Question 3 of 7: Point and areal estimates of precipitation, and stream flow measurement

Problem 3: Point and areal estimates of 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) — Point versus areal estimates of precipitation (7 marks)

First difference — what is actually measured, and over what support volume. A point estimate is a direct physical measurement made by one instrument at one location: a standard Canadian rain gauge with a 200 mm orifice samples an area of about 0.031 m², and the depth it reports is simply the caught volume divided by that orifice area,

An areal estimate, by contrast, is not measured at all. It is inferred — a weighted average of several point measurements, taken as representative of a basin whose area may be 10⁸ times the orifice area. The extrapolation from 0.03 m² to hundreds of square kilometres is the fundamental step, and everything else follows from it.

Second difference — the estimating equations. The point estimate needs none beyond the ratio above; the areal estimate requires an explicit weighting scheme, and three are standard. The arithmetic mean weights all gauges equally and is defensible only where the network is dense and evenly spread and the terrain is flat:

The Thiessen polygon method weights each gauge by the fraction of basin area lying closer to it than to any other gauge, which handles an uneven network:

The isohyetal method interpolates contours of equal depth using judgement about orography and storm structure, then averages the mean depth of each inter-isohyet band weighted by its area, P̄ = Σ a_j P̄_j / A. It is the most accurate in mountainous terrain and the most subjective. Modern practice adds inverse-distance weighting, kriging and radar–gauge merging, all of which are the same weighted average with a different rule for the weights.

Third difference — the error structure and therefore the appropriate use. The error in a point estimate is instrumental and exposure-related: wind-induced undercatch (10 to 15 per cent for rain and often 30 to 50 per cent for snow, which is why Nipher shields are standard at Canadian stations), wetting and evaporation losses, and splash. The error in an areal estimate is dominated instead by sampling: whether the gauge network resolves the storm at all. A convective cell 4 km across can be missed entirely by a network with 20 km spacing. Areal averaging also systematically smooths the extreme, so the areal mean depth is always less than the maximum point depth for the same storm — the basis of the areal reduction factor applied to point design rainfall. Consequently the point estimate is what IDF curves and design storms are built from, while the areal estimate is what water balances, runoff volumes and reservoir inflows require.

Given. A basin is instrumented with four gauges, and one storm produces the depths and Thiessen areas below (assumed, representative values).

Find. The arithmetic-mean and Thiessen areal depths over the basin, and the corresponding storm volume.

Approach. Average the four point depths without weights, then with the Thiessen area weights, and convert the weighted depth to a volume over the basin area.

1. Take the unweighted mean. $$\bar{P}_{\text{arith}} = \frac{62 + 48 + 91 + 75}{4} = \frac{276}{4} = 69.0\ \text{mm}$$
2. Take the area-weighted mean. Applying the Thiessen weights, $$\bar{P}_{\text{Th}} = \frac{(62)(18) + (48)(26) + (91)(31) + (75)(25)}{100} = \frac{7060}{100} = \boxed{70.6\ \text{mm}}$$ The two differ by only 1.6 mm here because no gauge is grossly over- or under-represented; the Thiessen value is nevertheless the defensible one, since it is the wettest gauge (C, 91 mm) that commands the largest area and the arithmetic mean does not know that.
3. Convert the areal depth to a volume. Depth times area gives $$V = \bar{P}_{\text{Th}}\,A = (0.0706\ \text{m})(100\times10^{6}\ \text{m}^2) = 7.06\times10^{6}\ \text{m}^3$$ which is the quantity a water balance or a reservoir inflow calculation actually needs, and which no point measurement could have supplied.

Part (ii) — Stream stage and the rating curve after a storm (7 marks)

The two are not alternative ways of measuring flow; they are the two halves of one measurement system, and the contrast between them is the point of the question.

Stage is the elevation of the water surface above an arbitrary local datum, recorded by a float-and-counterweight recorder in a stilling well, by a pressure transducer or bubbler, or by a non-contact radar sensor. It is directly and continuously measured, cheaply, automatically and at short intervals, and it can be telemetered in real time. Following a storm the stage record is what actually exists: a continuous trace showing the rise, the crest and the recession. Its virtue is completeness in time. Its limitation is that stage by itself is not a hydrologically useful quantity — it is a level, not a flow, and the same level on two different rivers, or on the same river after a channel change, corresponds to entirely different discharges.

The rating curve is the stage–discharge relation for the particular gauging section, the transfer function that converts the measured stage into the discharge that is wanted. It is not measured continuously; it is established by a programme of individual discharge measurements — typically current-meter gaugings spread across the full range of flows, sometimes supplemented by acoustic-Doppler transects — each of which pairs one discharge with the stage prevailing at the time. The paired points are then fitted with a power law of the form

where h₀ is the stage of zero flow at the section and the exponent b typically falls between 1.5 and 2.5, reflecting the hydraulic control — near 1.5 for a weir-like section, nearer 2.5 for a wide channel under uniform-flow control.

Given. A Water Survey of Canada gauge has the following assumed, representative rating and stage record from one storm.

Find. The discharge before the storm and at the crest, and the sensitivity of the computed discharge to an error in the stage reading.

Approach. Substitute each stage into the fitted power law, then differentiate the power law logarithmically to obtain the relative discharge error produced by a given stage error.

1. Convert the pre-storm stage. $$Q = 12.5\,(2.80 - 0.35)^{2.10} = 12.5\,(2.45)^{2.10} = 82.1\ \text{m}^3/\text{s}$$
2. Convert the crest stage. $$Q_{\text{crest}} = 12.5\,(4.20 - 0.35)^{2.10} = 12.5\,(3.85)^{2.10} = \boxed{212\ \text{m}^3/\text{s}}$$ A stage rise of 1.40 m, half of the pre-storm depth above the control, has multiplied the discharge by 2.58 — the strong non-linearity that makes the rating curve indispensable and makes eyeballing a flow from a level impossible.
3. Assess the sensitivity to stage error. Differentiating the power law logarithmically, $$\frac{\mathrm{d}Q}{Q} = b\,\frac{\mathrm{d}h}{h - h_0} = 2.10 \times \frac{0.03}{2.45} = 0.0257$$ so a 30 mm error in stage produces a 2.6 per cent error in discharge at that stage. The same absolute stage error matters progressively less as the river rises, because h − h₀ grows — which is why the largest rating errors at flood stage come not from reading the stage but from extrapolating the curve.

Where the comparison bites after a storm. Three practical issues distinguish the two during a flood. First, the crest stage is usually higher than any stage at which the rating has ever been calibrated, so the flood discharge is obtained by extrapolating the curve; the resulting peak may carry an uncertainty of 10 to 25 per cent even though the stage is known to millimetres. Second, an unsteady flood wave produces a looped rating: on the rising limb the water-surface slope is steeper than the bed slope and the discharge at a given stage exceeds the steady-flow value, while on the falling limb it is less, so the single-valued curve under-estimates the rising limb and over-estimates the falling one. Third, the control can shift — scour, deposition, debris, ice cover or vegetation growth all change the relation, so the curve must be re-verified by gauging after every major event and the record adjusted by a shift correction. Stage, in short, is reliable and continuous but not the quantity wanted; the rating curve supplies the quantity wanted but is periodic, fitted, extrapolated and perishable.

Part (iii) — Conventional stream flow measurement: challenges and approximations (6 marks)

The method considered. The conventional field method is the velocity–area method using a current meter, in its mid-section form (WMO Manual on Stream Gauging; ISO 748). The cross-section is divided into 20 to 30 vertical panels; at each vertical the depth is sounded and the velocity measured with a propeller or electromagnetic current meter, either at 0.6 of the depth below the surface (the single-point method) or as the mean of readings at 0.2 and 0.8 of the depth (the two-point method). The discharge attributed to each vertical is the product of its mean velocity, its depth and a width extending half-way to each neighbouring vertical, and the total discharge is the sum:

Given. One gauging on a small river returns the following assumed, representative field notes.

Find. The gauged discharge, the flow area and the section mean velocity.

Approach. Sum the panel discharges by the mid-section equation, sum the panel areas separately, and divide one by the other.

1. Sum the panel discharges. Each panel contributes v_i d_i times its width: $$Q = 1.008 + 3.375 + 4.131 + 1.815 = \boxed{10.33\ \text{m}^3/\text{s}}$$
2. Obtain the flow area and the mean velocity. $$A = \sum d_i\,\Delta b_i = (0.8 + 1.5 + 1.7 + 1.1)(3) = 15.3\ \text{m}^2, \qquad \bar{v} = \frac{Q}{A} = \frac{10.33}{15.3} = 0.675\ \text{m/s}$$ The mean velocity is a useful check: values far outside 0.3 to 3 m/s for a small river, or a section mean far from the largest panel velocity, usually signal a metering or bookkeeping error.

Two challenges in the field. The first is access and safety at the flows that matter most. Wading is impossible above roughly 1 m depth or 1 m/s velocity, so high flows must be gauged from a bridge, a cableway or a boat, in conditions of floating debris, ice and, on many Canadian rivers, remote access with limited daylight; the practical result is that ratings are least well calibrated exactly where they are most needed. The second is that the flow is changing while it is being measured. A complete gauging takes 45 to 60 minutes, during which a flood stage may rise by several tenths of a metre, so the measurement does not correspond to any single stage and must be assigned to a mean stage. Related field difficulties are non-uniform and oblique flow at the section, which requires an angle correction, and a mobile bed that changes the section between soundings.

Two approximations built into the method. The first is that the velocity at 0.6 of the depth equals the mean velocity in that vertical. This follows from assuming a logarithmic velocity profile, for which the depth-averaged velocity does occur near 0.6d; the assumption degrades where the profile is distorted by ice cover, dense weed, a very rough bed or a shallow depth, which is why the two-point method at 0.2d and 0.8d is preferred whenever the depth exceeds about 0.75 m. The second is that each measured vertical velocity is representative of a rectangular panel extending half-way to each neighbour: the mid-section equation replaces the true, curved velocity and bed geometry with a piecewise-constant one, and its error falls as the number of verticals rises, which is why the standards require each panel to carry no more than about 5 to 10 per cent of the total discharge. A third approximation, always present, is that the flow is treated as steady over the duration of the gauging.