Water Supply System Design — Population Forecasting & Demand
Population forecasting methods, per capita water demand, fire demand, system components, storage design, and distribution network analysis — with IS 1172 standards and GATE CE worked examples
Last Updated: April 2026
- Design population is forecasted 30–40 years ahead using arithmetic, geometric, incremental increase, or logistic methods — the most tested GATE CE environmental engineering calculation after BOD.
- Per capita water demand (IS 1172): 135 LPCD for domestic use in cities > 1 million; 70 LPCD for rural areas.
- Total demand = domestic + industrial + commercial + institutional + fire fighting + losses (15–25%).
- Fire demand: Kuichling’s formula Qf = 3182√P (P in thousands); Freeman’s Qf = 1136(P/5 + 10).
- Peak factors: maximum daily demand = 1.8 × average daily; maximum hourly demand = 1.5 × average hourly = 2.7 × average daily.
- Storage: balancing reservoir volume = difference between cumulative supply and demand; overhead tank = 1/3 of daily demand.
- Distribution system: grid-iron (most reliable), branching, ring (circular) — Hardy Cross method for pipe network analysis.
1. Components of a Water Supply System
| Component | Function |
|---|---|
| Source of supply | Surface water (rivers, reservoirs, lakes) or groundwater (wells, tube wells, springs) |
| Intake works | Withdraw water from source; include intake tower, screens, pumps |
| Pumping station (intake) | Lift raw water from source to treatment plant (raw water pumping) |
| Treatment plant | Remove impurities; produce water meeting IS 10500 |
| Clear water sump (CWS) | Storage of treated water at treatment plant; provides contact time for disinfection |
| Transmission main | Carry treated water from WTP to distribution reservoirs; large diameter, no house connections |
| Storage reservoirs | Balance hourly/daily demand fluctuations; overhead tanks (OHTs) or ground-level service reservoirs |
| Distribution network | Deliver water from service reservoirs to consumers; network of pipes, valves, meters |
| Appurtenances | Fire hydrants, air valves, washout valves, pressure relief valves, meters |
2. Population Forecasting Methods
The design of a water supply system is based on the future population the system must serve. Population is forecasted to the end of the design period (typically 30 years for transmission mains; 15 years for treatment plants) using historical census data.
2.1 Arithmetic Increase Method
Pn = P₀ + n × ΔPavg
where:
- Pn = population after n decades (10-year periods)
- P₀ = present population
- ΔPavg = average increase per decade = (Plast – Pfirst)/(number of decades)
- n = number of decades into future
Use: Old/mature cities with slow, steady growth; steady absolute increase each period
Limitation: Underestimates for growing cities; overestimates for declining cities
2.2 Geometric Increase Method (Geometric Growth)
Pn = P₀ × (1 + r/100)ⁿ
where r = average percentage rate of increase per decade (%)
r = average of [(P₂–P₁)/P₁ × 100] per decade
Alternatively: log Pn = log P₀ + n × log(1 + r/100)
Use: Growing cities, new towns, areas with rapid urbanisation
Limitation: Overestimates for mature cities (assumes growth rate stays constant forever)
2.3 Incremental Increase Method
Combines arithmetic and geometric elements:
Pn = P₀ + n × ΔPavg + n(n+1)/2 × Δ²Pavg
where Δ²Pavg = average of second differences (acceleration/deceleration of growth)
Procedure:
- List census populations at equal intervals
- Compute first differences (increments per decade)
- Compute second differences (change in increment per decade)
- Average first differences = ΔPavg; Average second differences = Δ²Pavg
Use: Most cities; accounts for acceleration or deceleration in growth rate
2.4 Decremental Rate of Growth Method
Used for cities approaching saturation (stable/declining cities). The growth rate decreases each decade by a constant amount. Not commonly tested in GATE CE.
2.5 Logistic (S-Curve) Method
Population follows an S-curve (sigmoid) — rapid growth early, slowing as the city approaches saturation population Ssat:
Pt = Ssat / (1 + m × e–nt)
where m and n are constants determined from historical data (requires 3 data points)
Use: Mature cities with identifiable saturation population; accurate long-range forecasting
2.6 Graphical Method
Plot historical population data and extend the trend curve graphically. Subjective but useful when mathematical methods are inconsistent. Can also use a “comparison city” whose growth pattern at a similar stage of development is projected onto the study city.
2.7 Summary — Method Selection
| Method | Best For | Tendency |
|---|---|---|
| Arithmetic | Stable, mature cities | Underestimates for growing cities |
| Geometric | Rapidly growing cities | Overestimates for mature cities |
| Incremental Increase | Most Indian cities (mixed growth) | Most balanced; widely used in India |
| Logistic | Cities approaching saturation | Most accurate long-term but complex |
3. Water Demand — Per Capita and Categories
3.1 Per Capita Water Demand (IS 1172:1993)
| Category | Per Capita Demand (LPCD) | Notes |
|---|---|---|
| Domestic (cooking, drinking, bathing, washing, toilet flushing) | 135 | For cities > 1 million; includes garden watering |
| Domestic (smaller towns) | 100–120 | Lower usage; fewer flush toilets |
| Rural | 40–70 | IS 1172: 40 LPCD minimum; Jal Jeevan Mission: 55 LPCD FHTC norm |
| Industrial | 45–450 | Highly variable by industry type; 45 LPCD general provision in city demand |
| Commercial | 20–45 | Offices, hotels, markets |
| Institutional | Variable | Hospitals: 340–450 LPCD per bed; Schools: 45–135 LPCD |
| Losses and waste (unaccounted for water) | 15–25% of total | Pipe leakage, metering errors; typical NRW in Indian cities = 30–50% |
3.2 Total Per Capita Demand
Total demand = domestic + institutional + commercial + industrial + losses
For a typical Class I Indian city: total ≈ 200–270 LPCD gross
IS 1172 specifies: for Class I cities (population > 100,000) → 150–200 LPCD net supply
For Class II cities (50,000–100,000) → 120–150 LPCD
Commonly used in GATE CE problems: 135 LPCD or 150 LPCD for domestic demand
4. Peak Demand Factors
Water demand is not uniform — it varies by hour of day, day of week, and season. The distribution system and storage must be designed for peak demand, not just average demand.
Average daily demand (ADD): Qavg = Annual demand / 365
⭐ Maximum daily demand (MDD):
MDD = 1.8 × ADD (IS 1172)
Occurs during summer, festivals, or periods of high demand
⭐ Maximum hourly demand (MHD):
MHD = 1.5 × average hourly demand on maximum day
= 1.5 × (MDD/24) = 1.5 × 1.8 × ADD/24
= 2.7 × ADD/24
or: MHD = 2.7 × average hourly demand (on an annual basis)
4.1 Design Demand for Different Components
| System Component | Design Demand | Reason |
|---|---|---|
| Intake works, treatment plant, pumps (raw water) | Maximum daily demand (MDD) | Sized to treat entire day’s peak supply in 24 hours |
| Transmission main (WTP to reservoir) | Maximum daily demand | Carries daily supply at one rate |
| Distribution mains | Maximum hourly demand | Must supply peak hour demand simultaneously |
| Storage reservoir (overhead tank) | Balances MDD supply with MHD demand | Stores during low demand, supplies during peak |
| Fire fighting | MDD + fire demand simultaneously | Pipes must carry both normal and fire flows |
5. Fire Demand
Fire demand is the additional water supply required to fight a major fire while maintaining minimum pressure (7 m or 10 psi) at all hydrants simultaneously. Though infrequent, fire demand can significantly exceed normal demand and must be accommodated in system design.
5.1 Fire Demand Formulas
⭐ Kuichling’s formula (most widely used in India):
Qf = 3182 √P
where Qf = fire demand (litres/minute); P = population in thousands
Freeman’s formula:
Qf = 1136.5 (P/5 + 10)
where Qf in litres/minute; P in thousands
National Board of Fire Underwriters (NBFU) formula:
Qf = 4637 √P (1 – 0.01√P)
where Qf in litres/minute; P in thousands
Minimum fire demand requirements (CPHEEO):
Small towns (< 50,000): 1100–2200 L/min (or as specified by fire code)
Large cities (> 200,000): 5500–9000 L/min
Duration: minimum 4–6 hours of fire fighting
Fire flow storage = Qf × duration
6. Design Period
The design period is the number of years for which a water supply component is designed — the time horizon over which the system should adequately serve the forecast population without major upgrades.
| Component | Design Period (Years) | Reason |
|---|---|---|
| Dams, barrages, major structures | 50–100 | Very long life; expensive to modify |
| Transmission mains, distribution mains (large) | 30 | Long life; costly to upsize later |
| Treatment plant structures | 15–30 | Structures last longer; equipment shorter |
| Pumps, electrical equipment | 10–15 | Mechanical/electrical life; technology changes |
| Distribution pipes (small) | 15–30 | Moderately long life; some upsizing expected |
7. Sources of Water
7.1 Surface Water Sources
| Source | Characteristics | Treatment Required |
|---|---|---|
| Rivers (perennial/seasonal) | High turbidity, variable quality; monsoon variation; BOD from sewage; temperature variation | Full treatment (coagulation, sedimentation, filtration, chlorination) |
| Impounding reservoirs (dams) | Lower turbidity (settled); algal growth risk (eutrophication); consistent quality; controlled release | Coagulation, filtration, chlorination; algae removal if necessary |
| Lakes/natural ponds | Moderate turbidity; seasonal algal bloom; bird/wildlife contamination risk | Coagulation, filtration, chlorination |
7.2 Groundwater Sources
| Source | Characteristics | Treatment Required |
|---|---|---|
| Open dug well | Shallow; vulnerable to surface contamination; seasonal water table variation | Disinfection minimum; often nothing in rural India |
| Tube well / bore well | Deep, protected; usually low turbidity; potential excess iron, manganese, fluoride, arsenic, nitrate | Aeration (Fe, Mn removal); defluoridation if needed; disinfection |
| Infiltration gallery | Horizontal collector near riverbank; bank filtration provides pre-treatment | Minimal treatment needed; disinfection only |
| Spring | Natural groundwater emergence; usually clean but check for contamination | Disinfection; check for minerals |
8. Storage — Balancing Reservoir and Overhead Tank
8.1 Service Reservoir (Balancing Reservoir)
A service reservoir (ground-level or elevated) stores water to balance the difference between the rate of supply from the treatment plant (uniform pumping) and the variable demand throughout the day.
⭐ Volume of balancing storage:
Vbalance = Maximum (cumulative demand – cumulative supply) over a 24-hour cycle
Determined graphically using a mass diagram (Rippl’s diagram) or tabular method:
Plot cumulative demand and cumulative supply vs time; the maximum vertical difference = required storage
Typical balancing storage: 1/3 to 1/2 of daily demand (for uniform pumping)
8.2 Overhead Tank (OHT)
Typical OHT capacity = 1/3 of daily demand
(Provides gravity pressure for distribution; positioned to maintain 7–10 m pressure at all consumer points)
OHT height above ground = depends on distribution area topography + minimum residual pressure required (7 m)
Height of OHT (HOHT): HOHT = highest consumer elevation + minimum pressure + friction head loss in supply main
8.3 Emergency Storage
In addition to balancing storage, 1–2 days of emergency supply is stored to cover pump failures, pipeline breaks, or treatment plant shutdowns. CPHEEO recommends: total storage = balancing storage + fire storage + emergency storage ≈ 1–2 days of average daily demand.
9. Distribution Network — Types and Analysis
9.1 Types of Distribution Systems
| System | Description | Advantage | Disadvantage | Use |
|---|---|---|---|---|
| Dead-end (tree) system | Main pipe with sub-mains and branches; no loops | Simple design; minimum pipe length; easy to isolate | Stagnation at dead ends; no alternate supply if main fails; poor pressure at ends | Rural areas; small towns; temporary supply |
| Ring (circular) system | Single loop around the distribution area; fed from one end | Better pressure; can isolate sections; alternate supply path | More pipe; limited flexibility for expansion | Medium-size towns |
| Grid-iron (interlaced/interconnected) | Pipes laid in a grid pattern; multiple loops; every node has multiple supply paths | Best reliability; alternate supply in any pipe failure; uniform pressure; no dead ends | Most pipe (highest cost); complex analysis; repair requires isolating area | All cities; recommended for Indian urban water supply |
9.2 Hardy Cross Method (Pipe Network Analysis)
The Hardy Cross method (see Civil_45 — Pipe Flow) solves the pipe network for flow distribution and head losses iteratively. For water distribution networks, the same method applies with head loss computed using Hazen-Williams formula instead of Darcy-Weisbach:
Hazen-Williams formula:
V = 0.8492 × C × R0.63 × S0.54
or: hf = 10.67 × L × Q1.852 / (C1.852 × D4.87)
where C = Hazen-Williams roughness coefficient; D = pipe diameter (m); L = length (m); Q = flow (m³/s)
Typical C values: new cast iron = 130; old cast iron = 100; concrete = 120; new PVC = 150
Hardy Cross correction: ΔQ = –Σ(hf)/[1.852 × Σ(hf/Q)]
(Compare with Darcy-Weisbach correction: ΔQ = –Σ(hf)/[2 × Σ(hf/Q)])
9.3 Pressure in Distribution System
Minimum residual pressure at consumer end: 7 m head (IS 1172)
Maximum pressure: 70 m head (to prevent pipe bursts and leakage)
Pressure at fire hydrant: minimum 1.05 kg/cm² = 10.3 m head during fire fighting
10. Worked Examples (GATE CE Level)
Example 1 — Arithmetic Increase Method (GATE CE 2022 type)
Problem: Census data for a city is as follows: 1991 = 1,00,000; 2001 = 1,20,000; 2011 = 1,45,000; 2021 = 1,70,000. Forecast the population for 2041 and 2051 using the arithmetic increase method.
Decade-wise increments:
1991–2001: 1,20,000 – 1,00,000 = 20,000
2001–2011: 1,45,000 – 1,20,000 = 25,000
2011–2021: 1,70,000 – 1,45,000 = 25,000
Average increment ΔPavg = (20,000 + 25,000 + 25,000)/3 = 70,000/3 = 23,333/decade
Forecast for 2041 (2 decades from 2021):
P₂₀₄₁ = 1,70,000 + 2 × 23,333 = 1,70,000 + 46,667 = 2,16,667 ≈ 2.17 lakh
Forecast for 2051 (3 decades from 2021):
P₂₀₅₁ = 1,70,000 + 3 × 23,333 = 1,70,000 + 70,000 = 2,40,000 = 2.40 lakh
Answer: P₂₀₄₁ ≈ 2.17 lakh; P₂₀₅₁ = 2.40 lakh
Example 2 — Geometric Increase Method (GATE CE 2021 type)
Problem: Using the same census data as Example 1, forecast the 2041 and 2051 populations using the geometric increase method.
Percentage growth rates per decade:
1991–2001: (20,000/1,00,000) × 100 = 20.0%
2001–2011: (25,000/1,20,000) × 100 = 20.83%
2011–2021: (25,000/1,45,000) × 100 = 17.24%
Average r = (20.0 + 20.83 + 17.24)/3 = 58.07/3 = 19.36%/decade
Forecast for 2041 (2 decades from 2021, n = 2):
P₂₀₄₁ = 1,70,000 × (1 + 19.36/100)² = 1,70,000 × (1.1936)²
= 1,70,000 × 1.4247 = 2,42,199 ≈ 2.42 lakh
Forecast for 2051 (3 decades from 2021, n = 3):
P₂₀₅₁ = 1,70,000 × (1.1936)³ = 1,70,000 × 1.7003 = 2,89,051 ≈ 2.89 lakh
Comparison: Geometric (2.42, 2.89 lakh) > Arithmetic (2.17, 2.40 lakh) — geometric always gives higher estimates for growing cities.
Answer: P₂₀₄₁ ≈ 2.42 lakh; P₂₀₅₁ ≈ 2.89 lakh (geometric method)
Example 3 — Incremental Increase Method (GATE CE 2020 type)
Problem: Using the same census data (1991: 1.00L, 2001: 1.20L, 2011: 1.45L, 2021: 1.70L), forecast 2031 population using the incremental increase method.
First differences (increments):
Δ₁ = 20,000; Δ₂ = 25,000; Δ₃ = 25,000
ΔPavg = (20,000 + 25,000 + 25,000)/3 = 23,333
Second differences (increments of increments):
δ₁ = Δ₂ – Δ₁ = 25,000 – 20,000 = 5,000
δ₂ = Δ₃ – Δ₂ = 25,000 – 25,000 = 0
Δ²Pavg = (5,000 + 0)/2 = 2,500
Forecast for 2031 (n = 1 decade from 2021):
P₂₀₃₁ = P₂₀₂₁ + n × ΔPavg + n(n+1)/2 × Δ²Pavg
= 1,70,000 + 1 × 23,333 + 1(2)/2 × 2,500
= 1,70,000 + 23,333 + 1 × 2,500
= 1,70,000 + 23,333 + 2,500 = 1,95,833 ≈ 1.96 lakh
Answer: P₂₀₃₁ ≈ 1.96 lakh (incremental increase method)
Example 4 — Design Water Demand and Daily Volume (GATE CE type)
Problem: A town has a design population of 2,50,000 in the year 2051. Per capita demand = 135 LPCD (domestic); add 25% for industrial, commercial, and institutional demands; and 15% for losses. Find (a) average daily demand, (b) maximum daily demand, and (c) maximum hourly demand.
Given: P = 2,50,000; domestic LPCD = 135
Total per capita demand:
Non-domestic demand = 25% of domestic = 0.25 × 135 = 33.75 LPCD
Sub-total = 135 + 33.75 = 168.75 LPCD
Losses = 15% of sub-total = 0.15 × 168.75 = 25.31 LPCD
Total per capita = 168.75 + 25.31 = 194.06 LPCD ≈ 194 LPCD
(a) Average daily demand (ADD):
ADD = 2,50,000 × 194 L/person/day = 4,85,00,000 L/day = 48.5 MLD
(b) Maximum daily demand (MDD):
MDD = 1.8 × ADD = 1.8 × 48.5 = 87.3 MLD
(c) Maximum hourly demand (MHD):
MHD = 2.7 × ADD/24 = 2.7 × 48.5/24 = 2.7 × 2.021 = 5.46 MLD/h = 5.46 × 10⁶/24 = 2,27,500 L/h
Or: MHD = 2.7 × 48.5 MLD/24 h = 131.0/24 = 5.46 MLD per hour
Answer: ADD = 48.5 MLD; MDD = 87.3 MLD; MHD = 5.46 MLD/hour
Example 5 — Fire Demand (GATE CE type)
Problem: A town has a population of 1,00,000. Find the fire demand using (a) Kuichling’s formula and (b) Freeman’s formula. Express in litres per minute.
Given: P = 1,00,000 = 100 × 10³; P in thousands = 100
(a) Kuichling’s formula:
Qf = 3182 √P = 3182 × √100 = 3182 × 10 = 31,820 L/min
(b) Freeman’s formula:
Qf = 1136.5 × (P/5 + 10) = 1136.5 × (100/5 + 10) = 1136.5 × (20 + 10) = 1136.5 × 30 = 34,095 L/min
For design, use the higher value: 34,095 L/min ≈ 34,100 L/min (Freeman’s governs)
Answer: Kuichling = 31,820 L/min; Freeman = 34,095 L/min; Design with higher = 34,095 L/min
Example 6 — Balancing Storage Volume (GATE CE type)
Problem: A town has an average daily demand of 5 MLD. The treatment plant pumps at a uniform rate for 8 hours per day. What is the minimum balancing storage volume needed in the service reservoir?
Given: ADD = 5 MLD; pumping for 8 h/day at uniform rate
Pumping rate:
Qpump = 5 MLD / 8 h = 0.625 MLD/h = 625 kL/h (during 8 hours)
During non-pumping 16 hours: Qpump = 0
Demand assumption: Assume average demand distributed uniformly throughout 24 hours
Average demand = 5 MLD/24 h = 208.3 kL/h
Storage required analysis:
During pumping 8 h: Supply = 625 kL/h; Demand = 208.3 kL/h; Surplus = 416.7 kL/h
Total surplus = 8 × 416.7 = 3333 kL
During non-pumping 16 h: Supply = 0; Demand = 208.3 kL/h; Deficit = 208.3 kL/h
Total deficit = 16 × 208.3 = 3333 kL
The surplus during pumping (stored) must equal deficit during non-pumping: 3333 kL = 3333 kL ✓
Required storage = 3333 kL = 3.33 ML = 3,333 m³
= 3333/5000 × 100 = 66.7% of daily demand
(Matches theoretical: uniform 8-hour pumping needs 2/3 of daily demand as storage)
Answer: Balancing storage = 3333 kL = 3.33 ML (≈ 2/3 of daily demand)
11. Common Mistakes
Mistake 1 — Using n = Number of Years Instead of Number of Decades in Population Forecasting
Error: Forecasting 30 years ahead with n = 30 (years) instead of n = 3 (decades) in Pn = P₀ + n × ΔP.
Root Cause: Census data in India is collected every 10 years — so one “period” in population forecasting formulas is one decade (10 years), not one year. The increment ΔP is computed per decade.
Fix: In all population forecasting formulas: n = number of decades to forecast, not years. If forecasting 30 years from 2021 (to 2051): n = 3 decades. Verify by checking that ΔPavg is in persons/decade — if it’s 20,000 per decade and you use n = 30 years, you’ll get 30 × 20,000 = 600,000 extra people instead of 3 × 20,000 = 60,000.
Mistake 2 — Confusing Maximum Daily Demand with Maximum Hourly Demand
Error: Designing distribution mains for MDD (1.8 × ADD) instead of MHD (2.7 × ADD/24).
Root Cause: Both are “peak demands” but for different purposes. Treatment plants and transmission mains are sized for MDD (peak in a day). Distribution mains must supply the maximum demand at any hour, which is MHD (much higher per unit time than MDD per unit time).
Fix: Treatment plant: sized for MDD. Distribution main capacity: sized for MHD. MHD in flow rate = 2.7 × ADD/24 hours. This is critical — a distribution main designed for MDD would be undersized and cause pressure drops during peak morning/evening hours.
Mistake 3 — Using Arithmetic Mean Instead of Average Percentage for Geometric Method
Error: Computing r = (r₁ + r₂ + r₃)/3 (arithmetic average) when geometric mean is more appropriate for the geometric method.
Root Cause: The geometric growth method ideally uses the geometric mean of growth rates: rgeo = [(1+r₁)(1+r₂)(1+r₃)]1/3 – 1. However, in practice, most GATE CE problems use the arithmetic average of percentage growth rates, which is an acceptable approximation.
Fix: For GATE CE: use arithmetic average of percentage growth rates (r₁ + r₂ + r₃)/n for simplicity unless the problem specifies geometric mean. The error is small (1–3%) for growth rates below 20% per decade. If the problem asks for geometric mean, compute: ravg = [(Plast/Pfirst)1/n – 1] × 100%.
Mistake 4 — Applying Kuichling’s Formula Without Checking Units (P in Thousands)
Error: Substituting P = 1,00,000 (full population) into Qf = 3182√P instead of P = 100 (in thousands).
Root Cause: In Kuichling’s formula, P must be in thousands. Using P = 1,00,000 gives Qf = 3182 × 316.2 = 10,06,000 L/min — impossibly high and clearly wrong.
Fix: Always convert: if population = 1,00,000 → P in thousands = 100. Qf = 3182 × √100 = 31,820 L/min. Similarly, Freeman’s formula also uses P in thousands.
Mistake 5 — Not Adding Losses When Computing Total Demand
Error: Computing design supply = population × domestic LPCD, ignoring the 15–25% unaccounted-for water (UFW) losses.
Root Cause: Students compute the net demand (water reaching consumers) but forget that the supply system must actually deliver more than net demand — the extra accounts for pipe leakage, metering errors, illegal connections, and operational losses.
Fix: Total supply = net demand / (1 – loss fraction). Or: add losses as a percentage on top. If net demand = 135 LPCD and losses = 15%: gross supply = 135/0.85 = 158.8 LPCD, or 135 + 0.15 × 135 = 155.25 LPCD (slightly different depending on whether % is on gross or net).
12. Frequently Asked Questions
Q1. Which population forecasting method should be used for a rapidly growing Indian city, and why?
For rapidly growing Indian cities — particularly medium-sized cities (2–10 lakh population) receiving significant rural-urban migration, IT/manufacturing employment centres, or newly declared municipal areas — the geometric increase method typically gives more realistic estimates because it assumes the percentage growth rate remains constant. If the city grew at 20% per decade historically (e.g., from 2 lakh to 2.4 lakh to 2.88 lakh), the geometric method correctly projects this proportional growth forward. The arithmetic method, by contrast, assumes the same absolute number is added each decade regardless of current size — which underestimates growth for expanding cities. However, for long-range forecasting (30–50 years), the geometric method often overestimates by projecting compound growth indefinitely — which no city sustains. In practice, Indian engineers use the incremental increase method as a compromise: it acknowledges changing growth rates and typically gives an intermediate forecast between arithmetic and geometric. For master planning purposes, multiple methods are applied and the professional judgement governs which is most credible given local economic and demographic conditions.
Q2. Why is the maximum daily demand factor 1.8 and the maximum hourly factor 2.7 in IS 1172?
These factors represent observed patterns of water use variation in Indian cities. The maximum daily demand of 1.8 × average daily reflects the fact that summer days (peak domestic use — filling overhead tanks, garden watering, cooling), festival days, or post-maintenance resumption periods can demand up to 80% more than the annual average. This factor was derived from metered demand studies across Indian water utilities and incorporated into IS 1172. The maximum hourly factor of 2.7 × average hourly represents the sharp morning peak (6–9 AM) and evening peak (6–9 PM) seen in Indian cities when all households use water simultaneously for bathing and cooking — creating a demand surge 1.5× above the already elevated maximum day demand rate. The 1.5 and 1.8 factors combine to give 2.7 (1.5 × 1.8 = 2.7). These factors are averages across many cities; actual values can range from 1.5 to 2.5 for daily peaking and 1.2 to 2.0 for hourly peaking depending on the city size, supply hours per day, and consumer storage behaviour. Cities with intermittent supply (which most Indian cities have — 4–8 hours per day rather than 24 hours) show much higher instantaneous peaks because consumers rush to fill storage vessels when water is available.
Q3. What is non-revenue water (NRW) and why is it so high in Indian water utilities?
Non-revenue water (NRW) is the difference between the water supplied into the distribution system and the water billed to consumers. It consists of two components: real losses (physical leakage from pipes, storage tanks, service connections — typically 60–70% of NRW in Indian cities) and apparent losses (unauthorised consumption from illegal connections and meter under-registration — typically 30–40%). Indian water utilities average 30–50% NRW — significantly higher than the 10–20% seen in well-managed systems. The primary causes in India are: old distribution pipes (asbestos cement, cast iron pipelines from the 1940s–1970s still in service in many cities) with high corrosion and break rates; intermittent supply creating pressure transients that accelerate leakage; poor metering infrastructure (many unmetered connections, inaccurate old meters); and underfunded operations and maintenance departments that cannot afford systematic leak detection. The Atal Mission for Rejuvenation and Urban Transformation (AMRUT) specifically funds NRW reduction programmes in Indian cities — including pipe replacement, smart meters, district metering areas (DMA), and 24×7 water supply demonstration projects which paradoxically reduce NRW by maintaining stable positive pressure (preventing vacuum-induced contamination ingress) and making leaks easier to detect and locate.
Q4. How are fire demand provisions integrated with the regular water distribution system in Indian towns?
Fire demand is integrated into the distribution system design through three mechanisms. First, the distribution pipe sizes are verified to carry normal maximum hourly demand plus fire flow simultaneously without pressure dropping below the minimum 7 m head. The pipe diameter is increased at critical nodes where fire demand would otherwise cause inadequate pressure. Second, dedicated fire hydrants are located at maximum 100 m spacing in commercial/industrial areas and 150 m spacing in residential areas, connected to mains of minimum 150 mm diameter. Third, a portion of the service reservoir storage is reserved as fire storage: Qf (litres/minute) × duration (typically 4 hours) = fire reserve volume. For a city of 1 lakh population: Qf = 31,820 L/min × 240 min = 76.4 lakh litres = 7.64 ML. This fire reserve is not normally used for regular supply and is maintained at full capacity. The National Building Code (NBC 2016) and CPHEEO Manuals provide detailed specifications for fire hydrant locations, flow testing, and pressure requirements. Most Indian municipalities have historically under-provisioned fire water supply, which is a significant contributing factor to the severity of urban fires.