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

Question 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.

(i) Two main differences between river routing and reservoir flood routing (7 marks)

Difference 1 — the storage-outflow relation, and hence the number of parameters. In reservoir routing the water surface is horizontal, so storage depends on outflow alone: $S = f(O)$ is single-valued, and continuity by itself closes the problem. In a river reach the water surface is sloped and the slope is steeper on the rising limb than on the falling limb, so the same discharge at the downstream end corresponds to more storage on the rise than on the fall. Storage therefore depends on both ends of the reach, which the Muskingum method expresses as $$S = K\,[\,X I + (1 - X) O\,]$$ with $X$ weighting inflow against outflow ($X = 0$ recovers the reservoir case, and $X$ is typically 0.2 to 0.3 in a natural channel). River routing consequently needs two parameters where reservoir routing needs none beyond the reservoir's own survey.

Difference 2 — what happens to the wave, and hence what the answer is used for. A reservoir attenuates strongly and translates hardly at all: the peak outflow may be a small fraction of the peak inflow, and it occurs at the instant of maximum storage, where the inflow and outflow hydrographs cross. A river reach mainly translates the wave — the peak arrives later, having travelled at the flood-wave celerity — and attenuates it only modestly through the wedge storage available in the channel and on the flood plain. Reservoir routing is therefore used to size spillways and set dam freeboard; river routing is used to forecast arrival time and peak stage at a downstream community, and to combine tributary hydrographs at their confluences with the correct relative timing.

(ii) Uniformly progressive wave and reservoir fluctuation (7 marks)

A uniformly progressive wave (also called a monoclinal or translatory wave) is a flood wave that moves downstream through a channel at a constant celerity with its shape essentially unchanged, so that an observer travelling with the wave sees a steady profile. It is the idealisation of a flood moving down a long, uniform reach, and its celerity exceeds the mean water velocity — for a wide channel, $c \approx \tfrac{5}{3}\,V$ under the Manning relation — which is why a flood peak overtakes the water that was in the reach when it began.

Reservoir fluctuation is the rise and fall of a body of water whose surface remains essentially horizontal, so that the whole pool changes level together rather than a wave travelling through it. There is no translation to speak of; the storage change is a horizontal slice across the surface area, and the outflow at any moment is fixed by the level alone. Every level-pool routing computation is a description of reservoir fluctuation.

The impact of a uniformly progressive wave on reservoir storage is the point of the question, and it is the phenomenon of wedge storage. When a translatory wave enters the upper end of a reservoir or a long pool, the water surface is no longer horizontal: it is tilted upward towards the inflow, because the wave has arrived there and not yet at the dam. The storage held under that tilt is a wedge, positive on the rising limb (inflow exceeds outflow, and the extra volume is stored in the sloping upper reach) and negative on the falling limb (outflow exceeds inflow, and the wedge drains out). Three consequences follow. The total storage at a given outflow is greater during the rise than during the recession, so a plot of storage against outflow traces a loop rather than a curve, and the level-pool assumption $S = f(O)$ is violated. The peak level at the upstream end of the pool is higher than the level-pool computation predicts, which matters for backwater flooding and for freeboard on upstream dykes. And because part of the flood volume is temporarily stored in the wedge rather than passed to the outlets, the outflow peak is further delayed — a real attenuation, but one that a level-pool routing attributes to the wrong mechanism. Where the wedge is significant, the reach must be routed as a channel (Muskingum, or a full dynamic model) up to the point where the pool is genuinely horizontal, and level-pool routing applied only from there.

(iii) The law of continuity and the mean outflow (6 marks)

Given. Every hydrologic routing method — Muskingum, level-pool, Puls — rests on the same conservation statement, and the illustrative routing interval below uses the solver's own representative numbers to show how the mean outflow is extracted from it.

Find. The equation form of the law of continuity, and the mean and end-of-period outflow it predicts for the tabulated interval.

Approach. Write continuity in differential form, integrate it over the routing period with the trapezoidal rule, and rearrange for the mean outflow.

1. State the law of continuity. For a reach or a reservoir treated as a control volume, the rate of change of storage equals inflow minus outflow: $$I - O = \frac{dS}{dt}$$ This is the hydrologic (storage) equation, and it contains no hydraulics whatever — it is a statement of mass conservation and therefore holds for every routing method regardless of how each method closes the storage term.
2. Integrate over the routing period. Multiplying by $dt$ and integrating from $t_1$ to $t_2 = t_1 + \Delta t$, with the trapezoidal rule applied to both hydrographs over an interval short enough that each is nearly linear: $$\frac{I_1 + I_2}{2}\,\Delta t - \frac{O_1 + O_2}{2}\,\Delta t = S_2 - S_1 = \Delta S$$ Every term is now a volume over the interval, which is the form used in every routing table.
3. Solve for the mean outflow. The second term is by definition the mean outflow during the period, so rearranging gives it directly: $$\bar{O} = \frac{O_1 + O_2}{2} = \frac{I_1 + I_2}{2} - \frac{\Delta S}{\Delta t}$$ Substituting the tabulated values, $$\bar{O} = \frac{12.0 + 18.0}{2} - \frac{36\,000}{7200} = 15.0 - 5.0 = \boxed{10.0\ \text{m}^3\text{/s}}$$ In words: the mean outflow during the period is the mean inflow less the rate at which storage is filling. That single sentence is the whole of hydrologic routing; the methods differ only in how they supply $\Delta S$.
4. Recover the end-of-period outflow and start the next step. Since the mean is the average of the two ends, $$O_2 = 2\bar{O} - O_1 = 2(10.0) - 8.0 = 12.0\ \text{m}^3\text{/s}$$ and $O_2$ becomes $O_1$ for the following interval, so the computation marches forward. Note the signature of storage: with inflow exceeding outflow throughout, storage is still rising, so the outflow peak has not yet been reached. If instead $S_1 = 250\,000$ m³, then $S_2 = 286\,000$ m³, consistent with the assumed gain.
5. Note how each method closes the equation. Continuity alone has two unknowns, $O_2$ and $S_2$, so a second relation is always needed: level-pool routing supplies $S = f(O)$ from the reservoir curves, Muskingum supplies $S = K[XI + (1 - X)O]$ for a channel, and a dynamic model replaces the closure with the momentum equation. This is why the question can say that all the methods are based on continuity — the differences are entirely in the closure.