--- title: "Clark's method flood hydrograph, 157 km2, K=8 h" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:clark-157km2-k8" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/clark-157km2-k8 site: BCT Engineering --- # Clark's method flood hydrograph, 157 km2, K=8 h Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2079 Bhadra, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/clark-157km2-k8 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question A drainage basin has area of 157 km$^2$, storage constant = 8h and time of concentration = 7h and has the following inter isochrone area distribution data. | Time (h) | 0-1 | 1-2 | 2-3 | 3-4 | 4-5 | 5-6 | 6-7 | |---|---|---|---|---|---|---|---| | Inter isochrone area (km$^2$) | 10 | 38 | 20 | 45 | 32 | 10 | 2 | Route the flood hydrograph. ## Short answer Clark's method: translation inflow I=2.78A_i/Δt per cm, then linear reservoir routing with C0=0.0588, C2=0.8824. For 1 cm excess rain, the IUH peak outflow is 38.40 m³/s at 6 h, and the 1-h unit hydrograph peak is 37.12 m³/s at 7 h. ## Answer #### Data and method Area of basin $A = 157\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 8$ h, $\Delta t = 1$ h. The strips add up to 157 km², equal to the given area. **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 $$ O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t} $$ $$ \begin{aligned} C_0 &= \frac{1}{2(8)+1} = 0.0588\\ C_2 &= \frac{2(8)-1}{2(8)+1} = 0.8824\end{aligned} $$ 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 $1$-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) | 1-h UH (m³/s) | |---|---|---|---| | 0 | 0.00 | 0.00 | 0.00 | | 1 | 27.80 | 1.64 | 0.82 | | 2 | 105.64 | 9.29 | 5.46 | | 3 | 55.60 | 17.68 | 13.49 | | 4 | 125.10 | 26.23 | 21.96 | | 5 | 88.96 | 35.74 | 30.99 | | 6 | 27.80 | 38.40 | 37.07 | | 7 | 5.56 | 35.85 | 37.12 | | 8 | 0.00 | 31.96 | 33.90 | | 9 | 0.00 | 28.20 | 30.08 | | 10 | 0.00 | 24.88 | 26.54 | | 11 | 0.00 | 21.95 | 23.42 | | 12 | 0.00 | 19.37 | 20.66 | | 13 | 0.00 | 17.09 | 18.23 | | 14 | 0.00 | 15.08 | 16.09 | | 15 | 0.00 | 13.31 | 14.19 | | 16 | 0.00 | 11.74 | 12.52 | | 17 | 0.00 | 10.36 | 11.05 | | 18 | 0.00 | 9.14 | 9.75 | | 19 | 0.00 | 8.07 | 8.60 | | 20 | 0.00 | 7.12 | 7.59 | | 21 | 0.00 | 6.28 | 6.70 | | 22 | 0.00 | 5.54 | 5.91 | | 23 | 0.00 | 4.89 | 5.21 | | 24 | 0.00 | 4.31 | 4.60 | | 25 | 0.00 | 3.81 | 4.06 | | 26 | 0.00 | 3.36 | 3.58 | | 27 | 0.00 | 2.96 | 3.16 | | 28 | 0.00 | 2.61 | 2.79 | | 29 | 0.00 | 2.31 | 2.46 | | 30 | 0.00 | 2.04 | 2.17 | | 31 | 0.00 | 1.80 | 1.92 | Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.8824$ times the previous one (the inflow is zero after $t = 7$ h). #### Flood hydrograph The routed outflow is the flood hydrograph for 1 cm of excess rain: IUH peak = **38.40 m³/s per cm** at t = 6 h, and 1-h unit hydrograph peak = **37.12 m³/s** at t = 7 h. Multiply the ordinates by the excess rainfall depth (cm) to get the flood for a given storm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With K = 8 h and dt = 1 h, what are C0 and C2? A. 0.0588 and 0.8824 B. 0.0625 and 0.875 C. 0.0588 and 0.9412 D. 0.1176 and 0.8824 Answer: A. C0 = 1/17 = 0.0588 and C2 = 15/17 = 0.8824. (Key point: Coefficients) 2. What is the inflow from the 38 km^2 strip? A. 105.64 m3/s B. 27.80 m3/s C. 55.60 m3/s D. 10.56 m3/s Answer: A. 2.78 x 38 / 1 = 105.64 m3/s per cm. (Key point: Inflow for second strip) 3. After t = 7 h when inflow is zero, how does the outflow change each hour? A. Falls by 0.8824 each hour in absolute terms B. Each ordinate equals 0.8824 times the previous one C. Drops to zero immediately D. Stays constant Answer: B. With zero inflow, O2 = C2 O1, an exponential recession. (Key point: Recession) 4. What is the IUH peak? A. 38.40 m3/s per cm at 6 h B. 37.12 m3/s per cm at 7 h C. 35.85 m3/s per cm at 7 h D. 26.23 m3/s per cm at 4 h Answer: A. The O column peaks at 38.40 at 6 h; 37.12 is the 1-h UH peak. (Key point: Result) ## Related in Engineering Hydrology - [Clark IUH, 157 km2 basin](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aiuh-157km2) — Engineering Hydrology, chapter 7 - [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× - [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× - [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'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, 128 km2 basin](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aiuh-128km2) — Engineering Hydrology, chapter 7