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16-Civ-B4 Engineering Hydrology · Undated paper

Question 6 of 7: Statistical methods of frequency and probability analysis applied to precipitation and floods

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2019, 16-Civ-B4 Engineering Hydrology, three hours’ duration, closed book with one candidate-prepared two-sided aid sheet (8½″ × 11″) and one approved Casio or Sharp calculator whose model designation must be written in the work book. Seven problems are printed, each divided into sub-parts (i), (ii) and (iii). Page-1 Note 4 states that any five (5) questions constitute a complete paper and that only the first five answers appearing in the work book will be marked; Note 5 weights every problem at twenty (20) points, for a maximum of one hundred (100) points. All seven problems are solved here, because this document is a study resource rather than a timed sitting. The sub-part mark values quoted below are the printed marginal marks, which on this sitting agree exactly with the page-6 marking scheme (7/7/6, 7/7/6, 7/7/6, 8/6/6, 6/7/7, 6/6/8, 10/5/5).

Reference texts. V. T. Chow, D. R. Maidment and L. W. Mays, Applied Hydrology (hydrologic cycle, unit hydrographs, reservoir and channel routing, frequency analysis); L. W. Mays, Water Resources Engineering, 3rd ed. (stormwater management, detention design, reservoir operation); W. Viessman and G. L. Lewis, Introduction to Hydrology, 5th ed. (precipitation measurement, areal averaging, streamflow gauging); P. B. Bedient, W. C. Huber and B. E. Vieux, Hydrology and Floodplain Analysis, 5th ed. (hydrograph analysis, urban hydrology, hydrologic modelling); C. W. Fetter, Applied Hydrogeology, 4th ed. (Darcy’s law and aquifer flow); V. T. Chow, Open-Channel Hydraulics (1959) (flood-wave propagation, Saint-Venant equations). Canadian practice references: Environment and Climate Change Canada Engineering Climate Datasets (short-duration rainfall IDF curves) and the Water Survey of Canada HYDAT archive; WMO Manual on Stream Gauging and ISO 748 (velocity–area gauging and stage–discharge ratings); Transportation Association of Canada Guide to Bridge Hydraulics and the provincial highway drainage manuals (culvert and roadside-drainage design); Canadian Dam Association Dam Safety Guidelines (inflow design flood and flood routing).

Check — every number below is the solver’s own illustrative value. All seven problems on this sitting are discussion questions; the paper supplies no numerical data whatever. Where a short calculation appears below it is there to demonstrate the method being asked about, and its inputs are declared explicitly in a Given line as assumed, representative Canadian values. They are not exam data. A candidate who assumed different but reasonable values and carried them through consistently would earn the same marks, and the examiner’s marks here are awarded for the explanation, the governing equation and the stated assumptions.

Problem 6: Statistical methods of frequency and probability analysis applied to precipitation and floods (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (i) — Generating and using an IDF curve (6 marks)

How the curve is generated. An intensity–duration–frequency curve is built from a long record of short-duration rainfall at a recording gauge, in Canada from the Environment and Climate Change Canada Engineering Climate Datasets, which publish IDF statistics for some six hundred stations. Four steps produce it. First, a set of standard durations is fixed — 5, 10, 15 and 30 minutes and 1, 2, 6, 12 and 24 hours. Second, the record is scanned with a moving window of each duration in turn, and for each year the greatest depth falling within that window is extracted, giving one annual maximum series per duration. Third, each series is fitted with an extreme-value distribution, almost always the Gumbel (EV1) distribution in Canadian practice, and the quantile for each required return period is read off; dividing the depth by the duration converts it to an intensity. Fourth, the resulting points are smoothed with an empirical relation of the form

$$i = \frac{a}{(t_d + b)^{c}}$$

where i is intensity in mm/h, td is duration in minutes, and a, b and c are fitted constants, one set per return period. Plotting one curve per return period on logarithmic axes gives the familiar family.

How it is used to predict peak runoff. The design storm duration is set equal to the time of concentration of the catchment, on the argument that this is the shortest storm that allows the entire catchment to contribute simultaneously and therefore produces the greatest peak. Entering the curve at td = tc and reading up to the curve for the design return period gives the design intensity, which is then converted to a peak discharge by the rational method,

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

with Qp in m3/s, i in mm/h and A in hectares.

Given. A municipal drainage design uses the following assumed, representative values.

Illustrative IDF parameters and catchment (assumed values, not exam data)
IDF relation, 10-year return periodi = 850/(td + 6.0)0.78, i in mm/h, td in minutes
Catchment area, A12 ha
Time of concentration, tc30 min
Runoff coefficient, C (residential)0.55

Find. The 10-year design intensity at the catchment’s time of concentration, and the resulting peak discharge.

Approach. Evaluate the fitted IDF relation at a duration equal to the time of concentration, then substitute into the rational formula.

  1. Read the design intensity from the curve. At td = tc = 30 min, $$i = \frac{850}{(30 + 6.0)^{0.78}} = \frac{850}{16.365} = 51.9\ \text{mm/h}$$ For comparison the same relation gives 79.1 mm/h at 15 minutes and 32.4 mm/h at 60 minutes, which is the steep duration dependence that makes the choice of tc so consequential.
  2. Convert to a peak discharge. Substituting into the rational method, $$Q_p = \frac{(0.55)(51.94)(12)}{360} = \boxed{0.952\ \text{m}^3/\text{s}}$$ a value consistent with a 750 mm storm sewer at typical grades, which is the practical output of the whole exercise.
Problem 6(i) — typical intensity–duration–frequency curves01530456090120050100150Storm duration, tᵈ (min)Rainfall intensity, i (mm/h)T = 2 yrT = 10 yrT = 100 yr10-yr, 30-min: 51.9 mm/hi = a / (tᵈ + b)ᶜone curve per return period T
Intensity–duration–frequency curves fitted to the illustrative gauge. Each curve is one return period; the design point is found by entering at a duration equal to the time of concentration and reading up to the required return period. The marked point is the 10-year, 30-minute intensity of 51.9 mm/h used in the rational-method calculation.

Part (ii) — Three important elements of a flood-frequency method (6 marks)

First element — the data series and its qualification. Frequency analysis operates on an annual maximum series: the single largest instantaneous peak discharge in each water year, one value per year. (The alternative is a partial-duration or peaks-over-threshold series, preferable for short records and for return periods below about ten years.) The series must satisfy four conditions before any distribution is fitted, and stating them is what earns the mark: the values must be independent of one another, identically distributed so that the record is homogeneous in cause, stationary so that no trend from regulation, land-use change or climate contaminates it, and random. Records affected by a dam constructed mid-series, or mixing snowmelt and rainfall-generated floods, violate these and must be split or adjusted.

Second element — the probability distribution and its fitted parameters. A theoretical extreme-value distribution is fitted to the series so that the record can be extrapolated beyond its own length. The Gumbel (EV1) distribution and the log-Pearson type III distribution are the standards in North American practice, with the generalised extreme value distribution increasingly preferred; parameters are estimated by the method of moments, by L-moments or by maximum likelihood. Alongside the fitted line, the observed data are plotted at empirical probabilities given by a plotting-position formula, most commonly Weibull,

$$P_X = \frac{m}{n + 1}, \qquad T_X = \frac{n + 1}{m}$$

where m is the rank of the event in descending order and n is the record length. Comparing the plotted points with the fitted line is the goodness-of-fit test that justifies the choice of distribution.

Third element — the frequency-factor equation and its confidence limits. Any of these distributions can be written in the general form introduced by Chow,

$$Q_T = \bar{Q} + K_T\,s$$

in which Q̄ and s are the mean and standard deviation of the series (or of its logarithms, for log-Pearson III) and KT is a frequency factor that depends on the distribution, the return period and, for log-Pearson III, the skew coefficient. This equation is what actually delivers the design flood, and it should always be accompanied by confidence limits, because the sampling error in a quantile estimated from thirty years of record and extrapolated to a hundred-year return period is large — commonly ± 25 per cent or more.

A caution the question invites. Flood-frequency analysis predicts the magnitude of the annual peak discharge; it does not by itself predict the time to peak. Timing comes from the basin’s response characteristics — the time of concentration, the lag and the unit hydrograph — applied to a design storm of the same return period, or from a seasonal analysis of the dates on which the annual maxima have historically occurred, which on Canadian rivers usually identifies the spring freshet. An answer that claims to obtain the peak time from a Gumbel fit is wrong, and saying so explicitly is worth a mark.

Part (iii) — Exceedance probability, return period and the 25-year flood (8 marks)

The general equation relating the probability of exceedance in any one year to the return period is

$$P_X = \frac{1}{T_X} \qquad\text{equivalently}\qquad T_X = \frac{1}{P_X}$$

The return period is therefore the average interval between exceedances over a very long record, not a guaranteed spacing: a 25-year flood has a 4 per cent chance of being equalled or exceeded in every year, independently of what happened last year, and two of them can occur in consecutive years without contradicting the definition. The probability that the event is exceeded at least once during a design life of N years — the encounter probability or hydrologic risk — follows from the binomial distribution as

$$R = 1 - \left(1 - \frac{1}{T_X}\right)^{N}$$

Given. A gauged river has the following assumed, representative record statistics.

Illustrative flood record statistics (assumed values, not exam data)
Length of the annual maximum series, n40 years
Mean annual peak discharge, Q̄420 m3/s
Standard deviation of annual peaks, s145 m3/s
Fitted distributionGumbel (EV1)
Design life of the structure, N50 years

Find. The annual exceedance probability of the 25-year flood, its magnitude, and the risk of encountering it during a 50-year design life.

Approach. Invert the return period to an exceedance probability, evaluate the Gumbel frequency factor for that return period, apply the frequency-factor equation, and finally apply the binomial risk formula.

  1. Convert return period to annual exceedance probability. $$P_X = \frac{1}{T_X} = \frac{1}{25} = 0.04 = 4\ \text{per cent per year}$$
  2. Evaluate the Gumbel frequency factor. For the EV1 distribution, $$K_T = -\frac{\sqrt{6}}{\pi}\left[\,0.5772 + \ln\!\left(\ln\frac{T}{T-1}\right)\right] = -0.7797\left[\,0.5772 + \ln\!\left(\ln\frac{25}{24}\right)\right] = 2.044$$
  3. Apply the frequency-factor equation. $$Q_{25} = \bar{Q} + K_{25}\,s = 420 + (2.044)(145) = \boxed{716\ \text{m}^3/\text{s}}$$ The corresponding flood level is then obtained by entering the site’s rating curve, or a backwater computation, with this discharge — frequency analysis yields a discharge, and the conversion to a stage is a hydraulic step, not a statistical one.
  4. Assess the risk over the design life. $$R = 1 - (1 - 0.04)^{50} = 1 - 0.130 = 0.870$$ so there is an 87 per cent chance that the 25-year flood will be equalled or exceeded at least once in fifty years. This is the number that justifies designing long-lived infrastructure to a return period well above its design life, and it is the single most useful thing to be able to quote to a non-technical client.

What data are required, and for how long. The analysis needs an annual maximum series of instantaneous peak discharge at the site, or at a gauge on the same river close enough for a drainage-area transfer to be defensible. In Canada this comes from the Water Survey of Canada HYDAT archive, which publishes annual instantaneous peaks alongside daily means; the instantaneous peak, not the daily mean, is required, because on a small basin the daily mean can be half the true peak. Ancillary data are the record of any regulation or diversion, so the series can be naturalised or truncated, and the dates of the annual maxima so that mixed snowmelt and rainfall populations can be identified.

On duration, the practical rules are these. A minimum of 10 years is required for any analysis at all, 25 to 30 years is the accepted minimum for a defensible design estimate, and the estimate should not be extrapolated much beyond about twice the record length. For a 25-year flood a record of 25 to 30 years is therefore adequate and one of 40 years, as assumed above, is comfortable; the same record would not support a credible 200-year estimate without regional pooling. Where the record is short, the standard remedies are a regional frequency analysis that pools data from hydrologically similar gauged basins, or a peaks-over-threshold analysis that extracts several events per year rather than one.

Problem 6 — results of the illustrative calculations
QuantitySymbolValue
10-year, 30-minute design intensityi51.9 mm/h
Rational-method peak dischargeQp0.952 m3/s
Annual exceedance probability of the 25-year floodPX0.04 (4 % per year)
Gumbel frequency factor for T = 25 yrK252.044
25-year peak dischargeQ25716 m3/s
Risk of exceedance in a 50-year design lifeR0.870 (87 %)
Weibull return period of the largest event in 40 yearsTX41 years