--- title: "Time area method for runoff hydrograph" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:time-area-method" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/time-area-method site: BCT Engineering --- # Time area method for runoff hydrograph Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2071 Chaitra, 8 marks - Video: [How to Correctly Apply the Time-Area Method in MSMA Manual Saliran Mesra Alam Malaysia JPS Malaysia](https://www.youtube.com/watch?v=Cl62Qfb2QUM) — MSMA Channel - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/time-area-method - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question Explain in detail the time area method for estimating runoff hydrograph. ## Short answer The time-area method computes direct runoff hydrograph by convolving effective rainfall blocks with inter-isochrone areas, ignoring storage: Qn = 2.78 ∑ rᵳ Aₙ₋₊ᵳᵳ/Δ t (area in km², r in cm, Δ t in h). It overestimates peaks; Clark improved it by adding linear reservoir routing to account for storage. ## Answer The **time-area method** estimates the direct runoff hydrograph at the outlet of a catchment from effective rainfall, taking into account only the **translation** of runoff to the outlet and ignoring storage effects. #### Basic idea - Lines of equal travel time to the outlet (**isochrones**) are drawn at interval $\Delta t$: $t_1, t_2, \dots, t_c$. - The area between two consecutive isochrones is $A_1, A_2, \dots, A_N$. Rain falling in $A_1$ reaches the outlet in the first interval, in $A_2$ in the second interval, and so on. The last strip takes time $t_c$ (time of concentration). - The plot of cumulative area against time (or the histogram of strip areas) is the **time-area diagram**. #### Time-area histogram ```text Area A3 | ____ A2 | ___| |___ A1 | | | | | | | | | |___ +-+----+----+----+---> t 0 Δt 2Δt 3Δt tc ``` #### Procedure 1. Prepare the isochrone map and find the strip areas $A_1, \dots, A_N$. 2. Convert the effective rainfall hyetograph into blocks $r_1, r_2, \dots, r_M$ (cm or mm) of duration $\Delta t$. 3. The discharge at the end of interval $n$ is the sum of contributions of every rainfall block over its corresponding strip: $$ Q_n = \frac{10^4}{3600\,\Delta t}\sum_{i=1}^{n} r_i\,A_{n-i+1}\quad (A \text{ in km}^2, r \text{ in cm}) $$ i.e. $Q_n = 2.78 \sum r_i A_{n-i+1}/\Delta t$ for $r$ in cm and $\Delta t$ in h. For example, $Q_1 = 2.78\,r_1A_1/\Delta t$, $Q_2 = 2.78\,(r_1A_2 + r_2A_1)/\Delta t$, $Q_3 = 2.78\,(r_1A_3 + r_2A_2 + r_3A_1)/\Delta t$. 4. Plot $Q_n$ against time to get the direct runoff hydrograph. The calculation is a convolution of the rainfall blocks with the area strips (reversed order), and the base flow is added afterwards. #### Limitations - Storage (channel and catchment) is ignored, so peaks are overestimated and the hydrograph is too steep. - Isochrones are hard to draw accurately and the velocity is assumed constant, independent of flow depth and rainfall intensity. - Uniform rainfall over each strip is assumed. Clark improved the method by routing the result through a linear reservoir to include storage. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the time-area method consider? A. Only storage B. Only translation, ignoring storage C. Infiltration only D. Evaporation only Answer: B. Storage is ignored. (Key point: Assumption) 2. What are isochrones? A. Lines of equal travel time to the outlet B. Lines of equal rainfall C. Lines of equal elevation D. Lines of equal infiltration Answer: A. Areas between them form the time-area diagram. (Key point: Isochrones) 3. Which expression gives the discharge at end of interval 2? A. Q2 = 2.78 (r1 A1 + r2 A2)/dt B. Q2 = 2.78 r2 A2/dt C. Q2 = r1 + r2 D. Q2 = 2.78 (r1 A2 + r2 A1)/dt Answer: D. Rainfall blocks are convolved with reversed area strips. (Key point: Discharge formula) 4. A main limitation is: A. It underestimates peaks B. It needs no map C. Storage is ignored so peaks are overestimated D. It includes storage twice Answer: C. Clark added a linear reservoir to fix this. 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