Water Resources HubWater topic 15 of 18 — free theory
Water Distribution Systems
Demand estimation, storage sizing, pressure zones, network analysis, and water age — how a drinking-water system is planned and checked.
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Demand estimation — average day and peaking factors
Everything starts with the average-day demand (ADD): population times per-capita use. Pipes, pumps, and tanks are then sized for the peaks, expressed as multipliers on the ADD rather than on each other.
ADD = population × per-capita use
| per-capita use | typical residential ≈ 300–450 L/person/day in exam problems (includes commercial/industrial share when stated) |
Qmax-day = PFmd × ADD Qpeak-hour = PFph × ADD
| PFmd | max-day peaking factor, typically 1.5–2.0 |
| PFph | peak-hour peaking factor, typically 2.5–4.0 |
| Both factors multiply the ADD — never chain them (peak-hour is not PFph × max-day). |
Max-day demand sizes supply and treatment; peak-hour demand sizes the distribution piping and sets the minimum-pressure check.
Storage sizing — three components
A distribution tank is not one volume but three stacked together: equalization to ride out the daily demand swing, fire storage for the design fire, and an emergency reserve.
Vtotal = Veq + Vfire + Vemerg
| Veq | equalization storage ≈ 20–25% of max-day demand (rule of thumb), or computed from the diurnal demand curve |
| Vfire | fire storage = fire flow × required duration (e.g. 90 L/s × 2.5 h) |
| Vemerg | emergency reserve ≈ 25% of (Veq + Vfire) |
Standard practice pairs the fire flow with max-day demand, not peak-hour — sizing for fire + peak-hour would build a tank nobody can justify.
The rigorous way to find equalization storage is the mass-curve method: plot cumulative inflow (pumps or treatment plant output) against cumulative demand over 24 hours. The required storage is the maximum cumulative surplus minus the maximum cumulative deficit — the tank absorbs the difference. The worked example below walks through it.
Pressure zones
Pressure in a gravity-fed zone is set by the tank's water level minus the elevation of the customer. When terrain varies too much for one tank to serve everyone, the system is split into zones, each with its own tank or pressure-reducing valves (PRVs).
p/γ = HGL − z p = γ(HGL − z)
| HGL − z | pressure head at the customer = hydraulic grade line elevation minus ground elevation |
| γ | unit weight of water, 9.81 kN/m³ |
Common design targets: minimum ≈ 35 psi (≈ 240 kPa) at peak-hour demand; maximum ≈ 80–100 psi to protect plumbing and limit leakage.
Zone boundaries are drawn where the elevation difference between high and low customers would otherwise push pressures outside that band. The tank overflow elevation fixes the zone's HGL; everything downstream of a PRV lives in the lower zone.
Network analysis — what the software enforces
Whether you iterate by hand (Hardy Cross) or let a solver do it, a pipe network must satisfy two conditions at once:
Node: ΣQin − ΣQout = qdemand Loop: ΣhL = 0
| Hardy Cross | hand method: guess loop flows, apply the correction ΔQ = −ΣhL / (n Σ|hL/Q|), iterate until ΣhL ≈ 0 |
| n | head-loss exponent: ≈ 2 for Darcy-Weisbach (with f fixed), 1.852 for Hazen-Williams |
Programs such as EPANET solve the same node-loop equations with a gradient algorithm and extend them over time (extended-period simulation) to track tank levels and water age. The software does not add physics — it just enforces continuity and energy everywhere, simultaneously.
Water age and quality basics
Water age is the average time water spends in the system before reaching a tap — essentially the detention time in tanks and slow-moving mains. Older water has lower disinfectant residual and more disinfection by-products, so tanks are designed for turnover: regular cycling between high and low levels, inlet/outlet separation for mixing, and occasionally booster chlorination on the far side of a zone.
PE depth: zone design and extended-period thinking
On PE problems, a pressure zone is defined by its HGL, and the tank overflow elevation is that HGL (minus small losses). When a question gives you a tank overflow at El. 180 m and a customer at El. 120 m with 6 m of head loss at peak flow, the available pressure head is 180 − 120 − 6 = 54 m — no pump curve needed.
Extended-period simulation is how designers verify a tank actually turns over: the model steps through the diurnal curve, filling the tank at night and draining it through the peaks, and reports the oldest water age anywhere in the zone. If the model shows a tank that never drops more than a metre, that tank is oversized for equalization and will have a water-age problem.
PE trap: booster pumps inside a distribution zone add head to the local HGL — downstream pressures rise by the pump's head at that flow, which can push a low-elevation customer over the maximum pressure limit. Check the maximum-pressure case (low demand, high tank level) as well as the minimum.
Worked example Equalization storage from a diurnal demand pattern
Given:
- Town of 12,000 people at 380 L/person/day; max-day peaking factor 1.8.
- Supply pumps run at a constant rate for 24 h; demand pattern per 6-hour block: 10%, 35%, 35%, 20% of max-day demand.
Solution:
- ADD = 12,000 × 0.380 = 4,560 m³/day. Max-day = 1.8 × 4,560 = 8,208 m³/day. Pump rate = 8,208/24 = 342 m³/h, so each 6-h block receives 2,052 m³.
- Block demands: 0.10 × 8,208 = 820.8; 0.35 × 8,208 = 2,872.8; 2,872.8; 0.20 × 8,208 = 1,641.6 m³.
- Cumulative storage change per block: +1,231.2, +410.4, −410.4, 0 m³. Maximum cumulative surplus = 1,231.2 m³; maximum deficit = 410.4 m³.
- Equalization storage = 1,231.2 − (−410.4) = 1,641.6 m³.
Answer: Equalization storage ≈ 1,640 m³ (about 20% of max-day demand — consistent with the rule of thumb).