--- title: "Storm hydrograph from 6-h UH, 15 cm storm" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:storm-hydrograph-6h-uh-15cm" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/storm-hydrograph-6h-uh-15cm site: BCT Engineering --- # Storm hydrograph from 6-h UH, 15 cm storm Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2079 Bhadra, 8 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/storm-hydrograph-6h-uh-15cm - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question The ordinates of 6-h UH of a basin are given as under. | Time (h) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | |---|---|---|---|---|---|---|---|---|---| | Ordinates (m$^3$/s) | 0 | 6 | 12 | 25 | 10 | 8 | 6 | 3 | 0 | Derive the storm hydrograph due to a 3-h storm with a total rainfall of 15 cm. Assume initial loss of 0.50 cm and $\Phi$-index of 1 cm/h. Take base flow of 4 m$^3$/s. ## Short answer Effective rainfall is 15-0.5-(1×3)=11.5 cm. Derive a3-h UH from the6-h UH using the S-curve method. Convolve 11.5 cm with the3-h UH and add base flow of4 m³/s. The storm hydrograph peaks at 297 m³/s at 15 h and returns to base flow at 45 h. ## Answer The storm lasts 3 h while the UH is for 6 h, so the **3-h UH is derived by the S-curve method** and then applied to the rainfall excess. #### Step 1: Rainfall excess Initial loss = 0.5 cm, then infiltration at φ = 1 cm/h during the 3-h storm. $$ ER = 15 - 0.5 - (1\times 3) = 11.5 \text{ cm} $$ #### Step 2: 3-h UH from the S-curve The S-curve at 6-h points is the running sum of the UH ordinates (0, 6, 18, 43, 53, 61, 67, 70, 70). Draw a smooth S-curve through them, read it at 3-h intervals, lag it by 3 h, and take the difference: $$ S(t) = U_6(t) + S(t-6), \qquad U_3(t) = \frac{6}{3}\,[S(t)-S(t-3)] $$ #### Step 3: Storm hydrograph $$ Q(t) = 11.5\,U_3(t) + 4 $$ | t (h) | 6-h UH | S(t) | S(t−3) | 3-h UH = 2[S(t)−S(t−3)] | 11.5 × UH | Q = +4 | |---|---|---|---|---|---|---| | 0 | 0 | 0.0 | 0.0 | 0.00 | 0.0 | 4.0 | | 3 | – | 2.4 | 0.0 | 4.75 | 54.6 | 58.6 | | 6 | 6 | 6.0 | 2.4 | 7.25 | 83.4 | 87.4 | | 9 | – | 11.0 | 6.0 | 9.95 | 114.4 | 118.4 | | 12 | 12 | 18.0 | 11.0 | 14.05 | 161.6 | 165.6 | | 15 | – | 30.7 | 18.0 | 25.48 | 293.1 | 297.1 | | 18 | 25 | 43.0 | 30.7 | 24.52 | 281.9 | 285.9 | | 21 | – | 48.7 | 43.0 | 11.35 | 130.5 | 134.5 | | 24 | 10 | 53.0 | 48.7 | 8.65 | 99.5 | 103.5 | | 27 | – | 57.3 | 53.0 | 8.51 | 97.8 | 101.8 | | 30 | 8 | 61.0 | 57.3 | 7.49 | 86.2 | 90.2 | | 33 | – | 64.4 | 61.0 | 6.71 | 77.2 | 81.2 | | 36 | 6 | 67.0 | 64.4 | 5.29 | 60.8 | 64.8 | | 39 | – | 69.0 | 67.0 | 4.00 | 46.0 | 50.0 | | 42 | 3 | 70.0 | 69.0 | 2.00 | 23.0 | 27.0 | | 45 | – | 70.0 | 70.0 | 0.00 | 0.0 | 4.0 | The S-curve becomes constant at 70 m³/s (sum of the 6-h UH ordinates). **Answer: peak of the storm hydrograph ≈ 297.1 m³/s at t = 15 h; it returns to the base flow of 4 m³/s at t = 45 h.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Rainfall excess for the 3-h storm (15 cm, initial loss 0.5 cm, phi = 1 cm/h) is: A. 11.5 cm B. 14.5 cm C. 12.0 cm D. 10.5 cm Answer: A. ER = 15 - 0.5 - 1 x 3 = 11.5 cm. (Key point: Rainfall excess) 2. Why is the S-curve method used in this problem? A. The base flow is unknown B. The storm duration (3 h) differs from the given UH duration (6 h) C. The rainfall is non-uniform D. To find the phi-index Answer: B. A 6-h UH cannot be applied directly to a 3-h storm, so a 3-h UH is derived by S-curve. (Key point: Why S-curve) 3. The ordinates of the storm hydrograph are obtained as: A. Q = 15 x U3(t) + 4 B. Q = 11.5 x U6(t) - 4 C. Q = 11.5 x U3(t) + 4 D. Q = 11.5 x U3(t) x 4 Answer: C. Multiply the 3-h UH by rainfall excess in cm and add the 4 m3/s base flow. (Key point: Storm hydrograph equation) 4. The S-curve at 6-h points is obtained by: A. Multiplying UH ordinates by 6 B. Subtracting successive UH ordinates C. Taking the average of UH ordinates D. Running sum of the 6-h UH ordinates, giving a constant 70 m3/s at the end Answer: D. S(t) = U6(t) + S(t-6), so values are 0, 6, 18, 43, 53, 61, 67, 70, 70. (Key point: S-curve construction) ## Close variants set in other exams - [Storm hydrograph from 6-h UH, 3-hr storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Astorm-hydrograph-6h-uh-3hr-storm) — Engineering Hydrology, chapter 5, asked 2× ## Related in Engineering Hydrology - [2-hr UH and flood hydrograph from storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-2h-from-4h-three-storms) — Engineering Hydrology, chapter 5 - [Flood hydrograph from 2-h UH, 5.2 cm rainfall](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-from-2h-uh-5-2cm) — Engineering Hydrology, chapter 5 - [2-h UH by S-curve from storm hydrograph](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Ascurve-2h-uh-4successive-hrs) — Engineering Hydrology, chapter 5 - [Flood hydrograph from 6-h UH, three storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-6h-three-storms-30cumec) — Engineering Hydrology, chapter 5 - [Flood hydrograph from 6-hr UH, two storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-hydrograph-3-3-5-5cm) — Engineering Hydrology, chapter 5 - [Flood hydrograph from 6-hr hydrograph, 3.3 and 5.5 cm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-hydrograph-770km2-6hr) — Engineering Hydrology, chapter 5 ## Related in other subjects - [AVL tree for 24, 12, 8, 15, 35, ...](https://www.bctengineer.com/question.md?key=semester_4%2Fdata_structure_and_algorithm%3Aavl-build-24-12-8) — Data Structure and Algorithm, chapter 6