Question 4 of 7: Areal precipitation and stream-flow measurement

Problem 4: Areal precipitation and stream-flow measurement (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) — Isohyetal and Thiessen-polygon methods (7 marks)

Rain gauges measure precipitation at points, but every hydrologic calculation needs the average depth over an area. Both methods answer the same question — how should each point measurement be weighted? — and both produce a weighted mean of the form $\bar P = \sum w_j P_j$ with $\sum w_j = 1$; they differ in how the weights are found.

Thiessen polygon method. The gauge locations, including gauges just outside the basin, are plotted on a map. Adjacent gauges are joined by straight lines, and the perpendicular bisector of each connecting line is drawn; the bisectors intersect to enclose each gauge in a polygon containing every point that is closer to that gauge than to any other. The polygon areas $A_j$ falling inside the basin are measured (today by GIS, historically by planimeter), and the areal average is

The weights $A_j/A$ depend only on the geometry of the network, so they are computed once and reused for every storm, which makes the method fast, objective and reproducible — its principal practical virtues. Its limitations are that it takes no account of topography, so a valley gauge is credited with the rainfall of the mountain slope beside it; it produces a discontinuous field, with rainfall jumping abruptly at a polygon boundary; and the whole polygon network must be redrawn whenever a gauge is added or a gauge record is missing for a particular storm. It is generally more accurate than a simple arithmetic mean whenever the gauges are unevenly spaced.

Isohyetal method. The observed depths are plotted at the gauge locations and lines of equal precipitation depth — isohyets — are interpolated between them, exactly as elevation contours are interpolated between spot heights. The essential difference from the Thiessen method is that the analyst is free, and expected, to bend the isohyets in accordance with knowledge that the gauge data alone do not contain: orographic lift on a windward slope, a rain shadow, the observed track of a convective cell, or radar imagery. The area $a_i$ between each pair of successive isohyets is then measured, and the average depth in that strip is taken as the mean of the two bounding isohyets, giving

Because it admits meteorological and topographic judgement, the isohyetal method is the most accurate of the standard techniques for a mountainous basin — which is to say, for most of British Columbia — and it is the only one of the two that represents the rainfall as a continuous field. Its costs are that it is laborious, that it must be repeated for every storm rather than computed once, and that it is subjective: two analysts drawing isohyets from the same gauge data will not obtain identical answers, so the result is only as good as the analyst’s knowledge of the local storm climatology. A dense gauge network reduces, but does not eliminate, that subjectivity.

A short numerical comparison makes the weighting explicit. Suppose three gauges in a 40 km² basin record 62, 48 and 75 mm, and that their Thiessen polygons occupy 12, 18 and 10 km² respectively.

Given. $P = \{62,\,48,\,75\}$ mm with Thiessen areas $A = \{12,\,18,\,10\}$ km², total basin area 40 km² (assumed, illustrative values).

Find. The Thiessen-weighted areal average, compared with the arithmetic mean.

1. Form the area-weighted sum. $$\sum A_j P_j = 12(62) + 18(48) + 10(75) = 744 + 864 + 750 = 2358\ \text{km}^2\!\cdot\!\text{mm}$$
2. Divide by the total area. $$\bar P = \frac{2358}{40} = \boxed{58.95\ \text{mm}}$$ The arithmetic mean of the same three readings is $(62+48+75)/3 = 61.67$ mm, which is 2.7 mm higher because it gives the driest gauge — the one with the largest polygon — too little weight. That difference is the entire point of areal weighting.

Part (ii) — Stream stage, the rating curve, and a flood alert system (7 marks)

Stream stage is the elevation of the water surface at a gauging section, measured above an arbitrary local datum called the gauge datum, and denoted $G$ or $h$. It is not the depth of flow and not an elevation above sea level unless the gauge datum has been levelled to a geodetic datum, which at Water Survey of Canada stations means CGVD2013. Stage is what is actually and continuously measured — formerly by staff gauge and float-and-counterweight recorder, now almost universally by pressure transducer, bubbler or non-contact radar sensor logging at 5- to 15-minute intervals and telemetered by satellite or cellular link. It is measured because it is cheap and continuous, whereas discharge is expensive and intermittent.

A rating curve, or stage-discharge relation, is the calibration that converts the continuously measured stage into the discharge that is actually wanted. It is the empirical relation $Q = f(G)$ for one specific section, and its usual algebraic form is the power law

where $G_0$ is the stage of zero flow — the gauge reading at which discharge would just cease, essentially the elevation of the section control — and $a$ and $b$ are fitted constants. The exponent $b$ is typically 1.5 for a section controlled by a weir-like crest and 2.0 to 2.5 for a channel-controlled section, which is no accident: those are the exponents that weir and Manning hydraulics predict.

How the rating is derived. A gauging station is first sited where a stable natural or artificial control — a bedrock riffle, a gravel bar, a constriction, or a built weir or flume — fixes the relation between stage and discharge and keeps it single-valued. Then, on many separate occasions spread across the range of flows, a discharge measurement is made concurrently with a stage reading. The standard technique is the velocity-area or mid-section method: the section is divided into 20 to 30 vertical panels, the depth is sounded in each, the mean velocity in each vertical is measured with a current meter or acoustic Doppler instrument at 0.6 of the depth (or as the average of readings at 0.2 and 0.8 depth), and the discharge is $Q = \sum v_i a_i$ over the panels. Modern practice increasingly uses a boat-mounted acoustic Doppler current profiler traversing the section. Each measurement yields one $(G,\,Q)$ pair. Twenty or more such pairs are plotted, usually on log-log paper where the power law becomes the straight line $\log Q = \log a + b\log(G - G_0)$, and $a$, $b$ and $G_0$ are fitted by least squares. Because floods are rare and dangerous, the upper end of the curve almost always has to be extrapolated above the highest measured gauging, which is done with hydraulic reasoning — a slope-area computation using Manning’s equation at surveyed high-water marks, a step-backwater model of the reach, or the theoretical weir exponent — rather than by blind curve extension. The rating is then re-checked after every significant flood, because scour or deposition at the control shifts it; the correction applied between check gaugings is called a shift, and a rating that has moved materially is replaced and dated.

Using the rating in a flood alert system. The rating is what turns a stage sensor into a forecasting instrument. The system is built as follows. Discharges of engineering significance are identified first — the bankfull discharge, the flow at which a low-lying road or intake is overtopped, the flow at which a floodplain development is threatened, and the regulatory design flood — and each is converted through the rating, read backwards into a corresponding stage. Those stages become the alert thresholds: a watch level, a warning level and an evacuation level, expressed in the units the telemetered gauge actually reports. The telemetry system then compares each transmitted stage against the thresholds automatically and issues the appropriate notification, so no discharge computation is needed in real time. Two refinements make it genuinely useful. First, the rate of rise is monitored as well as the absolute stage, because a rapidly rising limb at a low stage may be more alarming than a steady high stage. Second, upstream gauges are used as leading indicators: the travel time of a flood wave between an upstream and a downstream station, established from past events or from routing, buys warning lead time, so an upstream threshold exceedance triggers a downstream alert several hours in advance. Coupled to a rainfall-runoff model driven by radar and quantitative precipitation forecasts, the same rating converts forecast discharge back into forecast stage and hence into forecast inundation extent. In Canada this is the structure of provincial river-forecast centres, including the BC River Forecast Centre, operating on Water Survey of Canada real-time stage data.

Given. A station rating $Q = 12.5\,(G - 0.35)^{2.1}$ with $Q$ in m³/s and $G$ in m, and a telemetered stage of $G = 2.85$ m (assumed, illustrative constants).

Find. The discharge corresponding to the reported stage.

1. Form the effective head above the stage of zero flow. $$G - G_0 = 2.85 - 0.35 = 2.50\ \text{m}$$
2. Apply the rating equation. $$Q = 12.5\,(2.50)^{2.1} = 12.5 \times 6.850 = \boxed{85.6\ \text{m}^3/\text{s}}$$ Read in the other direction, a warning threshold set at 60 m³/s corresponds to a stage of $G = 0.35 + (60/12.5)^{1/2.1} = 2.46$ m, and that is the number programmed into the telemetry alarm.

Part (iii) — Peak-flow frequency analysis for the 100-year event (6 marks)

The historical rating-curve data provide, for every year of record, a continuous discharge series obtained by applying the rating to the recorded stage. Frequency analysis converts that record into a statement about the magnitude of rare events.

Assumptions, which must be stated because the whole method rests on them. The annual maximum flows are (1) independent from year to year, which is why the annual maximum series is used rather than a partial-duration series; (2) identically distributed, that is, drawn from one unchanging population — which requires that the basin be free of trend from urbanisation, forest harvesting, regulation by a new dam, or climate change, and that the rating has been maintained so the discharges are comparable through time; (3) randomly sampled, with no systematic omission of high years; and (4) adequately described by the chosen probability distribution. A record of at least 20 to 30 years is normally required to estimate a 100-year event with acceptable confidence, and the extrapolation should not extend much beyond about twice the record length without explicit caution.

1. Assemble the annual maximum series and rank it. Extract the single largest instantaneous peak discharge in each water year from the rated record, giving $n$ values, and rank them from largest ($m = 1$) to smallest ($m = n$).
2. Assign plotting positions and return periods. The Weibull plotting position gives the exceedance probability and hence the return period of each ranked observation, $$P(Q \ge Q_m) = \frac{m}{n+1}, \qquad T = \frac{1}{P} = \frac{n+1}{m}$$ so the second-largest peak in a 45-year record has $P = 2/46 = 0.0435$ and $T = 23.0$ years. Plotting positions are used to display the data, not to extrapolate — a 45-year record can never yield a 100-year point directly.
3. Fit a probability distribution and extrapolate with a frequency factor. Fit a distribution suitable for extremes — the Gumbel (Extreme Value Type I) or the log-Pearson Type III, the latter being the standard in much of North American practice — by the method of moments or of L-moments. Chow’s general frequency equation then expresses any quantile as $$Q_T = \bar Q + K_T\,s$$ where $\bar Q$ and $s$ are the sample mean and standard deviation of the annual maxima and $K_T$ is the frequency factor for the chosen distribution and return period. For the Gumbel distribution, $$K_T = -\frac{\sqrt{6}}{\pi}\left[0.5772 + \ln\ln\!\left(\frac{T}{T-1}\right)\right]$$
4. Evaluate the 100-year quantile. With an assumed record of $n = 40$ years giving $\bar Q = 420$ m³/s and $s = 130$ m³/s, the frequency factor for $T = 100$ years is $$K_T = -\frac{\sqrt{6}}{\pi}\left[0.5772 + \ln\ln\!\left(\frac{100}{99}\right)\right] = 3.137$$ and therefore $$Q_{100} = 420 + 3.137(130) = \boxed{828\ \text{m}^3/\text{s}}$$
5. State the result probabilistically and attach confidence limits. The 100-year flood is the discharge with an annual exceedance probability of $1/100 = 0.01$, or one percent in any year — not an event that occurs once a century. The probability that it is equalled or exceeded at least once during a 25-year project life is $$R = 1-(1-0.01)^{25} = 0.222$$ that is, 22 percent, which is the number that matters for design. Confidence limits on $Q_{100}$ should be computed from the standard error of the fitted quantile and reported alongside it; for a 40-year record they are typically wide, on the order of plus or minus 15 to 25 percent.