Question 6 of 7: Frequency and probability analysis of precipitation and floods

Question 6: Frequency and probability analysis of 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.

(i) Using a 50-year IDF curve to predict peak runoff, and two qualifications for the culvert (7 marks)

Given. A watershed is to be assigned a 50-year design peak flow from the local IDF family; the illustrative catchment used below is the solver's own, with area 25 ha, a time of concentration of 45 minutes and mixed residential land use. Find. The procedure that converts a 50-year IDF curve into a peak discharge, and two qualifications on using that discharge to size a highway culvert.

Approach. Establish the time of concentration, enter the 50-year IDF curve at that duration to obtain the design intensity, and combine with area and a runoff coefficient in the Rational Method — then state where that answer stops being adequate.

1. Compute the time of concentration for the watershed. This is the travel time from the hydraulically most remote point to the outlet, obtained from an overland-flow formula (Kirpich, Kerby, or the SCS velocity method) applied segment by segment along the longest flow path and summed. It is the single most influential quantity in the calculation, because it selects the point on the IDF curve. For the illustration, $t_c=45$ min $=0.75$ h.
2. Enter the 50-year IDF curve at a storm duration equal to the time of concentration. An IDF family is a set of curves, one per return period, of average intensity against duration; the 50-year curve is the one whose intensities have a 2 per cent annual exceedance probability. Reading it at $t_d=t_c$ gives the design intensity. Using the paper's own relation as the illustrative 50-year curve, $$i=7.0-0.2(0.75)=6.85\ \text{mm/h}$$ In Canadian practice the curve itself comes from the Environment and Climate Change Canada IDF file for the nearest long-record station, which fits a Gumbel distribution to the annual maximum series for each duration.
3. Select the runoff coefficient and apply the Rational Method. For mixed residential development take $C=0.45$; then $$Q=\frac{C\,i\,A}{360}=\frac{0.45\times 6.85\times 25}{360}=\boxed{0.214\ \text{m}^3/\text{s}}$$ For catchments beyond roughly 100 ha the Rational Method is replaced by a unit-hydrograph or continuous-simulation model, in which case the same IDF reading is used to build a design hyetograph and the hydrograph is generated by convolution as in Problem 3(ii). Where the design curve is increased for climate change — many Canadian municipalities now apply an intensity uplift of roughly 10 to 20 per cent, or use the ECCC IDF_CC tool — the uplift is applied at this step.

Qualification 1 — the frequency of the flow is not the frequency of the rain, and the culvert must be checked against a larger event. The Rational Method assumes the 50-year rainfall produces the 50-year flow, which requires that antecedent moisture, snowpack and storm movement all behave typically; it also ignores any storage upstream. More importantly, a highway culvert is a structure whose failure closes the road and can wash out the embankment, so Canadian practice sizes the barrel for the design flood but checks the headwater at a larger event — commonly 100-year for a major highway, with an explicit allowance for climate change — and verifies that the overtopping flow does not destroy the fill. The engineer should therefore report the 50-year flow together with the check-flood headwater, not the 50-year flow alone.

Qualification 2 — the peak discharge is only the first input to a culvert design, and hydraulic and environmental controls usually govern the final size. The barrel size follows from the allowable headwater depth under inlet or outlet control, whichever is more restrictive, and that is set by the embankment height, the upstream property that would be flooded and the highway agency's freeboard rule. On top of that, debris and ice blockage must be allowed for (a partially blocked culvert has a much steeper headwater curve), fish passage requirements under the federal Fisheries Act commonly dictate an embedded or open-bottom structure with a minimum width regardless of hydraulic need, and outlet velocity must be checked against scour with energy dissipation provided where required. A culvert sized from the peak flow alone will frequently be too small once these are applied. It is also worth recording explicitly that the Rational Method gives a peak rate and no volume, so it cannot be used if any storage or detention is contemplated.

(ii) How a 50-year flood can happen twice in five years (7 marks)

The agent's difficulty is entirely in the name. A "50-year flood" is not a flood that happens every fifty years, and nothing in its definition prevents two of them in five years, or two in the same spring. The term is shorthand for a flood magnitude, defined by a probability, not a schedule.

The definition is this. Rank the largest flood observed in each year over a long record. The 50-year flood is the discharge that is equalled or exceeded, on average, in one year out of fifty — that is, the discharge whose annual exceedance probability is

$$p=\frac{1}{T}=\frac{1}{50}=0.02=2\ \text{per cent per year}$$

Modern practice prefers the term 2 per cent AEP flood precisely because it does not invite the agent's misreading. The right mental model is a two-per-cent chance every single year, reset each January: the river has no memory of last year's flood, and experiencing one does not use up a quota or make the next one less likely, any more than rolling a six makes the next six less likely.

The arithmetic is then straightforward. If annual maxima are independent, the number of exceedances in $n$ years follows a binomial distribution, and the probability of exactly $k$ exceedances is

$$P(k \text{ in } n)=\binom{n}{k}p^{k}(1-p)^{\,n-k}$$

For $p=0.02$ and $n=5$ years, the chance of at least one exceedance is

$$P(\text{at least one})=1-(1-0.02)^{5}=1-0.9039=\boxed{0.0961\approx 9.6\ \text{per cent}}$$

and the chance of exactly two, which is what the agent actually saw, is

$$P(2\text{ in }5)=\binom{5}{2}(0.02)^{2}(0.98)^{3}=10(0.0004)(0.9412)=0.00376\approx 0.38\ \text{per cent}$$

So the event is uncommon — about one chance in 266 for any particular river over any particular five years — but it is not rare in the sense of being suspicious. The key point for an insurer is that this calculation applies to one river over one five-year window. Canada has thousands of gauged rivers; a 0.38 per cent chance evaluated across a thousand of them predicts a handful of such doubles somewhere in the country in every five-year period. Seeing one is expected, not evidence that the frequency analysis is broken.

Three further points complete the explanation, and each matters commercially. First, the return period is the average interval over an infinitely long stationary record, so intervals between exceedances are highly variable; back-to-back years are perfectly consistent with a 50-year magnitude. Second, the relevant number for a property owner is not the annual probability but the probability over the period of exposure: over a 25-year mortgage the chance of at least one 50-year flood is $1-(0.98)^{25}=0.397$, roughly two chances in five, and over a 30-year exposure it is 0.455. That is the figure that should drive an insurance or mitigation decision, and it is far more alarming than "one in fifty" sounds. Third, the estimate itself is uncertain and possibly drifting. A 50-year flood estimated from a 40-year gauge record carries wide confidence limits, so what is called the 50-year flood today may be revised when the record lengthens; and urbanisation upstream, channelisation, or a changing climate can make the series non-stationary, so that a discharge which genuinely was the 2 per cent AEP flood thirty years ago may now be the 4 per cent flood. Two exceedances in five years should prompt a re-analysis of the record, but it is not by itself proof that anything has changed.

(iii) Reading a distribution and its statistics off the probability plot (6 marks)

[Figure not reproduced: The source probability plot, redrawn with a fitted log-normal distribution. The median at 50 per cent exceedance gives the log-mean; the 84.1 and 15.9 per cent readings are one log-standard-deviation either side of it. See the official exam paper.]

A probability plot is a graph of the observed data against the exceedance probability assigned to each observation, and it does two jobs at once: it identifies which distribution fits, and it gives the parameters of that distribution without any algebra.

How the plot is constructed. The $n$ annual rainfall totals are ranked from largest ($m=1$) to smallest ($m=n$), and each is assigned a plotting position — the Weibull formula

$$P(X\ge x_m)=\frac{m}{n+1}$$

is the usual choice in Canadian practice, with Cunnane, Gringorten or Blom used for particular distributions. Each observation is then plotted against its plotting position, which is exactly the scatter of open circles in the source figure.

How the distribution is identified. The trick is the choice of axis scales. If the probability axis is transformed to the standard normal variate $z=\Phi^{-1}(1-P)$ — normal probability paper — then a normally distributed sample plots as a straight line. If in addition the data axis is logarithmic — log-normal probability paper — then a log-normally distributed sample plots as a straight line. The source plot uses linear axes for both, which is why the data trace the familiar reversed S-curve rather than a line; that S-shape by itself already tells you the sample is not uniformly distributed, but it does not identify the distribution. Replotting on normal and on log-normal paper and choosing the one on which the points fall closest to a straight line is the graphical goodness-of-fit test, and the departure of the upper tail from the line is exactly where the design values live, so it is the part to inspect most carefully. Annual rainfall totals are bounded below by zero and mildly right-skewed, which is why the log-normal is the natural first candidate; annual flood peaks, being more strongly skewed, usually call for the Gumbel or log-Pearson III distribution instead.

How the statistics are read off. Once the plot is linear, two readings fix the distribution completely, because a straight line has two parameters. Taking $Y=\ln X$ so that $Y$ is normal with mean $\mu$ and standard deviation $\sigma$:

1. Read the median at 50 per cent exceedance to obtain the log-mean. The 50 per cent point is the median of $X$ and also the median — hence the mean — of $Y$, so $\mu=\ln x_{50}$. From the plot, $x_{50}\approx 2030$ mm, giving $$\mu=\ln 2030=7.6158\ (\ln \text{mm})$$
2. Read the 84.1 and 15.9 per cent exceedance values to obtain the log-standard deviation. Those two probabilities are one standard deviation either side of the mean on a normal variate, so the difference in $\ln x$ between them spans $2\sigma$: $$\sigma=\tfrac{1}{2}\ln\!\left(\frac{x_{15.9}}{x_{84.1}}\right) =\tfrac{1}{2}\ln\!\left(\frac{2500}{1650}\right)=\boxed{0.2078}$$ Equivalently, and more robustly, fit a least-squares line of $\ln x$ against $z$ over all the points: the intercept is $\mu$ and the slope is $\sigma$.
3. Convert the log-space parameters back to the statistics of the rainfall itself. For a log-normal variable, $$\bar X=\exp\!\left(\mu+\tfrac{\sigma^{2}}{2}\right)=2074.3\ \text{mm},\qquad s_X=\bar X\sqrt{e^{\sigma^{2}}-1}=435.6\ \text{mm}$$ a coefficient of variation of 0.210. Note that the mean exceeds the median of 2030 mm, as it must for a right-skewed distribution.
4. Use the fitted line to estimate design quantiles. The rainfall with return period $T$ follows from $x_T=\exp(\mu+z_T\sigma)$ with $z_T=\Phi^{-1}(1-1/T)$. For $T=50$ years, $z=2.054$ and $x_{50\text{yr}}=3110$ mm; for $T=100$ years, $z=2.326$ and $x_{100\text{yr}}=3292$ mm. Both lie beyond the largest plotted observation, so they are extrapolations and should be reported with that caveat and with confidence limits.