16-Civ-B4 Engineering Hydrology · December 2017

Question 2 of 7: Runoff Hydrographs, Unit Hydrographs and Conceptual Models

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

Question 2: Runoff Hydrographs, Unit Hydrographs and Conceptual Models (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.

2(i) — Graphical explanation of hydrograph generation from a unit hydrograph (6 marks)

The question offers a choice between a graphical and a point-form explanation. The graphical route is taken here, because the whole content of the unit-hydrograph method is visible in one figure: the storm hyetograph is chopped into blocks, each block generates a scaled and lagged copy of the unit hydrograph, and the copies are added.

Given. A three-block effective-rainfall hyetograph and the watershed’s one-hour unit hydrograph, tabulated below.

Find. The storm hydrograph ordinates, and a graphical demonstration of how they arise.

Approach. Convolve the effective-rainfall blocks with the unit hydrograph, then check the answer by comparing the volume under the computed hydrograph with the volume of effective rainfall.

1. State the convolution as a sum. If $P_m$ is the depth of the $m$-th one-hour block and $U_j$ is the $j$-th unit-hydrograph ordinate, the direct-runoff ordinate at step $n$ is $$Q_n = \sum_{m} P_m \, U_{n-m+1}$$ Each term is one block’s scaled, lagged response; the sum is the superposition.
2. Scale and lag each block. The 12 mm block produces $12U_j$ starting at $t=0$; the 25 mm block produces $25U_j$ starting at $t=1$ h; the 8 mm block produces $8U_j$ starting at $t=2$ h. These are the three dashed curves in the figure.
3. Add the columns. At $t = 3$ h, for example, $$Q_3 = (12)(15) + (25)(20) + (8)(8) = 180 + 500 + 64 = \boxed{744\ \text{m}^3/\text{s}}$$ which is the peak of the storm hydrograph.
4. Add the baseflow. The unit hydrograph describes direct runoff only. Adding a baseflow of 20 m³/s, taken as constant through the event, gives a total peak of $744 + 20 = \boxed{764\ \text{m}^3/\text{s}}$.
5. Check the volume. The unit-hydrograph ordinates sum to 56 m³/s per mm, so one millimetre of effective rainfall corresponds to $(56)(3600) = 201\,600$ m³, which implies a watershed area of $201\,600 / 0.001 = 201.6$ km². The computed direct-runoff ordinates sum to 2520 m³/s, giving $(2520)(3600) = 9.072 \times 10^6$ m³, and the effective rainfall of 45 mm on 201.6 km² also gives $9.072 \times 10^6$ m³. The volumes agree exactly, which is the correct arithmetic check on any convolution.

Read graphically, the figure carries the whole idea: the shape of the response is a fixed property of the basin, the storm supplies only the scaling and the timing, and the peak occurs where the largest rainfall block coincides with the peak of the unit hydrograph. That is why an intense burst late in a storm, arriving on an already-wetted basin, is so much more dangerous than the same depth spread evenly.

2(ii) — Deriving, applying and assuming (7 marks)

(a) Generating a unit hydrograph for a watershed. The derivation reverses the convolution, and needs a gauged basin. Select an isolated storm that produced a single-peaked hydrograph, that fell reasonably uniformly over the basin, and whose duration is short compared with the basin’s time of concentration. Separate the baseflow from the recorded hydrograph, usually by a straight line from the start of the rise to the point on the recession where the flow returns to the pre-storm decay, and leave the direct-runoff hydrograph. Integrate that direct-runoff hydrograph over time and divide by the watershed area to obtain the depth of effective rainfall the storm produced — not the depth that fell, but the depth that ran off. Divide every direct-runoff ordinate by that depth. The result is the unit hydrograph for the storm’s effective duration. Repeat for several storms of similar duration and average, aligning the peaks before averaging so that the mean does not flatten the peak, then adjust the averaged ordinates so that the unit volume is exactly recovered. Where the required duration differs from the derived one, the S-curve construction converts between durations. On an ungauged basin, a synthetic unit hydrograph — Snyder, SCS dimensionless, or Clark — is fitted from measurable basin characteristics instead.

(b) Using a unit hydrograph to size a highway culvert. The design sequence is: choose the return period the road classification requires; read the design rainfall depth and distribution for the critical duration from the regional IDF curves; convert that rainfall to effective rainfall using an infiltration or curve-number loss model with a conservative antecedent-moisture assumption; convolve the effective-rainfall blocks with the basin’s unit hydrograph to obtain the design hydrograph; add baseflow; and size the structure so that the resulting headwater does not overtop the road or exceed the allowable inundation upstream. A worked check follows.

Given. A highway crossing on a 4.2 km² catchment. The 100-year design storm yields 62 mm of effective rainfall in one hour; the basin’s one-hour unit hydrograph peaks at 0.28 m³/s per mm; baseflow is 0.40 m³/s. The candidate structure is a 2.4 m diameter corrugated-steel pipe, 30 m long, Manning $n = 0.024$, entrance-loss coefficient $K_e = 0.5$, operating in outlet control under an available head of $H = 1.8$ m.

Find. The design discharge and whether the candidate barrel passes it.

Approach. Take the design peak from the unit hydrograph, then compute the full-flow outlet-control capacity of the barrel and compare.

1. Take the design peak from the unit hydrograph. For a single one-hour block the convolution collapses to a product: $$Q_{p,\text{DRO}} = P_e \, U_p = (62)(0.28) = 17.36\ \text{m}^3/\text{s}$$ Adding baseflow gives the design discharge $$Q_{\text{design}} = 17.36 + 0.40 = \boxed{17.76\ \text{m}^3/\text{s}}$$
2. Compute the barrel geometry. For the circular pipe flowing full, $$A = \frac{\pi D^2}{4} = \frac{\pi (2.4)^2}{4} = 4.524\ \text{m}^2, \qquad R = \frac{D}{4} = 0.600\ \text{m}$$
3. Assemble the loss coefficients. The friction term in the outlet-control equation is $$\frac{2gn^2L}{R^{4/3}} = \frac{2(9.81)(0.024)^2(30)}{(0.600)^{4/3}} = \frac{0.3390}{0.5060} = 0.670$$
4. Compute the outlet-control capacity. With entrance, friction and velocity heads, $$Q_{\text{cap}} = A\sqrt{\frac{2gH}{1 + K_e + 2gn^2L/R^{4/3}}} = 4.524\sqrt{\frac{(2)(9.81)(1.8)}{1 + 0.5 + 0.670}} = \boxed{18.25\ \text{m}^3/\text{s}}$$
5. Compare. Since $18.25 > 17.76$ m³/s, the 2.4 m barrel passes the design flood within the allowable headwater, with about three per cent spare. The next smaller commercial size would not, so 2.4 m is the correct selection.

(c) Two general assumptions behind the unit-hydrograph technique. The first is linearity and time invariance: the response is assumed proportional to the effective rainfall depth and independent of when the rainfall occurs, so that doubling the rainfall doubles every ordinate and responses to successive blocks simply superpose. Real basins are not linear — the effective velocity of the flood wave increases with stage, so large events peak earlier and more sharply than proportionality predicts, and a unit hydrograph derived from moderate storms will under-predict a rare one. The second is uniformity: the effective rainfall is assumed uniform in intensity over its duration and uniform in depth over the whole watershed. The assumption fails for convective cells on a large basin, and it fails badly when the storm moves along the basin axis, which can either concentrate or spread the runoff arrival far beyond what the stationary-storm unit hydrograph gives. A third assumption worth naming is that the basin itself is unchanged between derivation and application — a unit hydrograph derived before urbanisation, forest harvest or a major fire no longer describes the basin.

2(iii) — Calibrating a conceptual watershed model, and two limitations (7 marks)

A conceptual model represents the watershed as a small set of interconnected storages — interception, soil moisture, an upper fast-draining zone, a lower slow-draining zone, and a channel routing element — each drained by simple functional relations with parameters that are not directly measurable. HBV, the Sacramento soil-moisture accounting model used in Canadian and United States forecasting, GR4J and HEC-HMS in its soil-moisture accounting mode are all of this family. Because the parameters are conceptual rather than physical, they must be inferred from observed behaviour, which is what calibration means.

A workable calibration method — split-sample calibration against an objective function. The steps are as follows. First, assemble a continuous record of precipitation, temperature and observed discharge, and divide it into a calibration period and an independent validation period, each containing a representative range of wet and dry years and at least one large flood. Second, fix by inspection or by measurement every parameter that can be fixed — basin area, the degree-day melt factor from snow-course data, the routing time from observed travel times — so that only genuinely free parameters remain; a conceptual model with more than about six free parameters cannot be identified from one discharge record. Third, choose an objective function that matches the intended use. For peak rainfall-runoff simulation the Nash–Sutcliffe efficiency computed on discharge (or a weighted combination of Nash–Sutcliffe on peaks and percent bias on volume) is standard, because squaring the errors weights the large flows the design is about. Fourth, optimise the free parameters over the calibration period using an automatic search — the shuffled complex evolution algorithm is the usual choice because the response surface has many local optima — with manual review of the resulting hydrographs at each stage. Fifth, and non-negotiably, run the calibrated parameter set unchanged over the validation period and judge the model on that result. A model that fits the calibration period and fails the validation period has been fitted to noise.

Given. Six observed and simulated flood peaks from the validation period, in m³/s: observed 42, 118, 265, 190, 96, 51; simulated 38, 131, 240, 205, 88, 55.

Find. The Nash–Sutcliffe efficiency and the percent bias.

1. Compute the mean of the observations. $$\bar{Q}_o = \frac{42+118+265+190+96+51}{6} = \frac{762}{6} = 127.0\ \text{m}^3/\text{s}$$
2. Form the two sums of squares. $$\sum (Q_o - Q_s)^2 = 16+169+625+225+64+16 = 1115$$ $$\sum (Q_o - \bar{Q}_o)^2 = 7225+81+19\,044+3969+961+5776 = 37\,056$$
3. Evaluate the efficiency. $$\text{NSE} = 1 - \frac{\sum (Q_o - Q_s)^2}{\sum (Q_o - \bar{Q}_o)^2} = 1 - \frac{1115}{37\,056} = \boxed{0.970}$$
4. Check the bias. $$\text{PBIAS} = 100\,\frac{\sum Q_s - \sum Q_o}{\sum Q_o} = 100\,\frac{757-762}{762} = -0.66\%$$

An efficiency of 0.970 with a bias under one per cent would be regarded as a good calibration for peak simulation. Note that the efficiency is measured against the variance of the observations, so a value of zero means the model is no better than always predicting the mean flow, and a negative value means it is worse than that.

Limitation 1 — parameter non-uniqueness (equifinality). Many different parameter sets reproduce the observed discharge about equally well, because a single outflow series carries far less information than the number of free parameters requires. The calibrated values are therefore not physical properties of the basin and cannot be transferred to a neighbouring basin, nor safely used to simulate a changed basin. The practical consequence is that the model’s predictive uncertainty is much wider than its calibration fit suggests, which is why formal uncertainty methods such as GLUE report a range of behavioural parameter sets rather than one optimum.

Limitation 2 — the model is only valid inside the range it was calibrated over. A conceptual model calibrated on floods up to, say, a twenty-year event is being extrapolated when it is used to estimate the two-hundred-year design flood, and the storages and thresholds that were never exercised during calibration are unconstrained. The same limitation applies in time: a model calibrated on the historical record assumes stationarity of both climate and land use, so a changing snow regime or a new subdivision invalidates it. A third limitation worth stating is the model’s dependence on input quality — a lumped conceptual model fed by a sparse rain-gauge network inherits the areal-precipitation error directly, and no amount of parameter tuning can recover from a storm the gauges missed.