--- title: "Clark's 2-h IUH, 110 km2" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:clark-2h-iuh-110km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/clark-2h-iuh-110km2 site: BCT Engineering --- # Clark's 2-h IUH, 110 km2 Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2076 Asoj, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/clark-2h-iuh-110km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question A drainage basin has the following characteristics: Area = 110 km$^2$, time of concentration = 18 h, storage time constant = 12 h and inter-isochrone area distribution as below: | Travel time (h) | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 | 14-16 | 16-18 | |---|---|---|---|---|---|---|---|---|---| | Inter-isochrone area (km$^2$) | 3 | 9 | 20 | 22 | 16 | 18 | 10 | 8 | 4 | Determine the Clark's 2h-IUH for this catchment. ## Short answer Inflow from strips I=2.78A_i/2; route with K=12h, C0=0.0769, C2=0.8462. The 2-h IUH (2-h UH) obtained by averaging consecutive IUH ordinates peaks at 13.85 m³/s at 16 h. ## Answer #### Data and method Area of basin $A = 110\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 12$ h, $\Delta t = 2$ h. The table has 9 strips of 2 h each (total 18 h = $t_c$). **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 $$ O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t} $$ $$ \begin{aligned} C_0 &= \frac{2}{2(12)+2} = 0.0769\\ C_2 &= \frac{2(12)-2}{2(12)+2} = 0.8462\end{aligned} $$ 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 | 0.32 | 0.16 | | 4 | 12.51 | 1.55 | 0.94 | | 6 | 27.80 | 4.42 | 2.99 | | 8 | 30.58 | 8.23 | 6.32 | | 10 | 22.24 | 11.02 | 9.63 | | 12 | 25.02 | 12.96 | 11.99 | | 14 | 13.90 | 13.96 | 13.46 | | 16 | 11.12 | 13.74 | 13.85 | | 18 | 5.56 | 12.91 | 13.32 | | 20 | 0.00 | 11.35 | 12.13 | | 22 | 0.00 | 9.60 | 10.48 | | 24 | 0.00 | 8.13 | 8.87 | | 26 | 0.00 | 6.88 | 7.50 | | 28 | 0.00 | 5.82 | 6.35 | | 30 | 0.00 | 4.92 | 5.37 | | 32 | 0.00 | 4.17 | 4.54 | | 34 | 0.00 | 3.53 | 3.85 | | 36 | 0.00 | 2.98 | 3.25 | | 38 | 0.00 | 2.52 | 2.75 | | 40 | 0.00 | 2.14 | 2.33 | | 42 | 0.00 | 1.81 | 1.97 | | 44 | 0.00 | 1.53 | 1.67 | | 46 | 0.00 | 1.29 | 1.41 | | 48 | 0.00 | 1.09 | 1.19 | | 50 | 0.00 | 0.93 | 1.01 | | 52 | 0.00 | 0.78 | 0.86 | | 54 | 0.00 | 0.66 | 0.72 | Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.8462$ times the previous one (the inflow is zero after $t = 18$ h). #### Clark's 2-h IUH (2-h unit hydrograph) The 2-h unit hydrograph is the last column of the table, obtained by averaging two successive IUH ordinates. Its peak is **13.85 m³/s** at t = 16 h; the IUH peak is 13.96 m³/s per cm at t = 14 h. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For K = 12 h and dt = 2 h, what are C0 and C2? A. 0.0909 and 0.8182 B. 0.0769 and 0.8462 C. 0.0476 and 0.9048 D. 0.1 and 0.9 Answer: B. C0 = 2/26 = 0.0769, C2 = 22/26 = 0.8462. (Key point: Coefficients) 2. What is the inflow from the 3 km^2 first strip with dt = 2 h? A. 2.78 x 3/2 = 4.17 m^3/s B. 3/2.78 = 1.08 m^3/s C. 2.78 x 3 x 2 = 16.7 m^3/s D. 3 x 12 = 36 m^3/s Answer: A. I = 2.78 A/dt = 4.17. (Key point: Strip inflows) 3. Are all strips used? A. No, only up to 12 h (K) B. No, only the first four strips C. Only strips with area above 10 km^2 D. Yes, nine 2-h strips equal t_c = 18 h Answer: D. The table spans 18 h equal to t_c. (Key point: Strips) 4. What is the peak of the 2-h UH? A. 13.96 m^3/s at 14 h B. 30.58 m^3/s at 8 h C. 13.85 m^3/s at 16 h D. 12.91 m^3/s at 18 h Answer: C. Averaging IUH ordinates 13.96 and 13.74 gives 13.85 at 16 h. (Key point: Peak of 2-h UH) ## Close variants set in other exams - [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 Engineering Hydrology - [Clark IUH, 128 km2 basin](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aiuh-128km2) — 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, 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 - [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 - [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