Environmental Engineering covers the planning, design and operation of systems that protect public health and the environment — from water demand estimation and treatment, through sewer design and wastewater treatment, to solid waste management and air & noise pollution control. Every design formula, IS code limit, treatment-process diagram, solved example and exam-pattern table is included.
After studying this chapter you will be able to:
Prerequisite: Irrigation Engineering (canal hydraulics and well-discharge principles developed there carry over directly into water-supply conduit and groundwater design here). Leads to: Surveying, which provides the field measurement and levelling techniques needed to lay out water supply and sewerage networks.
| Source Type | Examples | Quality | Suitability |
|---|---|---|---|
| Surface water | Rivers, lakes, tanks, reservoirs | Variable; turbid, may have bacteria | Large cities after treatment; cheapest to develop |
| Ground water | Wells, springs, infiltration galleries, tube wells | Generally better; may have hardness, iron | Rural and small towns; often requires only disinfection |
| Rainwater / Rooftop harvesting | Roof collection, check dams | Generally clean initially | Supplementary; arid zones |
| Sea water (desalination) | SWRO, MSF plants | High TDS (35,000 mg/L) | Coastal cities; expensive |
| Type of Use | IS 1172 Per Capita (lpcd) | Remarks |
|---|---|---|
| Domestic | 135 lpcd (cities >1 lakh) | Drinking, cooking, bathing, flushing; 70% is indoor |
| Industrial | 50 lpcd (allowance) | Highly variable; 45–450 kL/tonne of product |
| Commercial / Institutional | 20 lpcd | Hotels, hospitals, offices, schools |
| Fire demand | As per Kuichling's formula | Not added to average; governs pipe sizing at hydrants |
| Losses / Thefts | 15% of total | Leakage, metering error, illegal use |
| Total design | 170–200 lpcd (urban India) | Sum of above with safety factor |
Water demand is never constant — it varies by season, day, and hour. Design must accommodate the maximum hourly demand, not just the average, or pressure will fail during peak consumption periods.
The design period is the future period for which a component is designed to be adequate — 30 years for dams and large headworks, 15–20 years for treatment plants, and 10 years for distribution networks (since distribution is cheaper to expand incrementally). Population at the end of the design period must be forecast using arithmetic, geometric, incremental-increase, or logistic (S-curve) methods, chosen based on the city's growth pattern and available census history.
Fire demand is a short-duration but very high instantaneous flow requirement, governed by empirical formulae (Kuichling's, Freeman's, or the IS formula) based on population. It is not added directly to the average demand but instead governs the minimum pipe sizing needed to deliver adequate pressure and flow to hydrants during a fire event, typically for a duration of 4–8 hours per IS 3764.
Average Daily Demand: \( ADD = \dfrac{\text{Annual consumption}}{365} \)
Maximum Daily Demand: \( MDD = 1.8\times ADD \)
Maximum Hourly Demand: \( MHD = 1.5\times MDD = 2.7\times ADD \)
Minimum Hourly Demand \( = 0.5\times ADD \); Peak factor \( = \dfrac{MHD}{ADD} = 2.7 \)
Arithmetic growth: \( P_n = P_0 + n\times r \)
Geometric growth: \( P_n = P_0\times(1+\dfrac{r}{100})^n \)
Logistic (S-curve): \( P = \dfrac{P_s}{1+m\,e^{-kn}} \)
Kuichling's: \( Q = 3182\sqrt{P} \) (L/min; P in thousands)
Freeman's: \( Q = \dfrac{1136\,(P/5+10)}{10} \)
IS formula: \( Q = 100\sqrt{P} \) (m³/hr; P in thousands)
Given: A city has an average daily demand of \(15{,}000\ \dfrac{\text{m}^3}{\text{day}}\). Find the maximum hourly demand.
Solution:
\( MHD = 2.7\times ADD = 2.7\times15{,}000 = 40{,}500\ \dfrac{\text{m}^3}{\text{day}} \)
Answer: \(MHD = 40{,}500\ \dfrac{\text{m}^3}{\text{day}}\).
Given: A town has a population of \(64{,}000\). Estimate the fire demand using Kuichling's formula.
Solution:
\( Q = 3182\sqrt{P} = 3182\sqrt{64} = 3182\times8 = 25{,}456\ \text{L/min} \)
Answer: \(Q\approx25{,}456\ \text{L/min} \approx 1527\ \dfrac{\text{m}^3}{\text{hr}}\).
Given: A city's current population is \(80{,}000\) with an average decadal growth rate of \(15\%\). Forecast the population after 20 years (2 decades) using geometric growth.
Solution:
\( P_n = P_0(1+\dfrac{r}{100})^n = 80{,}000\times(1.15)^2 = 80{,}000\times1.3225 = 105{,}800 \)
Answer: Forecast population \(\approx105{,}800\).
Fig. 1.1 — Daily demand variation: hourly consumption rises above the average during peak hours, reaching a maximum hourly demand of about 2.7 times the average daily demand.
Water conduits are analysed using different equations depending on flow type. Manning's equation applies to gravity (open-channel or partly-full) flow. The Hazen-Williams equation is the standard for pressurised water distribution mains, using an empirical roughness coefficient \(C\) that is easy to look up for common pipe materials. The Darcy-Weisbach equation, using a friction factor from the Moody chart, is the most rigorous (dimensionally consistent) head-loss formula and is preferred for precise engineering analysis.
| Material | Advantages | Disadvantages | Use |
|---|---|---|---|
| Cast Iron (CI) | Strong, long life, joints easy | Heavy, brittle, corrodes over time | Distribution mains; 75–600 mm dia |
| Ductile Iron (DI) | Strong + ductile; better than CI | Expensive | Modern distribution; replacing CI |
| Steel (MS) | Light, strong, any size | Corrodes; needs lining or coating | Transmission mains; large dia |
| PVC / CPVC | Light, corrosion-proof, cheap | Limited temperature/pressure range | Small dia distribution; house connections |
| HDPE | Flexible, good chemical resistance | UV degradation if exposed | Rural supply; sliplined rehabilitation |
| Asbestos Cement (AC) | Smooth, light, cheap | Brittle; health concerns (asbestos) | Now largely phased out |
| Pre-stressed Concrete | Large dia; suitable for gravity mains | Heavy; joints critical | Large transmission mains >600 mm |
The tree/dead-end system branches out from a main pipe with no interconnections; it is simple and cheap but the dead ends cause stagnation and sediment buildup, and any single pipe failure cuts off supply downstream. The ring/grid system interconnects pipes into loops, so water can reach any point from multiple directions — eliminating stagnation and maintaining supply even if one section is shut for repair, at the cost of higher construction cost and the need for iterative network analysis (Hardy-Cross). The radial/circular system divides a city into zones, each fed by a circular main from a central point, giving uniform pressure and easy fault detection in planned cities.
Looped (ring/grid) networks cannot be solved directly since flow splits are unknown at each junction. The Hardy-Cross iterative method starts by assuming a set of pipe flows that satisfy continuity at every junction, then computes a flow correction for each loop based on the imbalance of head losses around that loop, and repeats until the corrections become negligibly small (typically 3–5 iterations).
A service (distribution) reservoir provides three components of storage: balancing storage to absorb the mismatch between the (constant) supply rate from the treatment plant and the (fluctuating) hourly demand; fire reserve, held specifically for fire-fighting duration; and emergency/breakdown reserve, held against unplanned outages in supply. The elevated tank's height above the service area must be enough to deliver a minimum residual pressure head to the highest and farthest consumer after accounting for pipe friction losses.
\( V = \dfrac{1}{n}R^{2/3}S^{1/2} \), \( Q = \dfrac{A}{n}R^{2/3}S^{1/2} \)
n: CI pipe 0.012–0.014; PVC 0.009–0.011; concrete 0.013–0.015
\( V = 0.8492\,C\,R^{0.63}S^{0.54} \)
\( Q = 0.2785\,C\,D^{2.63}S^{0.54} \)
C: CI new=130, old CI=100, steel=120, PVC=140–150
\( h_f = \dfrac{fLV^2}{2gD} \)
f = friction factor (Moody chart); typical 0.01–0.05
\( \Delta Q = -\dfrac{\Sigma h_L}{n\,\Sigma|h_L/Q|} \)
\(h_f = rQ^n\) (n=1.85 for HW, 2 for DW); convergence when \(\Delta Q < 0.001\ \dfrac{\text{m}^3}{\text{s}}\)
Storage = balancing + fire reserve + emergency reserve
Balancing storage \(\approx \dfrac{1}{3}\) daily demand; Emergency reserve \(\approx 25\%\) daily demand
Given: A new CI pipe (\(C=130\)) of diameter \(D=0.3\ \text{m}\) has a hydraulic slope \(S=0.004\). Find the discharge.
Solution:
\( Q = 0.2785\times130\times0.3^{2.63}\times0.004^{0.54} \)
\( 0.3^{2.63}\approx0.0605 \); \( 0.004^{0.54}\approx0.0487 \)
\( Q = 0.2785\times130\times0.0605\times0.0487\approx0.1067\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q\approx0.107\ \dfrac{\text{m}^3}{\text{s}}\).
Given: Water flows at \(V=1.5\ \text{m/s}\) through a \(D=0.2\ \text{m}\) pipe, length \(L=500\ \text{m}\), \(f=0.02\). Find the head loss.
Solution:
\( h_f = \dfrac{fLV^2}{2gD} = \dfrac{0.02\times500\times1.5^2}{2\times9.81\times0.2} = \dfrac{22.5}{3.924}\approx5.73\ \text{m} \)
Answer: \(h_f\approx5.73\ \text{m}\).
Given: A town's average daily demand is \(6000\ \text{m}^3\). Estimate the balancing storage.
Solution:
Balancing storage \(\approx \dfrac{1}{3}\times6000 = 2000\ \text{m}^3\)
Answer: Balancing storage \(\approx2000\ \text{m}^3\) (fire and emergency reserves must be added separately for total capacity).
Fig. 2.1 — Distribution layouts: Tree system (dead-ends; simple but unreliable) versus Ring/Grid system (interconnected loops; reliable; requires Hardy-Cross analysis).
| Aquifer Type | Description | Water Table | Example |
|---|---|---|---|
| Unconfined (phreatic) | Water table exposed to atmosphere; water table is upper boundary | Varies freely with seasons | Alluvial plains; dug wells |
| Confined (artesian) | Sandwiched between impervious layers; water under pressure | Piezometric surface > top of aquifer | Deep tube wells; artesian wells |
| Semi-confined (leaky) | Confining layer is semi-permeable; leakage from adjacent aquifer | Piezometric varies with leakage | Alluvial basins with clay lenses |
| Perched aquifer | Local impervious lens above main water table | Localised; above regional WT | Hillside springs; seasonal |
Darcy's law states that the flow rate of water through a porous medium is proportional to the hydraulic gradient and the cross-sectional area, governed by a proportionality constant called the hydraulic conductivity. It is valid only for laminar flow through the pores (low Reynolds number), and does not hold for coarse gravel or fractured rock, where velocities are high enough to produce turbulent flow.
When a well is pumped, water flows radially inward toward it, and the water table (unconfined) or piezometric surface (confined) is drawn down into a funnel shape called the cone of depression. The radius of influence \(R\) is the distance beyond which the pumping has no measurable effect on the water level. For a confined aquifer, the drawdown analysis uses the aquifer thickness \(b\) directly since the aquifer itself stays fully saturated even as the piezometric surface drops.
The Dupuit-Thiem equation for a fully-penetrating well in an unconfined aquifer under steady-state pumping relates discharge to the square of the difference between the undisturbed head and the pumped water level in the well. The analogous Thiem equation for a confined aquifer uses the linear head difference (not squared) because the saturated thickness of a confined aquifer doesn't change with pumping — only the pressure (piezometric head) does.
Steady-state (Thiem) equations assume the cone of depression has stopped growing, which is only true after very long pumping durations. The Theis equation instead models the time-dependent (transient) drawdown at any point using the well function \(W(u)\), which depends on the aquifer's storage coefficient in addition to transmissivity. For small values of the dimensionless parameter \(u\), the Cooper-Jacob approximation simplifies the well function to a straight-line semi-log relationship, making field data analysis much easier.
| Well Type | Depth | Dia | Method | Yield |
|---|---|---|---|---|
| Open / Dug well | 10–20 m | 1–6 m | Manual/mechanical excavation | Low; 50–500 L/hr |
| Tube well (deep) | 60–300 m | 100–450 mm | Rotary/cable tool drilling | High; 100–3000 m³/day |
| Infiltration gallery | 5–10 m | — | Horizontal perforated pipe near riverbank | Medium; induced recharge |
| Spring collection | At surface | — | Gravity collection of natural spring | Variable; gravity-fed |
| Collector (Ranney) well | 15–30 m | Large caisson | Radial horizontal laterals | Very high; 1000–50,000 m³/day |
\( Q = K\,i\,A \)
K=hydraulic conductivity; i=hydraulic gradient=dh/dL
\( Q = \dfrac{\pi K(H^2-h^2)}{\ln(R/r_w)} \)
Sichardt: \( R = 3000(H-h)\sqrt{K} \)
\( Q = \dfrac{2\pi Kb(H-h)}{\ln(R/r_w)} = \dfrac{2\pi T(H-h)}{\ln(R/r_w)} \)
\(T = Kb\) (transmissivity, m²/day)
\( s = \dfrac{Q}{4\pi T}W(u) \), \( u = \dfrac{r^2 S}{4Tt} \)
Cooper-Jacob (small u): \( s = \dfrac{2.303\,Q}{4\pi T}\log\!\left(\dfrac{2.25\,Tt}{r^2 S}\right) \)
Given: A soil sample has hydraulic conductivity \(K=0.02\ \text{cm/s}\), cross-sectional area \(A=100\ \text{cm}^2\), and hydraulic gradient \(i=0.05\). Find the discharge.
Solution:
\( Q = K\,i\,A = 0.02\times0.05\times100 = 0.1\ \dfrac{\text{cm}^3}{\text{s}} \)
Answer: \(Q=0.1\ \dfrac{\text{cm}^3}{\text{s}}\).
Given: An unconfined aquifer has \(K=40\ \text{m/day}\), \(H=25\ \text{m}\), \(h=15\ \text{m}\), \(R=200\ \text{m}\), well radius \(r_w=0.15\ \text{m}\). Find the discharge.
Solution:
\( Q = \dfrac{\pi\times40\times(25^2-15^2)}{\ln(200/0.15)} = \dfrac{\pi\times40\times400}{\ln(1333.3)} = \dfrac{50{,}265}{7.195}\approx6987\ \dfrac{\text{m}^3}{\text{day}} \)
Answer: \(Q\approx6987\ \dfrac{\text{m}^3}{\text{day}}\).
Given: A confined aquifer of thickness \(b=20\ \text{m}\) has hydraulic conductivity \(K=15\ \text{m/day}\). Find the transmissivity.
Solution:
\( T = Kb = 15\times20 = 300\ \dfrac{\text{m}^2}{\text{day}} \)
Answer: \(T=300\ \dfrac{\text{m}^2}{\text{day}}\).
Fig. 3.1 — Unconfined well under steady pumping: the cone of depression forms between the static water table and the pumped level in the well, extending out to the radius of influence R.
| Parameter | Unit | Permissible (IS 10500) | Desirable | Test Method |
|---|---|---|---|---|
| Turbidity | NTU | 5 NTU | <1 NTU | Nephelometer; Jackson turbidimeter |
| Colour | Hazen units | 15 HU | 5 HU | Visual comparison; spectrophotometer |
| Taste & Odour | TON (threshold) | Unobjectionable | — | Dilution-to-threshold method |
| Temperature | °C | 10–25°C (max 45°C) | 10–20°C | Thermometer |
| Total Dissolved Solids (TDS) | mg/L | 500 mg/L | — | Gravimetric (105°C evaporation) |
| Parameter | Permissible Limit (IS 10500:2012) | Health Effect if Exceeded |
|---|---|---|
| pH | 6.5 – 8.5 | Corrosion (<6.5); scaling (>8.5) |
| Total Hardness | 300 mg/L (max 600) | Scale in pipes/boilers; soap wastage; not harmful directly |
| Fluoride | 1.0 mg/L (max 1.5) | Dental fluorosis >1.5; skeletal fluorosis >3–6 |
| Nitrate (as NO₃) | 45 mg/L | Methaemoglobinaemia (blue baby syndrome) in infants |
| Arsenic | 0.01 mg/L | Skin cancer, arsenicosis (chronic); WHO 0.01 mg/L |
| Iron | 0.3 mg/L | Staining, taste; no direct health effect at low levels |
| Chloride | 250 mg/L (max 1000) | Salty taste; corrosion of pipes |
| Chlorine (residual) | 0.2 mg/L (min) | Below → bacterial growth; above 0.5 → taste/odour |
| Sulphate | 200 mg/L (max 400) | Laxative effect; concrete attack |
| Lead | 0.01 mg/L | Neurotoxin; cumulative; children most vulnerable |
Coliform organisms (with E. coli as the specific index organism) serve as indicator bacteria for faecal contamination — their presence signals a risk of pathogenic contamination even though coliforms themselves are mostly harmless. IS 10500 requires zero coliforms in 100 mL of treated water. The Most Probable Number (MPN) method uses statistical inference from multiple-dilution tube tests to estimate the coliform count, while the Membrane Filter (MF) technique physically filters a sample and counts colonies grown on selective media directly.
Temporary hardness (from calcium and magnesium bicarbonates) can be removed simply by boiling, which drives off CO₂ and precipitates the carbonates. Permanent hardness (from sulphates and chlorides of calcium and magnesium) survives boiling and requires chemical softening. Both are conventionally expressed as an equivalent concentration of CaCO₃, allowing different hardness-causing ions to be compared on a common scale. The Langelier Saturation Index compares actual pH to the theoretical saturation pH to indicate whether water tends to deposit scale or corrode pipes.
As chlorine dose increases from zero, it first reacts with ammonia present in the water to form chloramines (combined residual), and the measured residual rises slowly. Beyond a certain dose, the chloramines are oxidised and destroyed, and the residual actually dips — this is the breakpoint. Beyond the breakpoint, any further chlorine added remains as free residual chlorine, rising linearly with dose. Operating beyond the breakpoint ensures the required free residual (minimum 0.2 mg/L) reaches the consumer's tap.
A Water Quality Index condenses several individual water-quality parameters into a single dimensionless number, giving a quick overall rating of a source's suitability (e.g. "excellent", "good", "poor"). It is computed as a weighted aggregation of the sub-index values of each parameter.
A rising WQI on the NSF scale means better water; on the weighted-arithmetic scale used by many Indian studies a lower value means better quality — always note which convention is in use.
Hardness (as CaCO₃) = ion concentration (mg/L) \(\times \dfrac{50}{\text{eq. wt of ion}}\)
\(Ca^{2+}\): eq. wt = 20; \(Mg^{2+}\): eq. wt = 12; \(HCO_3^-\): eq. wt = 61
\( LSI = pH_{actual} - pH_s \)
LSI > 0 → scaling; LSI < 0 → corrosive; LSI = 0 → stable
Chlorine demand = chlorine added − residual chlorine
Breakpoint dose \(\approx 7.6\times\)(NH₃-N concentration)
Given: A water sample has \(Ca^{2+}=40\ \text{mg/L}\) and \(Mg^{2+}=12\ \text{mg/L}\). Find the total hardness as CaCO₃.
Solution:
\( \text{Ca hardness} = 40\dfrac{\times50}{20} = 100\ \text{mg/L as CaCO}_3 \)
\( \text{Mg hardness} = 12\dfrac{\times50}{12} = 50\ \text{mg/L as CaCO}_3 \)
Answer: Total hardness \(= 100+50 = 150\ \text{mg/L as CaCO}_3\).
Given: Raw water has an ammonia-nitrogen concentration of \(0.5\ \text{mg/L}\). Estimate the breakpoint chlorine dose.
Solution:
\( \text{Breakpoint dose} \approx 7.6\times0.5 = 3.8\ \text{mg/L} \)
Answer: Approximately \(3.8\ \text{mg/L}\) of chlorine must be dosed to reach breakpoint, after which a free residual can be maintained.
Given: A water sample has \(pH_{actual}=8.2\) and calculated saturation \(pH_s=7.6\). Interpret the water's tendency.
Solution:
\( LSI = 8.2-7.6 = +0.6 \)
Answer: Since \(LSI>0\), the water has a scaling tendency (will tend to deposit CaCO₃ scale in pipes rather than corrode them).
Fig. 4.1 — Breakpoint chlorination curve: residual rises as chloramines form, dips sharply at the breakpoint as they are destroyed, then rises linearly as free residual chlorine beyond the breakpoint.
A conventional water treatment plant treats raw water through a fixed sequence of unit processes: Screening & Intake → Aeration (optional) → Coagulation & Flocculation → Sedimentation → Filtration (rapid or slow sand) → Disinfection (chlorine) → Storage & Distribution. Each stage removes progressively finer contaminants, so a later stage functioning correctly depends on the earlier stages having done their job (e.g., filtration works best on already-flocculated, low-turbidity water).
Aeration removes dissolved gases (CO₂, H₂S, volatile organics), oxidises dissolved iron and manganese (Fe²⁺→Fe³⁺, Mn²⁺→MnO₂) so they can be removed in the following sedimentation step, and reduces taste and odour. Spray aerators are simple and effective for taste/odour; cascade aerators (stepped weirs) increase dissolved oxygen and strip CO₂; diffused-air aerators bubble air through the water; packed-tower aerators use counter-current air flow and are the most efficient for volatile organic compound removal.
Coagulation destabilises negatively-charged colloidal particles (1–1000 nm) by adding a coagulant salt — alum is most common, optimally effective at pH 6.5–7.5 where its hydroxide floc is least soluble; ferric salts work better at low pH and low temperature than alum. A jar test determines the optimum coagulant dose and pH in the laboratory before full-scale dosing. Flocculation follows with gentle agitation to aggregate the destabilised micro-flocs into larger, settle-able macro-flocs, characterised by the velocity gradient × detention time product (Camp number).
Sedimentation removes settle-able solids under gravity. The key design parameter is the overflow rate (surface loading rate) — any particle with a settling velocity exceeding the overflow rate is completely removed, regardless of tank depth (Hazen's theory), which is why plain sedimentation uses a lower overflow rate than coagulation-sedimentation (heavier, faster-settling flocs allow a higher rate). Stokes' law gives the theoretical settling velocity of a discrete spherical particle as a function of the density difference and particle diameter.
| Parameter | Slow Sand Filter (SSF) | Rapid Sand Filter (RSF) |
|---|---|---|
| Filtration rate | 0.1–0.4 m/hr (0.1–0.2 usual) | 4–12 m/hr (5 m/hr typical) |
| Media | Sand (0.15–0.35 mm, UC<2) | Sand (0.45–0.70 mm) on gravel |
| Schmutzdecke (biofilm) | Critical; biological action; vital | Not present; purely physical |
| Turbidity of influent | ≤5 NTU (needs pre-treatment if higher) | Can handle higher (after coagulation) |
| Backwash | Scraping and washing; 1–2 months | Backwashing every 24–72 hrs (upward flow) |
| Area needed | Very large (low rate) | Much smaller |
| Cost | Low operating; high land cost | High capital (coagulation needed); lower land |
| Coliform removal | 98–99% | Lower; relies on disinfection |
The schmutzdecke — a biological film that grows on the SSF sand surface — is what actually does most of the purification in a slow sand filter (biological action, not just straining), which is why SSF achieves excellent coliform removal without any prior coagulation, unlike RSF which relies purely on physical straining and needs disinfection to finish the job.
Chlorination (gas Cl₂, hypochlorite solutions, or bleaching powder) is the most common disinfection method. Chick's law models disinfection as first-order die-off of organisms with time. The CT concept — concentration × contact time — captures the trade-off between disinfectant dose and required contact time to achieve a given level of pathogen inactivation, and is used to compare different disinfectants (chlorine, ozone, UV) on a common basis. Ozone and UV are powerful but leave no residual, so they are typically used together with a small chlorine dose to maintain protection throughout the distribution system.
\( G = \sqrt{\dfrac{P}{\mu V}} \)
Camp number \(G\times t = 10^4\)–\(10^5\) for flocculation
\( q_s = \dfrac{Q}{A_s} \); particles with \(v_s>q_s\) fully removed
\( v_s = \dfrac{g(\rho_s-\rho_w)d^2}{18\mu} \)
Plain: 12,000–18,000 L/m²/day; Coagulated: 30,000–40,000
Filtration rate: 5 m/hr (range 4–6); Sand depth: 600–900 mm
Backwash rate: 12–15 m/hr for 5–10 min; expansion 25–50%
\( N = N_0\,e^{-kt} \)
CT = constant for a given log inactivation
Given: A discrete particle of diameter \(d=0.05\ \text{mm}\), density \(\rho_s=2650\ \text{kg/m}^3\), settles in water (\(\rho_w=1000\ \text{kg/m}^3\), \(\mu=1.0\times10^{-3}\ \text{Pa·s}\)). Find the settling velocity.
Solution:
\( v_s = \dfrac{9.81\times(2650-1000)\times(0.05\times10^{-3})^2}{18\times1.0\times10^{-3}} = \dfrac{9.81\times1650\times2.5\times10^{-9}}{0.018} \)
\( v_s \approx \dfrac{4.05\times10^{-5}}{0.018} \approx 2.25\times10^{-3}\ \text{m/s} = 2.25\ \text{mm/s} \)
Answer: \(v_s\approx2.25\ \text{mm/s}\).
Given: A plant treats \(Q=5000\ \dfrac{\text{m}^3}{\text{day}}\), designed at an overflow rate of \(q_s=15{,}000\ \dfrac{\text{L/m}^2}{\text{day}}\) (plain sedimentation). Find the required surface area.
Solution:
\( A_s = \dfrac{Q}{q_s} = 5{,}000{,}000\ \text{L/day} / 15{,}000\ \dfrac{\text{L/m}^2}{\text{day}} = 333.3\ \text{m}^2 \)
Answer: \(A_s\approx333\ \text{m}^2\).
Given: A disinfectant has rate constant \(k=0.5\ \text{min}^{-1}\). Find the time needed to reduce organism count to 1% of initial (99% inactivation).
Solution:
\( N/N_0 = 0.01 = e^{-kt} \Rightarrow t = \dfrac{\ln(100)}{0.5} = \dfrac{4.605}{0.5} = 9.21\ \text{min} \)
Answer: \(t\approx9.2\ \text{minutes}\) of contact time required.
Fig. 5.1 — Water treatment plant flow sequence: Screening → Aeration → Coagulation/Flocculation → Sedimentation → Filtration → Disinfection → Distribution.
| System | Description | Advantage | Use |
|---|---|---|---|
| Separate system | Two separate pipes: one for sewage, one for storm water | Smaller sewage treatment plant; no dilution | New planned developments; most modern systems |
| Combined system | Single pipe carries both sewage and storm water | Lower initial cost; single network | Old cities; can cause overflow in heavy rain |
| Partially separate | Some storm water (from roofs) enters sewer | Compromise; reduces surcharge | Older cities modifying to separate |
The Dry Weather Flow (DWF) is the sewage generated from domestic and industrial use with no rainfall contribution, typically taken as about 80% of the water supplied (the return-flow fraction). Since sewage flow fluctuates hour-to-hour just like water demand, sewers are sized for a peak factor applied to DWF — a higher factor (×3) for small lateral sewers where fluctuations are more pronounced, and a lower factor (×2) for large interceptor sewers where flows from many sources average out. Storm water quantity, where relevant, is estimated by the rational method using a runoff coefficient that reflects the catchment's imperviousness.
Sewers are designed using Manning's equation, exactly as for gravity water conduits, but with two specific velocity constraints unique to sewage: a minimum self-cleansing velocity (to prevent solids from settling and blocking the pipe) and a maximum velocity (to prevent erosion/abrasion of the sewer lining from suspended grit at excessive speeds). A subtlety of circular pipe hydraulics is that the maximum discharge does not occur when the pipe runs completely full, but at about 94% of full depth — because near the very top, a small increase in depth adds disproportionately more wetted perimeter (and hence friction) than flow area.
The ratio of actual discharge to full-pipe discharge (\(\dfrac{Q}{Q_{full}}\)) and actual velocity to full-pipe velocity (\(V/V_{full}\)) both vary non-monotonically with the depth ratio \(\dfrac{d}{D}\) — each reaching a maximum before the pipe is completely full. Sewers should be designed so the anticipated peak flow does not exceed \(Q_{max}\) (i.e., stays below \(\dfrac{d}{D}=0.94\)), ensuring the pipe never needs to run under pressure.
| Appurtenance | Purpose | Spacing / Size |
|---|---|---|
| Manhole | Access for inspection, cleaning, rodding | Every 30–50 m on straight; at every change of direction, dia, gradient; junction |
| Drop connection | House drain enters sewer at much higher elevation | When incoming pipe invert is >0.6 m above sewer invert |
| Flushing tank | Periodic flushing of flat-gradient small sewers | Self-priming; at dead ends; automatic |
| Catch basin | Trap grit and floating material from storm drains | At road inlets; prevents grit entering sewer |
| Inverted siphon | Carries sewage under an obstruction (road, river) under pressure | Velocity ≥ 1.0 m/s to prevent deposition; minimum 2 barrels |
| Ventilation shaft | Releases sewer gases; prevents septic conditions | At every 150–300 m; at siphon ends |
\( Q_{sewage} = \dfrac{80}{100}\times Q_{water\ supply} \)
Peak factor: \(3\times DWF\) (small sewers); \(2\times DWF\) (interceptors)
\( V = \dfrac{1}{n}R^{2/3}S^{1/2} \), \( Q = \dfrac{A}{n}R^{2/3}S^{1/2} \)
n: RCC=0.013; glazed stoneware=0.011; PVC=0.009
\( V_{min} = 0.75\ \text{m/s} \) (self-cleansing, IS); \( V_{max} = 3.0\ \text{m/s} \) (erosion)
\(Q_{max}\) at \(\dfrac{d}{D}=0.94\) (\(=1.08\,Q_{full}\)); \(V_{max}\) at \(\dfrac{d}{D}=0.81\)
\( Q_{storm} = \dfrac{C\,I\,A}{360} \) (Q in m³/s, I in mm/hr, A in ha)
Given: A town's design water supply is \(200\ \text{lpcd}\) for a population of \(50{,}000\). Estimate the design (peak) sewage flow for a small sewer.
Solution:
\( Q_{water} = 200\times50{,}000 = 10{,}000{,}000\ \text{L/day} = 10{,}000\ \dfrac{\text{m}^3}{\text{day}} \)
\( DWF = 0.8\times10{,}000 = 8{,}000\ \dfrac{\text{m}^3}{\text{day}} \)
\( Q_{peak} = 3\times8{,}000 = 24{,}000\ \dfrac{\text{m}^3}{\text{day}} \)
Answer: Design peak sewage flow \(\approx24{,}000\ \dfrac{\text{m}^3}{\text{day}}\).
Given: A RCC sewer (\(n=0.013\)) has hydraulic radius \(R=0.15\ \text{m}\) and slope \(S=0.002\). Check if the self-cleansing velocity is achieved.
Solution:
\( V = \dfrac{1}{0.013}\times0.15^{2/3}\times0.002^{1/2} = 76.9\times0.282\times0.0447\approx0.97\ \text{m/s} \)
Answer: \(V\approx0.97\ \text{m/s} > 0.75\ \text{m/s}\), so self-cleansing velocity is achieved and the pipe is safely below the 3.0 m/s erosion limit.
Given: A paved catchment of area \(A=8\ \text{ha}\), runoff coefficient \(C=0.9\), design rainfall intensity \(I=60\ \text{mm/hr}\). Find the storm water discharge.
Solution:
\( Q_{storm} = \dfrac{0.9\times60\times8}{360} = \dfrac{432}{360} = 1.2\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q_{storm}=1.2\ \dfrac{\text{m}^3}{\text{s}}\).
Fig. 6.1 — Hydraulic elements of a circular sewer pipe: Q_max occurs at 94% full depth (Q=1.08 Q_full); V_max occurs at 81% depth. Design so peak flow stays below Q_max.
Sewage is approximately 99.9% water and only 0.1% solids (dissolved, suspended, colloidal) — yet that tiny solid fraction is what makes sewage a pollutant, since it comprises organic matter, inorganic salts, and disease-causing microorganisms.
| Parameter | Typical Range | Significance |
|---|---|---|
| BOD₅ (5-day BOD at 20°C) | 150–300 mg/L | Organic pollution indicator; O₂ demand for biodegradation |
| COD (Chemical Oxygen Demand) | 250–600 mg/L | Total oxidisable matter (biodegradable + non-biodeg.); COD > BOD always |
| Suspended Solids (SS) | 100–350 mg/L | Settles; causes sludge; reduces clarity |
| Dissolved Oxygen (DO) | 0–1 mg/L (septic) | Low/zero DO → anaerobic decomposition; H₂S odour |
| pH | 6.5–8.0 (fresh); acidic when septic | Controls biological treatment |
| Nitrogen (total) | 20–85 mg/L as N | Nutrient; causes eutrophication; TKN + NO₃-N |
| Phosphorus | 4–15 mg/L as P | Nutrient for algae; eutrophication |
| Coliforms (MPN) | 10⁶–10⁹/100 mL | Pathogen indicator; faecal contamination |
BOD measures the oxygen consumed by microorganisms as they biodegrade organic matter, following approximately first-order kinetics — the remaining unoxidised BOD decays exponentially with time. Since the standard 5-day test only captures a fraction of the total (ultimate) oxygen demand, the ratio \(\dfrac{BOD_5}{BOD_u}\) is typically 0.65–0.75 for domestic sewage. COD, which oxidises both biodegradable and non-biodegradable matter chemically, is always higher than BOD₅; a COD/BOD₅ ratio above about 2.5 signals a significant non-biodegradable fraction, a red flag for biological treatment feasibility.
When sewage (with its high BOD, low DO) is discharged into a river, dissolved oxygen initially falls as bacteria consume it to oxidise the organic load faster than re-aeration from the atmosphere can replenish it — reaching a minimum at the critical point — then recovers downstream as the BOD load is progressively consumed and re-aeration begins to dominate. The Streeter-Phelps equation models this "oxygen sag curve" as the combined effect of two competing first-order processes: deoxygenation (BOD exertion) and re-aeration.
SVI measures how well activated sludge settles — a sludge with good settleability (SVI 80–120 mL/g) compacts efficiently in the secondary clarifier, while a "bulking" sludge (SVI >200 mL/g), often caused by filamentous bacteria outcompeting floc-formers, settles poorly and can be washed out of the clarifier, degrading effluent quality. The related Food-to-Microorganism (F/M) ratio expresses the organic loading rate per unit of active biomass, and is a key operational control parameter distinguishing extended aeration (low F/M) from conventional activated sludge (higher F/M) plants.
\( BOD_t = L_0(1-e^{-k_d t}) \)
Temperature correction: \( k_d(T) = k_d(20^\circ C)\times\theta^{T-20} \), \(\theta=1.047\)
\( D_t = \dfrac{k_d L_0}{k_r-k_d}\left(e^{-k_d t}-e^{-k_r t}\right)+D_0 e^{-k_r t} \)
\( t_c = \dfrac{1}{k_r-k_d}\ln\!\left[\dfrac{k_r}{k_d}\left(1-\dfrac{D_0(k_r-k_d)}{k_d L_0}\right)\right] \)
\( D_c = \dfrac{k_d}{k_r}L_0\,e^{-k_d t_c} \)
\( SVI = \dfrac{\text{settled sludge volume (mL/L)}}{\text{MLSS (mg/L)}}\times1000 \)
\( \dfrac{F}{M} = \dfrac{\text{BOD input}}{MLVSS\times\text{aeration volume}} \)
Given: A sewage sample has \(BOD_5=180\ \text{mg/L}\), \(k_d=0.23\ \text{day}^{-1}\) (base-e). Find the ultimate BOD.
Solution:
\( 180 = L_0(1-e^{-0.23\times5}) = L_0(1-e^{-1.15}) = L_0(1-0.3166) = 0.6834\,L_0 \)
\( L_0 = \dfrac{180}{0.6834} \approx 263.4\ \text{mg/L} \)
Answer: \(L_0\approx263\ \text{mg/L}\).
Given: A 1-litre sample of mixed liquor with MLSS = 3000 mg/L settles to a volume of 300 mL after 30 minutes. Find the SVI and classify the sludge.
Solution:
\( SVI = \dfrac{300}{3000}\times1000 = 100\ \text{mL/g} \)
Answer: \(SVI=100\ \text{mL/g}\) — within the 80–120 mL/g range, indicating good-settling sludge.
Given: \(k_d\) at 20°C is \(0.23\ \text{day}^{-1}\). Find \(k_d\) at 25°C (\(\theta=1.047\)).
Solution:
\( k_d(25) = 0.23\times1.047^{5} = 0.23\times1.258 \approx 0.289\ \text{day}^{-1} \)
Answer: \(k_d(25^\circ C)\approx0.289\ \text{day}^{-1}\) — higher temperature speeds up biodegradation.
Fig. 7.1 — Streeter-Phelps DO sag curve: dissolved oxygen falls below saturation after the sewage outfall, reaching a minimum at the critical point t_c, then recovers as re-aeration overtakes deoxygenation downstream.
| Method | Description | BOD Removal | Best for |
|---|---|---|---|
| Irrigation (Land treatment) | Sewage applied to agricultural land; plants uptake nutrients | 85–95% | Nutrient-rich effluent; agriculture |
| Rapid infiltration | High rate loading in basins; percolates through soil | 90–99% | Recharge of groundwater |
| Overland flow | Sheet flow over grass terraces; collected in drain | 80–90% | Slow-draining soils |
| Subsurface application | Tile drains or absorption trenches | 90–99% | Small installations; septic tanks |
Discharging sewage into a flowing river relies on dilution — mixing with the much larger river flow — combined with the river's natural self-purification capacity. IS 2306 requires a minimum dilution of 1:20 (one part sewage to twenty parts river flow) for pathogenic safety, and separately requires that the outfall not push river DO below 4 mg/L. Self-purification itself proceeds through five simultaneous mechanisms: physical dilution and sedimentation, biological oxidation of organics by aerobic bacteria, photo-oxidation of pathogens by sunlight, and atmospheric re-aeration restoring dissolved oxygen.
A septic tank is a two-stage (anaerobic digestion + clarification) underground unit used for individual dwellings not connected to a public sewer. IS 2470 specifies a minimum capacity of 1000 litres for 1–5 users, with 180 litres added for each additional user beyond 5 — sized for a minimum 24-hour liquid detention time. The tank's effluent, still containing significant pathogens and nutrients, must be followed by a soak pit or absorption trench for subsurface percolation, and accumulated sludge requires periodic desludging every 1–3 years.
Treated wastewater reused for irrigation must meet standards that depend on the crop and exposure risk: unrestricted irrigation (including food crops eaten raw) requires the strictest limits, while restricted irrigation (non-food crops) permits higher BOD and coliform levels since human exposure is lower. Separately, CPCB sets discharge standards for effluent released to land or inland surface water, which are generally less strict than the unrestricted-irrigation reuse standards since dilution and further natural attenuation are assumed in the receiving environment.
\( D = \dfrac{Q_{river}+Q_{sewage}}{Q_{sewage}} \)
IS 2306: minimum dilution 1:20 for pathogenic safety
1–5 users: 1000 L; each additional user: +180 L
Detention time \(\ge24\ \text{hr}\); L:B:D = 2:1:1 to 4:1:1
Unrestricted irrigation: BOD ≤10, SS ≤20, Coliforms ≤1000/100mL
CPCB inland surface water: BOD ≤30, SS ≤100, pH 5.5–9
Given: A river carries \(Q_{river}=8\ \dfrac{\text{m}^3}{\text{s}}\), and sewage of \(Q_{sewage}=0.3\ \dfrac{\text{m}^3}{\text{s}}\) is discharged into it. Check whether the IS 2306 minimum dilution requirement is met.
Solution:
\( D = \dfrac{8+0.3}{0.3} = \dfrac{8.3}{0.3} \approx 27.7 \)
Answer: Dilution ratio \(\approx1:27.7\), which exceeds the IS 2306 minimum of 1:20 — the discharge is acceptable on the dilution criterion.
Given: A household of 8 people needs a septic tank. Find the minimum required capacity per IS 2470.
Solution:
First 5 users: 1000 L. Additional users: \(8-5=3\), so \(3\times180 = 540\ \text{L}\)
\( \text{Total} = 1000+540 = 1540\ \text{L} \)
Answer: Minimum capacity \(=1540\ \text{litres}\).
Given: A river with DO = 7 mg/L receives a sewage discharge that is estimated to reduce river DO by 2.5 mg/L at the critical point. Check compliance with IS 2306's DO requirement.
Solution:
Resulting DO \(= 7-2.5 = 4.5\ \text{mg/L}\)
Answer: Since \(4.5\ \text{mg/L} > 4\ \text{mg/L}\) (the IS 2306 minimum), the discharge is acceptable, though only with a small margin — any additional loading would risk non-compliance.
Fig. 8.1 — Septic tank cross-section: two-stage chamber (anaerobic digestion and settling, then clarification) followed by a soak pit for subsurface percolation of the effluent.
| Level | Processes | BOD Removal | SS Removal |
|---|---|---|---|
| Preliminary | Screening, grit removal, comminution | 5–10% | 5–10% |
| Primary | Plain sedimentation; skimming of floatables | 25–40% | 50–70% |
| Secondary (biological) | Activated sludge; trickling filter; waste stabilisation pond | 80–95% | 80–95% |
| Tertiary (advanced) | Nutrient removal (N, P); MF/UF; RO; disinfection | >95% | >95% |
In the ASP, raw sewage first passes through a primary clarifier, then enters an aeration tank where it is mixed with a suspension of active microorganisms (activated sludge) and aerated by blowers, allowing the biomass to metabolise the dissolved and suspended organic matter. The mixture then flows to a secondary clarifier, where the biomass settles out — most of it is returned to the aeration tank as Return Activated Sludge (RAS) to maintain the required biomass concentration, while the excess (Waste Activated Sludge, WAS) is removed to control sludge age.
A trickling filter passes sewage over a fixed bed of media (crushed stone or plastic) on which a biological slime layer grows, oxidising organic matter as the liquid trickles through. Unlike the ASP's suspended-growth biomass, the trickling filter is an attached-growth system. Design uses empirical formulae such as the NRC formula, which relates BOD removal efficiency to organic loading and a recirculation factor — recirculating part of the effluent improves distribution and wetting of the media and dilutes strong influent.
| Pond Type | DO | Depth | HRT | Removal |
|---|---|---|---|---|
| Anaerobic pond | Zero | 2–5 m | 5–30 days | 50–70% BOD |
| Facultative pond | Aerobic top; anaerobic bottom | 1–2 m | 5–30 days | 70–90% BOD |
| Maturation pond | Aerobic throughout | 0.5–1.5 m | 5–10 days | Pathogen removal (log reduction) |
WSPs form a natural treatment train — anaerobic ponds handle the heaviest organic load first, facultative ponds provide the bulk of BOD removal through a combined aerobic-anaerobic mechanism, and maturation ponds (fully aerobic, shallow, sunlit) focus specifically on pathogen die-off before final discharge.
Sludge removed from primary and secondary clarifiers must itself be treated before disposal: thickening concentrates it, digestion (anaerobic, in two stages — hydrolysis/acidogenesis then methanogenesis) stabilises it and produces biogas, dewatering reduces its volume for handling, and final disposal is via agricultural land application, landfill, or incineration.
Industrial wastewater often needs additional steps beyond a conventional municipal sewage train — equalization to smooth out flow and load variability, neutralization to bring pH into a biologically tolerable range, and specific tertiary steps (activated carbon, UV/ozone) tailored to the industry's particular contaminants. Zero Liquid Discharge (ZLD) is mandated by CPCB/MoEF for water-intensive, high-pollution-load industries such as textile, distillery, and paper mills.
MLSS: 2000–4000 mg/L; \(MLVSS=0.7\text{–}0.8\times MLSS\)
SRT: 5–15 days (conventional); HRT: 4–8 hours
\(\dfrac{F}{M} = \dfrac{BOD_{in}}{MLVSS\times HRT}\): 0.2–0.5 kg BOD/kg MLVSS/day
\( E = \dfrac{1}{1+0.4432\sqrt{BOD/(VF)}} \)
\( F = \dfrac{1+R}{(1+R/10)^2} \) (R = recirculation ratio)
Biogas: 60–70% CH₄, 30–40% CO₂
Gas production: 0.5–1.0 m³/kg VS destroyed; SRT 20–30 days (mesophilic, 35°C)
Given: BOD load entering an aeration tank is \(500\ \text{kg/day}\); MLVSS \(=2500\ \text{mg/L}\); aeration tank volume \(=1000\ \text{m}^3\). Find the F/M ratio.
Solution:
Biomass mass \(= MLVSS\times V = 2500\times10^{-3}\ \text{kg/m}^3\times1000\ \text{m}^3 = 2500\ \text{kg}\)
\( \dfrac{F}{M} = \dfrac{500}{2500} = 0.2\ \text{kg BOD/kg MLVSS/day} \)
Answer: \(\dfrac{F}{M}=0.2\), at the lower end of the conventional ASP range.
Given: A trickling filter operates at recirculation ratio \(R=2\). Find the recirculation factor F.
Solution:
\( F = \dfrac{1+2}{(1+2/10)^2} = \dfrac{3}{1.44} \approx 2.08 \)
Answer: \(F\approx2.08\), which increases the effective BOD removal efficiency in the NRC formula compared to no recirculation (\(R=0\), \(F=1\)).
Given: An anaerobic digester destroys \(2000\ \text{kg}\) of volatile solids per day, with a gas yield of \(0.75\ \text{m}^3\) per kg VS destroyed. Estimate the daily biogas volume and methane volume.
Solution:
Biogas \(=2000\times0.75=1500\ \dfrac{\text{m}^3}{\text{day}}\)
Methane (at 65% average) \(=1500\times0.65=975\ \dfrac{\text{m}^3}{\text{day}}\)
Answer: Total biogas \(\approx1500\ \dfrac{\text{m}^3}{\text{day}}\), of which \(\approx975\ \dfrac{\text{m}^3}{\text{day}}\) is methane.
Fig. 9.1 — Activated Sludge Process: raw sewage → Primary clarifier → Aeration tank (biomass + air) → Secondary clarifier → treated effluent; RAS maintains MLSS in the aeration tank, WAS controls sludge age.
Clean, dry air is roughly 78% nitrogen, 21% oxygen, 0.93% argon, and 0.04% CO₂ (rising steadily), with trace gases making up the remainder. Anthropogenic activity adds a further set of pollutants — particulate matter (PM₁₀, PM₂.₅), SO₂, NOx, CO, ozone (secondary, formed photochemically), lead, and volatile organic compounds — each with its own dominant source (combustion, vehicles, industry) and health/environmental effect.
| Pollutant | Primary Source | Health Effect | NAAQS Limit (24-hr) |
|---|---|---|---|
| Particulate Matter (PM₁₀) | Dust, construction, vehicles, industries | Respiratory disease; PM₂.₅ reaches alveoli | 100 µg/m³ (PM₁₀); 60 µg/m³ (PM₂.₅) |
| SO₂ | Combustion of S-containing coal; power plants; refineries | Acid rain (H₂SO₄); respiratory irritant | 80 µg/m³ (24-hr; annual 50) |
| NO₂ / NOx | High-temperature combustion; vehicles | Smog; acid rain (HNO₃); lung damage | 80 µg/m³ |
| CO | Incomplete combustion; vehicles | Binds haemoglobin (COHb); asphyxiation | 2 mg/m³ (8-hr) |
| Ozone (O₃) | Photochemical reaction (NOx + VOC + sunlight) | Lung irritant; eye irritation; crop damage | 100 µg/m³ (8-hr) |
| Lead (Pb) | Leaded fuel (now banned); battery industry; smelters | Neurotoxin; cognitive impairment | 0.5 µg/m³ (24-hr) |
| VOCs / Hydrocarbons | Vehicles; evaporation; industrial solvents | Precursor to photochemical smog | — |
| CO₂ / CH₄ | Fossil fuels; agriculture; waste | Greenhouse effect; climate change | No NAAQS (GHG) |
Particulate control devices span a range of efficiency and cost. Gravity settling chambers and cyclone separators remove only the coarser fraction cheaply. Fabric filters (baghouses) and electrostatic precipitators (ESPs) achieve very high efficiency down to sub-micron particle sizes — ESPs use a corona discharge to charge particles electrostatically before collecting them on oppositely-charged plates. Wet scrubbers additionally remove gaseous pollutants (SO₂, HCl) alongside particulates by direct contact with a liquid spray.
The Gaussian plume model predicts the ground-level concentration of a pollutant downwind of a stack, assuming the plume spreads in a Gaussian (bell-curve) distribution both horizontally and vertically as it travels downwind. The dispersion coefficients \(\sigma_y\) and \(\sigma_z\) grow with downwind distance and depend on atmospheric stability — more unstable conditions (strong solar heating, low wind) disperse the plume faster and reduce peak ground-level concentration, while stable conditions (clear nights, low wind) trap the plume and can produce higher ground-level concentrations closer to the source.
| Class | Stability | Condition | Dispersion |
|---|---|---|---|
| A | Extremely unstable | Strong insolation, low wind (<2 m/s) | Rapid, wide dispersion |
| B | Unstable | Moderate insolation | Good dispersion |
| C | Slightly unstable | Slight insolation | Moderate |
| D | Neutral | Overcast or windy (>6 m/s) | Average; most common |
| E | Slightly stable | Clear night, moderate wind | Poor dispersion |
| F | Stable | Clear night, low wind | Very poor; fumigation possible |
Sound pressure level is expressed on a logarithmic decibel scale referenced to the threshold of human hearing. Because decibels are logarithmic, sound levels from multiple sources cannot simply be added arithmetically — combining two equal sources only raises the level by 3 dB, not doubling it, since a 3 dB increase already represents a doubling of sound energy. The equivalent sound level \(L_{eq}\) is the single steady-level value that carries the same total sound energy as the actual, fluctuating noise over the measurement period, and is the standard metric used in noise regulations.
| Zone | Day (6am–10pm) dB(A) | Night (10pm–6am) dB(A) |
|---|---|---|
| Industrial | 75 | 70 |
| Commercial | 65 | 55 |
| Residential | 55 | 45 |
| Silence zone (hospitals, schools) | 50 | 40 |
Prolonged exposure above 85 dB carries a risk of permanent hearing damage (NIPTS — Noise-Induced Permanent Threshold Shift), which characteristically shows up first as a notch at 4000 Hz in an audiogram — a distinctive diagnostic signature of occupational noise exposure rather than age-related hearing loss.
\( C(x,0,0) = \dfrac{Q}{\pi\sigma_y\sigma_z U}\exp\!\left(\dfrac{-H^2}{2\sigma_z^2}\right) \)
Q=emission rate; U=wind speed; H=effective stack height
\( L = 20\log_{10}(\dfrac{P}{P_{ref}}) \), \(P_{ref}=20\ \mu\text{Pa}\)
Addition: \( L_{total} = 10\log_{10}\!\left(\Sigma10^{L_i/10}\right) \); two equal sources: \(L+3\ \text{dB}\)
\( L_{eq} = 10\log_{10}\!\left[\dfrac{1}{T}\int_0^T\left(\dfrac{P}{P_{ref}}\right)^2dt\right] \)
Given: Two machines each produce a sound level of \(80\ \text{dB}\) individually. Find the combined sound level when both operate together.
Solution:
\( L_{total} = 10\log_{10}(10^{8}+10^{8}) = 10\log_{10}(2\times10^8) = 10\times8.301 = 83.01\ \text{dB} \)
Answer: \(L_{total}\approx83\ \text{dB}\) (i.e., \(80+3\ \text{dB}\)), NOT \(160\ \text{dB}\).
Given: A sound has a measured pressure of \(P=2\ \text{Pa}\). Find the SPL.
Solution:
\( L = 20\log_{10}(\dfrac{2}{20\times10^{-6}}) = 20\log_{10}(100{,}000) = 20\times5 = 100\ \text{dB} \)
Answer: \(L=100\ \text{dB}\).
Given: Three sources produce \(70\ \text{dB}\), \(73\ \text{dB}\), and \(68\ \text{dB}\) individually. Find the combined level.
Solution:
\( L_{total} = 10\log_{10}(10^{7}+10^{7.3}+10^{6.8}) = 10\log_{10}(10{,}000{,}000+19{,}953{,}000+6{,}310{,}000) \)
\( = 10\log_{10}(36{,}263{,}000) \approx 10\times7.56 = 75.6\ \text{dB} \)
Answer: \(L_{total}\approx75.6\ \text{dB}\), dominated by (but higher than) the loudest individual source.
Fig. 10.1 — Gaussian plume dispersion: the plume spreads horizontally and vertically as it travels downwind, following a Gaussian distribution at each cross-section; ground-level concentration first rises then falls with downwind distance.
| Type | Sources | Examples | Characteristics |
|---|---|---|---|
| Municipal Solid Waste (MSW) | Household, commercial, institutional | Food waste, paper, plastic, glass, metals | Mixed; 50–60% biodegradable in India |
| Industrial solid waste | Factories, power plants | Fly ash, slag, packaging, process waste | May contain hazardous substances |
| Biomedical waste | Hospitals, labs, clinics | Sharps, body fluids, infected material | Hazardous; special handling (BMW Rules 2016) |
| Hazardous waste | Chemical industries; EV batteries | Solvents, heavy metals, pesticides | Toxic; special treatment and landfill (Hazardous Waste Rules) |
| Construction & demolition (C&D) | Building demolition, excavation | Concrete rubble, steel, brick, soil | Inert; large volume; recycling possible |
| E-waste | Consumer electronics | Computers, mobiles, batteries | Contains heavy metals (Pb, Hg, Cd, Cr); E-waste Rules 2022 |
Typical Indian MSW is dominated by biodegradable organics (40–60%, mainly food waste), with the remainder split among inert material (dirt/ash, 15–20%), paper, plastic, glass, and metals. Per capita generation ranges 0.3–0.6 kg/person/day in urban India, and the calorific value is relatively low (1000–2500 kcal/kg) due to high moisture and inert content — a key constraint on waste-to-energy feasibility, since low calorific value waste needs supplementary fuel to sustain combustion.
ISWM follows a waste management hierarchy that prioritises interventions in order of environmental preference: Reduce waste generation at the source first (most effective, since it avoids the waste entirely); then Reuse items as-is; then Recycle materials that can't be reused directly; then Recover energy from residual waste that can't be recycled (RDF, waste-to-energy, biogas); and only Dispose (landfill) of what remains after all higher options are exhausted.
| Method | Description | Advantage | Disadvantage |
|---|---|---|---|
| Open dumping | Waste thrown in open without treatment | None (cheap short-term) | Illegal; vermin, leachate, gas; prohibited under SWM Rules 2016 |
| Sanitary landfill | Compacted waste in lined cells; covered daily; leachate collected; gas managed | All waste types; low cost per tonne | Land intensive; 30+ year liability |
| Composting | Aerobic decomposition of organics; produces compost | Reduces waste; produces useful product | Needs source segregation; market for compost |
| Vermicomposting | Worms (Eisenia) decompose organics | High N compost; decentralised | Sensitive to temperature; lower capacity |
| Incineration (WTE) | Combustion at high temperature; energy recovered | 90%+ volume reduction; energy | Air pollution; expensive; needs high calorific waste |
| Biomethanation | Anaerobic digestion; biogas + compost | Energy + compost; closed system | Needs organics segregation |
| Pyrolysis / Gasification | Thermal decomposition without air | Handles mixed waste; energy rich products | Complex; expensive; emerging in India |
A sanitary landfill has several essential engineered components: a liner system (HDPE geomembrane, often combined with compacted clay of low permeability) to prevent leachate from contaminating groundwater; a leachate collection system to capture and treat the polluted liquid draining through the waste; a gas collection system to capture landfill gas (roughly equal parts methane and CO₂) since accumulated gas is both a fire/explosion hazard and a potent greenhouse gas if released untreated; and daily cover to control odour, vectors, and windblown litter.
\( L = \dfrac{P\times A\times R_c}{1000} \) (m³/day)
P=precipitation (mm/day); A=area (m²); R_c=runoff coefficient (0.3–0.5, covered landfill)
Gas: ~50% CH₄ + 50% CO₂
Generated: 1–5 m³/tonne refuse over 10–30 years
HDPE geomembrane: 1.5 mm minimum
Compacted clay: \(K\le10^{-7}\ \text{cm/s}\), 60 cm thick
Given: A covered landfill cell has area \(A=20{,}000\ \text{m}^2\), average daily precipitation \(P=8\ \text{mm/day}\), runoff coefficient \(R_c=0.4\). Find the leachate generation rate.
Solution:
\( L = \dfrac{8\times20{,}000\times0.4}{1000} = \dfrac{64{,}000}{1000} = 64\ \dfrac{\text{m}^3}{\text{day}} \)
Answer: \(L=64\ \dfrac{\text{m}^3}{\text{day}}\) of leachate must be collected and treated.
Given: A landfill receives \(500\ \text{tonnes/day}\) of refuse, with an expected gas yield of \(3\ \text{m}^3\) per tonne over its generation life. Estimate the total lifetime gas volume from one day's waste.
Solution:
\( \text{Gas volume} = 500\times3 = 1500\ \text{m}^3 \) (total, released over 10–30 years)
Answer: \(\approx1500\ \text{m}^3\) of landfill gas will eventually be generated from one day's waste input.
Given: A city generates MSW that is 55% biodegradable food waste with low calorific value (1200 kcal/kg) after removing recyclables. Recommend a disposal method for the biodegradable fraction.
Solution: Low calorific value (well below the ~1500 kcal/kg threshold typically needed for self-sustaining incineration) makes waste-to-energy incineration inefficient without supplementary fuel. The high biodegradable fraction is well suited to biological treatment instead.
Answer: Composting or biomethanation (anaerobic digestion) is the more suitable choice, converting the biodegradable fraction into compost and/or biogas rather than attempting incineration.
Fig. 11.1 — Sanitary landfill cross-section: HDPE + compacted-clay composite liner and leachate collection at the base, compacted waste cells with daily cover, and gas vents releasing landfill gas.
\( MDD=1.8\times ADD \); \( MHD=2.7\times ADD \)
\( Q=3182\sqrt{P} \) (L/min, P in thousands)
\( Q=0.2785\,C\,D^{2.63}S^{0.54} \)
\( \Delta Q=-\dfrac{\Sigma h_L}{n\Sigma|h_L/Q|} \)
\( Q=\dfrac{\pi K(H^2-h^2)}{\ln(R/r_w)} \)
\( Q=\dfrac{2\pi T(H-h)}{\ln(R/r_w)} \), \(T=Kb\)
\( v_s=\dfrac{g(\rho_s-\rho_w)d^2}{18\mu} \)
\( V_{min}=0.75\ \text{m/s} \), \(V_{max}=3.0\ \text{m/s}\)
\( BOD_t=L_0(1-e^{-k_d t}) \)
\( D_t=\dfrac{k_dL_0}{k_r-k_d}(e^{-k_dt}-e^{-k_rt})+D_0e^{-k_rt} \)
\( SVI=\dfrac{\text{settled vol (mL/L)}}{MLSS(\text{mg/L})}\times1000 \)
\( C(x,0,0)=\dfrac{Q}{\pi\sigma_y\sigma_zU}\exp\!\left(\dfrac{-H^2}{2\sigma_z^2}\right) \)
\( L_{total}=10\log_{10}(\Sigma10^{L_i/10}) \); equal sources: \(L+3\ \text{dB}\)
| Chapter | GATE Focus | ESE Focus | SSC JE Focus |
|---|---|---|---|
| Water demand | Per capita values; peak factors; population methods | Detailed demand calculations; fire demand; design period | Per capita; MDD and MHD values |
| Conduits & distribution | Hazen-Williams numericals; Hardy-Cross concept | Full network analysis; storage design | Pipe materials; layout types |
| Ground water & wells | Numerical: unconfined/confined well equations; \(T=Kb\) | Theis method; Thiem; test analysis; Sichardt R | Types of aquifers; formula recognition |
| Water quality | IS 10500 key limits; hardness types; BOD kinetics; Chick's law | Jar test; coagulant chemistry; Langelier index; breakpoint Cl | IS 10500 key limits only |
| Water treatment | SSF vs RSF rates; sedimentation overflow rate; disinfection CT | Tank design calculations; filter media specs; G×t for flocculation | Sequence of treatment; types of filters |
| Sewer design | Manning eq; self-cleansing velocity; Q at d/D=0.94; sewer types | Full sewer design; storm water; rational method | Manning equation; sewer materials; manhole spacing |
| Sewage quality | BOD-COD; Streeter-Phelps; DO sag; SVI; F/M | Full river quality analysis; critical DO calculations | BOD definition; COD concept |
| Sewage disposal | Dilution factor; IS 2306; septic tank sizing | Land disposal design; irrigation reuse standards | Septic tank basics; disposal methods |
| Sewage treatment | ASP design parameters; SRT; HRT; MLSS; biogas CH₄% | Complete STP design; ETP; ZLD; NRC formula for TF | Types of treatment; primary vs secondary |
| Air & noise pollution | Gaussian plume formula; NAAQS limits; stability classes; dB addition | Stack design; control device selection; photochemical smog | Pollutant sources and effects; NAAQS limits; noise zones |
| Solid waste | 3R hierarchy; sanitary landfill components; leachate | Landfill design; gas collection; EIA | Types of waste; disposal methods |
Q1. A city's average daily demand is \(20{,}000\ \dfrac{\text{m}^3}{\text{day}}\). Find the maximum daily demand and maximum hourly demand.
Q2. An unconfined aquifer has \(K=30\ \text{m/day}\), \(H=20\ \text{m}\), \(h=12\ \text{m}\), \(R=250\ \text{m}\), \(r_w=0.1\ \text{m}\). Find the well discharge.
Q3. A water sample has \(BOD_5=200\ \text{mg/L}\) with \(k_d=0.23\ \text{day}^{-1}\). Find the ultimate BOD \(L_0\).
Q4. A household with 10 members needs a septic tank sized per IS 2470. Find the minimum capacity.
Q5. Three equal noise sources each produce 65 dB. Find the combined sound level.