--- title: "Clark IUH, 172 km2" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:iuh-172km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/iuh-172km2 site: BCT Engineering --- # Clark IUH, 172 km2 Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2074 Asoj, 4 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/iuh-172km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question A drainage basin has the following characteristics: Area = 172 km$^2$, storage constant = 10 hour, time of concentration = 8 hour. The inter-isochrone area distributions are as follows: | Travel time (hr) | 0-1 | 1-2 | 2-3 | 3-4 | 4-5 | 5-6 | 6-7 | 7-8 | |---|---|---|---|---|---|---|---|---| | Inter-isochrone area (km$^2$) | 12 | 40 | 26 | 36 | 28 | 18 | 8 | 4 | Determine the IUH for this catchment. ## Short answer Time-area inflows from inter-isochrone areas are computed as I=2.78Aᵳ/Δ t and routed through a linear reservoir with C0=0.0476, C2=0.9048. Peak IUH ordinate is 32.87 m³/s per cm at 7 h. ## Answer #### Data and method Area of basin $A = 172\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 10$ h, $\Delta t = 1$ h. The strips add up to 172 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 $$ O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t} $$ $$ \begin{aligned} C_0 &= \frac{1}{2(10)+1} = 0.0476\\ C_2 &= \frac{2(10)-1}{2(10)+1} = 0.9048\end{aligned} $$ 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 | 33.36 | 1.59 | 0.79 | | 2 | 111.20 | 8.32 | 4.95 | | 3 | 72.28 | 16.27 | 12.29 | | 4 | 100.08 | 22.92 | 19.59 | | 5 | 77.84 | 29.21 | 26.07 | | 6 | 50.04 | 32.52 | 30.87 | | 7 | 22.24 | 32.87 | 32.69 | | 8 | 11.12 | 31.32 | 32.09 | | 9 | 0.00 | 28.87 | 30.10 | | 10 | 0.00 | 26.12 | 27.50 | | 11 | 0.00 | 23.63 | 24.88 | | 12 | 0.00 | 21.38 | 22.51 | | 13 | 0.00 | 19.35 | 20.36 | | 14 | 0.00 | 17.50 | 18.42 | | 15 | 0.00 | 15.84 | 16.67 | | 16 | 0.00 | 14.33 | 15.08 | | 17 | 0.00 | 12.96 | 13.65 | | 18 | 0.00 | 11.73 | 12.35 | | 19 | 0.00 | 10.61 | 11.17 | | 20 | 0.00 | 9.60 | 10.11 | | 21 | 0.00 | 8.69 | 9.14 | | 22 | 0.00 | 7.86 | 8.27 | | 23 | 0.00 | 7.11 | 7.49 | | 24 | 0.00 | 6.43 | 6.77 | | 25 | 0.00 | 5.82 | 6.13 | | 26 | 0.00 | 5.27 | 5.54 | | 27 | 0.00 | 4.77 | 5.02 | | 28 | 0.00 | 4.31 | 4.54 | | 29 | 0.00 | 3.90 | 4.11 | | 30 | 0.00 | 3.53 | 3.71 | | 31 | 0.00 | 3.19 | 3.36 | | 32 | 0.00 | 2.89 | 3.04 | | 33 | 0.00 | 2.61 | 2.75 | | 34 | 0.00 | 2.36 | 2.49 | | 35 | 0.00 | 2.14 | 2.25 | | 36 | 0.00 | 1.94 | 2.04 | | 37 | 0.00 | 1.75 | 1.84 | | 38 | 0.00 | 1.58 | 1.67 | Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.9048$ times the previous one (the inflow is zero after $t = 8$ h). #### Result The IUH of the catchment is given by the column $O$. Peak IUH ordinate = **32.87 m³/s per cm** at t = 7 h. The 1-h unit hydrograph (last column) peaks at 32.69 m³/s at t = 7 h. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For K = 10 h, dt = 1 h, C0 and C2 are: A. 0.0909 and 0.8182 B. 0.0476 and 0.9048 C. 0.1 and 0.9 D. 0.0476 and 0.0476 Answer: B. C0 = 1/21 and C2 = 19/21. (Key point: Coefficients) 2. Inflow from the 12 km^2 first strip (dt = 1 h) is: A. 33.36 m^3/s B. 4.32 m^3/s C. 120 m^3/s D. 12 m^3/s Answer: A. 2.78 x 12/1 = 33.36. (Key point: Inflow) 3. Peak IUH ordinate is: A. 111.2 m^3/s at 2 h B. 31.32 at 8 h C. 29.21 at 5 h D. 32.87 m^3/s per cm at t = 7 h Answer: D. Routed outflow peaks at 7 h as 32.87. (Key point: Peak) ## Close variants set in other exams - [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× ## Related in Engineering Hydrology - [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 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'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 ## Related in other subjects - [Heating constant and 1-hour rating (40 kVA)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aoil-temperature-40kva-rating) — Electric Machine Design, chapter 2, asked 2× - [Surge impedance and travel time of 400 km line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Asurge-impedance-220kv-400km) — High Voltage Engineering, chapter 3