--- title: "Clark IUH, 180 km2 basin" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:iuh-180km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/iuh-180km2 site: BCT Engineering --- # Clark IUH, 180 km2 basin Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2081 Baisakh, 8 marks; 2076 Chaitra, 8 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/iuh-180km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question A drainage basin has the following characteristics: Area = 180 km$^2$, Storage constant = 10 hrs, Time of concentration = 8 hours. The inter isochrones area distribution are as follows: | Travel time (hr) | 0-1 | 1-2 | 2-3 | 3-4 | 4-5 | 5-6 | 6-7 | 7-8 | |---|---|---|---|---|---|---|---|---| | Inter isochrones area (km$^2$) | 14 | 38 | 28 | 36 | 30 | 20 | 10 | 6 | Determine the instantaneous unit hydrograph (IUH) for this catchment. ## Short answer Compute inflow I=2.78A_i/1h for each strip, then route through linear reservoir with K=10h, C0=0.0476, C2=0.9048. The instantaneous unit hydrograph peaks at 34.43 m³/s per cm at 7 h. ## Answer #### Data and method Area of basin $A = 182\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 10$ h, $\Delta t = 1$ h. The strips add up to 182 km² (the stated area is 180 km²); the tabulated strip areas are used as given, the 1 % difference being negligible. **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 | 38.92 | 1.85 | 0.93 | | 2 | 105.64 | 8.56 | 5.21 | | 3 | 77.84 | 16.48 | 12.52 | | 4 | 100.08 | 23.39 | 19.93 | | 5 | 83.40 | 29.90 | 26.64 | | 6 | 55.60 | 33.67 | 31.78 | | 7 | 27.80 | 34.43 | 34.05 | | 8 | 16.68 | 33.27 | 33.85 | | 9 | 0.00 | 30.90 | 32.08 | | 10 | 0.00 | 27.95 | 29.43 | | 11 | 0.00 | 25.29 | 26.62 | | 12 | 0.00 | 22.88 | 24.09 | | 13 | 0.00 | 20.70 | 21.79 | | 14 | 0.00 | 18.73 | 19.72 | | 15 | 0.00 | 16.95 | 17.84 | | 16 | 0.00 | 15.33 | 16.14 | | 17 | 0.00 | 13.87 | 14.60 | | 18 | 0.00 | 12.55 | 13.21 | | 19 | 0.00 | 11.36 | 11.95 | | 20 | 0.00 | 10.28 | 10.82 | | 21 | 0.00 | 9.30 | 9.79 | | 22 | 0.00 | 8.41 | 8.85 | | 23 | 0.00 | 7.61 | 8.01 | | 24 | 0.00 | 6.89 | 7.25 | | 25 | 0.00 | 6.23 | 6.56 | | 26 | 0.00 | 5.64 | 5.93 | | 27 | 0.00 | 5.10 | 5.37 | | 28 | 0.00 | 4.61 | 4.86 | | 29 | 0.00 | 4.17 | 4.39 | | 30 | 0.00 | 3.78 | 3.98 | | 31 | 0.00 | 3.42 | 3.60 | | 32 | 0.00 | 3.09 | 3.25 | | 33 | 0.00 | 2.80 | 2.94 | | 34 | 0.00 | 2.53 | 2.66 | | 35 | 0.00 | 2.29 | 2.41 | | 36 | 0.00 | 2.07 | 2.18 | | 37 | 0.00 | 1.87 | 1.97 | | 38 | 0.00 | 1.70 | 1.79 | 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 = **34.43 m³/s per cm** at t = 7 h. The 1-h unit hydrograph (last column) peaks at 34.05 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 and dt = 1 h, what are C0 and C2? A. C0 = 0.0909 and C2 = 0.8182 B. C0 = 0.0476 and C2 = 0.9048 C. C0 = 0.1 and C2 = 0.9 D. C0 = 0.5 and C2 = 0 Answer: B. C0 = 1/21 = 0.0476, C2 = 19/21 = 0.9048. (Key point: Routing coefficients) 2. The strip areas sum to 182 km^2 but the stated area is 180 km^2. What was done? A. Tabulated strip areas were used as given, the 1% difference being negligible B. The strips were scaled to 180 C. The last strip was dropped D. The calculation was declared impossible Answer: A. The answer used the tabulated areas as given. (Key point: Area discrepancy) 3. How is the inflow for the first strip (14 km^2, dt = 1 h) found? A. 14 / 2.78 = 5.0 m^3/s B. 14 x 10 = 140 m^3/s C. 14 / 1 = 14 m^3/s D. 2.78 x 14 / 1 = 38.9 m^3/s Answer: D. I = 2.78 A/dt = 38.92 m^3/s. 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