Question 4 of 7: Channel Routing and Dam-Break Flood-Wave Behaviour

Question 4: Channel Routing and Dam-Break 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.

The paper labels the second sub-part “(iii)”; the page-6 marking scheme confirms Problem 4 has two sub-parts of 10 marks each. The printed labelling is retained below.

(i) A channel-routing method, worked example, and two limitations (10 marks)

Given. A river reach to be routed with the Muskingum method, calibrated from a past flood to $K = 2.0$ h and $X = 0.20$, routed at a time step of $\Delta t = 1.0$ h. At the start of the step the inflow is 10 m³/s, one hour later it is 30 m³/s, and the outflow at the start of the step is 10 m³/s.

Find. The routing coefficients and the outflow at the end of the step, and the two conditions under which the method should not be used.

Approach. The Muskingum method is hydrologic routing: it enforces the storage equation on the reach and closes it with a storage relation, giving an explicit three-coefficient recursion.

Channel routing answers the question “given the hydrograph entering a reach, what hydrograph leaves it?” The reach stores water on the rising limb and releases it on the falling limb, so the outflow peak is lower, later and broader than the inflow peak. The Muskingum method models this with a storage that depends on a weighted combination of inflow and outflow:

where K is the travel time through the reach and X weights inflow against outflow ($X = 0$ gives a level-pool reservoir, $X = 0.5$ gives pure translation with no attenuation; natural rivers fall in 0.1–0.3). Substituting this into the discrete continuity equation over a step gives the working recursion.

1. Form the common denominator and the three coefficients. $$D = K - KX + 0.5\,\Delta t = 2.0 - 0.4 + 0.5 = 2.10\ \text{h}$$ $$C_0 = \frac{-KX + 0.5\Delta t}{D} = \frac{0.10}{2.10} = 0.0476, \quad C_1 = \frac{KX + 0.5\Delta t}{D} = \frac{0.90}{2.10} = 0.4286, \quad C_2 = \frac{K - KX - 0.5\Delta t}{D} = \frac{1.10}{2.10} = 0.5238$$ As a check on the algebra the coefficients must sum to unity: $0.0476 + 0.4286 + 0.5238 = 1.0000$, which they do — this is simply the requirement that a steady inflow route to an equal steady outflow.
2. Check the stability window before routing. The step must satisfy $2KX \le \Delta t \le 2K(1-X)$, that is $0.8 \le 1.0 \le 3.2$ h. The chosen one-hour step is admissible; a step below 0.8 h would return a negative $C_0$ and produce a spurious initial dip in the outflow hydrograph.
3. Route one step. With $I_1 = 10$, $I_2 = 30$ and $Q_1 = 10$ m³/s, $$Q_2 = C_0 I_2 + C_1 I_1 + C_2 Q_1 = 0.0476(30) + 0.4286(10) + 0.5238(10)$$ $$\boxed{Q_2 = 1.43 + 4.29 + 5.24 = 10.95\ \text{m}^3/\text{s}}$$ The inflow tripled over the hour while the outflow rose by less than 10 percent: the reach absorbed the difference into storage, which is precisely the attenuation the method exists to represent. Repeating the recursion hour by hour, with $Q_2$ becoming the next step’s $Q_1$, produces the whole outflow hydrograph.

Limitation 1 — the parameters are not physical and must be calibrated on a past flood of similar magnitude. K and X are obtained by plotting weighted storage against weighted discharge for an observed inflow–outflow pair and choosing the X that makes the loop collapse to a line. On an ungauged reach they cannot be obtained at all without the Muskingum–Cunge variant, and even on a gauged reach they are strictly valid only near the calibration discharge, because travel time changes once the flood leaves the main channel and spreads across the floodplain.

Limitation 2 — it is a hydrologic method, so it carries no momentum equation and cannot represent backwater, steep waves, or unsteady effects. Muskingum routing knows nothing about water-surface slope, downstream boundaries, tides, tributary confluences or hydraulic structures; it cannot produce a stage, only a discharge; and it fails where the flood wave is steep or where flow reverses. Any of those conditions calls for hydraulic routing with the full Saint-Venant equations (HEC-RAS unsteady, MIKE 11) instead.

(iii) Predicting the dam-break wave, and measures for the town (10 marks)

Given. A large dam impounding a reservoir with a depth at the dam of $h_0 = 30$ m fails instantaneously. A town lies on the river bank 15 km downstream. The downstream channel is initially taken as dry, wide, horizontal and frictionless for the first-order estimate.

Find. The method of predicting the discharge, velocity and wave height along the river, a first-order estimate of the arrival time and severity at the town, and the risk-reduction measures to recommend.

Approach. Bound the problem analytically with the Ritter dam-break solution to establish the order of magnitude and the warning time, then state the modelling programme that produces the numbers a design risk assessment can actually use.

1. Establish the wave celerity from the reservoir depth. The undisturbed wave speed in still water of depth $h_0$ is $c_0 = \sqrt{g h_0}$, and the Ritter solution places the front of a dam-break wave on a dry bed at twice that speed: $$c_0 = \sqrt{9.81 \times 30} = 17.16\ \text{m/s}, \qquad \boxed{c_{\text{front}} = 2\sqrt{g h_0} = 34.31\ \text{m/s}}$$
2. Convert the celerity into a warning time for the town. Over the 15 km to the town, $$t = \frac{L}{c_{\text{front}}} = \frac{15\,000\ \text{m}}{34.31\ \text{m/s}} = 437\ \text{s} = 7.3\ \text{min}$$ This is the shortest credible arrival time — channel friction, a wet initial bed and floodplain storage all slow the front — and it is the single most important number in the risk assessment, because it decides whether evacuation is even conceivable.
3. Obtain the depth, velocity and unit discharge at the dam site. The Ritter solution is self-similar, and at the dam section itself it gives fixed fractions of the reservoir depth: $$h = \tfrac{4}{9}h_0 = 13.33\ \text{m}, \qquad v = \tfrac{2}{3}\sqrt{g h_0} = 11.44\ \text{m/s}, \qquad q = \tfrac{8}{27} h_0 \sqrt{g h_0} = 152.5\ \text{m}^3/\text{s per metre of width}$$ On a 200 m wide valley that is of order 30 000 m³/s — one to two orders of magnitude above any natural flood the town has experienced, which is why dam-break hazard is assessed separately from flood-frequency hazard.
4. Replace the idealisation with a real prediction. The analytical result assumes an instantaneous complete breach, a horizontal frictionless prismatic channel and a dry bed, none of which holds. The design assessment runs a breach-formation model (breach width, side slopes and formation time from the CDA or the Froehlich regression relations) to generate the outflow hydrograph, routes it with the full one-dimensional Saint-Venant equations through a surveyed cross-section series in HEC-RAS unsteady flow — two-dimensional where the valley opens out and the flow spreads through the town — with Manning’s n calibrated on a known flood, and outputs at every cross-section the peak discharge, peak depth and velocity, the depth–velocity product that governs building and human stability, and the time of first flooding. Sensitivity runs on breach width, formation time and roughness bracket the result, and the outputs become an inundation map with arrival-time contours.

Measures to recommend to the town. They fall in the standard order of preference. Prevent and detect: strengthen the dam-safety programme itself — surveillance, instrumentation, an emergency drawdown capability, and an automatic breach-detection and alarm link from the dam to the town, since with a 7-minute front there is no time for a human decision chain. Warn: a legislated Emergency Preparedness Plan under the CDA guidelines, with siren and mass-notification coverage of the inundation zone, signed evacuation routes leading laterally out of the valley rather than along it, annual exercises, and arrival-time contours published so residents know their own margin. Avoid the hazard: use the inundation map in the municipal official plan to prohibit new residential, institutional and critical-facility development in the high depth–velocity zone, and relocate or flood-proof the hospital, school, water treatment plant and emergency services if they lie within it. Resist what cannot be avoided: dykes and floodwalls sized on the modelled depth only where the depth–velocity product is low enough that a levee is credible, elevation of existing structures above the modelled level, protection of the bridge crossings whose loss would strand the evacuation, and anchoring of fuel tanks and hazardous storage that would otherwise turn a flood into a contamination event. Recover: mutual-aid agreements, financial protection, and a post-event plan.