--- title: "Construct 3-h UH and flood hydrograph, 524.88 ha" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:uh-3h-triangular-524-88ha" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-3h-triangular-524-88ha site: BCT Engineering --- # Construct 3-h UH and flood hydrograph, 524.88 ha Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2066 Magh (old course), 16 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-3h-triangular-524-88ha - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question Unit hydrograph of 3h duration from a catchment area of 524.88 ha has peak discharge value after 9 hours from the start of the storm. It has half of the peak discharge value at 3 hours and 15 hours after the start of storm respectively. If time base ($t_b$) is 30 hours, constitute the unit hydrograph and calculate the flood hydrograph from the following rainfall data. Assume the Phi index ($\phi$) as 0.333 cm/hour and base flow as 20 cumecs. | Time (hours) | 0 | 6 | 9 | 12 | |---|---|---|---|---| | Cumulative Rainfall (cm) | 0 | 12 | 12 | 23 | ## Short answer Catchment area 524.88 ha yields UH volume 52488 m³; triangular shape with peak at 9 h gives Qp = 1.08 m³/s. 3-h UH ordinates computed. Rainfall excess after φ = 0.333 cm/h: 5, 5, 0, 10 cm in successive 3-h blocks. Superposition with baseflow 20 m³/s gives peak flood 35.7 m³/s at 18 h. ## Answer #### Step 1: Shape of the 3-h UH Given: $t_b$ = 30 h, peak at 9 h, half-peak at 3 h and 15 h. So the UH is a straight-line polygon through (0, 0), (3, ½$Q_p$), (9, $Q_p$), (15, ½$Q_p$) and (30, 0). ```text Q Qp | /\ | / \ Qp/2 | / \_ | / \__ 0 +--+-----+-----+------+---- t (h) 0 3 9 15 30 ``` #### Step 2: Peak ordinate from the area A UH has 1 cm of runoff over the catchment: $$ V = 524.88\text{ ha}\times 10^4 \times 0.01 \text{ m} = 52\,488 \text{ m}^3 $$ Area under the polygon (in units of $Q_p\cdot$h): $$ \tfrac12(3)(0.5) + \tfrac12(0.5+1)(6) + \tfrac12(1+0.5)(6) + \tfrac12(15)(0.5) = 0.75+4.5+4.5+3.75 = 13.5 $$ $$ Q_p \times 13.5 \times 3600 = 52\,488 \ \Rightarrow\ Q_p = 1.08 \text{ m}^3/\text{s} $$ #### Step 3: 3-h UH ordinates (3-h interval) | t (h) | 0 | 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 | 30 | |---|---|---|---|---|---|---|---|---|---|---|---| | Fraction of Q_p | 0.000 | 0.500 | 0.750 | 1.000 | 0.750 | 0.500 | 0.400 | 0.300 | 0.200 | 0.100 | 0.000 | | UH (m³/s per cm) | 0.000 | 0.540 | 0.810 | 1.080 | 0.810 | 0.540 | 0.432 | 0.324 | 0.216 | 0.108 | 0.000 | #### Step 4: Rainfall excess (3-h blocks) Cumulative rain: 12 cm at 6 h, still 12 cm at 9 h (no rain 6 – 9 h), 23 cm at 12 h. Rain is taken uniform inside each interval. | Block | Rain (cm) | Loss φt = 0.333 × 3 (cm) | Excess (cm) | |---|---|---|---| | 0 – 3 h | 6 | 1.0 | 5.0 | | 3 – 6 h | 6 | 1.0 | 5.0 | | 6 – 9 h | 0 | 0 | 0 | | 9 – 12 h | 11 | 1.0 | 10.0 | #### Step 5: Flood hydrograph $$ Q(t) = 5\,U(t) + 5\,U(t-3) + 0\cdot U(t-6) + 10\,U(t-9) + 20 $$ | t (h) | UH (m³/s) | 5 cm × UH | 5 cm × UH | 10 cm × UH | Base | Q (m³/s) | |---|---|---|---|---|---|---| | 0 | 0.00 | 0.0 | 0.0 | 0.0 | 20.0 | 20.0 | | 3 | 0.54 | 2.7 | 0.0 | 0.0 | 20.0 | 22.7 | | 6 | 0.81 | 4.1 | 2.7 | 0.0 | 20.0 | 26.8 | | 9 | 1.08 | 5.4 | 4.1 | 0.0 | 20.0 | 29.5 | | 12 | 0.81 | 4.1 | 5.4 | 5.4 | 20.0 | 34.9 | | 15 | 0.54 | 2.7 | 4.1 | 8.1 | 20.0 | 34.9 | | 18 | 0.43 | 2.2 | 2.7 | 10.8 | 20.0 | 35.7 | | 21 | 0.32 | 1.6 | 2.2 | 8.1 | 20.0 | 31.9 | | 24 | 0.22 | 1.1 | 1.6 | 5.4 | 20.0 | 28.1 | | 27 | 0.11 | 0.5 | 1.1 | 4.3 | 20.0 | 25.9 | | 30 | 0.00 | 0.0 | 0.5 | 3.2 | 20.0 | 23.8 | | 33 | 0.00 | 0.0 | 0.0 | 2.2 | 20.0 | 22.2 | | 36 | 0.00 | 0.0 | 0.0 | 1.1 | 20.0 | 21.1 | | 39 | 0.00 | 0.0 | 0.0 | 0.0 | 20.0 | 20.0 | **Answer: peak flood ≈ 35.66 m³/s at t = 18 h** (the flood hydrograph returns to the base flow of 20 m³/s at t = 39 h). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What runoff depth does the UH represent, and its volume for 524.88 ha? A. 1 m, giving 5,248,800 m3 B. 3 cm, giving 157,464 m3 C. 1 mm, giving 5,249 m3 D. 1 cm, giving 52,488 m3 Answer: D. 524.88 ha = 5,248,800 m2; x 0.01 m = 52,488 m3. (Key point: UH volume) 2. How is Qp found from the polygon? A. Area under polygon (13.5 Qp.h) x 3600 = 52,488, so Qp = 1.08 m3/s B. Qp = 52,488 / 3600 = 14.58 m3/s C. Qp = 52,488 / 30 = 1749.6 D. Qp is the half-peak value 0.54 Answer: A. Volume equals area under UH; 13.5 x 3600 Qp = 52,488. (Key point: Peak from area) 3. With phi = 0.333 cm/h, what are the 3-h excess blocks? A. 6, 6, 0, 11 cm B. 5, 5, 0, 10 cm C. 5, 5, 5, 10 cm D. 12, 0, 11 cm Answer: B. Blocks 6, 6, 0, 11 cm each lose 1.0 cm, except the zero-rain block (no loss to subtract). (Key point: Excess blocks) 4. What superposition gives the flood? A. Q = 5U(t) + 5U(t-3) + 10U(t-6) + 20 B. Q = 20U(t) + 20 C. Q = 5U(t) + 5U(t-3) + 0U(t-6) + 10U(t-9) + 20 D. Q = 5U(t) + 5U(t-6) + 10U(t-12) + 20 Answer: C. Blocks are 3 h apart; the dry 6-9 h block contributes zero. 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