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Gumbel 500-year flood from two risk-based floods

Short answer

Risk 63.6% → T=50 yr (4200 m³/s); risk 39.5% → T=100 yr (5800 m³/s). Using Gumbel line slope 2291.6, the 500-year discharge is 9497 m³/s.

The project life of a headworks is 50 years. The flood discharges at the risks 63.6% and 39.5% are 4200 cumecs and 5800 cumecs respectively, which is derived from long term historical data using Gumbel distribution. Calculate the discharge from 500 years return period. Plot these three discharges on Gumbel paper.

Step 1: Return periods from the risk

Life n = 50 years; T=11−(1−R)1/nT=\dfrac{1}{1-(1-R)^{1/n}}.

  • R = 63.6%: T=11−(0.364)1/50=49.98≈50T=\dfrac{1}{1-(0.364)^{1/50}}=49.98\approx50 years, so 4200 m³/s is the 50-year flood.
  • R = 39.5%: T=11−(0.605)1/50=100.00≈100T=\dfrac{1}{1-(0.605)^{1/50}}=100.00\approx100 years, so 5800 m³/s is the 100-year flood.

Step 2: Gumbel line

xT=xˉ+Kσ=a+b yTx_T=\bar x+K\sigma=a+b\,y_T is a straight line in the reduced variate yT=−ln⁡[ln⁡(T/(T−1))]y_T=-\ln[\ln(T/(T-1))].

T (yr)yTy_TxTx_T (m³/s)
503.9024200
1004.6005800
5006.214?
b=5800−42004.600−3.902=2291.6 m3/sb=\frac{5800-4200}{4.600-3.902}=2291.6 \text{ m}^3/\text{s} x500=5800+2291.6 (6.214−4.600)=9497 m3/sx_{500}=5800+2291.6\,(6.214-4.600)=9497 \text{ m}^3/\text{s}

Answer: the 500-year discharge ≈ 9497 m³/s.

Step 3: Plotting on Gumbel paper

Use arithmetic graph paper with a uniform yTy_T axis (labelled with the return period T at the positions y = 3.902, 4.600 and 6.214) and an arithmetic discharge axis. Plot the three points (T = 50, 4200), (T = 100, 5800) and (T = 500, 9497); they lie on one straight line.

 Q (m3/s)
  9497|                              o  T=500
  5800 |                 o  T=100
  4200 |         o  T=50
       +----+----+----+----+----+----+--- y_T
            3    4    5    6    7

✓ Answer checked by Claude Sonnet 5.5

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