Question 5 of 7: Channel or river routing and flood wave behaviour

Problem 5: Channel or river routing and flood wave behaviour (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) — Three key steps in applying a channel routing method (6 marks)

The method taken here is the Muskingum method, the standard hydrologic routing procedure for a river reach.

Step 1 — Define the reach and choose the routing interval. Divide the river into reaches short enough that the flood wave travel time through each is comparable to the routing interval, and select Δt to satisfy the stability requirement 2KX ≤ Δt ≤ 2K(1 − X); a Δt outside that band produces negative coefficients and physically impossible oscillating outflows. Identify the inflow hydrograph at the upstream end and the initial outflow at the downstream end.

Step 2 — Determine the routing parameters. Muskingum represents reach storage as the sum of a prism proportional to outflow and a wedge proportional to the difference between inflow and outflow:

where K is the travel time of the flood wave through the reach and X is a dimensionless weighting factor between 0 and 0.5. On a gauged reach both are found by trial: plotting S against [XI + (1 − X)O] for a past flood and selecting the X that collapses the loop into a straight line, whose slope is K. On an ungauged reach, K is estimated as the reach length divided by the flood-wave celerity, and X is taken near 0.2 for a natural channel (0 gives pure reservoir behaviour, 0.5 gives pure translation with no attenuation).

Step 3 — Compute the coefficients and march the routing equation. Combining the storage relation with continuity gives

with D = 2K(1 − X) + Δt, and the coefficients must sum to unity, which is the arithmetic check. Stepping through the inflow ordinates then produces the outflow hydrograph, from which the routed peak and its time of occurrence are read directly.

Given. A 60 km reach of river is routed with the following assumed, representative values.

Find. The routed outflow hydrograph, the magnitude and time of the routed peak, and hence the time for the flood to peak at the downstream town.

Approach. Check the stability band, compute the three Muskingum coefficients, apply the routing equation step by step, then read the peak and compare its timing with the inflow peak.

1. Check the routing interval against the stability band. With K = 8 h and X = 0.20, $$2KX = 3.2\ \text{h} \;\le\; \Delta t = 4\ \text{h} \;\le\; 2K(1-X) = 12.8\ \text{h}$$ so Δt = 4 h is admissible and no coefficient will be negative.
2. Compute the coefficients. With D = 12.8 + 4 = 16.8 h, $$C_0 = \frac{4 - 3.2}{16.8} = 0.0476, \qquad C_1 = \frac{4 + 3.2}{16.8} = 0.4286, \qquad C_2 = \frac{12.8 - 4}{16.8} = 0.5238$$ and the check C₀ + C₁ + C₂ = 1.000 is satisfied exactly.
3. Route the hydrograph. Applying O₂ = C₀I₂ + C₁I₁ + C₂O₁ from the initial condition gives outflows of 20.0, 21.9, 43.9, 87.7, 91.7, 75.4 and 55.4 m³/s at 0, 4, 8, 12, 16, 20 and 24 h. The routed peak is $$O_{\max} = \boxed{91.7\ \text{m}^3/\text{s at } t = 16\ \text{h}}$$ against an inflow peak of 140 m³/s at t = 8 h — an attenuation of 34.5 per cent and a translation of 8 h.
4. Answer the timing question. The flood therefore takes 8 h to travel the 60 km reach, so a peak observed at the upstream gauge at 08:00 will reach the town at about 16:00; the implied flood-wave celerity is $$c = \frac{L}{K} = \frac{60\ \text{km}}{8\ \text{h}} = 7.5\ \text{km/h} = 2.08\ \text{m/s}$$ which is the number an emergency-management agency needs, since it converts an upstream reading into a warning time.

Part (ii) — Two simplifying assumptions, and hydraulic versus hydrologic methods (7 marks)

First assumption — reach storage is a single-valued function of inflow and outflow. Hydrologic routing methods replace the true, spatially distributed storage in the reach with an algebraic expression such as the Muskingum prism-plus-wedge relation S = K[XI + (1 − X)O]. This asserts that the whole water-surface profile through the reach is determined by just two numbers, the discharges at its two ends. It is a good approximation while the wave is long and gentle relative to the reach, and it breaks down where the profile is controlled by something else — a downstream lake, a tidal boundary, a bridge constriction or a confluence producing backwater.

Second assumption — the routing parameters are constant through the flood. K and X are treated as fixed, which requires the reach to be prismatic and its geometry and roughness to be stage-invariant, with no significant lateral inflow along it. In reality the flood-wave celerity changes markedly once flow spills onto a floodplain, where the effective K may double, so a single parameter pair calibrated on a moderate flood misrepresents an extreme one. Variable-parameter forms such as Muskingum–Cunge, which recompute K and X from the channel geometry and the discharge at each step, exist precisely to relax this assumption.

The main difference between hydraulic and hydrologic methods. Hydraulic routing solves both governing equations of unsteady open-channel flow — the Saint-Venant system, comprising continuity and momentum:

as partial differential equations in distance and time, numerically, over a surveyed set of cross-sections. It therefore produces depth and velocity everywhere at every instant, reproduces backwater, looped ratings, tidal and structure effects, and can represent subcritical and supercritical transitions. Its costs are cross-section survey data, roughness calibration, computational effort and the possibility of numerical instability. HEC-RAS in unsteady mode and MIKE 11 are the familiar implementations, and the kinematic-wave, diffusion-wave and dynamic-wave models are successive truncations of the momentum equation.

Hydrologic routing, by contrast, discards the momentum equation entirely and retains only lumped continuity, dS/dt = I − O, closed by an empirical storage–discharge relation. It treats the whole reach as a single element and produces only the outflow hydrograph at its downstream end. It is cheap, robust, needs no cross-section survey, and is entirely adequate for the great majority of design-flood work — but it cannot represent backwater, cannot give a stage anywhere, and cannot be used where the flow is controlled from downstream.

Part (iii) — Continuity plus level-pool routing as a model of flood routing (7 marks)

Lumped flood-wave routing rests on exactly two ingredients: the continuity equation, which is exact, and a storage–discharge functional relationship, which is not. Level-pool routing is the special case in which that relationship is single-valued — the outflow depends on the storage alone, O = f(S) — and it is the cleanest illustration of how the pair works together.

Continuity for the routing element states that the rate of change of stored water equals inflow minus outflow, which written over one interval and with the time derivative replaced by a central difference gives

This single equation contains two unknowns at the end of the step, S₂ and O₂, so it cannot be solved alone; the storage–discharge relation supplies the second equation and closes the system. Rearranging so that both unknowns appear in one combined variable yields the storage-indication form,

whose right-hand side is entirely known at the start of the step. Because a level pool has a unique S(H) from the bathymetry and a unique O(H) from the outlet hydraulics, the quantity 2S/Δt + O can be tabulated once against O; each step is then a table look-up rather than an iteration. Marching from the initial condition to the end of the inflow hydrograph produces the outflow hydrograph, the storage history and the maximum pool level.

Applied to a river reach rather than a reservoir, the same machinery becomes Muskingum routing, with the wedge term KX(I − O) added to the prism term to acknowledge that a river’s water surface is not horizontal and that its storage therefore depends on inflow as well as outflow. Setting X = 0 in the Muskingum relation recovers level-pool routing exactly, which is the useful way to remember the connection: a reservoir is a reach whose wedge storage has vanished.

Assumptions in the procedure. The water surface in the routing element is horizontal, so storage is a function of one level only; the outflow is a unique single-valued function of that storage, which requires the outlet to be free-discharging and not submerged by a downstream tailwater; the inflow varies linearly within each interval, which is what justifies the trapezoidal average (I₁ + I₂)/2; the routing interval is short enough to resolve the rising limb, conventionally Δt ≤ T_p/5; and losses within the element — seepage, evaporation, bank storage — together with any lateral inflow along it are negligible over the event. Where any of these fails, in particular where a downstream backwater makes the stage–discharge relation looped, a hydraulic (Saint-Venant) solution is required instead.