--- title: "Why flow routing, time-area method of routing" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:flow-routing-time-area" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flow-routing-time-area site: BCT Engineering --- # Why flow routing, time-area method of routing Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2079 Bhadra, 6 marks - Video: [STUDENT 22 - FLOW ROUTING TECHNIQUES(HYDROLOGICAL ROUTING)](https://www.youtube.com/watch?v=Vb9koCj-ans) — Kate Lyka Mae Rosal - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flow-routing-time-area - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question Why is flow routing required? Explain Time-Area method of Routing. ## Short answer Flow routing determines outflow hydrographs for design of spillways, flood forecasting, and protection works. The Time-Area method uses isochrones to translate excess rain to the outlet, with outflow Qn = (2.78/Δt) Σ r_i A_{n-i+1}, accounting only for translation (no storage), yielding a hydrograph that is too peaked. ## Answer #### Why flow routing is required Flow routing finds the outflow hydrograph from a reach, reservoir, or catchment for a given inflow, accounting for storage. It is required for: - designing spillways, reservoir capacity, and flood-control works (the outflow peak controls the design); - flood forecasting and warning at downstream points (arrival time and peak); - designing flood protection works such as embankments, bridges, and culverts along a river; - studying how a flood changes (attenuation and lag) as it moves downstream; - obtaining a catchment hydrograph from rainfall, as in the time-area method. #### Time-area method of routing The time-area method routes excess rain over the catchment surface to the outlet, considering only **translation**. 1. Draw isochrones at interval $\Delta t$ and find the strip areas $A_1, A_2, \dots, A_N$ ($N\Delta t = t_c$). ```text Outlet *---- A1 (0 - dt) *---- A2 (dt - 2dt) *---- A3 (2dt - 3dt) ... up to tc ``` 2. Divide the effective rainfall into blocks $r_1, r_2, \dots$ of duration $\Delta t$ (cm). 3. Rain on strip $A_i$ reaches the outlet after time $i\Delta t$ at a rate $2.78\,r\,A_i/\Delta t$ (m³/s). The ordinate at time $n\Delta t$ is the sum of the contributions of all blocks: $$ Q_n = \frac{2.78}{\Delta t}\sum_{i=1}^{n} r_i\,A_{n-i+1} $$ For example, $Q_1 = 2.78\,r_1A_1/\Delta t$, $Q_2 = 2.78\,(r_1A_2 + r_2A_1)/\Delta t$. 4. Plot $Q_n$ against time for the direct runoff hydrograph; add base flow for the total flow. Since storage is neglected, the hydrograph is too peaked. Clark's method corrects this by routing the result through a linear reservoir. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is flow routing needed? A. To design spillways, flood control works and to forecast floods downstream B. To measure rainfall C. To compute evaporation D. To calibrate rain gauges Answer: A. Routing gives outflow peak, attenuation and lag, which control design and forecasting. (Key point: Why routing) 2. In time-area routing, when does rain on strip A_i reach the outlet? A. Immediately B. After time i x dt C. After tc always D. After 2 x dt Answer: B. Each strip lies i intervals of travel time from the outlet. (Key point: Strips and travel time) 3. What is Q3 in the time-area method? A. 2.78(r1A3 + r2A2 + r3A1)/dt B. 2.78(r1A1 + r2A2 + r3A3)/dt C. 2.78 r3 A3/dt D. 2.78(r1 + r2 + r3)A1/dt Answer: A. Block i uses strip A(n-i+1), so r1 pairs with A3, r2 with A2, r3 with A1. (Key point: Ordinate formula) 4. Why does the result need Clark's correction? A. Storage is neglected so the hydrograph is too peaked B. Rainfall is too small C. Strips are too large D. Baseflow is ignored Answer: A. Clark's method routes the time-area result through a linear reservoir to add the storage effect. 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