--- title: "2-hr UH from sketch, net rain 5 in/hr" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:uh-2hr-sketch-5in-cfs" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-2hr-sketch-5in-cfs site: BCT Engineering --- # 2-hr UH from sketch, net rain 5 in/hr Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2071 Chaitra, 6+3+5 marks - Video: [Unit Hydrograph concept-derivation and Numericals](https://www.youtube.com/watch?v=FbPqxFotoeI) — Chandrakant Jejurkar - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-2hr-sketch-5in-cfs - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question A 2-hr unit hydrograph for a basin is shown in the sketch. [Figure: 2-hr unit hydrograph, discharge (cfs) versus time (hr), t = 0 to 9. The curve rises from 0 at t = 0, passes about 200 cfs near t = 1 and 400 cfs near t = 2, peaks at 600 cfs at t = 3, then falls through about 400 cfs near t = 4 and 200 cfs near t = 5, and tails off to near zero by about t = 9. Note: in = inches; 12 inches = 1 feet; cfs = cubic feet/sec; a unit hydrograph is the response of catchment due to 1 inch effective rainfall over it.] (a) Determine the peak discharge (in cfs) for a net rain of 5.00 in./hr and a duration of 2 hr. (b) What is the total direct surface runoff (in inches) for the storm described in part (a)? (c) A different storm with a net rain of 0.50 in./hr lasts for 4 hr. What is the discharge at 8 p.m. if the rainfall started at 4 p.m.? ## Short answer For 5 in/hr for2hr, net rain is10 in, so peak discharge =10×600 =6000 cfs. Total direct runoff equals rainfall depth,10 in. For 0.5 in/hr for4 hr, net rain 2 in gives two 1-in blocks; at 8 p.m. (t=4 hr) Q =400+400=800 cfs. ## Answer A 2-hr UH gives the response to **1 inch of net rain in 2 hr** (intensity 0.5 in/hr). Ordinates are directly proportional to the net rain depth, and responses of successive blocks are added with a time lag (principle of proportionality and superposition). Read from the sketch (t, hr: Q, cfs): 0: 0; 1: 200; 2: 400; 3: 600 (peak); 4: 400; 5: 200; tail to about 0 at 9. #### (a) Peak discharge for 5.00 in/hr for 2 hr Duration is the same as the UH duration (2 hr), so just scale. $$ \begin{aligned} \text{net rain} &= 5.00 \times 2 = 10 \text{ in} \\ Q_{peak} &= 10 \times 600 = 6000 \text{ cfs} \end{aligned} $$ **Answer: 6000 cfs** (at t = 3 hr). #### (b) Total direct surface runoff Direct runoff depth equals net (effective) rain depth. $$ \text{Runoff} = 5.00 \text{ in/hr} \times 2 \text{ hr} = 10 \text{ in} $$ **Answer: 10 inches** over the basin. #### (c) Storm of 0.50 in/hr for 4 hr, discharge at 8 p.m. Net rain is 0.5 × 4 = 2 in, which is two successive 2-hr blocks of 0.5 in/hr × 2 hr = 1 in each. Each block gives one UH (multiplier 1); the second is lagged by 2 hr. Rain starts at 4 p.m., so 8 p.m. is t = 4 hr: $$ Q(4) = 1\times U(4) + 1\times U(4-2) = 400 + 400 = 800 \text{ cfs} $$ **Answer: 800 cfs at 8 p.m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The UH is for 1 in of net rain in 2 hr. For a 2-hr storm of 10 in net rain, how are ordinates found? A. Multiply every UH ordinate by 10 B. Multiply by 5 (the intensity only) C. Multiply by 2 (the duration only) D. Shift the UH by 10 hr Answer: A. Same duration, so ordinates scale directly with depth (principle of proportionality). (Key point: Proportionality) 2. What is the peak for 5 in/hr for 2 hr, given UH peak 600 cfs? A. 3000 cfs B. 6000 cfs C. 1200 cfs D. 600 cfs Answer: B. Net rain = 10 in, so Qpeak = 10 x 600 = 6000 cfs. (Key point: Peak discharge) 3. For 0.5 in/hr for 4 hr, how is the discharge at t = 4 hr obtained? A. 2 x U(4) = 800 cfs but without lag, since total is 2 in B. U(4) only = 400 cfs C. 1 x U(4) + 1 x U(2), two 1-in blocks with the second lagged by 2 hr, giving 800 cfs D. U(4) + U(4+2) = 600 cfs Answer: C. Each 2-hr block gives 1 in; the second is lagged 2 hr: 400 + 400 = 800 cfs. 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