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Gumbel 200-year flood from 50 and 100-year floods

Short answer

Using yT: b=(740-660)/(4.6001-3.9019)=114.58 m³/s. 200-yr flood = 660+114.58×(5.2958-3.9019)=820 m³/s.

From the analysis of available data on annual flood peaks of a small stream for a period of 35 years, 50 years and 100 years flood have been estimated to be 660 m3^3/s and 740 m3^3/s. Using Gumbel's method, estimate the 200 years flood for the stream. (Take Yn=0.54034Y_n = 0.54034 and Sn=1.12843S_n = 1.12843 for n = 35 years)

Gumbel's equation is linear in the reduced variate: xT=a+b yTx_T=a+b\,y_T, with yT=−ln⁡[ln⁡(T/(T−1))]y_T=-\ln[\ln(T/(T-1))]. The two given floods fix the line, so YnY_n and SnS_n are not needed for x200x_{200} (they are needed only to separate the line into xˉ\bar x and σ\sigma).

T (yr)yTy_TxTx_T (m³/s)
503.9019660
1004.6001740
2005.2958?
b=σSn=740−6604.6001−3.9019=114.58 m3/sb=\frac{\sigma}{S_n}=\frac{740-660}{4.6001-3.9019}=114.58 \text{ m}^3/\text{s} x200=660+114.58 (5.2958−3.9019)=820 m3/sx_{200}=660+114.58\,(5.2958-3.9019)=820 \text{ m}^3/\text{s}

(Check with the statistics: σ=b Sn=114.58×1.12843=129.3\sigma=b\,S_n=114.58\times1.12843=129.3 m³/s and xˉ=660−b(y50−Yn)=274.8\bar x=660-b(y_{50}-Y_n)=274.8 m³/s.)

Answer: 200-year flood ≈ 820 m³/s.

✓ Answer checked by Claude Sonnet 5.5

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