--- title: "Clark IUH, 157 km2 basin" subject: "Engineering Hydrology" chapter: 7 key: "semester_5/engineering_hydrology:iuh-157km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/iuh-157km2 site: BCT Engineering --- # Clark IUH, 157 km2 basin Engineering Hydrology · Chapter 7: Flow Routing - Asked in: 2068 Chaitra, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/iuh-157km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7 ## Question A drainage basin has area = 157 km$^2$, storage constant K = 9.5 h and time of concentration = 7 h. The following isochrones area distribution data are available. Determine the IUH of this catchment. | 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 | ## Short answer For each strip compute time–area inflow Iᵢ = 2.78 Aᵢ/Δt (m³/s per cm). Route through linear reservoir with K=9.5 h, Δt=1 h, C₀=0.05, C₂=0.9. Peak IUH ordinate is 33.86 m³/s per cm at t=6 h. ## Answer #### Data and method Area of basin $A = 157\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 9.5$ 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 $$ O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t} $$ $$ \begin{aligned} C_0 &= \frac{1}{2(9.5)+1} = 0.0500\\ C_2 &= \frac{2(9.5)-1}{2(9.5)+1} = 0.9000\end{aligned} $$ 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 | 1.39 | 0.69 | | 2 | 105.64 | 7.92 | 4.66 | | 3 | 55.60 | 15.19 | 11.56 | | 4 | 125.10 | 22.71 | 18.95 | | 5 | 88.96 | 31.14 | 26.92 | | 6 | 27.80 | 33.86 | 32.50 | | 7 | 5.56 | 32.15 | 33.01 | | 8 | 0.00 | 29.21 | 30.68 | | 9 | 0.00 | 26.29 | 27.75 | | 10 | 0.00 | 23.66 | 24.97 | | 11 | 0.00 | 21.29 | 22.48 | | 12 | 0.00 | 19.16 | 20.23 | | 13 | 0.00 | 17.25 | 18.21 | | 14 | 0.00 | 15.52 | 16.39 | | 15 | 0.00 | 13.97 | 14.75 | | 16 | 0.00 | 12.57 | 13.27 | | 17 | 0.00 | 11.32 | 11.95 | | 18 | 0.00 | 10.18 | 10.75 | | 19 | 0.00 | 9.17 | 9.68 | | 20 | 0.00 | 8.25 | 8.71 | | 21 | 0.00 | 7.42 | 7.84 | | 22 | 0.00 | 6.68 | 7.05 | | 23 | 0.00 | 6.01 | 6.35 | | 24 | 0.00 | 5.41 | 5.71 | | 25 | 0.00 | 4.87 | 5.14 | | 26 | 0.00 | 4.38 | 4.63 | | 27 | 0.00 | 3.95 | 4.16 | | 28 | 0.00 | 3.55 | 3.75 | | 29 | 0.00 | 3.20 | 3.37 | | 30 | 0.00 | 2.88 | 3.04 | | 31 | 0.00 | 2.59 | 2.73 | | 32 | 0.00 | 2.33 | 2.46 | | 33 | 0.00 | 2.10 | 2.21 | | 34 | 0.00 | 1.89 | 1.99 | | 35 | 0.00 | 1.70 | 1.79 | | 36 | 0.00 | 1.53 | 1.61 | Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.9000$ times the previous one (the inflow is zero after $t = 7$ h). #### Result The IUH of the catchment is given by the column $O$. Peak IUH ordinate = **33.86 m³/s per cm** at t = 6 h. The 1-h unit hydrograph (last column) peaks at 33.01 m³/s at t = 7 h. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With K = 9.5 h and dt = 1 h, what are C0 and C2? A. C0 = 0.0526, C2 = 0.8947 B. C0 = 0.0500, C2 = 0.9000 C. C0 = 0.1000, C2 = 0.8000 D. C0 = 0.0500, C2 = 0.9500 Answer: B. C0 = 1/(2*9.5+1) = 0.05 and C2 = 18/20 = 0.90; they satisfy 2C0 + C2 = 1. (Key point: Routing coefficients) 2. What is the inflow for the first strip of 10 km^2 with dt = 1 h? A. 27.8 m3/s B. 2.78 m3/s C. 278 m3/s D. 10 m3/s Answer: A. I = 2.78 x 10 / 1 = 27.8 m3/s per cm. (Key point: Inflow ordinates) 3. What check should the routing coefficients satisfy? A. C0 = C2 B. C0 + C0 + C2 = 1 C. C0 - C2 = 1 D. C2 = 2 C0 Answer: B. Since C0 applies to both I1 and I2, 2C0 + C2 must equal 1. (Key point: Check on coefficients) 4. Where does the IUH ordinate peak, and what is it? A. 33.86 m3/s per cm at t = 6 h B. 33.01 m3/s per cm at t = 7 h C. 125.10 m3/s per cm at t = 4 h D. 31.14 m3/s per cm at t = 5 h Answer: A. The outflow column O peaks at 33.86 at 6 h; 33.01 at 7 h is the 1-h UH peak. 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