Water Resources HubWater topic 18 of 18 — free theory
Flood Frequency & Reservoir Operations
Return periods, the Log-Pearson III distribution, Rippl storage sizing, and why a routed flood peak is always lower and later.
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Return period and exceedance probability
A “T-year flood” is the flow whose annual exceedance probability is 1/T — a 100-year flood has a 1% chance of being equalled or exceeded in any given year. Over a project life of n years, the chance of seeing it at least once is:
P = 1T Riskn = 1 − (1 − 1T)n
| P | annual exceedance probability |
| T | return period, years (an average recurrence interval, not a schedule) |
| Riskn | probability of at least one exceedance in n years |
When n is much smaller than T, Riskn ≈ n/T is a decent approximation — but for n near T, use the full formula. A 100-year flood over a 50-year mortgage is not 50%: it is about 39.5%.
Log-Pearson III — the standard for flood frequency
US practice fits annual peak flows with the Log-Pearson Type III distribution (Bulletin 17C): take the logarithms of the annual peaks, compute their mean, standard deviation, and skew, then
log QT = (mean of logs) + K · (std. dev. of logs)
| QT | flood magnitude with return period T |
| K | frequency factor — depends on T and on the skew of the logs |
Skew is the interesting part. Zero skew collapses LP3 to the log-normal case; positive skew stretches the upper tail, so rare-flood estimates come out larger than log-normal would give; negative skew compresses it. On the exam, read the skew first — it tells you which way the tail leans.
Reservoir storage — the Rippl (mass curve) method
Plot cumulative inflow against time and cumulative demand on the same axes. Whenever demand outruns inflow, the reservoir makes up the difference; the required active storage is the largest cumulative deficit drawn during the critical dry spell. Surpluses that arrive before the dry spell spill away and don’t count — only the drawdown matters.
The flip side is firm yield: the maximum constant demand the reservoir can sustain through the driest period on record, given its storage. More storage buys more firm yield — up to the mean inflow, beyond which no storage helps.
Flood routing — storage attenuates the peak
Routing a flood through reservoir storage (level-pool routing) is continuity in action:
I − O = ΔSΔt
| I, O | inflow and outflow rates |
| ΔS/Δt | rate of change of reservoir storage |
While inflow exceeds outflow the pool rises and stores water, so the outflow peak must come later and lower than the inflow peak. Storage doesn’t destroy flood volume — it reshapes it.
PE depth: storage-indication routing and spillway checks
Hand routing uses the storage-indication (Puls) form of continuity — stepping (2S/Δt + O) forward in time — but the exam usually tests the concepts rather than the full table: peak attenuation, lag, and the fact that outflow can never exceed inflow once the pool stops rising.
Pair this with trap efficiency from Topic 17: the same reservoir that tames the flood is quietly filling with sediment and losing the storage its routing depends on. A spillway sized for today’s storage–elevation curve deserves a second look once sedimentation is factored in.
PE trap: routing conserves volume (minus evaporation and seepage). Any answer showing less outflow volume than inflow volume — without losses — violates continuity.
Worked example 50-year risk and a Rippl storage check
Given:
- A home with a 50-year mortgage sits in the 100-year floodplain.
- A reservoir serves a constant demand of 8 million m³/month. Six-month inflows (million m³): 12, 9, 6, 4, 5, 10.
Solution:
- Risk of at least one 100-year flood in 50 years: Risk = 1 − (1 − 1/100)50 = 1 − 0.9950. Since 0.9950 ≈ 0.6050, Risk ≈ 1 − 0.6050 = 0.395 ≈ 39.5%.
- Rippl running balance (start full; demand − inflow each month): M1: 8−12 = −4 (surplus spills, reservoir stays full); M2: 8−9 = −1 (full); M3: 8−6 = 2 drawn (2); M4: 8−4 = 4 drawn (6); M5: 8−5 = 3 drawn (9); M6: 8−10 = −2 (drawn falls to 7).
- The deepest drawdown is 9 million m³ at the end of month 5 — that is the required active storage.
Answer: roughly a 2-in-5 chance (39.5%) of flooding during the mortgage; required active storage ≈ 9 million m³.