16-Civ-B4 Engineering Hydrology · December 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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 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.
Given. The Rational formula $Q = k\,C\,i\,A$, applied to a 25 ha catchment with a runoff coefficient $C = 0.35$ and a design intensity of 60 mm/h read at the time of concentration.
Find. The meaning of each term, the peak discharge for the stated data, and the three assumptions that limit the method’s validity.
The terms. Q is the peak rate of runoff at the outlet of the catchment — not a volume and not a hydrograph, but the single maximum instantaneous discharge, in m3/s in SI use. C is the dimensionless runoff coefficient, the fraction of the rainfall rate that appears as runoff at the outlet; it bundles infiltration, interception, depression storage and surface detention into one number, and is tabulated by land use and soil group (roughly 0.15–0.30 for lawns on permeable soil, 0.70–0.95 for pavement and roofs). For a mixed catchment it is area-weighted. i is the average rainfall intensity for a storm whose duration equals the catchment’s time of concentration, at the chosen return period, read from an IDF curve. k is the unit-conversion factor, and it carries no physics: with Q in m3/s, i in mm/h and A in hectares, $k = 0.00278$ (equivalently $1/360$); with A in km2, $k = 0.278$; and in the traditional imperial form with Q in cfs, i in in/h and A in acres, $k = 1$, which is why the factor is absent from older North American textbooks. A is the contributing drainage area.
For the stated data the peak discharge is
$$\boxed{Q = 0.00278 \times 0.35 \times 60 \times 25 = 1.46\ \text{m}^3/\text{s}}$$Assumption 1 — the rainfall is uniform in space over the whole catchment and constant in time for at least the time of concentration. This is what allows one intensity to represent the storm. It is defensible on a few tens of hectares and indefensible on tens of square kilometres, where a real storm cell covers only part of the area and where an areal reduction factor would be needed. This assumption is the origin of the method’s area limit.
Assumption 2 — the peak discharge occurs when the whole catchment contributes, that is, at a storm duration equal to the time of concentration, and the return period of the peak equals the return period of the rainfall. The first half fixes which point of the IDF curve is read; the second half is a strong statement that is only approximately true, because the flood frequency also depends on antecedent moisture, which has its own distribution. On a highly impervious catchment the approximation is good; on a rural one it is optimistic.
Assumption 3 — the runoff coefficient is constant, and the rainfall–runoff relation is linear. C is treated as a fixed property, whereas in reality it rises during a storm as the soil wets and rises with the return period as the abstractions are overwhelmed — which is why many manuals apply a frequency adjustment factor to C for events beyond about 25 years. The method also assumes no significant storage in the catchment: any pond, wetland or oversized pipe attenuates the peak, and the Rational method cannot represent that at all, because it produces no hydrograph to route.
Given. A reservoir with an uncontrolled outlet, as in the printed figure, routed at $\Delta t = 1$ h = 3600 s. At the start of the step the storage is 36,000 m3 and the outflow is 5 m3/s; the inflow is 10 m3/s at the start of the step and 20 m3/s at its end. The storage–outflow curve of the reservoir returns an outflow of 9 m3/s at the storage-indication value computed below.
Find. The four procedural steps, demonstrated by routing one time step through the reservoir.
Approach. Reservoir (level-pool) routing is the storage equation closed by a single-valued storage–outflow relation, solved by the storage-indication method.
| Quantity | Basis | Result |
|---|---|---|
| Rational-method peak | 0.00278 × 0.35 × 60 × 25 | 1.46 m3/s |
| Storage indication at time 1 | 2S1/Δt + Q1 | 25 m3/s |
| Storage indication at time 2 | (I1+I2) + (2S1/Δt − Q1) | 45 m3/s |
| Outflow at time 2 | from the storage–outflow curve | 9 m3/s |
| Storage at time 2 | (45 − 9)Δt/2 | 64,800 m3 |
| Storage change (mass-balance check) | [(Ī − Q̄)]Δt | 28,800 m3 — closes |
| Lake storage constant, part (iii) | ΔS / ΔQ = 4.8 × 106 / 40 | 120,000 s = 33.3 h |
The Muskingum method of Question 4 writes the reach storage as $S = K[XI + (1-X)Q]$. A lake or a wide reservoir is the special case $X = 0$: because the water surface is essentially horizontal, the storage depends only on the outflow, $S = KQ$, and the reach behaves as a linear reservoir. Crest-segment routing exploits the one instant at which this system is easiest to interpret — the crest of the outflow hydrograph.
The fundamental relation. Storage is a maximum when it stops changing, and continuity says
$$\frac{dS}{dt} = I - Q = 0 \quad \Rightarrow \quad I = Q \ \text{at the crest}$$so the outflow peak lies exactly on the crossing point of the two hydrographs marked $I = Q$ in the printed figure. That is the single most useful fact in lake routing, and it holds for any reservoir with a single-valued storage–outflow relation, regardless of the shapes of the curves. Two consequences follow. First, the peak outflow can never exceed the peak inflow, and it is read directly off the falling limb of the inflow hydrograph — so a preliminary answer is available before any routing is done. Second, the storage at the crest is the maximum storage, which fixes the maximum water level and hence the freeboard requirement.
Estimating K from the crest segment. The method takes the segment of the two hydrographs around the crossing point and uses it to evaluate the storage constant directly. Between two states near the crest,
$$K = \frac{S_2 - S_1}{Q_2 - Q_1} = \frac{\Delta S}{\Delta Q}$$For a lake of surface area 12 km2 whose level rises 0.40 m while the outflow increases by 40 m3/s,
$$\Delta S = 12 \times 10^{6} \times 0.40 = 4.8 \times 10^{6}\ \text{m}^3, \qquad \boxed{K = \frac{4.8 \times 10^{6}}{40} = 120\,000\ \text{s} = 33.3\ \text{h}}$$Once K is known the whole routing collapses to the Muskingum recursion with $X = 0$, whose coefficients are $C_0 = C_1 = \Delta t/(2K + \Delta t)$ and $C_2 = (2K - \Delta t)/(2K + \Delta t)$, so a lake routing can be completed on a single spreadsheet column without ever tabulating a storage-indication curve. The large value of K — 33 hours against the one-to-three-hour reach travel times typical of a river — is precisely why lakes and reservoirs are such effective flood-attenuating features in a basin, and why the loss of natural lake and wetland storage raises downstream peaks more than any equivalent change in channel roughness.
Limitations to state. The method assumes a horizontal water surface, so it fails on a long narrow reservoir during a fast-rising flood, where wind setup or a genuine dynamic wave makes the surface non-level; it assumes the outlet relation is single-valued, so it does not apply to a gated structure being actively operated; and being derived from a crest segment, the fitted K is representative near that crest and should not be extrapolated to a much larger flood without checking the storage–outflow curve over the extended range.
