--- title: "Time-area method with histogram, drawbacks" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:time-area-method-purpose-drawbacks" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/time-area-method-purpose-drawbacks site: BCT Engineering --- # Time-area method with histogram, drawbacks Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2070 Chaitra, 1+5+2 marks - Video: [10 Common Mistakes by Engineers in Applying MSMA: Video 5- Application of the Time-Area Method JPS](https://www.youtube.com/watch?v=u6kopSGaIEM) — 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-purpose-drawbacks - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question For what purpose is the time area method used? Explain the time area method using a time area histogram of a catchment and a set of effective rainfall hydrograph over it. Comment on its drawbacks. ## Short answer Used to compute direct runoff hydrograph from excess rainfall for ungauged catchments (basis of Clark IUH). Isochrones divide catchment into strips Aᵢ; discharge Qₙ = (2.78/Δt) Σ rᵢ A₍ₙ₋ᵢ₊₁₎. Drawbacks: neglects storage, assumes constant velocity and uniform rainfall, yields overly sharp peaks. ## Answer #### Purpose The time-area method is used to compute the direct runoff hydrograph at a catchment outlet for a known effective (excess) rainfall hyetograph, especially for ungauged catchments (it is also the base of Clark's IUH). #### The method Draw isochrones (lines of equal travel time to the outlet) at interval $\Delta t$. Let the areas between them be $A_1, A_2, \dots, A_N$ (the last at $t_c$). The time-area histogram: ```text Area (km2) A4 | ____ A3 | ___| | A2 | __| | |___ A1 || | | | | +-+--+---+----+---+--> t 0 Δt 2Δt 3Δt 4Δt ``` Let the effective rainfall be in blocks $r_1, r_2, \dots, r_M$ (cm) of duration $\Delta t$. - Rain $r_1$ falling on $A_1$ reaches the outlet in the first interval, on $A_2$ in the second interval, etc. - The discharge at the end of each interval is the sum of the flows from all blocks: $$ Q_n = \frac{2.78}{\Delta t}\sum_{i=1}^{n} r_i\,A_{n-i+1} $$ So: - $Q_1 = 2.78\,r_1A_1/\Delta t$ - $Q_2 = 2.78\,(r_1A_2 + r_2A_1)/\Delta t$ - $Q_3 = 2.78\,(r_1A_3 + r_2A_2 + r_3A_1)/\Delta t$ - and so on, until all rainfall blocks and strips have been used. Here $A$ is in km², $r$ in cm and $\Delta t$ in h. Plotting $Q_n$ against $t$ gives the direct runoff hydrograph (base flow is added to get the total flood). #### Drawbacks - It considers only translation and ignores storage in channels and on the catchment, so the computed peak is too high and the hydrograph too sharp. - Constant velocity is assumed, so travel times do not change with discharge or rainfall intensity. - Drawing isochrones accurately needs a good topographic map and field knowledge. - Rainfall and losses are taken to be uniform over each strip. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the time-area method mainly used for? A. Computing the direct runoff hydrograph at the outlet for a given effective rainfall, even for ungauged catchments B. Estimating infiltration capacity of soil C. Finding the average rainfall over a basin D. Routing a flood through a reach using storage Answer: A. The method converts effective rainfall into a direct runoff hydrograph using the time-area histogram, and suits ungauged catchments. (Key point: Purpose) 2. What do the isochrones used in the method represent? A. Lines of equal rainfall depth B. Lines of equal travel time to the outlet C. Lines of equal ground elevation D. Lines of equal infiltration loss Answer: B. Isochrones are lines of equal travel time to the outlet; the areas between them form the time-area histogram. (Key point: Isochrones) 3. For Q2 in the time-area method, which combination is correct? A. 2.78(r1A1 + r2A2)/dt B. 2.78(r1A2 + r2A1)/dt C. 2.78(r1A1)/dt only D. 2.78(r2A2 + r1A1 + r3A3)/dt Answer: B. Block r1 has by now reached the outlet from strip A2 while r2 contributes from the nearest strip A1, giving r1A2 + r2A1. (Key point: Ordinate computation) 4. Why is the hydrograph computed by the time-area method too peaked? A. Losses are over-estimated B. Rainfall is assumed to vary in space C. It considers only translation and ignores storage in channels and catchment D. It uses too small a time interval Answer: C. Ignoring storage attenuation gives a too-high peak and too-sharp hydrograph. 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