Question 2 of 7: Conceptual models, unit hydrographs and runoff hydrographs

Problem 2: Conceptual models, unit hydrographs and runoff hydrographs (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) — Conceptual versus physical models (10 marks)

A physical model is a tangible, usually scaled, replica of the prototype in which real water flows: a 1:40 sectional model of a spillway in a hydraulics laboratory flume, a distorted-scale estuary model, or a rainfall simulator over an instrumented soil plot. Its behaviour is produced by the same physics as the prototype, and it is designed by similitude — geometric, kinematic and dynamic similarity, with the governing dimensionless group (Froude number for free-surface flow, Reynolds number for closed-conduit or viscous flow) held equal between model and prototype. A conceptual model is a mathematical abstraction: the system is represented by a small number of idealised storage elements and transfer functions whose parameters are calibrated against observed records rather than measured directly. The Muskingum reach, the linear-reservoir cascade of Nash, the Stanford Watershed Model and its descendants (HSPF, HEC-HMS, HBV, GAWSER) are all conceptual models — a watershed is not really a chain of leaky buckets, but treating it as one reproduces the observed hydrograph well.

The two main differences follow from those definitions.

First difference — the nature of the representation and therefore of the errors. In a physical model the governing physics is present automatically and does not have to be written down; the modeller never has to decide how turbulence, air entrainment or a three-dimensional separation zone behaves, because the model fluid does it. What the modeller must worry about is scale effects: only one dimensionless group can be preserved at a time, so a Froude-scaled spillway model has too low a Reynolds number and exaggerates viscous and surface-tension effects, and the roughness of the model surface cannot be scaled down indefinitely. In a conceptual model the reverse holds. There are no scale effects at all — the model runs at prototype scale by construction — but every process that was not written into the equations is simply absent, and the parameters are lumped quantities with no directly measurable field counterpart, so they must be obtained by calibration and are not transferable to an ungauged basin without a regionalisation step. Physical-model error is dominated by similitude and instrumentation; conceptual-model error is dominated by structural inadequacy and parameter identifiability.

Second difference — cost, time and the range of conditions that can be explored. A physical model is expensive to build and instrument, occupies laboratory space, takes months to commission, and each experimental condition takes real time to run and to reach steady state. Consequently only a handful of design alternatives and a handful of flow conditions can be tested. A conceptual model, once coded and calibrated, costs almost nothing per run; a continuous simulation of fifty years of hourly record, or a Monte Carlo ensemble of ten thousand storm realisations, or a sensitivity sweep across every plausible value of a parameter, is routine. This is what makes conceptual models the only practical vehicle for probabilistic, long-record and climate-change analysis. Against that, a conceptual model can only reproduce behaviour of a kind that was anticipated when its equations were written, whereas a physical model can reveal a phenomenon nobody expected — a vortex forming at an intake, a hydraulic jump sweeping out of a stilling basin, cavitation damage on a chute.

Two further differences are worth a sentence each. The outputs differ in kind: a physical model yields spatially detailed, three-dimensional velocity and pressure fields directly observable by eye and by probe, while a conceptual model yields lumped time series at a few nodes. And the validation route differs: a physical model is validated by similitude argument and, ideally, by comparison with a prototype at one condition, while a conceptual model is validated by split-sample testing, calibrating on one period of record and verifying on an independent one.

Example where a conceptual model is preferred. Consider the design of a stormwater management pond for a new subdivision, where the requirement is to demonstrate that post-development peak flows for the 2-, 5-, 10-, 25-, 50- and 100-year events do not exceed pre-development peaks, and that the pond drains within 48 hours. A physical model of the catchment is inconceivable — there is no way to scale infiltration, evapotranspiration, soil-moisture accounting or the spatial pattern of a storm over several square kilometres, and no meaningful similitude criterion exists for a rainfall-runoff process. A conceptual rainfall-runoff model such as HEC-HMS or SWMM, with SCS curve-number or Horton losses, a unit-hydrograph or kinematic-wave transform and level-pool routing through the pond, answers the question in an afternoon and can be re-run for every return period, for a range of antecedent moisture conditions, and for a climate-adjusted IDF curve. This is the ordinary case in urban drainage design in British Columbia and across Canada.

The converse case is equally clear and worth stating to show the boundary: the stilling basin at the toe of that pond’s emergency spillway, or the intake of a large hydroelectric plant, is precisely where a physical model earns its cost, because the local three-dimensional flow, air entrainment and cavitation potential are exactly the phenomena that no conceptual model contains.

Part (ii) — Deriving a unit hydrograph and building the design storm hydrograph (10 marks)

The question asks only for the method, so what follows is the procedure with its assumptions stated, together with the reasoning specific to the watershed shown.

Assumptions made, and stated as the question requires. (1) Each sub-area is treated as hydrologically homogeneous, with one curve number or one loss-rate parameter set and one time of concentration. (2) The unit-hydrograph theory is applied only to direct runoff, with baseflow separated first and added back at the end. (3) The three unit-hydrograph postulates below hold within each sub-area. (4) The design storm is areally uniform over the whole watershed, which is defensible at this size but would need an areal-reduction factor over a large basin. (5) The pond serving sub-area B is a level-pool reservoir, that is, its water surface is horizontal so that storage and outflow are both single-valued functions of stage. (6) Channel travel times between the sub-area outlets and the watershed outlet are known or can be estimated.

Deriving the unit hydrograph. A unit hydrograph of specified duration $D$ is the direct-runoff hydrograph produced by one unit depth (1 cm, or 10 mm in Canadian practice) of effective rainfall falling uniformly over the whole sub-area at a uniform rate during the time $D$. It is derived from observed rainfall and streamflow records for the sub-area, or for a hydrologically similar gauged basin, in the following steps.

1. Select suitable storm events from the record. Choose isolated storms of roughly uniform intensity, of a duration close to the desired $D$ (a common rule is $D$ between one-fifth and one-third of the basin lag), that fall over the whole sub-area and produce a well-defined single-peaked hydrograph. Several such events are needed, because a unit hydrograph derived from one storm is not reliable.
2. Separate baseflow from the observed hydrograph. Apply a consistent separation rule to every event — a straight line from the start of the rise to the point on the recession at $N = 0.83\,A^{0.2}$ days after the peak ($A$ in km²), or the constant-slope or concave method. Consistency between events matters more than the choice of rule. The remainder is the direct-runoff hydrograph.
3. Compute the volume of direct runoff and express it as a depth. Integrating the direct-runoff hydrograph over time gives the volume, and dividing by the sub-area gives the depth of effective rainfall that produced it: $$P_e = \frac{1}{A_{sub}}\int Q_d\,\mathrm{d}t \;\approx\; \frac{\Delta t \sum Q_d}{A_{sub}}$$ The corresponding loss rate follows from the difference between total and effective rainfall and is used later to check the loss model.
4. Scale the ordinates to unit depth. Divide every direct-runoff ordinate by $P_e$ expressed in the chosen unit depth. This is where the proportionality postulate enters: $$U_j = \frac{Q_{d,j}}{P_e}$$ The resulting ordinates constitute the $D$-hour unit hydrograph for that sub-area, and their integral must return exactly the unit depth — a check that should always be made.
5. Average several storms and adjust the duration if required. Average the peak discharges and the times to peak of the individual unit hydrographs rather than averaging ordinates blindly, sketch a curve of that peak and timing, and adjust it so that the volume again equals unit depth. To convert to a different duration, superpose lagged copies (the S-curve method): build the S-curve by summing unit hydrographs lagged by $D$ indefinitely, then take the difference of two S-curves offset by the new duration $D'$ and multiply by $D/D'$.

Where no gauge exists on the sub-area — the normal situation for a new development — a synthetic unit hydrograph is used instead: Snyder’s method, the SCS dimensionless unit hydrograph with $t_p = 0.6\,t_c$ and $q_p = 2.08\,A/t_p$ in SI, or Clark’s time-area method. The parameters are estimated from measurable basin characteristics (area, main-channel length, slope, land use) and, where possible, regionalised from gauged neighbours in the Water Survey of Canada HYDAT record.

The three postulates that make all of this legitimate are: constancy of the time base (all storms of the same duration produce direct runoff over the same time base, regardless of intensity); proportionality (ordinates scale linearly with effective-rainfall depth); and superposition (the hydrograph from a sequence of rainfall pulses is the sum of the appropriately lagged individual responses). Together they say that the basin is a linear, time-invariant system.

Developing the design storm hydrograph for the whole watershed. The unit hydrograph converts effective rainfall to direct runoff; a design hydrograph therefore requires a design storm and a loss model in front of it, and routing behind it.

1. Select the design storm. Choose the return period from the applicable municipal or provincial criterion, read the intensities for the required durations from the Environment and Climate Change Canada IDF curve for the nearest station (adjusted for climate change with the IDF_CC tool where the approving authority requires it), and distribute the total depth in time using an accepted hyetograph — the Chicago distribution, the SCS Type II distribution, or the AES/Atmospheric Environment Service 12-hour distribution commonly used in Canada. The storm duration should at least equal the time of concentration of the whole watershed.
2. Convert design rainfall to effective rainfall on each sub-area separately. Apply a loss model — SCS curve number, Horton or Green–Ampt infiltration, or an initial-abstraction-plus-constant-loss scheme — using parameters appropriate to each sub-area. This is the step that distinguishes the developed sub-area from the undeveloped ones: B has a high impervious fraction, so its curve number is high, its initial abstraction small and its effective rainfall much larger for the same storm, while A and C retain the higher losses of undeveloped ground.
3. Convolve each sub-area’s unit hydrograph with its own effective-rainfall sequence. The discrete convolution equation is $$Q_n \;=\; \sum_{m=1}^{n\le M} P_m\,U_{\,n-m+1}$$ where $P_m$ are the effective-rainfall pulses of duration $D$ and $U_j$ the unit-hydrograph ordinates. This produces one direct-runoff hydrograph at the outlet of each of A, B and C.
4. Route the developed sub-area’s hydrograph through the stormwater pond. Sub-area B discharges to the pond, so its hydrograph is the inflow to a level-pool reservoir routing calculation. Combine the continuity equation with the pond’s stage-storage relation (from the grading plan) and its stage-discharge relation (from the outlet structure — orifice, weir and emergency spillway in series) using the storage-indication method described under Problem 3. The pond outflow is attenuated and delayed relative to its inflow, which is the whole purpose of the facility.
5. Route each sub-area hydrograph to the common outlet and superpose. Translate and attenuate each hydrograph along its channel to the watershed outlet using Muskingum, Muskingum–Cunge or kinematic-wave routing, then add the routed hydrographs ordinate by ordinate at a common time base. The peaks must be added at the correct clock times, not aligned at their individual peaks — sub-areas with different times of concentration peak at different moments, and the attenuation supplied by the pond may shift B’s peak so that it now coincides with, rather than precedes, the peak from A and C. It is entirely possible for a pond sized only on its own outflow to make the combined peak worse, and checking for this is a required part of the analysis.
6. Add baseflow and check the result. Restore the baseflow removed in the derivation step, check that the total runoff volume equals the effective rainfall depth times the contributing area, and compare the resulting peak with a regional-frequency or rational-method estimate as an order-of-magnitude test.

A miniature numerical demonstration of step 3 makes the convolution concrete. Suppose a sub-area has the 1-hour unit-hydrograph ordinates 0, 5, 12, 8, 3, 0 m³/s per 10 mm, and that the design storm yields effective rainfall of 10 mm in the first hour and 20 mm in the second, that is, pulses of 1.0 and 2.0 units.

Given. $U = \{0,\,5,\,12,\,8,\,3,\,0\}$ m³/s per 10 mm at 1 h intervals; effective rainfall pulses $P = \{1.0,\,2.0\}$ units of 10 mm.

Find. The direct-runoff hydrograph and its peak.

1. Apply the convolution equation ordinate by ordinate. With $Q_n = \sum_m P_m U_{n-m+1}$, $$\begin{aligned} Q_1 &= 1.0(0) = 0 \\ Q_2 &= 1.0(5) + 2.0(0) = 5 \\ Q_3 &= 1.0(12) + 2.0(5) = 22 \\ Q_4 &= 1.0(8) + 2.0(12) = \boxed{32\ \text{m}^3/\text{s}} \\ Q_5 &= 1.0(3) + 2.0(8) = 19 \\ Q_6 &= 1.0(0) + 2.0(3) = 6 \end{aligned}$$ so the storm hydrograph is 0, 5, 22, 32, 19, 6, 0 m³/s and peaks at 32 m³/s in the fourth hour — later than the unit-hydrograph peak, because the heavier second pulse dominates.
2. Check the volume. The total effective depth is 30 mm, or 3.0 units, and the unit hydrograph carries 28 m³/s·h per unit; 3.0 × 28 = 84 m³/s·h, which equals the sum of the storm ordinates, confirming that no volume was lost.