18.2 Water Distribution Pressure, Demand, and Storage

Key Takeaways

  • Node pressure equals the hydraulic grade line elevation minus node elevation, converted between feet of water and psi using 2.31.
  • Average day, maximum day, peak hour, and fire-flow demands answer different design questions and must not be interchanged.
  • Storage normally combines operational, equalization, fire, and emergency components, subject to local design criteria.
  • Tanks set or stabilize hydraulic grade; pumps, pressure-reducing valves, and pressure zones control where that grade is usable.
  • Looped networks improve reliability, but network answers still require continuity and headloss compatibility around each loop.
Last updated: June 2026

Distribution Systems as Hydraulic Grade Networks

The PE Civil WRE specification covers drinking-water distribution systems, present and future demands, and storage. Distribution problems look like pipe-loss calculations, but the decision being tested is service reliability: can the system deliver the required flow while holding usable pressure, acceptable water age, and an emergency reserve?

A distribution system is a connected set of sources, pumps, tanks, pressure zones, transmission mains, distribution mains, valves, hydrants, and service connections. At any node, pressure head equals the hydraulic grade line (HGL) elevation minus the ground or pipe elevation at that node. In U.S. customary units, pressure in psi equals pressure head in feet divided by 2.31 for water (because 1 psi = 2.31 ft of water). This single conversion is the fastest plausibility check on a distribution answer: 40-80 psi normal service translates to roughly 92-185 ft of head above the node.

Demand Terms

Demand termTypical meaningPE use
ADDAverage day demandAnnual planning baseline
MDDMaximum day demandSource, treatment, storage sizing
PHDPeak hour demandDistribution pressure and pipe capacity
Fire flowRequired hydrant flow for a durationFire storage and residual pressure
Future demandProjected demand at buildoutCapacity and phasing

Peak hour demand is not maximum day demand. A common chain is MDD = peaking factor x ADD, then PHD = peaking factor x MDD. If ADD is 2.0 MGD with MDD factor 1.8 and PHD factor 1.5, then MDD = 3.6 MGD and PHD = 5.4 MGD. MDD usually governs supply and treatment; PHD plus fire flow usually governs distribution mains, storage drawdown, and residual pressure at critical nodes. Apply each peaking factor to the correct base.

Pressure and Storage Workflow

  1. Convert population, unit use, industrial flow, and leakage to the requested demand basis.
  2. Identify the controlling case: average, maximum day, peak hour, fire flow, or pump-fill.
  3. Set the HGL from the tank water surface, reservoir, pump discharge head, or pressure-zone boundary.
  4. Subtract pipe and minor losses along the flow path to the node of interest.
  5. Convert remaining head above node elevation to pressure and compare with the criterion.
  6. Build storage as the sum of required components in the problem's units and duration.

Worked pressure example. An elevated tank sits at water surface 920 ft. A critical node is at elevation 760 ft, and headloss to it under PHD is 25 ft. HGL at the node is 920 - 25 = 895 ft; pressure head is 895 - 760 = 135 ft; pressure is 135 / 2.31 = 58 psi, which clears a typical 40 psi minimum. During a coincident fire-flow event the added flow increases headloss, drops the HGL, and the same node must still hold a residual (commonly 20 psi) at the hydrant.

Storage Components

ComponentPurposeCommon form
OperationalPrevents excessive pump cyclingVolume between control levels
EqualizationCovers hourly demand above supply ratePercentage or mass-curve volume
FireSupports required fire flow for a durationFlow x duration
EmergencyCovers outages, breaks, or source lossStated reserve or demand fraction

Elevated tanks and standpipes set the HGL directly from water-surface elevation. Ground storage needs pumps to create service pressure unless sited high enough. Pressure zones divide steep terrain so low areas avoid excessive pressure and high areas keep adequate pressure. Pressure-reducing valves (PRVs) lower downstream grade; booster pumps raise it.

Network Reasoning and Operations

A looped distribution network must satisfy both continuity at junctions (flow in equals flow out) and headloss compatibility around loops (the algebraic sum of headlosses around each closed loop is zero). You may not need a full Hardy Cross iteration on a six-minute item, but you should know the logic: flow redistributes until each loop balances. Closing a valve, adding fire flow, or taking a tank offline rewrites the network and can expose a low-pressure pocket.

Hardy Cross flow correction for a loop is delta-Q = -sum(hL) / [n x sum(hL/Q)], where n is the exponent on Q in the headloss law (about 1.85 for Hazen-Williams, 2 for Darcy-Weisbach). Sign convention (clockwise positive) must be consistent, and the correction is applied to every pipe in the loop until delta-Q is negligible.

Operationally, high headloss flags undersized pipe, a closed valve, tuberculation, high roughness, excessive velocity, or an overstated demand. Low chlorine residual or water-age complaints suggest oversized dead-end storage or poor turnover.

SymptomLikely causeFirst check
Low pressure at high node, peak hourUndersized main, tank drawdown, closed valveTrace HGL to that node
High pressure in low areaMissing or failed PRV / pressure zoneVerify PRV setpoint
Discolored or stale waterDead-end main, low turnoverFlushing and storage turnover
Cannot meet fire flow residualInsufficient main capacity or storageRecompute losses at fire flow

A PE answer should connect the calculation to the service goal: adequate pressure, adequate storage, and reliable flow under the specified condition, not just a number.

Test Your Knowledge

An elevated tank has a water surface elevation of 875 ft during a peak-hour condition. A junction is at elevation 730 ft, and headloss from the tank to the junction is 18 ft. What is the approximate pressure at the junction?

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Test Your Knowledge

A community has an average day demand of 1.8 MGD. Maximum day demand is 1.8 times average day. Required equalization storage is 20 percent of maximum day demand. Fire storage is 2,500 gpm for 2 hours, and emergency storage is 0.50 MG. What total storage is required?

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