--- title: "Clark's method flood hydrograph, 123 km2" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:clark-123km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/clark-123km2 site: BCT Engineering --- # Clark's method flood hydrograph, 123 km2 Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2079 Baisakh, 6 marks; 2075 Chaitra, 6 marks (asked 2 times) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/clark-123km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question A drainage basin has the following characteristics: Area = 123 km$^2$, time of concentration = 14 hr, storage constant = 10 h and inter-isochrone area distribution as below: | Travel time (hr) | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 | 14-16 | 16-18 | |---|---|---|---|---|---|---|---|---|---| | Inter-isochrone area (km$^2$) | 4 | 10 | 21 | 24 | 18 | 20 | 12 | 9 | 5 | Compute the flood hydrograph by using Clark's method. ## Short answer Translation inflow I = 2.78×A_i/Δt. Routing with linear reservoir (K=10 h, Δt=2 h, C0=0.0909, C2=0.8182) gives IUH peaking at 17.20 m³/s/cm at 14 h; 2-h unit hydrograph peaks at 16.96 m³/s at 16 h. Multiply ordinates by excess rain for flood hydrograph. ## Answer #### Data and method Area of basin $A = 123\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 10$ h, $\Delta t = 2$ h. The table has 9 strips of 2 h each (total 18 h), and all strips are used. Although $t_c$ is given as 14 h, the table extends to 18 h, so every strip of the table is used. **Step 1: Translation (time-area) inflow.** For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is $$ I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm} $$ **Step 2: Storage (linear reservoir) routing.** With $S = KQ$ and the continuity equation, the outflow is $$ O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t} $$ $$ \begin{aligned} C_0 &= \frac{2}{2(10)+2} = 0.0909\\ C_2 &= \frac{2(10)-2}{2(10)+2} = 0.8182\end{aligned} $$ Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1). #### Computation Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $2$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$. | t (h) | Inflow I (m³/s) | IUH ordinate O (m³/s/cm) | 2-h UH (m³/s) | |---|---|---|---| | 0 | 0.00 | 0.00 | 0.00 | | 2 | 5.56 | 0.51 | 0.25 | | 4 | 13.90 | 2.18 | 1.34 | | 6 | 29.19 | 5.70 | 3.94 | | 8 | 33.36 | 10.35 | 8.03 | | 10 | 25.02 | 13.78 | 12.07 | | 12 | 27.80 | 16.07 | 14.93 | | 14 | 16.68 | 17.20 | 16.63 | | 16 | 12.51 | 16.72 | 16.96 | | 18 | 6.95 | 15.45 | 16.09 | | 20 | 0.00 | 13.27 | 14.36 | | 22 | 0.00 | 10.86 | 12.07 | | 24 | 0.00 | 8.89 | 9.87 | | 26 | 0.00 | 7.27 | 8.08 | | 28 | 0.00 | 5.95 | 6.61 | | 30 | 0.00 | 4.87 | 5.41 | | 32 | 0.00 | 3.98 | 4.42 | | 34 | 0.00 | 3.26 | 3.62 | | 36 | 0.00 | 2.67 | 2.96 | | 38 | 0.00 | 2.18 | 2.42 | | 40 | 0.00 | 1.78 | 1.98 | | 42 | 0.00 | 1.46 | 1.62 | | 44 | 0.00 | 1.19 | 1.33 | | 46 | 0.00 | 0.98 | 1.09 | | 48 | 0.00 | 0.80 | 0.89 | Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.8182$ times the previous one (the inflow is zero after $t = 18$ h). #### Flood hydrograph The tabulated outflow is the flood hydrograph produced by 1 cm of excess rain: the IUH (column $O$) and the 2-h unit hydrograph (last column). The IUH peaks at **17.20 m³/s per cm** at t = 14 h, and the 2-h unit hydrograph peaks at **16.96 m³/s** at t = 16 h. For any storm, multiply the ordinates by the excess rainfall depth in cm (and superpose for successive blocks). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the translation inflow for each strip computed in Clark's method? A. I = A_i x dt B. I = 2.78 A_i / dt, for 1 cm of excess rain, A in km^2, dt in h C. I = 2.78 x A_i x dt D. I = A / K Answer: B. For 1 cm over area A_i in time dt, I = 2.78 A_i/dt m^3/s. (Key point: Inflow per strip) 2. For K = 10 h and dt = 2 h, what are C0 and C2? A. C0 = 0.0909 and C2 = 0.8182 B. C0 = 0.1 and C2 = 0.9 C. C0 = 0.0476 and C2 = 0.9048 D. C0 = 0.2 and C2 = 0.6 Answer: A. C0 = dt/(2K+dt) = 2/22 = 0.0909 and C2 = (2K-dt)/(2K+dt) = 18/22 = 0.8182. (Key point: Routing coefficients) 3. Which equation routes the translation hydrograph through the linear reservoir? A. O2 = C0 I2 + C1 I1 + C2 O1 with x = 0.2 B. O2 = I2 - K C. O2 = (I1 + I2)/2 D. O2 = C0 (I1 + I2) + C2 O1 Answer: D. The linear reservoir S = KO gives O2 = C0(I1 + I2) + C2 O1, and C0 + C0 + C2 = 1. (Key point: Routing equation) 4. How is the 2-h unit hydrograph obtained from the routed ordinates? A. Multiply IUH ordinates by 2 B. Take the IUH ordinate at twice the time C. Average two successive IUH ordinates, (O_n + O_n-1)/2 D. Subtract successive ordinates Answer: C. The dt-hour UH is the average of two successive IUH ordinates. (Key point: Unit hydrograph from IUH) ## Close variants set in other exams - [Clark's method flood hydrograph, 120 km2](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aclark-120km2) — Engineering Hydrology, chapter 7, asked 3× ## Related in Engineering Hydrology - [Clark's method flood hydrograph, 157 km2, K=8 h](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aclark-157km2-k8) — Engineering Hydrology, chapter 7 - [Clark IUH, 180 km2 basin](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aiuh-180km2) — Engineering Hydrology, chapter 7, asked 2× - [Clark IUH, 128 km2 basin](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aiuh-128km2) — Engineering Hydrology, chapter 7 - [Clark's 2-h IUH, 110 km2](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aclark-2h-iuh-110km2) — Engineering Hydrology, chapter 7, asked 2× - [Clark IUH, 157 km2 basin](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aiuh-157km2) — Engineering Hydrology, chapter 7 - [Clark IUH, 172 km2](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aiuh-172km2) — Engineering Hydrology, chapter 7, asked 2×