Flood Hydrology

Rational method

The rational method estimates the peak discharge from a small catchment, assuming that the peak occurs when the whole catchment contributes, i.e. when the rainfall duration equals the time of concentration $t_c$, and that the runoff rate equals the rainfall excess rate.

$$Q_p = \frac{C\,i\,A}{360}$$

where $Q_p$ is in m³/s, $i$ is the average rainfall intensity (mm/h) for duration $t_c$ and the required return period (from IDF curves), $A$ is the area in hectares, and $C$ is the runoff coefficient (ratio of peak runoff to rainfall, 0 to 1). In SI units with A in km²: $Q_p = 0.278\,C\,i\,A$.

Steps

1. Find $t_c$ (e.g. Kirpich: $t_c = 0.0195\,L^{0.77}S^{-0.385}$ min, L in m, S in m/m).
2. Take the rainfall intensity $i$ for duration $t_c$ and the design return period from the IDF curve.
3. Select $C$ from tables according to land use (e.g. 0.8 – 0.95 for paved areas, 0.1 – 0.3 for forest or flat cultivated land); use a weighted $C$ for mixed surfaces: $C=\sum C_iA_i/\sum A_i$.
4. Compute $Q_p$.

The rational hydrograph is a triangle or trapezoid; the method gives only the peak. It assumes the return period of the flood equals that of the rainfall.

Limitations

1. Applicable only to small catchments (generally up to about 50 km², best below 5 km²) because the rainfall is assumed uniform in space and time.
2. Gives only the peak discharge, not the flood volume or hydrograph shape.
3. $C$ is assumed constant, although it depends on soil moisture, storm intensity, slope and season (it increases with return period).
4. Ignores storage in the catchment and channels and the effect of antecedent moisture.
5. Assumes the duration of rain equals $t_c$ and that rainfall intensity is constant in that period.
6. $t_c$ itself is empirical; different formulae give widely different values.
7. Needs reliable local IDF curves, which are scarce for Nepal's hilly catchments.

Log Pearson Type III distribution

In this distribution, the logarithms of the annual flood peaks are assumed to follow the Pearson Type III (gamma-type) distribution, which has three parameters: the mean, standard deviation and coefficient of skew of the log series. It is recommended in the USA (US Water Resources Council) for flood frequency. When the skew of the logs is zero, it reduces to the log-normal distribution.

For a variate $z = \log_{10} x$, the magnitude at return period T is

$$z_T = \bar z + K_T\, S_z \qquad\Rightarrow\qquad x_T = 10^{\,z_T}$$

where $\bar z$ = mean of the log values, $S_z$ = standard deviation of the logs, and $K_T$ = frequency factor, which depends on T and on the skew coefficient $C_s$ of the logs (given in tables).

Procedure (N years of data, estimate for T > N)

1. Arrange the annual maximum flood series $x_1,\dots,x_N$ and convert each to $z_i = \log_{10}x_i$.
2. Compute the mean:

$$\bar z = \frac{1}{N}\sum z_i$$

3. Compute the standard deviation:

$$S_z = \sqrt{\frac{\sum (z_i-\bar z)^2}{N-1}}$$

4. Compute the coefficient of skew:

$$C_s = \frac{N\sum (z_i-\bar z)^3}{(N-1)(N-2)\,S_z^3}$$

5. For the required return period T and the computed $C_s$, read the frequency factor $K_T$ from the table of Pearson Type III (for $C_s = 0$, $K_T$ is the standard normal variate; Wilson–Hilferty approximation for other skews). Interpolate if necessary.
6. Compute $z_T = \bar z + K_T S_z$.
7. Flood of return period T: $x_T = 10^{z_T}$ m³/s.
8. (Optional) Plot the points with Weibull plotting positions $T=(N+1)/m$ on log-probability paper, draw the fitted line $x_T$ against T and extrapolate to the required T. Confidence limits can also be computed.

Because T > N, the result is an extrapolation of the fitted distribution; its reliability decreases when T is much larger than N (rule of thumb T ≤ 2N).

Design flood

Reliability = 0.85, so the risk $R=1-0.85=0.15$ in the life of n = 40 years.

$$R=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.85)^{1/40}\ \Rightarrow\ T=247 \text{ years}$$ $$y_T=-\ln\left[\ln\frac{T}{T-1}\right]=5.506, \qquad K=\frac{y_T-\bar y_n}{S_n}=\frac{5.506-0.5252}{1.0696}=4.657$$ $$x_T=\bar x+K\sigma=8520+4.657\times3900=26681 \text{ m}^3/\text{s}$$

Answer: recommended design flood ≈ 26681 m³/s (return period ≈ 247 years).

Gumbel's method: $x_T = \bar x + K\sigma$, with $K=\dfrac{y_T-\bar y_n}{S_n}$ and $y_T = -\ln\!\left[\ln\dfrac{T}{T-1}\right]$. Here $\bar y_n=0.536$, $S_n=1.1124$ (n = 30).

(i) 200-year and 1000-year floods

First find $\bar x$ and $\sigma$ from the two given floods.

$$\sigma = \frac{12100-9900}{5.104-3.653} = 1516.8 \text{ m}^3/\text{s}, \qquad \bar x = 9900 - 3.653\times 1516.8 = 4358.4 \text{ m}^3/\text{s}$$

Answer: $x_{200}$ ≈ 10849 m³/s and $x_{1000}$ ≈ 13046 m³/s.

(ii) Confidence limits

$$x_T' = x_T \pm f(c)\,S_e, \qquad S_e = \frac{b\,\sigma}{\sqrt{N}},\quad b=\sqrt{1+1.3K+1.1K^2}$$

(iii) Hydrologic risk for 12300 m³/s, n = 25 years

$$K=\frac{12300-4358.4}{1516.8}=5.236, \quad y_T=0.536+5.236\times 1.1124=6.360$$ $$T=\frac{1}{1-e^{-e^{-y_T}}} = 579 \text{ years}$$ $$\text{Risk}=1-\left(1-\frac1T\right)^{n}=1-\left(1-\frac{1}{579}\right)^{25} = 0.0423$$

Answer: T ≈ 579 years and the risk ≈ 4.2%.

(iv) Flood for 99% reliability over 25 years

Reliability = 0.99 means risk = 0.01.

$$1-\frac1T = (0.99)^{1/25} \Rightarrow T = 2488 \text{ years}$$ $$y_T=7.819, \quad K=6.547, \quad x_T = 4358.4 + 6.547\times1516.8 = 14289 \text{ m}^3/\text{s}$$

Answer: design peak flow ≈ 14289 m³/s (return period ≈ 2488 years).

The number of years N is not given; N = 30 is assumed because $S_n=1.1124$ is the Gumbel value for N = 30.

Return period

$$K=\frac{x_T-\bar x}{\sigma}=\frac{300-140}{50}=3.20$$ $$y_T = \bar y_n + K S_n = 0.5224 + 3.20\times 1.1124 = 4.082$$ $$T=\frac{1}{1-e^{-e^{-y_T}}} = \frac{1}{1-e^{-e^{-4.082}}} = 60 \text{ years}$$

Answer: return period of the design flood ≈ 60 years (probability of exceedance in any year = 0.0167).

90% confidence limit

$$b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(3.20)+1.1(3.20)^2}=4.053$$ $$S_e=\frac{b\,\sigma}{\sqrt N}=\frac{4.053\times 50}{\sqrt{30}}=37.00 \text{ m}^3/\text{s}$$ $$x_T' = x_T \pm f(c)S_e = 300 \pm 1.645\times37.00 = 300 \pm 60.9$$

Answer: 90% confidence limits are 239 m³/s to 361 m³/s (for N = 30).

(i) Design flood

Reliability = 0.85, so the risk $R=1-0.85=0.15$ in the life of n = 40 years.

$$R=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.85)^{1/40}\ \Rightarrow\ T=247 \text{ years}$$ $$y_T=-\ln\left[\ln\frac{T}{T-1}\right]=5.506, \qquad K=\frac{y_T-\bar y_n}{S_n}=\frac{5.506-0.5252}{1.0696}=4.657$$ $$x_T=\bar x+K\sigma=8520+4.657\times3900=26681 \text{ m}^3/\text{s}$$

Answer: recommended design flood ≈ 26681 m³/s (return period ≈ 247 years).

(ii) 80% confidence limits

$$b=\sqrt{1+1.3K+1.1K^2}=5.559, \qquad S_e=\frac{b\sigma}{\sqrt N}=\frac{5.559\times3900}{\sqrt{21}}=4731 \text{ m}^3/\text{s}$$ $$x_T'=26681\pm1.282\times4731=26681\pm6065$$

Answer: 80% confidence limits are 20615 m³/s and 32746 m³/s.

For N = 43 years the table gives $\bar y_n=0.5453$ and $S_n=1.1480$.

Step 1: Design return period

Reliability 85% means risk $R=1-0.85=0.15$ during the life n = 50 years.

$$R=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.85)^{1/50}\ \Rightarrow\ T=308 \text{ years}$$

Step 2: Reduced variate and frequency factor

$$y_T=-\ln\left[\ln\frac{T}{T-1}\right]=5.729$$ $$K=\frac{y_T-\bar y_n}{S_n}=\frac{5.729-0.5453}{1.1480}=4.515$$

Step 3: Design flood

$$x_T=\bar x+K\sigma=330+4.515\times187=1174 \text{ m}^3/\text{s}$$

Answer: recommended design flood ≈ 1174 m³/s (return period ≈ 308 years).

Frequency factor

$$K=\frac{x_T-\bar x}{\sigma}=\frac{350-140}{65}=3.231$$

Reduced variate

$$y_T=\bar y_n+K S_n=0.5299+3.231\times1.0876=4.044$$

Return period

$$T=\frac{1}{1-e^{-e^{-y_T}}}=\frac{1}{1-e^{-e^{-4.044}}}=57.5 \text{ years}$$

Answer: the return period of the 350 m³/s design flood ≈ 58 years (annual exceedance probability ≈ 0.0174).

Rational formula: $Q_p = 0.278\,C\,i\,A$ (Q in m³/s, i in mm/h, A in km²), where $i$ is the intensity for a duration equal to the time of concentration.

Step 1: Time of concentration (Kirpich)

$$\begin{aligned} t_c &= 0.0195\,L^{0.77}S^{-0.385} \\ &= 0.0195\,(6000)^{0.77}(0.013)^{-0.385} = 84.2 \text{ min} \end{aligned}$$

Step 2: Rainfall intensity for duration $t_c$

Cumulative rainfall from the start of the storm:

The depth in the first 84.2 min, by linear interpolation between 50 and 84 min (this is also the most intense 84.2-min window of the storm):

$$P = 40 + \frac{84.2-50}{84-50}(70-40) = 70.1 \text{ mm}$$ $$i = \frac{70.1}{84.2}\times 60 = 50.0 \text{ mm/h}$$

Step 3: Peak flow

$$\begin{aligned} Q_p &= 0.278\times 0.6\times 50.0 \times 100 \\ &= 833 \text{ m}^3/\text{s} \end{aligned}$$

Answer: peak flow ≈ 833 m³/s (note: a 100 km² catchment is large for the rational method, so this is a rough estimate).

Rational method

$$Q_p = \frac{C\,i\,A}{360}\quad(\text{A in ha, i in mm/h, Q in m}^3/\text{s})\qquad\text{or}\qquad Q_p=0.278\,C\,i\,A\ (A\text{ in km}^2)$$

i = intensity for duration $t_c$ and return period T; C = runoff coefficient.

• Limitations: only small catchments (up to about 50 km²); uniform rainfall and constant C assumed; gives only the peak; ignores storage and antecedent moisture; needs IDF curves.
• Appropriate use: design of urban drainage, culverts, small bridges, storm sewers and spillways of small catchments.

Empirical methods

Based on regional observations, the flood is a function of catchment area (and sometimes return period):

1. Dickens (north India, hilly regions): $Q = C_D\,A^{3/4}$ (A in km², $C_D$ = 11 – 14 for plains, 14 – 28 for hilly areas, up to 25 – 30 for the Himalayan foothills).
2. Ryves (south India): $Q = C_R\,A^{2/3}$ ($C_R$ = 6.8 – 10.1 depending on distance from the coast).
3. Inglis (Maharashtra, ghat areas): $Q = \dfrac{124A}{\sqrt{A+10.4}}$
4. Fuller (USA): $Q_T = C_f\,A^{0.8}(1+0.8\log T)$
5. Nepal (WECS/DHM method): $Q_2 = 1.8767\,(A_{<3000}+1)^{0.8783}$ and $Q_{100}=14.63\,(A_{<3000}+1)^{0.7342}$, where $A_{<3000}$ is the catchment area below 3000 m elevation (km²), used for rivers without records.

• Limitations: valid only for the region and range of area where the constants were derived; they ignore rainfall intensity, slope, shape and land use of the particular catchment; the constants are subjective; the return period is not known (except Fuller and WECS).
• Appropriate use: quick, approximate estimates for preliminary design or for checking other methods, mainly in ungauged basins; not for final design of major structures.

Final design of large projects should use flood-frequency analysis (Gumbel, log-Pearson) or the unit hydrograph method with the observed data.

In Gumbel's method $x_T=\bar x+K\sigma = a + b\,y_T$, which is a straight line in $y_T$ (the reduced variate). Using two known points gives the line, so $\bar y_n$ and $S_n$ are not needed.

$$y_T = -\ln\left[\ln\frac{T}{T-1}\right]$$

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

$$b=\frac{1020-850}{4.6001-3.9019}=243.48 \text{ m}^3/\text{s}$$ $$x_{500}=1020+243.48\,(6.2136-4.6001)=1413 \text{ m}^3/\text{s}$$

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

(i) Statistics (n = 10)

$$\bar x=\frac{\sum x}{n}=\frac{401.7}{10}=40.17 \text{ m}^3/\text{s}$$ $$s=\sqrt{\frac{\sum (x-\bar x)^2}{n-1}}=\sqrt{\frac{1016.92}{9}}=10.63 \text{ m}^3/\text{s}$$ $$C_s=\frac{n\sum (x-\bar x)^3}{(n-1)(n-2)\,s^3}=\frac{10\times6480.3}{9\times 8\times1201.1}=0.75$$

Answer: mean = 40.17 m³/s, standard deviation = 10.63 m³/s, skewness coefficient = 0.75 (positively skewed).

(ii) Probability of a 250 m³/s flood next year (Gumbel)

$$K=\frac{250-40.17}{10.63}=19.74$$ $$y_T=\bar y_n+K\sigma_n = 0.4952 + 19.74\times0.9497=19.24$$ $$P(X\ge 250)=1-e^{-e^{-y_T}}=1-e^{-e^{-19.24}} = 4.40e-09$$

The return period is $T=1/P$ ≈ 2.27e+08 years.

Answer: probability ≈ 4.40e-09 (about 4.4e-07%), i.e. practically zero. The value is far outside the range of the 10-year record (maximum 59.6 m³/s), so it is a very large extrapolation.

Exceedance probability

The exceedance probability $P(X\ge x_T)$ is the probability that an event of magnitude $x_T$ is equalled or exceeded in any one year. For example, a 100-year flood has an exceedance probability of $P=0.01$ (1%) each year.

Recurrence interval (return period)

The recurrence interval $T$ is the average number of years between events equal to or greater than a given magnitude. It is the reciprocal of the exceedance probability:

$$T=\frac{1}{P}$$

It is an average over a long record, not a fixed period: a 50-year flood may occur in two consecutive years. The probability that it occurs at least once in n years is $1-(1-1/T)^n$.

Frequency factor

The frequency factor $K_T$ is the number of standard deviations by which the event of return period T differs from the mean, so that

$$x_T=\bar x + K_T\,\sigma$$

It depends on the distribution used and on T (and, for Pearson III, on the skewness). For Gumbel's distribution $K_T=\dfrac{y_T-\bar y_n}{S_n}$ with $y_T=-\ln[\ln(T/(T-1))]$, and for the normal distribution it is the standard normal variate $z$. It is the common basis of Chow's general equation for frequency analysis.

Method. Rank the annual maxima, find the return period of each by a plotting-position formula, plot rainfall against T on semi-log paper (T on the log axis), fit a straight line and extend it to T = 50 years.

Weibull formula (N = 16 years of data, m = rank in descending order):

$$T=\frac{N+1}{m}=\frac{17}{m}, \qquad P=\frac1T$$

Plotting. Plot the 16 points (T on the logarithmic horizontal axis, 24-h rainfall on the arithmetic vertical axis). Gumbel-distributed data fall close to a straight line on this paper, with equation:

$$P_{24}=a+b\log_{10}T$$

A line fitted through the points gives $b=7.49$ cm per log-cycle and $a=7.33$ cm.

 P24 (cm)
  20 |                        o  <- T=50 (extrapolated)
  16 |                  o
  12 |          o  o
   8 |   o o o
   4 +--+------+-------+--------- T (yr, log)
     1.1      2       10      50

Reading at T = 50 years:

$$P_{24,50}= 7.33 + 7.49\times\log_{10}(50)=20.05 \text{ cm}$$

Answer: 24-h maximum rainfall for the 50-year return period ≈ 20 cm. As a check, Gumbel's analytical method with $\bar y_n=0.5157$, $S_n=1.0316$ (N = 16), mean = 10.31 cm and $\sigma$ = 2.59 cm gives 18.8 cm, which is of the same order.

(i) Peak flood for 0.4% probability of exceedance

$$T=\frac1P=\frac{1}{0.004}=250 \text{ years}$$ $$y_T=-\ln\left[\ln\frac{T}{T-1}\right]=-\ln\left[\ln\frac{250}{249}\right]=5.519$$ $$K=\frac{y_T-\bar y_n}{S_n}=\frac{5.519-0.556}{1.189}=4.174$$ $$x_T=\bar x+K\sigma=5561+4.174\times1718=12733 \text{ m}^3/\text{s}$$

Answer: $x_{250}$ ≈ 12733 m³/s.

(ii) 90% confidence limits

$$b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(4.174)+1.1(4.174)^2}=5.059$$ $$S_e=\frac{b\sigma}{\sqrt N}=\frac{5.059\times1718}{\sqrt{75}}=1004 \text{ m}^3/\text{s}$$ $$x_T'=x_T\pm f(c)S_e=12733\pm1.6\times1004=12733\pm1606$$

Answer: the 90% confidence limits are 11127 m³/s and 14339 m³/s.

(i) Design flood

Reliability = 0.87, so risk $R=0.13$ over n = 35 years.

$$\frac1T=1-(0.87)^{1/35}\ \Rightarrow\ T=252 \text{ years}$$ $$y_T=5.527,\qquad K=\frac{5.527-0.5402}{1.1285}=4.419$$ $$x_T=5460+4.419\times2950=18495 \text{ m}^3/\text{s}$$

Answer: predicted flood for the project ≈ 18495 m³/s (T ≈ 252 years).

(ii) Safety factor 1.5

$$Q_{adopted}=1.5\times18495=27743 \text{ m}^3/\text{s}$$

Safety margin = adopted − predicted flood = 27743 − 18495 = 9248 m³/s (that is 50% of the predicted flood).

Answer: adopt 27743 m³/s; safety margin = 9248 m³/s.

(iii) 95% confidence limits

$$b=\sqrt{1+1.3K+1.1K^2}=5.312,\qquad S_e=\frac{b\sigma}{\sqrt N}=\frac{5.312\times2950}{\sqrt{20}}=3504 \text{ m}^3/\text{s}$$ $$x_T'=18495\pm1.96\times3504=18495\pm6868$$

Answer: 95% confidence limits are 11627 m³/s and 25364 m³/s.

Gumbel's cumulative distribution (probability of non-exceedance):

$$F(x)=P(X\le x)=\exp\left[-e^{-\alpha(x-\beta)}\right]$$

Using the hint:

$$\alpha=\frac{1.28255}{3000}=4.2752e-04\ \text{per m}^3/\text{s},\qquad \beta=10000-0.48\times3000=8560 \text{ m}^3/\text{s}$$

Probability that the annual peak exceeds 15000 m³/s

$$\alpha(x-\beta)=4.2752e-04\times(15000-8560)=2.753$$ $$F(15000)=\exp(-e^{-2.753})=0.9383$$ $$P(X>15000)=1-F=0.0617$$

Answer: probability ≈ 0.0617 (about 6.17%), a return period of ≈ 16.2 years.

Discharge with exceedance probability 0.1

Exceedance 0.1 means $F=0.9$:

$$-e^{-\alpha(x-\beta)}=\ln 0.9\ \Rightarrow\ x=\beta-\frac{1}{\alpha}\ln(-\ln0.9)$$ $$x=8560-\frac{\ln(-\ln0.9)}{4.2752e-04}=8560-\frac{(-2.2504)}{4.2752e-04}=13824 \text{ m}^3/\text{s}$$

Answer: $Q_{0.1}$ ≈ 13824 m³/s (the 10-year flood).

Continuous and discrete random variables

Formulae for return period

With N years of record and m the rank of the event (largest = 1), the plotting position formulae are:

1. Weibull: $T=\dfrac{N+1}{m}$ (most commonly used)
2. California: $T=\dfrac{N}{m}$
3. Hazen: $T=\dfrac{2N}{2m-1}$ (i.e. $N/(m-0.5)$)

Other forms are Gringorten, $T=\dfrac{N+0.12}{m-0.44}$, and Blom. In terms of probability, $T=1/P$, where $P$ is the exceedance probability.

Method. Arrange the data in descending order, assign the Weibull return period $T=(N+1)/m$ with N = 20, plot flood peak against T on semi-log paper (T on the log axis) and read the flood at T = 30 years from the extended line.

Graph. Plot Q (arithmetic scale) against T (log scale). The points lie approximately on a straight line for the Gumbel type of data. A best-fit line through the points has the equation

$$Q=88.2+167.2\log_{10}T \quad(\text{m}^3/\text{s})$$

 Q (m3/s)
 300 |                    o   (x at T=30 line)
 250 |              o  o
 200 |         o
 150 |    o o o
 100 | o o
     +--+-----+------+-------- T (yr, log)
      1.05     2     10  30

Reading at T = 30 years:

$$Q_{30}=88.2+167.2\times\log_{10}(30)=335 \text{ m}^3/\text{s}$$

Answer: the annual peak flood with a 30-year return period ≈ 335 m³/s. (Check by Gumbel's equation with mean 155.6 and standard deviation 60.6 m³/s, using $\bar y_n=0.5236$ and $S_n=1.0628$: 319 m³/s; the graph estimate depends on the line drawn through the points.)

Gumbel's distribution (extreme value type I)

Gumbel considered a series of annual maxima, each the largest of many daily values, and showed that for a large sample the probability that a value $x$ is equalled or exceeded in a year is

$$P(X\ge x)=1-e^{-e^{-y}}$$

where $y$ is the reduced variate, a linear function of $x$:

$$y=a(x-u), \qquad a=\frac{1.2825}{\sigma},\qquad u=\bar x-0.45\sigma$$

($\bar x$ and $\sigma$ = mean and standard deviation of the annual maxima.) The non-exceedance probability is $F(x)=e^{-e^{-y}}$. It has a long right tail, so it suits floods and maximum rainfall.

Return period and reduced variate

Since $P=1/T$:

$$\frac1T=1-e^{-e^{-y_T}}\ \Rightarrow\ e^{-e^{-y_T}}=\frac{T-1}{T}$$

Taking natural logs twice:

$$y_T=-\ln\left[\ln\frac{T}{T-1}\right]$$

For large T, $y_T\approx\ln T$.

Derivation of the frequency factor

Chow's general equation: $x_T=\bar x+K\sigma$.

From $y=a(x-u)$, the value at return period T is $x_T=u+\dfrac{y_T}{a}$. For the infinite sample the mean of the reduced variate is $\bar y=0.5772$ (Euler's constant) and its standard deviation is $\sigma_y=1.2825$. Since $\bar x=u+\bar y/a$ and $\sigma=\sigma_y/a$:

$$x_T-\bar x=\frac{y_T-\bar y}{a}=\sigma\,\frac{y_T-\bar y}{\sigma_y}$$

Comparing with $x_T=\bar x+K\sigma$:

$$K=\frac{y_T-0.5772}{1.2825}$$

For a finite sample of N years the theoretical mean and standard deviation of $y$ are replaced by the reduced mean $\bar y_n$ and reduced standard deviation $S_n$ (tabulated against N):

$$\boxed{K=\frac{y_T-\bar y_n}{S_n}}\qquad\text{so}\qquad x_T=\bar x+K\sigma$$

with $\bar y_n\to0.5772$ and $S_n\to1.2825$ as $N\to\infty$. For example, T = 100 years gives $y_T=4.600$ and $K=(4.600-0.5772)/1.2825=3.14$ for a large sample.

Gumbel's equation $x_T=\bar x+K\sigma$ is linear in the reduced variate $y_T$, so two known floods fix the line and the other floods follow by proportion (no need for the sample size).

$$y_T=-\ln\left[\ln\frac{T}{T-1}\right]$$

Slope of the line:

$$b=\frac{x_{250}-x_{25}}{y_{250}-y_{25}}=\frac{550-90}{5.5195-3.1985}=198.20 \text{ m}^3/\text{s}$$ $$x_T=90+198.20\,(y_T-3.1985)$$

Answer: $x_{50}$ ≈ 229 m³/s, $x_{500}$ ≈ 688 m³/s, $x_{1000}$ ≈ 825 m³/s.

Method. Use the rational formula, with the time of concentration from Kirpich's empirical formula and the intensity from the maximum daily rainfall by the empirical reduction formula (IMD/Indian practice).

Data

A = 100 km², L = 60 km = 60 000 m, fall H = 20 m, C = 0.6, maximum daily (24 h) rainfall P = 200 mm.

Step 1: Slope and time of concentration

$$S=\frac{H}{L}=\frac{20}{60000}=3.333e-04$$ $$t_c=0.0195\,L^{0.77}S^{-0.385}=0.0195\,(60000)^{0.77}(3.333e-04)^{-0.385}=2032 \text{ min}=33.9 \text{ h}$$

Step 2: Rainfall intensity for duration $t_c$

$$i=\frac{P}{24}\left(\frac{24}{t_c}\right)^{2/3}=\frac{200}{24}\left(\frac{24}{33.9}\right)^{2/3}=6.62 \text{ mm/h}$$

Step 3: Peak flood

$$Q=0.278\,C\,i\,A=0.278\times0.6\times6.62\times100=110 \text{ m}^3/\text{s}$$

Answer: flood discharge ≈ 110 m³/s. (A 100 km² catchment is larger than the usual limit of the rational method, so the result is approximate.)

Gumbel's equation is linear in the reduced variate: $x_T=\bar x+K\sigma=a+b\,y_T$, with $y_T=-\ln[\ln(T/(T-1))]$. The two given floods fix the line.

$$b=\frac{620000-215000}{4.3757-(-0.2254)}=88023 \qquad\Rightarrow\qquad x_T=215000+88023\,[y_T-(-0.2254)]$$

(i) Probability of a flood ≥ 440 000

$$y=-0.2254+\frac{440000-215000}{88023}=2.3308$$ $$P(X\ge440000)=1-e^{-e^{-y}}=1-e^{-e^{-2.3308}}=0.0926$$

Answer: probability ≈ 0.0926 (about 9.3%; return period ≈ 10.8 years).

(ii) Flood with a 40-year recurrence interval

$$y_{40}=-\ln\left[\ln\frac{40}{39}\right]=3.6762$$ $$x_{40}=215000+88023\,(3.6762-(-0.2254))=558429 \text{ units}$$

Answer: $x_{40}$ ≈ 558429 units.

(iii) Probability of a flood ≥ 575 000 in the next 25 years

$$y=-0.2254+\frac{575000-215000}{88023}=3.8645$$ $$p=1-e^{-e^{-3.8645}}=0.02075 \quad(T=48.2 \text{ years})$$

The probability of at least one such flood in n = 25 years is

$$1-(1-p)^{25}=1-(1-0.02075)^{25}=0.4080$$

Answer: probability ≈ 0.4080 (about 40.8%).

Graphical (plotting-position) method for flood of return period T

1. Collect the annual maximum flood series for N years.
2. Arrange the floods in descending order and assign the rank m (m = 1 for the largest).
3. Compute the return period of each flood by a plotting-position formula, e.g. Weibull $T=\dfrac{N+1}{m}$ (or the exceedance probability $P=m/(N+1)$).
4. Choose suitable probability paper (semi-log, Gumbel, or log-normal paper) so that the data fall nearly on a straight line.
5. Plot the flood magnitude (vertical axis) against T or P (horizontal axis) for every point.
6. Draw the best-fit straight line (or smooth curve) through the points.
7. Extend the line (extrapolate) to the required return period T and read the flood magnitude from the vertical axis.
8. Report the flood, noting that reliability falls when T is much larger than N.

Method. Take $z=\log_{10}Q$, find its mean, standard deviation and skewness, then $z_T=\bar z+K_TS_z$ and $Q_T=10^{z_T}$.

n = 17.

$$\bar z=\frac{\sum z}{n}=\frac{58.1088}{17}=3.4182$$ $$S_z=\sqrt{\frac{\sum(z-\bar z)^2}{n-1}}=\sqrt{\frac{0.6735}{16}}=0.2052$$ $$C_s=\frac{n\sum(z-\bar z)^3}{(n-1)(n-2)S_z^3}=\frac{17\times(0.00774)}{16\times15\times0.00864}=0.063$$

Frequency factor for T = 50 years. From the table, $C_s=0$ gives 2.054 and $C_s=0.1$ gives 2.107. Interpolating for $C_s=0.063$:

$$K_T=2.054+\frac{0.063-0}{0.1-0}(2.107-2.054)=2.088$$

Flood magnitude.

$$z_{50}=\bar z+K_TS_z=3.4182+2.088\times0.2052=3.8465$$ $$Q_{50}=10^{3.8465}=7022 \text{ m}^3/\text{s}$$

Answer: the 50-year flood ≈ 7022 m³/s.

Step 1: Return periods from the risk

Risk over n = 50 years: $R=1-(1-1/T)^{n}$, so $T=\dfrac{1}{1-(1-R)^{1/n}}$.

• R = 63.58303%: $T=\dfrac{1}{1-(0.3641697)^{1/50}}=50.00$ years, so 4200 m³/s is the 50-year flood.
• R = 39.49939%: $T=\dfrac{1}{1-(0.6050061)^{1/50}}=100.00$ years, so this is the 100-year flood.

Step 2: Gumbel parameters

The average flood is the mean $\bar x=3500$ m³/s. For a long record, $\bar y_n=0.5772$ and $S_n=1.2825$.

$$y_{50}=-\ln[\ln(50/49)]=3.902,\qquad K_{50}=\frac{3.902-0.5772}{1.2825}=2.592$$ $$\sigma=\frac{x_{50}-\bar x}{K_{50}}=\frac{4200-3500}{2.592}=270.0 \text{ m}^3/\text{s}$$

Step 3: Discharges

Answer: discharge at 39.49939% risk (T = 100 years) ≈ 4347 m³/s; discharge for T = 500 years ≈ 4687 m³/s.

Step 4: Gumbel probability paper from ordinary graph paper

1. Draw the horizontal axis with uniform spacing for $y_T$ (e.g. 1 cm = 1 unit of y, range −1 to 7).
2. Mark the return periods at their $y_T$ positions (T = 2, 5, 10, 20, 50, 100, 200, 500, 1000 at $y_T$ = 0.367, 1.500, 2.250, 2.970, 3.902, 4.600, 5.296, 6.214, 6.907) and label them T. This makes the T scale non-uniform.
3. Draw the vertical axis for discharge (arithmetic scale, 3000 to 8000 m³/s).
4. Plot the points; Gumbel data fall on a straight line.

 Q (m3/s)
  4700 |                              o  T=500
  4347 |                 o  T=100
  4200 |         o  T=50
       +----+----+----+----+----+----+--- y_T
            3    4    5    6    7

The three plotted points (T = 50, 100, 500 years) lie on one straight line, which can be extended to any other T.