16-Civ-B4 Engineering Hydrology · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examination, May 2018, 16-Civ-B4 Engineering Hydrology, three hours’ duration, closed book with one two-sided candidate-prepared aid sheet (8½″ × 11″) and one approved Casio or Sharp calculator whose model designation must be written on the first inside left-hand sheet of the work book. 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 work book will be marked; Note 5 weights each problem at twenty (20) points, so the examinable total is 5 × 20 = 100 points. All seven problems are solved here, because this set is a study resource rather than a timed sitting. Sub-part mark values below are the printed ones from the page-6 marking scheme, which on this sitting is internally consistent — every problem’s sub-parts sum to twenty, and the marks printed in the page margins agree with the scheme.
Reference texts. V. T. Chow, D. R. Maidment and L. W. Mays, Applied Hydrology (hydrologic cycle, unit hydrographs, conceptual models, routing, frequency analysis); L. W. Mays, Water Resources Engineering, 3rd ed. (urban and highway drainage, stormwater management, reservoir operation); W. Viessman and G. L. Lewis, Introduction to Hydrology, 5th ed. (precipitation and streamflow measurement, hydrologic modelling); C. W. Fetter, Applied Hydrogeology, 4th ed. (Darcy’s law and aquifer hydraulics); V. T. Chow, Open-Channel Hydraulics (1959) (flood-wave propagation and unsteady flow). Canadian practice references: Environment and Climate Change Canada Engineering Climate Datasets (short-duration rainfall and IDF curves) and the Water Survey of Canada HYDAT archive; the WMO Manual on Stream Gauging (WMO-No. 1044) and ISO 1100-2 (stage–discharge ratings); the Transportation Association of Canada Guide to Bridge Hydraulics and provincial highway drainage manuals; provincial stormwater management planning and design manuals (e.g. Ontario MOECC 2003, British Columbia Stormwater Planning Guidebook); and the Canadian Dam Association Dam Safety Guidelines (inflow design flood and reservoir routing).
Check — which numbers come from the paper and which are the solver’s. This sitting supplies numerical data in only three places: Problem 2(iii) (the 7 ha suburban development, its 30-minute time of concentration and the IDF relation), Problem 5(iii) (the printed IDF chart, from which an intensity must be read graphically), and Problem 6(iii) (the 50-year return period and the 10-year exposure). Those three answers are computed from the paper’s own data. Every other number below appears inside a short illustrative example whose inputs are stated in an explicit Given line as the solver’s own representative Canadian values; they exist to make a discussion answer concrete and checkable, and they are not exam data. A candidate who assumed different but reasonable values and carried them through consistently would receive the same 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.
Difference 1 — how space is represented, and therefore what kind of equations are solved. A lumped model treats the whole watershed as a single unit. It accepts basin-average inputs, carries a small number of basin-average state variables (a soil-moisture store, a groundwater store), and solves ordinary differential or difference equations in time alone; the unit hydrograph with a curve-number loss model is the archetype. A distributed model divides the watershed into grid cells, hillslope elements or hydrologic response units, gives each one its own inputs and its own state variables, and solves the mass and momentum equations in space as well as in time, passing water from cell to cell. The state of a lumped model is one number per store for the whole basin; the state of a distributed model is a map.
Difference 2 — the nature of the parameters, and therefore the data and calibration burden. Lumped parameters are effective quantities: a single recession constant or a single curve number that reproduces the outlet hydrograph. They have no direct field meaning and must be calibrated against an observed record at the outlet, but there are few of them and the calibration is quick and stable. Distributed parameters are nominally physical — hydraulic conductivity, Manning roughness, soil depth, leaf area index — and can in principle be assigned from soil surveys, land-cover mapping and a digital elevation model, but there are thousands of them, the field data are never available at the model’s resolution, and the calibration suffers from equifinality, where many different parameter fields reproduce the same outlet hydrograph equally well. The computational cost differs by orders of magnitude, and a distributed model also requires spatially distributed rainfall (radar or a dense gauge network) to make use of its resolution.
When each is preferred. A lumped model is preferred when the watershed is small and reasonably homogeneous; when the only data available are a rain-gauge record and an outlet discharge record; when the question is a single design flow at a single point (a culvert, a bridge, a storm sewer); when long continuous simulation or Monte Carlo work makes runtime the constraint; and in operational real-time forecasting, where speed and robustness matter more than internal realism. A distributed model is preferred when the watershed is large and heterogeneous in soil, land cover, elevation or precipitation; when the question concerns an ungauged interior point or the relative contribution of sub-areas; when the effect of a spatial change — urbanisation, forest harvesting, a wildfire, a proposed reservoir — must be evaluated, since a lumped model has no way to represent where the change is; when storms are convective and move across the basin; and when spatially distributed outputs such as flood inundation maps or soil-moisture fields are themselves the deliverable. A common and defensible compromise is the semi-distributed model: lumped sub-basins linked by channel routing, which captures the spatial pattern that matters most at a fraction of the cost.
Reservoir routing answers the question “given the flood coming in, how high will the water rise and how much will go out?” It rests on one physical statement and one geometric one. The physical statement is the continuity equation for the pool:
$$\frac{dS}{dt}=I(t)-O(t)$$The geometric statement is that the water surface of a reservoir or lake is essentially horizontal, so that the storage and the outflow are both single-valued functions of the same variable, the water-surface elevation $H$. That is what makes the problem tractable: outflow becomes a unique function of storage, $O = f(S)$, with no dependence on the inflow, and the technique is accordingly called level-pool or storage-indication routing.
The two curves that carry the geometry are obtained before any routing is done. The elevation–storage curve $S(H)$ comes from the bathymetry: contour areas are planimetered or taken from a survey and integrated vertically. The elevation–discharge curve $O(H)$ comes from the hydraulics of the outlets — for a free overflow spillway of crest length $L$, $O = C_w L H^{3/2}$, with orifice or culvert equations added for low-level outlets and gates. Writing the continuity equation over a time step $\Delta t$ with the trapezoidal rule,
$$\frac{I_1+I_2}{2}-\frac{O_1+O_2}{2}=\frac{S_2-S_1}{\Delta t}$$and collecting the unknowns at the end of the step on the left gives the working form,
$$\left(\frac{2S_2}{\Delta t}+O_2\right)=(I_1+I_2)+\left(\frac{2S_1}{\Delta t}-O_1\right)$$Everything on the right is known at the start of the step. The single storage-indication curve of $(2S/\Delta t + O)$ against $O$, built once from the two elevation curves, then converts the computed left-hand side into $O_2$; and $O_2$ read back on the discharge curve gives the elevation, and the elevation read on the storage curve gives $S_2$. Stepping forward in time produces the whole outflow hydrograph and the whole elevation record, from which the maximum pool level, the required freeboard and the peak discharge to the downstream channel all follow.
Given. A worked step, with $\Delta t = 1$ h $=3600$ s, an initial storage of $0.50 \times 10^{6}\ \text{m}^3$, an initial outflow of $5.0\ \text{m}^3/\text{s}$, and inflows of $10$ and $30\ \text{m}^3/\text{s}$ at the ends of the step; the spillway is $L = 20$ m long with $C_w = 1.8$. Find. The storage-indication value entering the step, and the discharge when the head over the crest reaches 0.80 m.
Two properties of the result are worth stating because they are frequently examined. For an uncontrolled outlet the peak outflow occurs exactly where the falling limb of the inflow hydrograph crosses the outflow hydrograph, because that is the instant at which $dS/dt = 0$ and the storage — and therefore the level — is at its maximum. And the area between the two hydrographs up to that instant is the flood storage that the reservoir has had to find, which is the volume that fixes the required freeboard.
1. Bathymetry, giving the elevation–area–capacity relation. This is the single most important input, because it is the storage that does the attenuating. It is obtained by an echo-sounding or multibeam survey of the pool tied to a levelled shoreline survey (or, for the zone above normal pool, by LiDAR or photogrammetry), and reduced to a table of surface area and cumulative volume against elevation. Its predictive value is direct: an error in the capacity curve translates one-for-one into an error in the computed peak level and outflow. It must also be re-surveyed periodically, because sedimentation progressively destroys the lower storage — a reservoir routed on its as-built curve fifty years after construction will predict an attenuation it no longer provides. Suspended-sediment and delta survey data support that judgement.
2. Outlet rating and operating information, giving the elevation–discharge relation. The routing is only as good as the assumed $O(H)$. What is needed from the field is the surveyed crest elevation, length and profile of the spillway; the invert, size and condition of every low-level outlet, gate and culvert; and, critically, a field verification of the theoretical rating by current-metering the outflow at several observed heads, because the discharge coefficient of a real spillway differs from the textbook value through approach conditions, piers, debris, trash racks and submergence by tailwater. To this must be added the operating rule for any gated outlet — a gated reservoir does not follow $O = f(S)$ at all, and the routing must be driven by the operator’s rule curve instead.
Two further items are worth collecting where the budget allows, and either would be accepted as an answer: a continuous stage record at the pool plus inflow gauging on the main tributaries, which is the only way to calibrate and verify the routing against a real event; and the wind fetch and tailwater data needed for wind set-up, wave run-up and submergence corrections, which affect the freeboard the routing is ultimately used to justify.
