Question 4 of 7: Basics of hydrologic modelling, and reservoir and lake routing

Problem 4: Basics of hydrologic modelling, and reservoir and lake routing (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) — Lumped versus distributed hydrologic models (8 marks)

First difference — spatial discretisation. A lumped model treats the entire watershed as a single homogeneous unit: one rainfall input, one set of state variables, one outlet hydrograph. A distributed model divides the watershed into many computational elements — a regular grid, sub-basins, or hydrologic response units defined by soil, land cover and slope — and solves the water balance separately in each, routing the outputs from element to element through the drainage network. A semi-distributed model occupies the middle ground, lumping within sub-basins but distributing between them; HEC-HMS in its usual configuration is of this kind.

Second difference — the meaning and origin of the parameters. Lumped-model parameters are effective basin-average quantities: a single curve number, a single lag, a single storage coefficient. They have no directly measurable field counterpart at the basin scale and are obtained by calibration against an observed outlet hydrograph. Distributed-model parameters are assigned cell by cell from mapped data — soil surveys, digital elevation models, land-cover classifications — and in a fully physically based model such as MIKE SHE they are quantities that can in principle be measured in the field, such as saturated hydraulic conductivity and Manning’s n. In practice distributed parameters are still adjusted by regional multipliers, but the starting point is measurement rather than fitting.

Third difference — data demand, computational cost and the information produced. A lumped model needs a basin-average rainfall series and runs in seconds, but it produces one number per time step: the discharge at the outlet. A distributed model needs spatially resolved rainfall (radar or a dense network), terrain, soils and land cover, and its run time and calibration burden are orders of magnitude larger — but it produces internal state fields, so it can report the hydrograph at any interior point, map the areas generating runoff, and represent a change confined to part of the basin.

When each is better suited. A lumped model gives better accuracy per unit of effort on small to moderate basins, roughly below a few hundred square kilometres, that are physically homogeneous, where a good outlet flow record exists for calibration, where rainfall is spatially uniform over the basin (frontal or synoptic-scale systems), and where only the outlet hydrograph is required — the typical design-flood or reservoir-yield study. A distributed model is preferred where spatial variability genuinely controls the answer: large basins whose rainfall is never uniform, convective summer storms covering only part of the basin, snowmelt driven by elevation and aspect bands in mountainous terrain, basins with strong soil or land-use contrasts, studies of partial land-use change such as urban development or forest harvest, prediction at interior ungauged points, and any application needing the spatial pattern of flooding rather than a single hydrograph. The honest qualifier is that a distributed model only realises its advantage when the spatial input data are good; driven by a single rain gauge, it will not outperform a well-calibrated lumped model, and its extra parameters simply add uncertainty.

Part (ii) — Three key steps in reservoir or lake routing (6 marks)

Reservoir routing — also called level-pool or storage-indication routing — determines the outflow hydrograph and the storage history produced when a known inflow hydrograph passes through a reservoir whose outflow depends only on the water level. Three steps carry it out.

Step 1 — Assemble the storage–elevation–discharge relationships. From bathymetric survey or contour mapping, build the storage–elevation curve S(H) by planimetering the area within each contour and integrating upward. From the hydraulics of the outlet works, build the discharge–elevation curve O(H): a weir equation Q = C L H^3/2 for a spillway, an orifice equation Q = C_d A√(2gh) for a low-level outlet, summed over all outlets that are active at that level. Eliminating H between the two gives the single functional relation O = f(S) on which the whole method depends, usually tabulated as the storage-indication function 2S/Δt + O against O.

Step 2 — Write continuity in finite-difference form. The reservoir has no momentum equation to satisfy because the pool is treated as horizontal, so conservation of mass alone governs:

Collecting the unknowns at the end of the interval on the left gives the storage-indication form used in practice,

Step 3 — March forward in time. Beginning from the known initial storage and outflow, evaluate the right-hand side from quantities already known, enter the storage-indication table with that value to read the corresponding O₂, recover S₂, and repeat for the next interval. The routing interval must be short enough to resolve the rise of the inflow hydrograph — a common rule is Δt ≤ T_p/5 — and the results are the outflow hydrograph, the storage history plotted below, and the maximum water level, which fixes the required dam crest and freeboard.

The storage–time curve has one property worth stating explicitly because examiners test it: storage peaks at the instant the falling limb of the inflow hydrograph crosses the outflow hydrograph, that is, when I = O and hence dS/dt = 0. Since outflow is a monotonic function of storage in a level pool, the maximum outflow occurs at that same instant, which is why the peak of a reservoir outflow hydrograph always lies on the recession limb of the inflow hydrograph.

Part (iii) — Maximum storage required for a given inflow hydrograph (6 marks)

The maximum storage a reservoir must provide to hold the outflow down to an allowable release is the largest cumulative volume by which inflow exceeds that release. It is obtained either graphically from a mass curve or analytically by integrating the difference of the two hydrographs. The mass-curve (Rippl) construction plots cumulative inflow volume against time and superimposes a straight line whose slope is the allowable release; the greatest vertical departure of the mass curve above that line is the required storage. Equivalently and more directly,

taken between the two times at which the inflow hydrograph crosses the allowable release.

Given. A design storm on a watershed produces the triangular inflow hydrograph below, and the downstream channel can accept only a limited release (assumed, representative values).

Find. The maximum storage the reservoir must provide, and that storage as a fraction of the total inflow volume.

Approach. Find the two times at which the triangular inflow crosses the allowable release, integrate the excess between them by simple geometry, and compare the result with the total volume under the inflow triangle.

1. Locate the crossings. On the rising limb the inflow is I = (120/8)t = 15t, so I = 45 m³/s at $$t_1 = \frac{45}{15} = 3.0\ \text{h}$$ On the falling limb I = 7.5(24 − t), so I = 45 m³/s at t₂ = 24 − 6 = 18.0 h. Storage accumulates only between these two times.
2. Integrate the excess inflow. The region between the hydrograph and the release line is a triangle with apex at the inflow peak; taking it as two trapezoids minus the release rectangle, $$\int I\,\mathrm{d}t = \tfrac{1}{2}(45+120)(5) + \tfrac{1}{2}(120+45)(10) = 412.5 + 825 = 1237.5\ \text{m}^3/\text{s}\cdot\text{h}$$ $$S_{\max} = \left[1237.5 - (45)(15)\right](3600) = (562.5)(3600) = \boxed{2.03\times10^{6}\ \text{m}^3}$$
3. Express the storage as a fraction of the event volume. The whole inflow triangle carries $$V_{\text{in}} = \tfrac{1}{2}(120)(24)(3600) = 5.184\times10^{6}\ \text{m}^3, \qquad \frac{S_{\max}}{V_{\text{in}}} = 0.391$$ so holding the release to 37.5 per cent of the peak inflow costs storage equal to 39 per cent of the entire flood volume. That steep trade-off is the central economic fact of detention design and is worth quoting to a client who asks for a further reduction in release.

Assumptions made. Five, and each should be stated in an examination answer. First, the reservoir begins the event empty at the level of the control structure, so all of the computed volume is available; if it does not, the antecedent storage must be added. Second, the release is held at the allowable rate throughout the period of storage — a gated structure can approximate this, but an uncontrolled weir or orifice releases a rate that rises with head, which reduces the storage requirement somewhat and makes this estimate conservative. Third, the pool is level and losses over the event — seepage, evaporation, bank storage — are negligible over the day or so that matters. Fourth, there is no lateral inflow to the reach between the reservoir and the point being protected, and no downstream backwater affecting the outlet. Fifth, the inflow hydrograph used is the appropriate design event, both in return period and in shape; because the storage requirement depends on the volume under the hydrograph and not only on its peak, a long-duration event of lower peak often governs the storage even when a short intense one governs the spillway, and both must be tested.