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Distribution networks

Intermediate Water Supply Engineering Reviewed 24 Oct 2026 2 min read

The distribution network carries treated water from service reservoirs to every consumer, at enough pressure and in enough quantity, around the clock where possible. It is the largest part of a water scheme's capital cost and the source of most losses and complaints.

Layouts include dead-end (tree) systems, simple but prone to stagnant water and supply cuts when a pipe breaks; grid (looped) systems, where water reaches each point by more than one route, giving better pressures and reliability; and ring and radial systems for planned areas. Networks are divided into district metered areas so that flows can be measured and leaks found. Service reservoirs, elevated or ground-level, balance hourly demand against steady supply, store water for fire and emergencies, and set pressure.

Diagram: Indiacon original, not to scale.

Pipes are designed for peak hourly demand, which can be two to three times average, with minimum residual pressures specified by building height and velocities kept in a reasonable range to limit head loss and water hammer. Pipe materials include ductile iron, mild steel, HDPE, PVC and DI or steel with linings. Valves isolate sections, air valves release trapped air at high points, scour valves drain low points and fire hydrants are spaced along streets. Continuous pressurised supply, rather than intermittent supply, is the goal, because intermittent supply draws contaminated water into pipes when they empty.

Tip

Model the network with real demand patterns and calibrate it against field pressure readings.

Key formulas
  • Design flow = average demand × peak factor
  • Hazen-Williams: h_f = 10.67 L Q^1.852 / (C^1.852 D^4.87)
  • Balancing storage from mass curve of supply versus demand
  • Residual pressure ≥ code minimum at farthest node
Solved example

Problem. A zone has average demand 4 MLD with peak factor 2.5. A 350 mm main with C = 130 runs 1.8 km. Find peak flow and head loss.

  1. Peak = 4 × 2.5 = 10 MLD = 0.1157 m³/s.
  2. h_f = 10.67 × 1 800 × 0.1157^1.852 / (130^1.852 × 0.35^4.87).
  3. 0.1157^1.852 = 0.01844; 130^1.852 = 8 197; 0.35^4.87 = 0.00602.
  4. h_f = 10.67 × 1 800 × 0.01844 / (8 197 × 0.00602) = 354.2 / 49.35 = 7.2 m.

Answer. Peak about 116 L/s; head loss about 7.2 m

Quick quiz

1Why are looped networks preferred?

Better pressures and reliability, since water has more than one route.

2What is a district metered area?

A zone with measured inflow used to monitor demand and leakage.

3Why is intermittent supply harmful?

Empty pipes draw in contaminated water.

Code note
  • CPHEEO Manual on Water Supply and Treatment (India): Central government manual for planning, design and operation of public water supply.
  • IS 1172 (India): Basic requirements for water supply, drainage and sanitation, including per-capita demand.

Clauses and documents to read

  • CPHEEO Manual on Water Supply and Treatment (distribution system chapters, cited by name)
  • IS 1172:1993 (water requirements)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • S. K. Garg, Water Supply Engineering (Environmental Engineering Vol. I), Khanna Publishers, latest edition.
  • MWH (J. C. Crittenden et al.), Water Treatment: Principles and Design, Wiley, 3rd edition.
  • H. S. Peavy, D. R. Rowe, G. Tchobanoglous, Environmental Engineering, McGraw Hill, 1985.

Free and official sources

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