Sewage & Wastewater — Characteristics & Quantity Estimation

1. Definition and Sources of Sewage

Sewage (wastewater) is the used water from domestic, commercial, and industrial activities, plus any infiltrating groundwater and stormwater that enters the sewer system. It is a dilute mixture (99.9% water, 0.1% dissolved and suspended pollutants) carrying both inorganic and organic matter, microorganisms, and potentially hazardous chemicals.

1.1 Sources of Wastewater

2. Sewage Quantity Estimation

2.1 Dry Weather Flow (DWF)

2.2 Peak Sewage Flow

2.3 Wet Weather Flow (WWF)

In combined sewer systems, wet weather flow = DWF + stormwater runoff. Combined sewers are sized for the combined flow, but treatment plants cannot handle the full peak wet weather flow — combined sewer overflows (CSOs) are a major pollution problem. In separate systems, storm drains handle stormwater independently.

3. Storm Water Runoff — Rational Method

3.1 Runoff Coefficients (C)

3.2 Time of Concentration

4. Characteristics of Sewage

4.1 Physical Characteristics

4.2 Chemical Characteristics

4.3 Biological Characteristics

4.4 Septic Sewage

When sewage becomes anaerobic due to long travel time in sewers (especially in flat terrain or warm climates), it becomes “septic” — characterised by H₂S production (black colour, rotten egg odour), pH drop, and corrosion of concrete sewers. Prevention: adequate slope (velocity), ventilation at manholes, chlorination of influent, or forced aeration at critical points.

5. Types of Sewer Systems

6. Sewer Design — Manning’s Equation

Sewers in India are designed as open channels (gravity flow, not pressurised), using Manning’s formula — the same equation used for irrigation canals and storm drains.

6.1 Manning’s n Values for Sewers

6.2 Minimum Size Requirements (CPHEEO)

6.3 Sewer Gradient Design Rules

7. Self-Cleansing Velocity

Self-cleansing velocity is the minimum velocity required to keep the sewer pipes clean by preventing the deposition and accumulation of sediment (silt, grit, solids) on the sewer invert. If sewage flows too slowly, solids settle and clog the sewer, causing blockages and anaerobic decomposition.

7.1 Self-Cleansing Grade (Minimum Slope)

The minimum slope required to achieve self-cleansing velocity for a given sewer diameter. As sewer diameter increases, the hydraulic radius R = D/4 also increases, so a less steep slope is needed to achieve the same velocity. This is why large trunk sewers can run at flatter gradients than small lateral sewers.

8. Partial Flow in Circular Sewers

Sewers are designed for full-bore (running full) conditions for design calculations, but they rarely flow full — most of the time they carry only a fraction of their design capacity. The hydraulic properties change as the depth of flow changes.

9. Sewer Appurtenances

10. Worked Examples (GATE CE Level)

Example 1 — Sewer Flow Capacity (Manning’s Equation, GATE CE 2022 type)

Problem: A circular sewer of 450 mm diameter is laid at a slope of 1 in 200. Manning’s n = 0.013. Find the velocity and discharge when the sewer is (a) running full and (b) running half full.

Answer: Full: V = 1.27 m/s, Q = 201 L/s; Half: V = 1.27 m/s, Q = 101 L/s (same velocity at half and full flow)

Example 2 — Minimum Slope for Self-Cleansing (GATE CE type)

Problem: A 300 mm diameter concrete sewer (n = 0.013) must have a minimum velocity of 0.75 m/s when flowing full. Find the minimum slope required.

Answer: Minimum slope = 1 in 333 (S = 0.003)

Example 3 — Sewage Quantity Estimation (GATE CE type)

Problem: A town with a population of 2,00,000 has a water supply rate of 150 LPCD. Estimate (a) the average daily sewage flow (assume 80% return to sewers), (b) the peak sewage flow using Harmon’s formula.

Answer: Average daily sewage = 24 MLD; Peak sewage = 42.5 MLD

Example 4 — Rational Formula for Storm Drain (GATE CE type)

Problem: A 20 hectare urban catchment has a runoff coefficient C = 0.65. The time of concentration is 25 minutes and the rainfall intensity for a 10-year storm at 25-minute duration = 40 mm/hour. Find the peak runoff using the rational formula.

Answer: Peak runoff Q = 1.44 m³/s

Example 5 — Sewer Size Selection (GATE CE type)

Problem: Design a circular concrete sewer (n = 0.013) to carry a peak flow of 0.15 m³/s at a slope of 1 in 250. The sewer should flow at approximately 75–80% of full capacity. Find the required sewer diameter.

Answer: D = 450 mm circular concrete sewer (theoretical 428 mm → round up to standard 450 mm)

Example 6 — BOD Load from Sewage (GATE CE type)

Problem: A town with 1,00,000 population generates sewage at 110 LPCD with BOD₅ = 250 mg/L. Find (a) the daily sewage volume, (b) the total BOD load in kg/day, and (c) the per capita BOD contribution.

Answer: Q = 11 MLD; BOD load = 2750 kg/day; Per capita BOD = 27.5 g/capita/day

11. Common Mistakes

Mistake 1 — Assuming the Same Velocity at Full and Half-Full Flow as a Coincidence

Error: Stating that a half-full circular sewer “happens to have” the same velocity as when full, without understanding the mathematical reason.
Root Cause: At half-full, both area A and wetted perimeter P are exactly halved compared to full-bore conditions. Therefore R = A/P = (A_full/2)/(P_full/2) = A_full/P_full = R_full — hydraulic radius is unchanged. Since Manning’s velocity V = (1/n)R^2/3S^1/2 and R is the same, V is the same. This is not a coincidence — it is a mathematical consequence of the circular geometry.
Fix: Understand and state: “For a circular sewer at half-full depth, R_half = R_full = D/4, so velocity is identical to full-bore flow. Discharge is exactly half because area is halved.”

Mistake 2 — Using Total Water Supply as Sewage Quantity

Error: Taking sewage flow = 100% of water supply (135 LPCD) instead of 80% (108 LPCD).
Root Cause: Not all water supplied to a household is returned to the sewer — some is consumed (drinking, cooking), some evaporates from gardens, and some is lost to leakage. Typically, 80% returns as sewage in Indian cities.
Fix: Sewage = 80% × water supply. If water supply = 135 LPCD: sewage = 0.80 × 135 = 108 LPCD ≈ 110 LPCD. Always state this assumption in sewer design calculations.

Mistake 3 — Designing Sewers for Average Flow Without Checking Peak Flow

Error: Sizing the sewer pipe for average DWF without applying the peak factor, resulting in overflow during morning peak hours.
Root Cause: Sewage flows vary significantly (2–3× average during morning peaks), just as water demand varies. A sewer designed only for average flow will surcharge (flow under pressure) and potentially overflow at manholes during peak periods.
Fix: Design sewer for peak DWF = peak factor × average DWF. Use Harmon’s formula for peak factor. Also check minimum velocity at DWF (not just at peak) to ensure self-cleansing conditions are maintained throughout the day.

Mistake 4 — Using the Wrong Manning’s n for Sewers

Error: Using n = 0.025 (earthen channel value) for a concrete sewer, significantly underestimating capacity.
Root Cause: Manning’s n for a smooth concrete sewer = 0.013, which is much lower (smoother) than for earthen channels (0.025). Using 0.025 for a concrete sewer gives a velocity that is 0.025/0.013 × (adjusted) ≈ 1.9× too low — the pipe appears to require much larger diameter than actually needed.
Fix: Concrete sewer, vitrified clay pipe: n = 0.013. Brick sewer: n = 0.013–0.015. Corrugated metal: 0.021–0.025. PVC/HDPE: 0.009–0.011.

Mistake 5 — Applying the Rational Formula with Q in m³/s but Forgetting the /360 Conversion Factor

Error: Writing Q = C × i × A without the /360 factor, getting Q in mm·ha/hr instead of m³/s.
Root Cause: The rational formula Q = CiA has units: C (dimensionless) × i (mm/hr) × A (hectares). 1 mm/hr over 1 ha = 10,000 m² × 0.001 m/3600 s = 10,000/3,600,000 m³/s = 1/360 m³/s. So Q [m³/s] = C × i [mm/hr] × A [ha] / 360.
Fix: Q = CiA/360 (all units as above). If i is in cm/hour: Q = CiA/36. If i is in m/hour: Q = CiA × 10,000/3600 = CiA × 2.778. Always check units of the rainfall intensity given in the problem before applying the conversion constant.

12. Frequently Asked Questions

Q1. Why do separate sewer systems have lower effectiveness in Indian cities than intended?

The separate sewer system (sanitary sewer for sewage, storm drain for rainwater) is theoretically superior because it prevents dilution and overloading of treatment plants during storms and eliminates combined sewer overflows (CSOs). However, in Indian cities, the separate system often fails in practice because: illegal cross-connections are made by residents and contractors (sanitary drains connected to storm drains, which is common when storm drains are accessible and sanitary sewers are not), so stormwater dilutes sewage and untreated sewage enters drains; old combined sections remain in parts of the city that haven’t been upgraded; infiltration and inflow (I/I) of groundwater and stormwater through cracked pipes and open manholes is high; and storm drains receive sewage from slum settlements that lack sewer connections. A study of Indian STPs (Sewage Treatment Plants) found that many plants receive 40–60% more flow during monsoon than their design capacity — significantly above what infiltration alone explains, indicating substantial cross-connections. Effective operation of separate systems requires systematic I/I assessment using CCTV inspection of sewers, smoke testing of storm drains, and flow monitoring at sewer junctions to identify and seal cross-connections.

Q2. What is an inverted siphon in a sewer system, and when is it used?

An inverted siphon (also called a depressed sewer or sag pipe) is a section of sewer that dips below the normal gradient to pass under an obstacle — a river, a canal, a railway crossing, a road underpass, or a buried utility corridor. Unlike a true siphon (which relies on suction and can convey liquid above the hydraulic gradient), an inverted siphon works under pressure: the sewage is forced through the dip by the head difference between the upstream and downstream water levels in the upstream and downstream manholes. Design requirements include: minimum velocity of 1.0–1.5 m/s through the siphon legs to prevent sedimentation (silt readily settles in low-velocity pressurised pipes); provision of multiple parallel pipes (typically 2 or 3 legs of different diameters) so that the smaller legs handle DWF with high velocity while larger legs open during peak flow; provision of sluice gates to isolate individual legs for maintenance; a cleanout chamber at the bottom of the dip where sediment collects and can be flushed; and bypass provision for maintenance shutdown. Inverted siphons are avoided where possible due to maintenance difficulty — sewage accumulates sediment at the bottom, and the pressurised section cannot be inspected by CCTV without careful planning. They are nevertheless unavoidable in flat terrain cities like Chennai, Kolkata, and Mumbai where crossing waterways without raising the sewer above the obstacle grade is necessary.

Q3. How does Harmon’s peak factor formula work, and what are its limitations?

Harmon’s formula P_f = 1 + 14/(4 + √P) (P in thousands) gives the ratio of peak hourly sewage flow to average daily flow for a sewer section serving population P. The formula reflects the observation that smaller populations served by a sewer section show higher variability in flow (everyone uses water at the same time in the morning) — peak factors of 3–4 for small neighborhoods — while large sewer mains serving millions show smoother flows with peak factors of 1.5–2.0 because different neighborhoods reach their peaks at slightly different times, averaging out the fluctuations. The formula’s limitations are: it was originally calibrated on American data and may not fully represent Indian usage patterns (which show sharper morning peaks due to intermittent water supply and cultural bathing habits); it doesn’t account for industrial contributions that peak at different times; and it assumes sewage flow variation mirrors water supply variation, which is approximately true for residential areas but less accurate for commercial or industrial zones. CPHEEO guidelines allow peak factors up to 3.0 for small laterals and 2.0–2.5 for sub-mains in Indian design practice.

Q4. What is the significance of the grease trap in commercial and food service sewer connections?

Grease traps (also called grease interceptors) are small tanks installed in the drain lines of commercial kitchens, restaurants, hotels, and food processing facilities before they connect to the municipal sewer. Hot greasy wastewater from cooking enters the trap, where it cools; the solidified grease and fats float to the top and are retained, while cooler wastewater flows through to the sewer. Without grease traps, fats, oils, and grease (FOG) from cooking accumulate in sewer pipes, solidify on the pipe walls (especially in small diameter laterals), and cause blockages — the so-called “fatberg” problem now notorious in London, New York, and increasingly Indian cities like Mumbai and Delhi. FOG blockages are a major cause of sewer overflows (sewage backs up and overflows at manholes or into buildings) and require expensive high-pressure jetting to clear. IS 5600 and CPHEEO guidelines require grease traps on all food service connections to municipal sewers. In practice, enforcement is poor in India — most restaurants lack functional grease traps, contributing to the high sewer blockage rates in commercial areas. The SWM Rules 2016 and NMCG (National Mission for Clean Ganga) guidelines are strengthening requirements for grease trap installation as part of the Namami Gange programme’s sewer infrastructure upgradation.