only at the exact sampling moment. Suitable for consistent water quality sources (e.g., protected groundwater).
• Composite Sample: A combination of multiple grab samples collected at fixed time intervals (typically over $24\text{ hours}$) proportional to flow rate. Used for volatile or variable industrial and municipal wastewater streams.
• Preservation and Storage: Wastewater changes rapidly due to microbial activity. Samples must be analyzed immediately or stored in ice boxes. Chemical preservatives include chloroform, formaldehyde, and concentrated sulphuric acid ($\text{H}_2\text{SO}_4$). Crucial Rule: Chemical preservatives must never be added to samples designated for Biochemical Oxygen Demand (BOD) testing.

Biochemical Oxygen Demand (BOD)

BOD is the quantity of dissolved oxygen required by microorganisms to biochemically decompose biodegradable organic matter in sewage under aerobic conditions at a specified standard temperature ($20^\circ\text{C}$).

Applications of BOD Test Results:

1. Sizing wastewater treatment units.
2. Determining oxygen requirements for organic stabilization.
3. Evaluating treatment plant efficiency.
4. Measuring sewage strength.
5. Calculating dilution water requirements for river discharge.

Carbonaceous vs. Nitrogenous Stages of BOD

• First Stage BOD (Carbonaceous Demand): Biological breakdown of carbon compounds into $\text{CO}_2$, $\text{H}_2\text{O}$, and simple amine compounds. Lasts for $7\text{ to }10\text{ days}$.
• Second Stage BOD (Nitrogenous Demand): Oxidation of nitrogenous matter (ammonia, organic nitrogen) into nitrites ($\text{NO}_2^-$) and nitrates ($\text{NO}_3^-$) by nitrifying bacteria. Takes place after $8\text{ to }10\text{ days}$ and continues over extended periods.

Derivation of First-Stage BOD Equation

De-oxygenation rate is directly proportional to the organic matter remaining at time $t$ ($L_t$):

$\frac{dL_t}{dt} = -k' L_t$

Integrating from $t = 0$ (where $L_t = L_0$) to time $t$:

$\int_{L_0}^{L_t} \frac{dL_t}{L_t} = -k' \int_0^t dt \implies \ln\left(\frac{L_t}{L_0}\right) = -k' t$$\frac{L_t}{L_0} = e^{-k' t}$

Converting to Base-10 log format (where $k = \frac{k'}{2.303}$):

$\log_{10}\left(\frac{L_t}{L_0}\right) = -k t \implies L_t = L_0 10^{-kt}$

The BOD exerted ($y_t$) at time $t$ is the difference between ultimate initial BOD ($L_0$) and remaining BOD ($L_t$):

$y_t = L_0 - L_t = L_0 (1 - 10^{-kt})$

Temperature Correction for Rate Constant ($K$): $K_T = K_{20} (1.047)^{T - 20}$ Where $K_T$ is the rate constant at temperature $T^\circ\text{C}$ and $K_{20}$ is the rate constant at $20^\circ\text{C}$.

Chemical Oxygen Demand (COD)

COD is the quantity of oxygen consumed to chemically oxidize both biodegradable and non-biodegradable organic matter using a strong chemical oxidizing agent (potassium dichromate, $\text{K}_2\text{Cr}_2\text{O}_7$) in an acid medium.

• Fast analysis ($3\text{ hours}$ compared to $5\text{ days}$ for BOD).
• COD values are always higher than BOD values ($\text{COD} > \text{BOD}$).
• Used for process operational control, whereas BOD is used for regulatory standards.

Decomposition Processes

• Aerobic Decomposition: Breakdown carried out by aerobic and facultative bacteria in the presence of dissolved oxygen. End products are non-offensive and stable: Nitrates ($\text{NO}_3^-$), Carbon Dioxide ($\text{CO}_2$), Sulphate ($\text{SO}_4^{2-}$), Water ($\text{H}_2\text{O}$), and energy. Occurs in oxidation ponds, trickling filters, aerated lagoons.
• Anaerobic Decomposition (Putrefaction): Breakdown carried out by anaerobic and facultative bacteria in the complete absence of free dissolved oxygen. Oxygen is extracted from organic radical bounds ($\text{NO}_3^-$, $\text{SO}_4^{2-}$). End products are offensive gases and unstable solids: Hydrogen Sulphide ($\text{H}_2\text{S}$), Ammonia ($\text{NH}_3$), Methane ($\text{CH}_4$), Nitrogen ($\text{N}_2$). Occurs in septic tanks, sludge digesters.

Comparison: Aerobic vs. Anaerobic Decomposition

Characteristics of Sewage

1. Physical Characteristics

• Color: Fresh sewage is light brown. After $6\text{ hours}$ of aerobic breakdown, it becomes medium grey. Aged anaerobic sewage becomes dark grey or black.
• Odor: Fresh sewage is practically odorless. Stale sewage emits $\text{H}_2\text{S}$ (rotten egg odor). Causes loss of appetite, nausea, impaired respiration.
• Temperature: Typical wastewater temperatures are slightly higher than domestic water supplies. Biological activity increases up to $60^\circ\text{C}$. High temperatures decrease gas solubility (lower DO) and reduce fluid viscosity.
• Turbidity and Solids: Total solids ($TS$) are categorized into:
  • Colloidal Solids: Finely divided non-settleable particles; removed by coagulation/biological oxidation.
  • Suspended Solids: Settleable and non-settleable solids.
  • Dissolved Solids: Solutes that pass through filter media.

2. Chemical Characteristics

• pH: Fresh sewage is slightly alkaline ($\text{pH } 7.3\text{ to }7.5$). As organic acids form during decomposition, pH drops (acidic). Extreme pH ($< 7$ or $> 7$ significantly) indicates heavy industrial waste discharges.
• Solids Composition: Organic solids (carbon, hydrogen, oxygen) volatilize completely at $550^\circ\text{C}$ for $20\text{ minutes}$ (Volatile Solids). Inorganic solids (nitrogen, chlorine, salts) remain as ash (Fixed Solids).
• Nitrogen Content: Present as Organic Nitrogen, Free Ammonia ($\text{NH}_3$), Ammonium ($\text{NH}_4^+$), Nitrites ($\text{NO}_2^-$), and Nitrates ($\text{NO}_3^-$). High nitrogen accelerates aquatic algae growth.

3. Biological Characteristics

Sewage contains bacteria, protozoa, fungi, algae, and viruses. Bacteria are grouped by oxygen demand into Aerobic, Anaerobic, and Facultative.

Sewage Testing and Examination Methods

• Physical Tests:
  • Temperature: Ordinary thermometer ($^\circ\text{C}$).
  • Color: Tintometer.
  • Odor: Threshold Odor Number ($TON$).
  • Turbidity: Turbidity Rod or Jackson's Turbidimeter.
• Solids Laboratory Procedures:
  1. Total Solids ($TS$):      Evaporate known sample volume $V$ in crucible of initial weight $W_1$ at $130\text{ to }150^\circ\text{C}$ for $24\text{ hours}$. Weigh dry crucible ($W_2$).      $TS = \frac{W_2 - W_1}{V} \quad (\text{mg/L})$
  2. Total Volatile Solids ($TVS$) and Fixed Solids ($TFS$):      Ignite the dry residue from $\text{TS}$ testing in a muffle furnace at $500 \pm 50^\circ\text{C}$ for $2\text{ hours}$. Cool and weigh ($W_3$).      $TVS = \frac{W_2 - W_3}{V} \quad (\text{mg/L})$$TFS = TS - TVS \quad (\text{mg/L})$
  3. Total Suspended ($TSS$) and Dissolved Solids ($TDS$):      Filter volume $V$ through Whatman No. 44 filter paper. Evaporate filtrate at $130\text{ to }150^\circ\text{C}$ to determine $TDS = \frac{W_{2,\text{filt}} - W_{1,\text{filt}}}{V}$. Calculate $TSS = TS - TDS$.
  4. Settleable Solids ($SES$):      Measured using an Imhoff Cone (1-liter capacity). Allow sample to settle undisturbed for $2\text{ hours}$. Settleable volume read directly off cone bottom ($\text{mL/L}$).
• Chemical Oxygen Demand ($COD$) Test:   Add known excess standard potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) and concentrated sulphuric acid ($\text{H}_2\text{SO}_4$) with silver sulphate catalyst to sample. Boil for $2\text{ hours}$. Titrate remaining dichromate against standard ferrous ammonium sulphate ($FAS$).
• BOD Dilution Test Procedure:
  1. Measure initial Dissolved Oxygen ($DO_i$) of sample/dilution water.
  2. Dilute wastewater with aerated nutrient-enriched dilution water.
  3. Incubate diluted sample at $20^\circ\text{C}$ in the dark for $5\text{ days}$.
  4. Measure final Dissolved Oxygen ($DO_f$).
  5. Calculate depletion: $\text{DO depleted} = DO_i - DO_f$
  6. Compute BOD:      $BOD_5 = \text{DO depleted} \times \text{Dilution Ratio}$$\text{Dilution Ratio} = \frac{\text{Volume of Diluted Sample}}{\text{Volume of Undiluted Wastewater}}$

Design, Hydraulics, and Construction of Sewers

Design Criteria for Sewers

• Specific gravity of sewage is taken as $1.0$.
• Design Period: $25\text{ to }30\text{ years}$.
• Minimum sewer pipe diameter: $15\text{ cm}$; recommended minimum: $20\text{ cm}$. Commercially available in increments up to $3\text{ m}$.
• Sewer Gradients: Selected to maintain velocities between self-cleansing (minimum) and non-scouring (maximum). Slopes range from $1:100$ (steep) to $1:20$ (flat).
• Pipe design assumes gravity open-channel flow running full at maximum peak discharge.

Velocity Criteria

1. Self-Cleansing Velocity

The minimum flow velocity required to keep solid particles in suspension and prevent deposition/silting at the sewer invert. Calculated via Camp Shield's Formula:

$V_{\text{min}} = \sqrt{\frac{8 K (S_g - 1) g d}{f}}$

Where:

• $d$ = Particle diameter
• $g$ = Acceleration due to gravity ($9.81\text{ m/s}^2$)
• $S_g$ = Specific gravity of solids ($1.2$ for organic, $2.65$ for inorganic)
• $f$ = Darcy-Weisbach friction factor ($\approx 0.03$)
• $K$ = Dimensionless particle characteristic constant ($0.04\text{ to }0.06$)

Minimum Self-Cleansing Velocities for Separate Systems:

• Diameter $15\text{ to }25\text{ cm}$: $1.0\text{ m/s}$
• Diameter $30\text{ to }60\text{ cm}$: $0.75\text{ m/s}$
• Diameter $> 60\text{ cm}$: $0.60\text{ m/s}$
• Combined Sewer Minimum: $0.75\text{ m/s}$

2. Non-Scouring (Maximum) Velocity

The maximum permissible flow velocity that prevents structural erosion, scouring, and abrasion of the interior sewer lining.

Maximum Non-Scouring Velocities by Material:

• Earthen channels: $0.6\text{ to }1.2\text{ m/s}$
• Ordinary brick lining: $1.2\text{ to }2.5\text{ m/s}$
• Cement concrete: $3.0\text{ to }4.5\text{ m/s}$
• Stoneware: $3.5\text{ to }4.5\text{ m/s}$
• Cast iron pipe: $2.5\text{ to }3.0\text{ m/s}$
• Vitrified clay and salt glazed: $4.5\text{ to }5.0\text{ m/s}$

Hydraulic Formulas

• Discharge Equation: $Q = A V$

• Manning's Formula:   $V = \frac{1}{n} R^{2/3} S^{1/2}$   Where $R = \frac{A}{P}$ (Hydraulic radius), $S$ = slope, $n$ = Manning's rugosity coefficient.      Manning's Rugosity Coefficient ($n$):

  • Fine soil / Sand: $0.020\text{ to }0.035$
  • Grass lined: $0.035\text{ to }0.060$
  • Brick work: $0.013\text{ to }0.015$
  • Cast iron: $0.012\text{ to }0.013$
  • Cement concrete (good): $0.013$; Cement concrete (fair): $0.015$
  • Salt glazed stoneware (good): $0.012$; (fair): $0.015$
  • Plastic pipe: $0.011$

• Hazen-Williams Formula:   $V = 0.849 C_H R^{0.63} S^{0.54}$Coefficient $C_H$ Values: Vitrified clay = $110$; Brick = $100$; Cast iron (old) = $100$; Cast iron (new) = $130$; Concrete/Plastic/AC = 120\text{ to }150$.\n\n- **Chezy's Formula**: \n  V = C \sqrt{R S}\n  Using Manning relation C = \frac{1}{n} R^{1/6}.\n - *Kutter's Formula for C*:\n    C = \frac{23 + \frac{0.00155}{S} + \frac{1}{n}}{1 + \left(23 + \frac{0.00155}{S}\right) \frac{n}{\sqrt{R}}}\n - *Bazin's Formula for C*:\n    C = \frac{157.6}{1 + \frac{m}{\sqrt{R}}}\n    Where m = 0.11$(very smooth),$m = 0.21$(smooth brick/concrete),$m = 0.83$(poor brickwork),$m = 1.54$(good earth channel),$m = 2.36 (ordinary earth channel).\n\n### Partial Flow Hydraulics in Circular Sewers\n\nLet D$= Full pipe diameter,$d$= Partial depth of flow,$\phi = Angle subtended by flow surface at center (degrees).\n\n- **Depth ratio**: \n  \frac{d}{D} = \frac{1}{2} \left(1 - \cos\left(\frac{\phi}{2}\right)\right)\n- **Proportional Area (\frac{a}{A})**:\n  \frac{a}{A} = \frac{\phi}{360} - \frac{\sin\phi}{2\pi}\n- **Proportional Wetted Perimeter (\frac{p}{P})**:\n  \frac{p}{P} = \frac{\phi}{360}\n- **Proportional Hydraulic Radius (\frac{r}{R})**:\n  \frac{r}{R} = 1 - \frac{360 \sin\phi}{2\pi \phi}\n- **Proportional Velocity (\frac{v}{V})**:\n  \frac{v}{V} = \frac{n}{n_d} \left(\frac{r}{R}\right)^{2/3}\n- **Proportional Discharge (\frac{q}{Q})**:\n  \frac{q}{Q} = \frac{a}{A} \frac{v}{V} = \left(\frac{a}{A}\right) \left(\frac{n}{n_d}\right) \left(\frac{r}{R}\right)^{2/3}\n\n*Equal Self-Cleansing Tractive Force Condition*:\nTo provide equal self-cleansing at partial depth, tractive shear forces must be equal (\tau_s = \tau \implies \gamma r S_s = \gamma R S):\n\n\frac{S_s}{S} = \frac{R}{r}\nv_s = \frac{1}{n_d} r^{2/3} S_s^{1/2} = V \left(\frac{n}{n_d}\right) \left(\frac{r}{R}\right)^{1/6}\n\n### Sewer Shapes\n\n1. **Circular Sewers**: Most widely used. Maximum hydraulic efficiency (least perimeter for given area); uniform stress distribution; easy to manufacture. Max diameter up to 1.5\text{ m}. Unsuitable for combined sewers with large flow variations because self-cleansing velocity drops at low DWF.\n2. **Egg-Shaped Sewers**: Depth is 1.5 \times \text{width}. Smaller bottom radius and larger top arch. Maintains high velocity during low DWF. Highly efficient for combined sewers. Demerits: Tedious construction, lower structural stability than circular sewers, expensive.\n3. **Non-Circular Sewers**: Used for soft foundation soils or constrained headroom.\n - *Semi-Elliptical*: Excellent load distribution for large sewers (> 2\text{ m}).\n - *Parabolic (Delta)*: Large capacity, lower normal flow level.\n - *Horse-Shoe*: Heavy top arch load support; wide flat invert.\n - *Rectangular (Box)*: Simple formwork, used in restricted headroom, RCC construction.\n - *U-Shaped*: Good hydraulic properties, easy maintenance for conduits > 1\text{ m} wide.\n\n### Sewer Pipe Materials\n\n1. **Salt Glazed Stoneware**: Manufactured from vitrified clay/shale at 1200^\circ\text{C}$. Salt added at high temperature creates a smooth glazed surface. Diameters:$80\text{ to }1000\text{ mm}$; lengths$600\text{ to }900\text{ mm}. Socket and spigot joints. Highly resistant to abrasion and acid corrosion. Weak in tension, heavy, brittle.\n2. **Cement Concrete (PCC / RCC)**: Widely used for medium/large sewers. PCC used for diameter \le 60\text{ cm}$under low pressure ($< 15\text{ m} head); RCC used for large diameters/pressures.\n - *Crown Corrosion Phenomenon*: Bacterial reduction of sulphate compounds under anaerobic invert deposits generates \text{H}2\text{S} gas:\n     \text{SO}_4^{2-} + \text{Organic matter} \xrightarrow{\text{Bacteria}} \text{S}^{2-} + \text{H}_2\text{O} + \text{CO}_2\n     \text{S}^{2-} + 2\text{H}^+ \rightarrow \text{H}_2\text{S}\n     \text{H}_2\text{S}$escapes into the sewer atmosphere, condenses on the wet crown surface, and is oxidized by *Thiobacillus thiooxidans* into sulphuric acid ($\text{H}_2\text{SO}_4):\n     \text{H}_2\text{S} + 2 \text{O}_2 \xrightarrow{\text{Thiobacillus}} \text{H}_2\text{SO}_4\n     \text{H}_2\text{SO}_4$attacks concrete ($\text{CaCO}_3), causing crown structural disintegration.\n3. **Cast Iron (CI)**: High structural, tensile, and flexural strength. Used for high internal/external pressure, wet soil, traffic vibrations, river crossings, and inverted siphons.\n\n### Solved Engineering Problems\n\n#### Example 1\nPopulation = 80,000$, water supply =$145\text{ lpcd}$, return factor =$0.8$, Manning's$n_d = 0.013$, slope =$1:400$, peak factor =$3$. Find minimum circular sewer diameter running at$0.75$depth ($\frac{d}{D} = 0.75).\n\n*Solution*:\n1. **Design Discharge (q)**:\n   q = \frac{80000 \times 145 \times 0.8 \times 3}{24 \times 3600 \times 1000} = 0.334\text{ m}^3/\text{sec}\n2. **Hydraulic Geometry at d/D = 0.75**:\n   \cos\left(\frac{\phi}{2}\right) = 1 - 2\left(\frac{d}{D}\right) = 1 - 2(0.75) = -0.5 \implies \frac{\phi}{2} = 120^\circ \implies \phi = 240^\circ\n   \sin\phi = \sin(240^\circ) = -0.866\n   a = \frac{\pi D^2}{4} \left(\frac{240}{360} - \frac{-0.866}{2\pi}\right) = 0.63 D^2\n   p = \pi D \left(\frac{240}{360}\right) = 2.094 D\n   r = \frac{a}{p} = \frac{0.63 D^2}{2.094 D} = 0.301 D\n3. **Discharge Equation**:\n   q = \frac{1}{n_d} a r^{2/3} S^{1/2}\n   0.334 = \frac{1}{0.013} (0.63 D^2) (0.301 D)^{2/3} \left(\frac{1}{400}\right)^{1/2}\n   0.334 = 76.923 \times 0.63 D^2 \times 0.4492 D^{2/3} \times 0.05\n   0.334 = 1.088 D^{8/3} \implies D^{8/3} = 0.3069 \implies D = 0.64\text{ m}\n\n#### Example 2\nDesign circular combined sewer: Area = 60\text{ ha}$, Max velocity =$3.2\text{ m/s}$, flow time =$18\text{ min}$, entry time =$5\text{ min}$,$C = 0.55$, Population =$65,000$, water supply =$235\text{ lpcd}.\n\n*Solution*:\n1. **Storm Flow**:\n   t = 18 + 5 = 23\text{ min}\n   I = \frac{1020}{23 + 20} = 23.72\text{ mm/hr}\n   Q{\text{storm}} = \frac{0.55 \times 23.72 \times 60}{360} = 2.174\text{ m}^3/\text{sec}\n2. **Sanitary Flow**:\n   Q_{\text{sewage}} = \frac{65000 \times 235 \times 0.8}{24 \times 3600 \times 1000} = 0.1414\text{ m}^3/\text{sec}\n   Taking peak factor 1.25\implies Q = 0.1767\text{ m}^3/\text{sec}\n3. **Combined Flow**: Q = 2.174 + 0.1767 = 2.350\text{ m}^3/\text{sec}\n4. **Sewer Dimensions**:\n   Using Q = \frac{1}{n} A R^{2/3} S^{1/2} = 2.350\n   For \frac{1}{S} = 600$,$n = 0.012\n   Solving gives D = 1.35\text{ m}$, Area =$1.424\text{ m}^2$, Velocity =$1.649\text{ m/s}.\n\n#### Example 3\nFind full diameter of sanitary sewer: Population = 90,000$, supply =$180\text{ lpcd}$,$n = 0.025$, Slope =$1:500$,$R.F. = 0.8.\n\n*Solution*:\n1. **Average Sanitary Discharge**:\n   Q = \frac{90000 \times 180 \times 0.8}{24 \times 3600 \times 1000} = 0.15\text{ m}^3/\text{sec}\n2. **Velocity**: V = \frac{1}{0.025} \left(\frac{D}{4}\right)^{2/3} \left(\frac{1}{500}\right)^{1/2} = 0.713 D^{2/3}\n3. **Discharge Equation**:\n   Q = A V \implies 0.15 = \left(\frac{\pi D^2}{4}\right) (0.713 D^{2/3}) = 0.560 D^{8/3}\n   D^{8/3} = 0.2679 \implies D = 0.611\text{ m}\n\n#### Example 4\nCalculate size of circular sewer to carry Q = 0.6\text{ m}^3/\text{sec}$running half full ($d = D/2$). Slope =$1:1000$,$n = 0.013$.

Solution: For half-full flow: $a = \frac{\pi D^2}{8}$, $p = \frac{\pi D}{2}$, $r = \frac{D}{4}$. $Q = a V = \left(\frac{\pi D^2}{8}\right) \frac{1}{0.013} \left(\frac{D}{4}\right)^{2/3} \left(\frac{1}{1000}\right)^{1/2}$$0.6 = (0.3927 D^2) (76.923) (0.3968 D^{2/3}) (0.03162)$$0.6 = 0.3791 D^{8/3} \implies D^{8/3} = 1.5827 \implies D = 1.186\text{ m}$(Note: If solved with $\pi D^2/8 \times 0.969 D^{2/3} = 0.6 \implies D = 1.812\text{ m}$ under high-roughness parameter).

Example 5

Circular sewer $D = 1\text{ m}$, slope = $1:400$, $n = 0.012$. Find velocity and half-full discharge.

Solution: For half-full: $r = \frac{D}{4} = 0.25\text{ m}$, $a = \frac{\pi (1)^2}{8} = 0.3927\text{ m}^2$$V = \frac{1}{0.012} (0.25)^{2/3} \left(\frac{1}{400}\right)^{1/2} = \frac{1}{0.012} (0.3968) (0.05) = 1.653\text{ m/s}$$Q = a V = 0.3927 \times 1.653 = 0.649\text{ m}^3/\text{sec}$

Sewer Construction Procedure

1. Setting out Centerline: Centerline pegs are driven at $7.5\text{ m}$, $15\text{ m}$, or $30\text{ m}$ intervals starting from downstream (tail end).
   • Method 1 (Offset Line): Parallel line offset $2\text{ to }3\text{ m}$ from centerline to clear traffic.
   • Method 2 (Sight Rails): Vertical posts driven on both sides of trench supporting horizontal cross-rails.
2. Alignment and Gradient Control: Maintained using Boning Rods (Travelers) and sight rails. A cord stretched between consecutive sight rails represents an imaginary line parallel to the invert slope.
3. Excavation and Timbering: Trench excavated to minimum depth $0.9\text{ m}$ above sewer, width $0.6\text{ m}$. For depths $> 1.5\text{ to }2.0\text{ m}$, vertical or horizontal timber sheeting and struts prevent side cave-ins.
4. Laying and Jointing: Pipe bedded on concrete or firm soil, aligned with plumb bobs, and jointed (cement mortar, bitumen, rubber rings).
5. Sewer Testing:
   • Straightness & Obstruction Test: A smooth wooden ball (diameter $13\text{ mm}$ less than pipe internal diameter) is rolled down the invert, or light from a lamp at one end is viewed via a mirror at the opposite end.
   • Water Test (Exfiltration Test): Low end plugged, high end fitted with funnel/air vent. Filled with water under head. Loss after $30\text{ min}$ must not exceed $2\text{ Litres per cm diameter per km length}$.
   • Air Test: Used for large pipes. Test section pressurized to $100\text{ mm}$ water column. Pressure drop must remain within $75\text{ mm}$ water column over test interval.
6. Backfilling: Excavated earth free of stones is backfilled and compacted in layers after test approval.

Primary, Secondary, and Biological Sewage Treatment Processes

Wastewater treatment converts harmful organic and inorganic impurities into stable solids, yielding safe effluent.

Process Classification

1. Preliminary Processes: Screening (removes large floating solids), Grit Removal (removes heavy inorganic discrete solids), Skimming (removes fats, oils, grease).
2. Primary Treatment: Plain Sedimentation and Chemical Precipitation. Removes settleable organic solids ($50\%\text{ to }70\%$ TSS removal, $25\%\text{ to }40\%$ BOD removal).
3. Secondary (Biological) Treatment: Microbial decomposition of dissolved and colloidal volatile organic matter under aerobic or anaerobic conditions.
   • Suspended Growth Processes: Microorganisms suspended in liquid mix via aeration/agitation. (e.g., Activated Sludge Process, Aerated Lagoons, Anaerobic Digesters).
   • Attached Growth (Fixed-Film) Processes: Microorganisms attached to inert media (rock, slag, plastic). (e.g., Trickling Filters, Rotating Biological Contactors).
   • Combined Processes: Hybrid systems (e.g., Trickling filter with activated sludge).
4. Tertiary (Advanced) Treatment: Removal of specific recalcitrant compounds, nutrients (N, P), heavy metals, pathogens.

Primary Treatment Units

1. Racks and Screens

Traps coarse floating debris (plastics, rags, leaves, dead animals).

• Coarse Screens (Racks): Clear opening between bars $> 50\text{ mm}$ (typically $75\text{ to }150\text{ mm}$). Collects $\approx 7.5\text{ m}^3$ material per million $\text{m}^3$ sewage.
• Medium Screens: Bar spacing $20\text{ to }50\text{ mm}$. Collects $40\text{ to }120\text{ m}^3$ per million $\text{m}^3$ sewage.
• Fine Screens: Perforated plates or wire mesh with openings $< 20\text{ mm}$ ($0.7\text{ to }1.5\text{ mm}$). Removes up to $20\%$ TSS.

2. Grit Chamber

Long narrow channels designed to settle out inorganic discrete grit (sand, gravel, ash, eggshells; diameter $\ge 0.2\text{ mm}$, specific gravity $S_g \approx 2.65$) while keeping lighter organic particles in suspension.

• Settling Velocity ($V_s$) (Transition Zone formula for $d = 0.1\text{ to }1.0\text{ mm}$):      $V_s = 60.6 (S_g - 1) d \left(\frac{3T + 70}{100}\right)$   Where $V_s$ is in $\text{mm/s}$, $d$ in $\text{mm}$, $T$ in $^\circ\text{C}$.

• Surface Overflow Rate ($SOR$ or $V_0$): $V_0 = \frac{Q}{A_s} = \frac{Q}{B \times L}$. Efficiency $\eta = \frac{V_s}{V_0}$.

• Design Parameters:

  • Detention Time: $45\text{ to }90\text{ seconds}$.
  • Horizontal Flow Velocity ($V_h$): $0.15\text{ to }0.30\text{ m/s}$.
  • Critical Scouring Velocity ($V_{sc}$):     $V_{sc} = \sqrt{\frac{8 K (S_g - 1) g d_g}{f}}$     For inorganic grit ($S_g=2.65$, $K=0.04$, $f=0.03$): $V_h < V_{sc,\text{grit}}$.     For organic matter ($S_g=1.2$, $K=0.06$): $V_h > V_{sc,\text{org}}$ (prevents organic deposition).

3. Skimming Tank

Removes light floating substances (oil, grease, soap, vegetable debris).

• Compressed air diffusers at bottom ($0.2\text{ m}^3\text{ air per m}^3\text{ sewage}$) coagulate grease and float scum into side troughs.
• Surface Area Equation: $A = \frac{0.00622 q}{V_r}$, where $V_r = 0.25\text{ m/min}$ (minimum rising velocity).
• Detention Time: $3\text{ to }5\text{ minutes}$.

4. Primary Sedimentation Tank (PST)

Settleable organic solids removal under gravity.

• Settling Regimes:
  • Type 1 (Discrete Settling): Unhindered settling of non-flocculating particles (Grit chambers).
  • Type 2 (Flocculent Settling): Particles coalesce during settling, increasing mass and velocity (PST).
  • Type 3 (Hindered / Zone Settling): High solids concentration; particles settle as a contiguous mass matrix (Secondary Settling Tanks).
  • Type 4 (Compression Settling): Mechanical compaction of lower sludge matrix (Bottom of SST / Thickeners).
• PST Design Criteria:
  • Detention Period: $1.5\text{ to }3.0\text{ hours}$ (plain sedimentation); $2.0\text{ to }2.5\text{ hours}$ (coagulated).
  • Flow Velocity: $30\text{ to }60\text{ cm/min}$.
  • Tank Dimensions: Rectangular $L:B = 3:1\text{ to }5:1$ ($L \le 30\text{ to }100\text{ m}$, $B = 6\text{ to }10\text{ m}$); Circular Diameter $\le 60\text{ m}$.
  • Effective Liquid Depth: $\le 3.5\text{ m}$.
  • Surface Overflow Rate ($SOR$): Plain $12,000\text{ to }18,000\text{ L/day/m}^2$ ($12\text{ to }18\text{ m}^3/\text{m}^2/\text{day}$); Flocculated $24,000\text{ to }30,000\text{ L/day/m}^2$
  • Freeboard: $0.3\text{ to }0.5\text{ m}$.

Solved Mathematical Examples for Primary Units

Example 1 (Grit Chamber Design)

Flow = $55 \times 10^6\text{ L/day}$ ($0.636\text{ m}^3/\text{s}$), $S_g = 2.65$, $T = 27^\circ\text{C}$, particle diameter $d = 0.21\text{ mm}$, $K = 0.06$, $f = 0.03$. Design grit chamber.

Solution:

1. Detention time $t = 60\text{ seconds}$.
2. Volume $V = 0.636 \times 60 = 38.16\text{ m}^3$.
3. Settling Velocity ($V_s$):    $V_s = 60.6 (2.65 - 1) \times 0.21 \times \left(\frac{3(27) + 70}{100}\right) = 31.7\text{ mm/s} = 0.0317\text{ m/s}$
4. Depth $D = V_s \times t = 0.0317 \times 60 = 1.9\text{ m}$.
5. Total Plan Area $A = \frac{38.16}{1.9} = 20.1\text{ m}^2$.    Provide 2 units of area $10.5\text{ m}^2$ each. For width $B = 1.0\text{ m}$, Length $L = 10.5\text{ m}$.    Total depth = $1.9 + 0.25 \text{ (grit storage)} + 0.25 \text{ (freeboard)} = 2.4\text{ m}$.
6. Critical Velocity Checks:    $V_h = \frac{0.636 / 2}{1.0 \times 1.9} = 0.167\text{ m/s}$$V_{sc,\text{grit}} = \sqrt{\frac{8(0.06)(2.65 - 1)(9.81)(0.00021)}{0.03}} = 0.233\text{ m/s}$$V_{sc,\text{org}} = \sqrt{\frac{8(0.06)(1.2 - 1)(9.81)(0.00021)}{0.03}} = 0.081\text{ m/s}$    Check: $V_{sc,\text{org}} (0.081) < V_h (0.167) < V_{sc,\text{grit}} (0.233)$. (Design OK).

Example 2 (Circular PST Design)

Population = $80,000$, water supply rate = $150\text{ lpcd}$, return factor = $0.7$, Peak factor = $2.5$, detention time = $2\text{ hours}$, $SOR = 48\text{ m}^3/\text{m}^2/\text{day}$.

Solution:

1. Average Sewage = $0.7 \times 80000 \times 150 \times 10^{-3} = 8400\text{ m}^3/\text{day}$.
2. Peak Flow $Q_{\text{max}} = 2.5 \times 8400 = 21,000\text{ m}^3/\text{day}$.
3. Tank Volume $V = 21000 \times \frac{2}{24} = 1750\text{ m}^3$.
4. Plan Area $A = \frac{Q_{\text{max}}}{SOR} = \frac{21000}{48} = 437.5\text{ m}^2$$\frac{\pi d^2}{4} = 437.5 \implies d = 23.62\text{ m}$.
5. Liquid Depth $D = \frac{V}{A} = \frac{1750}{437.5} = 4.0\text{ m}$.    Total Depth = $4.0 + 0.3 \text{ (FB)} + 1.0 \text{ (Sludge)} = 5.3\text{ m}$.