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Reservoir planning

Intermediate Irrigation and Water Resources Engineering Reviewed 24 Oct 2026 2 min read

A reservoir stores water when it is plentiful to release it when it is scarce. Planning one means deciding how much storage is needed for a target supply, where to locate the dam, how to divide the storage among uses, and how long the reservoir will last before sediment fills it.

Storage is divided into zones. Dead storage, below the lowest outlet, holds sediment and cannot be released by gravity. Live storage, between the minimum drawdown level and the full reservoir level, supplies irrigation, drinking water and power. Above full reservoir level, the flood cushion up to the maximum water level absorbs part of the design flood, and freeboard above that protects the dam from waves.

Diagram: Indiacon original, not to scale.

The live storage needed for a given demand can be found from a mass curve, the cumulative inflow plotted against time, by drawing demand lines tangent to its peaks and measuring the largest gap below them. The sequent peak algorithm does the same in a table: it carries forward the cumulative deficit, resetting to zero whenever inflow refills the reservoir, and the largest value is the required storage. Longer records, including droughts, give more reliable answers, and evaporation and seepage losses are added.

Sedimentation reduces capacity every year. The trap efficiency of a reservoir, which depends on the ratio of capacity to annual inflow, estimates how much incoming sediment stays. Catchment treatment, sediment flushing and sluicing extend reservoir life.

Tip

Use the longest inflow record available and include the worst drought.

Key formulas
  • Sequent peak: K_t = max(0, K_(t−1) + D_t − I_t); required storage = max K_t
  • Live storage = storage at FRL − storage at minimum drawdown level
Common mistake

Ignoring evaporation, which can be large from wide shallow reservoirs.

Solved example

Problem. Monthly inflows to a reservoir are 10, 4, 2, 1, 6 and 15 Mm³, and the demand is 6 Mm³ every month. Find the storage needed by the sequent peak method (ignore losses).

  1. K starts at 0. Month 1: 0 + 6 − 10 = −4, so K = 0. Month 2: 0 + 6 − 4 = 2.
  2. Month 3: 2 + 6 − 2 = 6. Month 4: 6 + 6 − 1 = 11.
  3. Month 5: 11 + 6 − 6 = 11. Month 6: 11 + 6 − 15 = 2.
  4. Maximum K = 11 Mm³. In practice the record is run twice (or over many years) to confirm the cycle.

Answer. Required live storage 11 Mm³

Quick quiz

1What is dead storage?

Storage below the lowest outlet, reserved for sediment.

2What does the flood cushion do?

Absorbs part of the design flood between FRL and MWL.

3What does the sequent peak method give?

The storage needed to meet a demand over the inflow record.

Code note
  • IS 5477 (India): Methods for fixing the capacities of reservoirs (general, dead, live and surcharge storage).

Clauses and documents to read

  • IS 5477 (Parts 1 to 4): methods for fixing the capacities of reservoirs (general, dead, live and surcharge storage)

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, Irrigation Engineering and Hydraulic Structures, Khanna Publishers, latest edition.
  • G. L. Asawa, Irrigation and Water Resources Engineering, New Age International, latest edition.

Free and official sources

All references for this subject →

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