--- title: "Peak flow from 3-h UH, two successive storms" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:peak-flow-3h-uh-4h-3h" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/peak-flow-3h-uh-4h-3h site: BCT Engineering --- # Peak flow from 3-h UH, two successive storms Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2073 Shrawan, 10 marks - Video: [Unit Hydrograph Solved Problems | Engineering Hydrology](https://www.youtube.com/watch?v=nnuaxfeMvew) — APSEd - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/peak-flow-3h-uh-4h-3h - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question The 3 h unit hydrograph of a basin with an area of 20 km$^2$ at one hour interval are as given below: 0, 0.41, 1.38, 4, 7.72, 10.06, 9.24, 6.62, 4.57, 3.86, 2.76, 2.07, 1.38, 0.83, 0.41, 0 (m$^3$/s). If rainfall excess with intensity of 2.0 cm/h for a period of 4 h followed immediately by another 3 h storm with an intensity of 1 cm/h occurs on the basin, what is the peak flow produced by this rainfall and at what time after the commencement of rainfall would this peak flow occur? Assume baseflow is negligible. ## Short answer The 3-h UH was converted to a 1-h UH by the S-curve method, then convolved with rainfall excess: 2 cm/h for the first 4 h and 1 cm/h for the next 3 h. Peak flow is approximately 76.5 m³/s, occurring 7 h after commencement. ## Answer **Check of the UH:** $\sum U\times3600=55.31\times3600=199\,116$ m$^3$, which is 0.996 cm over 20 km$^2$, so it is a 1 cm, 3-h UH. The storm consists of 4 h at 2 cm/h followed by 3 h at 1 cm/h. These durations are not multiples of 3 h, so the 3-h UH is first converted to a **1-h UH by the S-curve method**. #### Step 1: S-curve $S(t)=U_3(t)+S(t-3)$ (response to a continuous 1 cm per 3 h = 0.333 cm/h). #### Step 2: 1-h UH $$ U_1(t)=\frac{3}{1}\left[S(t)-S(t-1)\right] $$ #### Step 3: Hourly effective rainfall Hours 1 to 4: 2 cm each; hours 5 to 7: 1 cm each. $$ Q(t)=\sum_{j=0}^{6} P_j\,U_1(t-j),\qquad P=(2,2,2,2,1,1,1)\ \text{cm} $$ | t (h) | $U_3$ | S(t) | $U_1$ | Q (m$^3$/s) | |---|---|---|---|---| | 0 | 0 | 0.00 | 0.00 | 0.00 | | 1 | 0.41 | 0.41 | 1.23 | 2.46 | | 2 | 1.38 | 1.38 | 2.91 | 8.28 | | 3 | 4 | 4.00 | 7.86 | 24.00 | | 4 | 7.72 | 8.13 | 12.39 | 48.78 | | 5 | 10.06 | 11.44 | 9.93 | 67.41 | | 6 | 9.24 | 13.24 | 5.40 | 75.30 | | 7 | 6.62 | 14.75 | 4.53 | 76.50 | | 8 | 4.57 | 16.01 | 3.78 | 70.44 | | 9 | 3.86 | 17.10 | 3.27 | 64.14 | | 10 | 2.76 | 17.51 | 1.23 | 53.34 | | 11 | 2.07 | 18.08 | 1.71 | 39.84 | | 12 | 1.38 | 18.48 | 1.20 | 28.53 | | 13 | 0.83 | 18.34 | 0.00 | 19.86 | | 14 | 0.41 | 18.49 | 0.00 | 14.10 | | 15 | 0 | 18.48 | 0.00 | 8.61 | | 16 | 0 | 18.34 | 0.00 | 4.14 | | 17 | 0 | 18.49 | 0.00 | 2.91 | | 18 | 0 | 18.48 | 0.00 | 1.20 | | 19 | 0 | 18.34 | 0.00 | 0.00 | The $U_1$ ordinates after 12 h are rounding noise and are taken as zero. The total ER is $2\times4+1\times3=11$ cm. **Answer: Peak flow $\approx 76.5$ m$^3$/s, occurring 7 h after the commencement of rainfall (at the end of the 7-h storm).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 3-h UH is converted to a 1-h UH because: A. Storm durations 4 h and 3 h are not both multiples of 3 h B. The area is small C. Flow is low D. Intensity is 2 cm/h Answer: A. 4 h is not a multiple of 3 h, so a 1-h UH is needed to represent the hourly rainfall. (Key point: Why 1-h UH) 2. U1(t) from the 3-h S-curve is: A. 3[S(t) - S(t-1)] B. (1/3)[S(t) - S(t-1)] C. 3[S(t) - S(t-3)] D. S(t) - S(t-1) Answer: A. S is for 1 cm per 3 h; scaling by 3/1 gives 1 cm per 1 h. (Key point: 1-h UH formula) 3. The hourly ER sequence P is: A. 2,2,2,2,1,1,1 cm B. 2,2,2,1,1,1 cm C. 8,3 cm D. 2,1 cm Answer: A. Four hours at 2 cm/h followed by three hours at 1 cm/h. (Key point: Hourly rainfall) 4. The peak flow and time are: A. 76.5 m^3/s at 7 h B. 75.3 m^3/s at 6 h C. 48.8 m^3/s at 4 h D. 70.4 m^3/s at 8 h Answer: A. From table, maximum Q = 76.5 at t = 7 h. 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