SME
Hydrology · Problem type
The 100-year flood by Gumbel’s method
Use this for “Yearly peak flows average 500 m³/s with a standard deviation of 150 m³/s. Estimate the 100-year flood by Gumbel’s method.”
Table
Return period (T)
(yr)
Design flood (xT)
(m³/s)
This is an example. Type your own values to start a new problem.
xT (m³/s)
=
x̄ (m³/s)
+
KT
×
s (m³/s)
T
Return period
example · yr
Formulas
KT
=
−0.7797
×
(0.5772
+
ln(ln(
T
T
− 1
)))
3.1367
=
−0.7797
×
(0.5772
+
ln(ln(
100
100 − 1
)))
xT
=
x̄
+
KT
×
s
970.5 = 500 + 3.1367 × 150
Assumptions
•
Yearly peak flows follow a Gumbel (extreme value type I) distribution.
•
This frequency factor is for a long record; short records use tables with slightly larger factors.
•
0.5772 is Euler’s constant, and √6 ÷ π = 0.7797 links the standard deviation to the Gumbel spread.
Step-by-step
WE KNOW
T = 100 yr, x̄ = 500 m³/s, s = 150 m³/s
FIND
frequency factor (KT), design flood (xT)
Step 1 · Find frequency factor (KT)
Use KT = −0.7797 × (0.5772 + ln(ln(T ÷ (T − 1))))
The longer the return period, the more standard deviations the flood sits above the mean.
KT
=
−0.7797
×
(0.5772
+
ln(ln(
100
100 − 1
)))
ln(
100
99
)
=
0.01005
ln(0.01005) = −4.6
0.5772 + (−4.6) = −4.023
−0.7797 × (−4.023) = 3.137
KT = 3.1367
Step 2 · Find design flood (xT)
Use xT = x̄ + KT × s
Start at the mean and go up KT standard deviations.
xT = 500 + 3.1367 × 150
xT = 500 + 470.505
xT
=
970.5
m³/
s
Check
Put the numbers back in. Both sides should match.
3.1367
=
−0.7797
×
(0.5772
+
ln(ln(
100
100 − 1
)))
✓
970.5 = 500 + 3.1367 × 150 ✓
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