16-Civ-B4 Engineering Hydrology · December 2013

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

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

Notes on this paper

Paper format. National Examinations, December 2013 — 98-Civ-B4 Engineering Hydrology. Three hours, closed book, one candidate-prepared two-sided 8½″ × 11″ aid sheet, and one approved Casio or Sharp calculator whose model must be declared. Seven problems are printed; page-1 Note 4 states that any five (5) questions constitute a complete paper and that only the first five answers appearing in the workbook are marked. Each problem carries twenty (20) points, so the examinable total is 5 × 20 = 100 points. The page-6 marking scheme breaks each problem into its sub-parts. All seven problems are solved here, because this set is a study resource rather than a timed sitting; the sub-part mark values shown below are the printed ones.

Reference texts. V. T. Chow, D. R. Maidment and L. W. Mays, Applied Hydrology (hydrologic cycle, unit hydrographs, routing, frequency analysis, infiltration); L. W. Mays, Water Resources Engineering, 3rd ed. (design application, rainfall–runoff, reservoir operation); W. Viessman and G. L. Lewis, Introduction to Hydrology, 5th ed. (measurement, areal precipitation, energy budget); V. T. Chow, Open-Channel Hydraulics (1959) (flood-wave propagation, dam-break and gradually varied unsteady flow); C. W. Fetter, Applied Hydrogeology, 4th ed. (Darcy’s law, hydraulic conductivity, recharge). Canadian practice references: Environment and Climate Change Canada / Water Survey of Canada HYDAT archive and the ECCC Engineering Climate Datasets (IDF curves and the IDF_CC climate-adjustment tool); the ISO 1100 / WMO Manual on Stream Gauging series as adopted by the Water Survey of Canada; the Canadian Dam Association Dam Safety Guidelines (inflow design flood and dam-break consequence classification); and the Transportation Association of Canada Guide to Bridge Hydraulics, 2nd ed.

Check — conventions used throughout this paper. A hydrologic year is taken as 365 days = 31 536 000 s unless a question says otherwise. Water density is 1000 kg/m³, gravitational acceleration is 9.81 m/s², and the latent heat of vaporisation of water is 2.45 MJ/kg at 20 °C. Several sub-parts ask for an explanation with an example rather than for the solution of stated data; in those cases a realistic Canadian data set is declared at the point of use and every number arising from it. Where the printed data are internally inconsistent — and Question 6(i) is such a case — the inconsistency is demonstrated arithmetically, the governing conservation requirement is stated, and the corrected reading actually used is declared, as page-1 Note 1 invites (“the candidate is urged to submit… a clear statement of any assumptions made”).

Question 2: Unit Hydrographs, Runoff 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.

(i) Three pieces of engineering information carried by the UH and the TRH (10 marks)

Given. Read from the printed figure: the unit hydrograph rises to a peak of about 400 L/s at $t \approx 0.90$ h and its base length is about 2.50 h; the total runoff hydrograph peaks at about 720 L/s at $t \approx 1.80$ h with a base length of about 5.00 h. Both curves are close enough to triangular that the area under each may be taken as one half base times peak.

Find. Three quantities the pair of curves supplies to an engineer who has to size and control the hydraulic system that receives the total runoff hydrograph.

[Figure not reproduced: Figure 2.1 — The printed unit hydrograph and total runoff hydrograph redrawn to scale. The lag between the two peaks and the ratio of the two areas are the two readings that carry the design information. See the official exam paper.]

Approach. Treat each curve as a triangle, obtain its volume, and interpret the peak, the volume ratio and the timing separately.

Volume under the unit hydrograph. With a base of 2.50 h = 9000 s and a peak of 400 L/s, $$V_{UH} = \tfrac{1}{2}\,Q_p\,t_b = \tfrac{1}{2}(400\ \text{L/s})(9000\ \text{s}) = 1.80 \times 10^{6}\ \text{L} = 1800\ \text{m}^3$$ This is the response of the catchment to one unit of effective rainfall — the catchment’s transfer function, independent of any particular storm.
Volume under the total runoff hydrograph. With a base of 5.00 h = 18 000 s and a peak of 720 L/s, $$V_{TRH} = \tfrac{1}{2}(720\ \text{L/s})(18\,000\ \text{s}) = 6.48 \times 10^{6}\ \text{L} = 6480\ \text{m}^3$$
The ratio of the two volumes is the effective rainfall of the storm. Because the unit hydrograph is by definition the response to one unit depth, $$\boxed{\frac{V_{TRH}}{V_{UH}} = \frac{6480}{1800} = 3.6\ \text{units of effective rainfall}}$$ and the peak ratio, $720/400 = 1.8$, is smaller than 3.6 — the signature of a storm spread over more than one unit-hydrograph duration, whose peaks superpose only partially.

Those three readings translate directly into the three pieces of engineering information the question asks for.

Information 1 — the design peak and hence the capacity of every hydraulic element. The TRH peak of 720 L/s is the flow the culvert, storm sewer, channel or pump station must convey, and the UH tells the engineer how that peak was produced: since the UH peak is 400 L/s per unit of effective rainfall, any storm depth can be converted into a peak by convolution without re-running a rainfall–runoff model. This is the property that makes the unit hydrograph a design tool rather than a description of one event.

Information 2 — the runoff volume, and therefore the storage needed to control the TRH. Capacity alone does not control a hydrograph; storage does. The 6480 m³ under the TRH is the water that must be either passed or held. If the receiving system can accept only 300 L/s, the volume that must be detained is the area of the TRH lying above that release rate, and the pond is sized from it. The UH makes this an explicit design loop: change the effective rainfall, reconvolve, re-read the volume above the permitted release.

Information 3 — the timing: lag, time to peak and time of concentration. The TRH peak occurs 0.9 h after the UH peak, and the UH time to peak of 0.90 h with a 2.50 h base is a direct measurement of the catchment’s response time. Timing governs whether upstream sub-catchments arrive together or in sequence, when a detention outlet must be open, how long a road crossing is overtopped, and how much warning time an operator has. It is also the quantity most changed by development: paving shortens the time to peak, sharpens the UH, and raises the peak even when the volume is unchanged.

Final results — Question 2(i)
Quantity	Basis	Result
Volume under the unit hydrograph	½ × 400 L/s × 9000 s	1800 m³
Volume under the total runoff hydrograph	½ × 720 L/s × 18 000 s	6480 m³
Effective rainfall of the storm	V_TRH / V_UH	3.6 units
Peak lag, UH to TRH	1.80 h − 0.90 h	0.90 h
Design conveyance capacity	TRH peak	720 L/s

(ii) Conceptual versus analytical hydrologic models (10 marks)

The hydrologic conceptual model. A conceptual model represents the catchment as a small set of interconnected stores — interception, soil moisture, an upper (fast) reservoir and a lower (slow) groundwater reservoir — linked by transfer functions that are physically motivated but not derived from the equations of flow. Its essential elements are: a defined structure of stores and fluxes; a small number of lumped parameters (store capacities, recession constants, split fractions) that are not measured in the field but calibrated against an observed record; a continuous water balance stepped through time; and an explicit objective function against which the calibration is judged. HBV, the Sacramento soil-moisture accounting model, and the tank models used across Canadian operational forecasting are of this type. The model is a bookkeeping device that has been taught to imitate one catchment.

The analytical hydrologic model. An analytical model starts from a governing equation with a closed-form or quasi-closed-form solution, and its parameters are physical quantities with independent meaning. Its essential elements are: a stated conservation law (mass, momentum, energy); a set of idealising assumptions strong enough to make the equation solvable (homogeneity, one-dimensionality, a simple boundary geometry); an exact solution valid within those assumptions; and parameters — hydraulic conductivity, storativity, Manning’s n, wave celerity — that can be measured directly. The Theis well-drawdown solution, Green–Ampt infiltration, the Ritter dam-break solution and the linear-reservoir hydrograph are analytical models. The model is a derivation, and it is exactly right about an idealised catchment rather than approximately right about a real one.

Four key areas of comparison
Area of comparison	Hydrologic conceptual model	Analytical hydrologic model
1. Basis and derivation	Structure invented to mimic observed behaviour; stores and transfers chosen for plausibility, not derived	Closed-form solution of a stated conservation equation under explicit idealising assumptions
2. Parameters and how they are obtained	Lumped, effective, and not directly measurable; obtained by calibration against a historical record, so they carry the errors of that record	Physically defined and independently measurable (K, S, n, celerity); no calibration record is required
3. Data demand and effort	High — needs a multi-year concurrent rainfall and streamflow record for calibration and a separate period for validation; runs cheaply once calibrated	Low — needs only the few physical parameters and the boundary conditions; can be evaluated by hand or on a spreadsheet
4. Range of valid use and failure mode	Reliable inside the range of conditions it was calibrated over; extrapolates poorly to extreme events or to a changed land use, and fails quietly (a plausible wrong number)	Reliable wherever its assumptions hold; fails visibly when they do not (homogeneity, one-dimensionality), and cannot represent a whole catchment’s continuous water balance

In practice the two are complementary rather than competing: an analytical solution supplies the process description inside one component of a conceptual model — Green–Ampt inside the infiltration store, a linear reservoir inside the baseflow store — and the conceptual framework supplies the continuous accounting the analytical solution cannot.

Topic. The unit hydrograph as a linear transfer function, and the two families of hydrologic model.

Key relations. Triangular hydrograph volume $V = \tfrac{1}{2} Q_p t_b$; convolution $Q_n = \sum_{m} P_m\,U_{n-m+1}$; effective rainfall $= V_{TRH}/V_{UH}$ when the UH is a unit response.

Why this works. The unit hydrograph rests on two assumptions — that the catchment response is proportional to the effective rainfall depth (linearity) and that it does not depend on when the rain falls (time invariance). Together these make the storm response a convolution, so a single measured response curve, scaled and shifted, reproduces the hydrograph of any storm. Reading the volume ratio backwards, as done above, simply inverts that statement: the total runoff volume divided by the unit response volume must be the number of units of effective rainfall that fell.

Common pitfall. Comparing the UH and TRH peaks and concluding that the storm delivered 1.8 units. Peaks superpose only partially when the storm lasts longer than the unit duration, so the volume ratio is the correct measure and it is larger. A second pitfall is forgetting that a unit hydrograph applies to direct runoff only — baseflow must be separated from the observed hydrograph before the UH is derived and added back after it is applied.

Source. Chow, Maidment & Mays, Applied Hydrology, Ch. 7 (unit hydrograph) and Ch. 1 (model classification); Viessman & Lewis, Introduction to Hydrology, Ch. 9.

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