--- title: "Direct runoff hydrograph analysis, 4250-acre basin" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:trapezoid-dro-4250-acre" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/trapezoid-dro-4250-acre site: BCT Engineering --- # Direct runoff hydrograph analysis, 4250-acre basin Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2072 Kartik, 4+3+3+4 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/trapezoid-dro-4250-acre - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question A hydrograph for a 4,250-acre basin is shown in the accompanying sketch. The given hydrograph actually appeared as a direct runoff hydrograph from the basin, caused by net rain falling at an intensity of 0.20 in./hr for a duration of 5 hr, beginning at t = 0. [Figure: direct surface runoff rate Q (cfs) versus time t (hr), t = 0 to 8. The rising limb passes about 200 cfs near t = 1 and about 700 cfs near t = 3, then the hydrograph flattens to a plateau at the peak $Q_p$ between t = 3 and t = 5. The falling limb passes about 400 cfs near t = 6 and about 200 cfs near t = 7 and reaches zero at t = 8.] Note: in = inches; cfs = cubic feet/sec; 12 inches = 1 foot; 1 acre = 43560 sq.ft. (a) Determine the excess release time of the basin. (b) What percentage of the drainage basin was contributing to direct runoff 4 hr after rain began (t = 4)? (c) Use your response to part (b) to determine $Q_p$ as shown in the sketch. Do not scale $Q_p$ from the drawing. (d) Note that rain continued to fall between t = 3 and t = 5. Why did the hydrograph form a plateau between t = 3 and t = 5, rather than continue to rise during these 2 hours? ## Short answer Excess release time is 3 h, the time of concentration. At t=4 h, 100% of the basin contributes. Peak discharge Qp = (0.20/12) ft/hr × 4250 acre × 43560 ft²/acre ÷ 3600 ≈ 857 cfs. The plateau occurs because after t_c = 3 h the entire basin contributes at equilibrium, keeping flow constant until rain stops. ## Answer The hydrograph is a trapezoid produced by constant net rain of 0.20 in/hr for 5 hr. The rise stops when the whole basin contributes (equilibrium), and the recession starts when rain stops. ```text Q Qp | ________ | /| |\ | / | | \ | / | | \ |/______|_______|______\___ t (hr) 0 3 5 8 rain 0 -> 5 hr ``` #### (a) Excess release time Rain stops at t = 5 hr and the direct runoff ends at t = 8 hr. The time taken for the last excess water to leave the basin is $$ t_{release} = 8 - 5 = 3 \text{ hr} $$ This equals the time of concentration, $t_c = 3$ hr (also the duration of the rising limb, 0 to 3 hr). #### (b) Contributing area at t = 4 hr Since $t = 4\text{ hr} > t_c = 3\text{ hr}$, water from every part of the basin, including the farthest point, has reached the outlet. So $$ \frac{A_{contributing}}{A} = 100\% $$ (At t < 3 hr the contributing fraction is t/3, e.g. 33% at t = 1 hr.) #### (c) Peak discharge $Q_p$ With the whole basin contributing, $Q_p$ = net rain intensity × area (equilibrium discharge). $$ \begin{aligned} Q_p &= i\,A = \frac{0.20}{12}\ \text{ft/hr} \times 4250 \times 43560\ \text{ft}^2 \\ &= 3,085,500 \text{ ft}^3/\text{hr} \\ &= \frac{3,085,500}{3600} = 857 \text{ cfs} \end{aligned} $$ **Answer: $Q_p$ ≈ 857 cfs** (the sketch values are only approximate readings). #### (d) Why the plateau between t = 3 and t = 5 hr After t = 3 hr (= $t_c$), the entire basin already contributes. Rain continues at the same constant intensity, so the inflow to the outlet from the basin is constant: the rate of runoff leaving equals the rate of net rain falling (equilibrium). No new area can be added to increase the flow, so the hydrograph stays flat at $Q_p$ until the rain stops at t = 5 hr. It then recedes to zero in another $t_c$ = 3 hr. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the trapezoidal hydrograph, rain stops at t = 5 hr and direct runoff ends at t = 8 hr. What is the excess release time? A. 3 hr, equal to the time of concentration B. 5 hr, equal to the rain duration C. 8 hr, equal to the base time D. 2 hr, the length of the plateau Answer: A. The last excess water takes 8 - 5 = 3 hr to leave, which equals tc. (Key point: Excess release time) 2. Why is 100% of the basin contributing at t = 4 hr? A. Because rain is still falling at t = 4 hr B. t = 4 hr exceeds tc = 3 hr, so water from the farthest point has reached the outlet C. Because the hydrograph is at its recession limb D. Because the net rain intensity doubles after 3 hr Answer: B. After tc every part of the basin, including the farthest, contributes. (Key point: Contributing area) 3. How is Qp found when the whole basin contributes? A. Qp = intensity x duration x area / tc B. Qp = intensity x area / 2, as for a triangle C. Qp = net rain intensity x basin area (about 857 cfs) D. Qp is read from the sketch at t = 3 hr Answer: C. At equilibrium outflow equals rainfall input: 0.20/12 ft/hr x 4250 x 43560 ft2 = 857 cfs. (Key point: Equilibrium peak discharge) 4. Why does the hydrograph stay flat between t = 3 and 5 hr although rain continues? A. Rain intensity decreases after t = 3 hr B. Infiltration increases and cancels the extra rain C. Storage in the channel stops the flow from rising D. Whole basin already contributes, so outflow equals the constant rain input and no new area can be added Answer: D. Equilibrium: constant intensity over the full area gives constant discharge until rain stops. (Key point: Plateau reason) ## Related in Engineering Hydrology - [2-hr UH from sketch, net rain 5 in/hr](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-2hr-sketch-5in-cfs) — Engineering Hydrology, chapter 5 - [Derive 4-hr UH from triangular direct runoff hydrograph](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Atriangular-dro-4hr-uh) — Engineering Hydrology, chapter 5 - [Phi-index of storm, 20 cm rain, 12 cm runoff](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-index-20cm-12cm) — Engineering Hydrology, chapter 3 - [Infiltration index from 8-hr rainfall, Bagmati Basin](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-index-bagmati-11-6cm) — Engineering Hydrology, chapter 3 - [Phi-index and rainfall excess duration, 0.340 Mm3 runoff](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-index-500ha-0-340mm3) — Engineering Hydrology, chapter 3, asked 2× - [Total rainfall, net runoff, W-index of 140-min storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-w-index-140min-storm) — Engineering Hydrology, chapter 3, asked 2× ## Related in other subjects - [Rate analysis: wash basin 800×400](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Arate-wash-basin) — Estimating and Costing, chapter 4 - [Rate analysis: 20 mm plaster 1:4 per 100 m²](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Arate-plaster-20mm-1-4) — Estimating and Costing, chapter 4, asked 3× - [Runoff from a 20 ha built-up area](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Arunoff-builtup-area-20ha) — Sanitary Engineering, chapter 2 - [Hydrograph and flow duration curve](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Ahydrograph-flow-duration) — Power Plant Design, chapter 3, asked 3×