--- title: "2-hr UH and flood hydrograph from storm" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:uh-2h-from-4h-three-storms" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-2h-from-4h-three-storms site: BCT Engineering --- # 2-hr UH and flood hydrograph from storm Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2075 Asoj, 12 marks - Video: [Unit Hydrograph to Flood Hydrograph || Derivation of Flood Hydrograph from Unit Hydrograph](https://www.youtube.com/watch?v=8ZXviXtKVdk) — Engineering School - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-2h-from-4h-three-storms - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question The ordinates of 4 hr unit hydrograph are given below. | Time (hr) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 | |---|---|---|---|---|---|---|---|---|---|---|---|---|---| | 4-hr UH ordinates (m$^3$/s) | 0 | 9 | 12 | 28 | 40 | 52 | 49 | 36 | 29 | 20 | 13 | 10 | 0 | The storm has successive 2 hr, 4 hr and 6 hr rainfall of 2.5, 8.0 and 9.0 cm respectively. $\phi$-index is of 0.15 cm/hr and base flow of 40 m$^3$/s. Determine the 2 hr UH and resulting flood hydrograph from above storm. ## Short answer Using the S-curve method, the 2-h UH ordinates at 0,2,…,24 h are 0, 15, 20, 34, 46, 50.5, 42.5, 32.5, 24.5, 16.5, 11.5, 5, 0 m³/s. The resulting flood hydrograph with 40 m³/s baseflow peaks at about 726 m³/s at 14 h. ## Answer #### Part 1: 2-h UH from the 4-h UH (S-curve method) $S(t)=U_4(t)+S(t-4)$, and $U_2(t)=\dfrac{4}{2}\left[S(t)-S(t-2)\right]$ (the S-curve lagged by the new duration, 2 h, and scaled by $4/2$). The raw S-curve points zig-zag because the ordinates of the given 4-h UH are irregular. A smooth S-curve is drawn through them (3-point smoothing of the ordinates, ending at the mean equilibrium value 149 m$^3$/s). Differences are taken from the smooth curve. | t (h) | $U_4$ | Raw $S$ | Smoothed $S$ | $U_2=2\,[S(t)-S(t-2)]$ | |---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | | 2 | 9 | 9 | 7.5 | 15 | | 4 | 12 | 12 | 17.5 | 20 | | 6 | 28 | 37 | 34.5 | 34 | | 8 | 40 | 52 | 57.5 | 46 | | 10 | 52 | 89 | 82.75 | 50.5 | | 12 | 49 | 101 | 104 | 42.5 | | 14 | 36 | 125 | 120.25 | 32.5 | | 16 | 29 | 130 | 132.5 | 24.5 | | 18 | 20 | 145 | 140.75 | 16.5 | | 20 | 13 | 143 | 146.5 | 11.5 | | 22 | 10 | 155 | 149 | 5 | | 24 | 0 | 143 | 149 | 0 | (The sum of the $U_2$ ordinates is 298, equal to the sum of the $U_4$ ordinates, so the volume of 1 cm is conserved.) #### Part 2: Effective rainfall ($\phi=0.15$ cm/h) | Storm | Rain (cm) | Loss $\phi t$ (cm) | ER (cm) | ER per 2 h block | |---|---|---|---|---| | 0-2 h | 2.5 | 0.30 | 2.2 | 2.2 | | 2-6 h | 8.0 | 0.60 | 7.4 | 3.7, 3.7 | | 6-12 h | 9.0 | 0.90 | 8.1 | 2.7, 2.7, 2.7 | #### Part 3: Flood hydrograph $Q(t)=40+\sum_j P_j\,U_2(t-2j)$ with $P=(2.2,3.7,3.7,2.7,2.7,2.7)$. | t (h) | $U_2$ | 2.2 U(t) | 3.7 U(t-2) | 3.7 U(t-4) | 2.7 U(t-6) | 2.7 U(t-8) | 2.7 U(t-10) | Q (m$^3$/s) | |---|---|---|---|---|---|---|---|---| | 0 | 0 | 0.0 | 0 | 0 | 0 | 0 | 0 | 40.0 | | 2 | 15 | 33.0 | 0.0 | 0 | 0 | 0 | 0 | 73.0 | | 4 | 20 | 44.0 | 55.5 | 0.0 | 0 | 0 | 0 | 139.5 | | 6 | 34 | 74.8 | 74.0 | 55.5 | 0.0 | 0 | 0 | 244.3 | | 8 | 46 | 101.2 | 125.8 | 74.0 | 40.5 | 0.0 | 0 | 381.5 | | 10 | 50.5 | 111.1 | 170.2 | 125.8 | 54.0 | 40.5 | 0.0 | 541.6 | | 12 | 42.5 | 93.5 | 186.9 | 170.2 | 91.8 | 54.0 | 40.5 | 676.9 | | 14 | 32.5 | 71.5 | 157.2 | 186.9 | 124.2 | 91.8 | 54.0 | 725.6 | | 16 | 24.5 | 53.9 | 120.2 | 157.2 | 136.4 | 124.2 | 91.8 | 723.8 | | 18 | 16.5 | 36.3 | 90.7 | 120.2 | 114.8 | 136.4 | 124.2 | 662.5 | | 20 | 11.5 | 25.3 | 61.1 | 90.7 | 87.8 | 114.8 | 136.4 | 555.9 | | 22 | 5 | 11.0 | 42.6 | 61.1 | 66.2 | 87.8 | 114.8 | 423.2 | | 24 | 0 | 0.0 | 18.5 | 42.6 | 44.6 | 66.2 | 87.8 | 299.5 | | 26 | 0 | 0 | 0.0 | 18.5 | 31.1 | 44.6 | 66.2 | 200.2 | | 28 | 0 | 0 | 0 | 0.0 | 13.5 | 31.1 | 44.6 | 129.1 | | 30 | 0 | 0 | 0 | 0 | 0.0 | 13.5 | 31.1 | 84.5 | | 32 | 0 | 0 | 0 | 0 | 0 | 0.0 | 13.5 | 53.5 | | 34 | 0 | 0 | 0 | 0 | 0 | 0 | 0.0 | 40.0 | | 36 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 40.0 | **Answer: The 2-h UH ordinates (0, 2, ..., 24 h) are 0, 15, 20, 34, 46, 50.5, 42.5, 32.5, 24.5, 16.5, 11.5, 5, 0 m$^3$/s. The flood hydrograph, including base flow of 40 m$^3$/s, peaks at about 726 m$^3$/s at 14 h.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 2-h UH from the 4-h S-curve is: A. U2 = 2[S(t) - S(t-2)] B. U2 = 0.5[S(t) - S(t-2)] C. U2 = 2[S(t) - S(t-4)] D. U2 = S(t) - S(t-2) Answer: A. Ratio 4/2 = 2 scales the difference. (Key point: 2-h UH formula) 2. For the 6 h storm of 9.0 cm with phi = 0.15 cm/h, ER is: A. 8.1 cm B. 7.4 cm C. 9.0 cm D. 8.85 cm Answer: A. 9.0 - 0.15 x 6 = 8.1 cm, split as 2.7 cm per 2 h block. (Key point: Effective rain) 3. The 2-6 h storm of 8 cm gives effective 7.4 cm; per 2 h block this is: A. 3.7 cm, 3.7 cm B. 7.4 cm C. 2.5 cm each D. 4 cm, 3.4 cm Answer: A. Uniform intensity gives equal ER per block. (Key point: Blocks) 4. The sum of U2 ordinates equals the sum of U4 ordinates (298) because: A. 1 cm volume is conserved B. Both are peaks C. Base flow equal D. Storm equal Answer: A. Both are UHs of 1 cm over the same area. 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