Watershed Runoff-Computation Procedure and Linear-Reservoir Routing: 22-Agric-B7 Principles of Hydrology, December 2016, Q3

Question 3: Watershed Runoff-Computation Procedure and Linear-Reservoir Routing (33.3 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.

Given. A watershed of seven labelled sub-watersheds (A–G) draining through four numbered channels to a single outlet (Part 3.1); a linear-reservoir storage–outflow relationship and a 12-hour inflow hydrograph (Part 3.2):

Time (hr)	0	1	2	3	4	5	6	7	8	9	10	11
Inflow (m³/s)	0	100	200	400	300	200	100	50	0	0	0	0

Reservoir constant $K=1.21\ \text{s}$ (as printed); routing interval $\Delta t=1\ \text{hr}=3600\ \text{s}$.

Find. (3.1) A procedure (steps or schematic) fixing the order of sub-watershed runoff computation, hydrograph combination and channel routing. (3.2) The routed outflow at t = 4 hr and t = 8 hr.

Approach. Part 3.1 is answered by method: work strictly from headwater sub-watersheds down to the outlet, computing, combining and routing in the sequence the channel network forces. Part 3.2 uses the storage-indication (modified Puls) routing method, exploiting the fact that $S=KO$ makes the storage-indication function linear in outflow so each step solves in closed form.

Part 3.1 — Runoff-computation procedure for the networked watershed. The figure shows a dendritic network with TWO independent headwater branches. On the upper branch, sub-watersheds F, G and D drain into Channel 1, and sub-watershed B enters where Channel 1 becomes Channel 2. On the lower branch, sub-watersheds E and C drain into Channel 3. Channels 2 and 3 meet at one confluence, where sub-watershed A also enters, and the combined flow runs down Channel 4 to the outlet. Because a downstream channel's inflow depends on everything upstream of it, the computation must proceed strictly upstream-to-downstream, one confluence at a time:
1. For EACH sub-watershed independently: from the design storm's total rainfall, subtract losses (interception, infiltration – e.g. by the SCS curve-number or Green–Ampt method) to get that sub-watershed's own rainfall excess.
2. Convolve each sub-watershed's rainfall excess with its OWN unit hydrograph (scaled to its own area and lag time) to obtain the LOCAL runoff hydrograph entering the network at that sub-watershed's own channel node.
3. Upper branch — at its head (Channel 1, fed directly by sub-watersheds F, G and D): sum the three local hydrographs, time-ordinate by time-ordinate, to get Channel 1's inflow hydrograph.
4. Route Channel 1's inflow hydrograph down its own reach to the Channel 1/Channel 2 confluence (e.g. Muskingum or kinematic-wave channel routing, using the reach's own length, slope and roughness).
5. At that confluence, ADD sub-watershed B's own local hydrograph (Step 2) to the routed Channel 1 hydrograph (Step 4) to obtain Channel 2's inflow.
6. Route Channel 2's hydrograph down its reach to the outlet confluence.
7. Lower branch, which shares no upstream node with Steps 3–6 and may therefore be computed in parallel with them: sum the local hydrographs of sub-watersheds E and C to obtain Channel 3's inflow, then route Channel 3 down its own reach to the same outlet confluence.
8. At that confluence, ADD three hydrographs at matching time ordinates — the routed Channel 2 hydrograph, the routed Channel 3 hydrograph, and sub-watershed A's own local hydrograph — to obtain Channel 4's inflow.
9. Route Channel 4's combined hydrograph to the outlet — the result is the design runoff hydrograph for the whole watershed.
The controlling rule is simply: never route or combine a hydrograph before every one of its upstream contributions has itself been computed, combined and routed to that point — the network's topology (a directed tree draining to one outlet) fixes the valid computation orders, and the two headwater branches are independent of one another until they meet above Channel 4, as the schematic below shows.

Computation order read off the source sketch: local hydrographs from sub-watersheds A–G are generated independently, then combined and routed upstream-to-downstream. Channels 1→2 and Channel 3 are independent branches that meet only at the Channel 4 confluence, where sub-watershed A also enters.

Part 3.2 — Storage-indication routing set-up. Continuity over one interval is $\tfrac{I_1+I_2}{2}\Delta t=(S_2-S_1)+\tfrac{O_1+O_2}{2}\Delta t$, rearranged into the standard storage-indication form $$I_1+I_2+\left(\frac{2S_1}{\Delta t}-O_1\right)=\left(\frac{2S_2}{\Delta t}+O_2\right)$$ Because $S=KO$ here, the right-hand "storage-indication" quantity is exactly linear in $O$: $\dfrac{2S}{\Delta t}+O=O\left(\dfrac{2K}{\Delta t}+1\right)$. With $K=1.21$ s and $\Delta t=3600$ s, $$c=\frac{2K}{\Delta t}+1=\frac{2(1.21)}{3600}+1=\boxed{1.000672}$$ so each step solves directly: $O_2=\left[I_1+I_2+X_1-2O_1\right]/c$, where $X_1=2S_1/\Delta t+O_1$ carries forward from the previous step. Starting the reservoir empty, $O_0=0$, $X_0=0$.
Part 3.2 — March the recursion forward from t = 0 to t = 8 hr. Applying $O_{n+1}=\left[I_n+I_{n+1}+X_n-2O_n\right]/c$ successively with the given hourly inflows:

t (hr) 0 1 2 3 4 5 6 7 8

Inflow, I (m³/s) 0 100 200 400 300 200 100 50 0

Outflow, O (m³/s) 0 99.9 200.0 399.9 300.2 199.9 100.2 49.8 0.2

Reading off the two requested times: $$O(t=4\ \text{hr})=\boxed{300.2\ \text{m}^3/\text{s}}$$ $$O(t=8\ \text{hr})=\boxed{0.2\ \text{m}^3/\text{s}}$$

t (hr)	0	1	2	3	4	5	6	7	8
Inflow, I (m³/s)	0	100	200	400	300	200	100	50	0
Outflow, O (m³/s)	0	99.9	200.0	399.9	300.2	199.9	100.2	49.8	0.2

Inflow hydrograph (blue) and routed outflow (red) — with K only 1.21 s against a 3600 s routing interval the two curves are visually almost identical.

Check: with $K=1.21$ s printed against a routing interval of 3600 s, the reservoir's storage $S=KO$ is only about 1/3000 of what one hour of flow would need, so the routing correctly returns essentially NO attenuation or lag — the outflow tracks the inflow almost exactly one interval later. This is solved literally as given; a source intending meaningful attenuation more plausibly meant $K=1.21$ HOURS $(=4{,}356\ \text{s})$, which would give $c=2(4{,}356)/3600+1=3.42$ and hence real attenuation and lag. Both readings use the identical method above; only $K$ (and hence $c$) would change.

Quantity	Value
Storage-indication coefficient, c	1.000672
Outflow at t = 4 hr	300.2 m³/s
Outflow at t = 8 hr	0.2 m³/s

Concept & Theory

Topic. Order-of-operations for distributed rainfall–runoff modelling over a channel network, and storage-indication (modified Puls) reservoir routing for a linear storage–outflow reservoir.

Key relations. Storage-indication continuity: $I_1+I_2+(2S_1/\Delta t-O_1)=(2S_2/\Delta t+O_2)$. Linear reservoir: $S=KO\ \Rightarrow\ 2S/\Delta t+O=O(2K/\Delta t+1)$.

Why this works. A dendritic drainage network is a directed acyclic graph rooted at the outlet; any valid computation order is simply a topological sort of that graph, which is why the rule reduces to "never process a node before all its upstream parents." Storage-indication routing rewrites the continuity + storage equations so that the ONLY unknown at each step ($O_2$) appears once, linearly, on one side — ordinarily this needs the reservoir's own storage–outflow (or elevation) curve to convert $2S/\Delta t+O_2$ back to $O_2$; the exceptional case here is a reservoir whose storage is ALREADY linear in outflow, which collapses that lookup into one algebraic division.

Common pitfall. Applying storage-indication routing without carrying $X_1=2S_1/\Delta t+O_1$ forward correctly between steps (it is NOT simply $2S_2/\Delta t+O_2$ recomputed from scratch); and, for the network procedure, combining hydrographs at a confluence before routing the upstream reach into it, or vice-versa — the reach travel time and the tributary's own hydrograph shape are both needed at the SAME point in time.

Source. Chow, Maidment & Mays, Applied Hydrology, Ch. 8 (hydrologic reservoir routing, storage-indication method) and Ch. 9 (distributed and network runoff routing).

Question 3 of 4: Watershed Runoff-Computation Procedure and Linear-Reservoir Routing

Question 3: Watershed Runoff-Computation Procedure and Linear-Reservoir Routing (33.3 marks)