--- title: "Clark's method flood hydrograph, 120 km2" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:clark-120km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/clark-120km2 site: BCT Engineering --- # Clark's method flood hydrograph, 120 km2 Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2078 Kartik, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/clark-120km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question A drainage basin has the following characteristics: Area = 120 km$^2$, time of concentration = 14 h, 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$^2$) | 3 | 10 | 20 | 26 | 18 | 18 | 13 | 8 | 4 | Compute the flood hydrograph by using Clark's method. ## Short answer For 2-h strips, inflow I=2.78A_i/2. Route using K=10h, C0=0.0909, C2=0.8182. The flood hydrograph (2-h UH from 1 cm excess) peaks at 16.74 m³/s at 16 h. ## Answer #### Data and method Area of basin $A = 120\ \text{km}^2$ (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 are used, although $t_c$ is stated as 14 h. **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 | 4.17 | 0.38 | 0.19 | | 4 | 13.90 | 1.95 | 1.17 | | 6 | 27.80 | 5.39 | 3.67 | | 8 | 36.14 | 10.22 | 7.81 | | 10 | 25.02 | 13.92 | 12.07 | | 12 | 25.02 | 15.94 | 14.93 | | 14 | 18.07 | 16.96 | 16.45 | | 16 | 11.12 | 16.53 | 16.74 | | 18 | 5.56 | 15.04 | 15.79 | | 20 | 0.00 | 12.81 | 13.93 | | 22 | 0.00 | 10.48 | 11.65 | | 24 | 0.00 | 8.58 | 9.53 | | 26 | 0.00 | 7.02 | 7.80 | | 28 | 0.00 | 5.74 | 6.38 | | 30 | 0.00 | 4.70 | 5.22 | | 32 | 0.00 | 3.84 | 4.27 | | 34 | 0.00 | 3.14 | 3.49 | | 36 | 0.00 | 2.57 | 2.86 | | 38 | 0.00 | 2.10 | 2.34 | | 40 | 0.00 | 1.72 | 1.91 | | 42 | 0.00 | 1.41 | 1.57 | | 44 | 0.00 | 1.15 | 1.28 | | 46 | 0.00 | 0.94 | 1.05 | | 48 | 0.00 | 0.77 | 0.86 | 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 **16.96 m³/s per cm** at t = 14 h, and the 2-h unit hydrograph peaks at **16.74 m³/s** at t = 16 h. Multiply by the excess rainfall (cm) for an actual storm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Though t_c is stated as 14 h, which strips are used? A. Only the first seven strips (14 h) B. All nine 2-h strips of the table (18 h) C. Only the largest strip D. Only strips above 10 km^2 Answer: B. The table extends to 18 h, so every strip is used. (Key point: Strips used) 2. What are C0 and C2 for K = 10 h, dt = 2 h? A. 0.0909 and 0.8182 B. 0.0476 and 0.9048 C. 0.1667 and 0.6667 D. 0.5 and 0.5 Answer: A. C0 = 2/22, C2 = 18/22. (Key point: Coefficients) 3. After inflow ceases at 18 h, how does the outflow behave? A. It drops to zero immediately B. It stays constant C. It rises again D. Each ordinate is C2 = 0.8182 times the previous one Answer: D. With zero inflow, O2 = C2 O1. (Key point: Recession) 4. What are the IUH and 2-h UH peaks? A. IUH 36.14 at 8 h; UH 25.02 at 10 h B. IUH 16.74 at 14 h; UH 16.96 at 16 h C. IUH 16.96 m^3/s per cm at 14 h; 2-h UH 16.74 m^3/s at 16 h D. Both peak at 18 h Answer: C. The routed IUH peaks at 14 h and the averaged UH peak is one step later. (Key point: Peaks) ## Close variants set in other exams - [Clark's method flood hydrograph, 123 km2](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aclark-123km2) — 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× ## Related in other subjects - [Time and space mean speed from 14 travel times](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Atime-space-mean-speed-14-vehicles) — Transportation Engineering II, chapter 1