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Gumbel flood for 500-year return period

Short answer

Using Gumbel's reduced variate, slope b = (1020-850)/(4.6001-3.9019) = 243.48 m³/s. Then 500-year flood = 1020 + 243.48×(6.2136-4.6001) ≈ 1413 m³/s.

For a river the estimated flood peaks for two return periods by the use of Gumbel's method are given below:

Return period (yrs)Peak flood (m3^3/s)
1001020
50850

What flood discharge in this river will have a return period of 500 yrs?

In Gumbel's method xT=xˉ+Kσ=a+b yTx_T=\bar x+K\sigma = a + b\,y_T, which is a straight line in yTy_T (the reduced variate). Using two known points gives the line, so yˉn\bar y_n and SnS_n are not needed.

yT=−ln⁡[ln⁡TT−1]y_T = -\ln\left[\ln\frac{T}{T-1}\right]
T (yr)yTy_TxTx_T (m³/s)
503.9019850
1004.60011020
5006.2136?

Slope of the line (equal to σ/Sn\sigma/S_n):

b=1020−8504.6001−3.9019=243.48 m3/sb=\frac{1020-850}{4.6001-3.9019}=243.48 \text{ m}^3/\text{s} x500=1020+243.48 (6.2136−4.6001)=1413 m3/sx_{500}=1020+243.48\,(6.2136-4.6001)=1413 \text{ m}^3/\text{s}

Answer: the 500-year flood ≈ 1413 m³/s.

✓ Answer checked by Claude Sonnet 5.5

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