22-Agric-B7 Principles of Hydrology · December 2015
Question 1 of 6: Unit Hydrograph Analysis — Watershed Area and Storm Convolution
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 04-Agric-B7, Principles of Hydrology. Three-hour, open-book exam; any non-communicating calculator is permitted. Format: five questions constitute a complete paper (the first five as they appear in the answer book are marked), each of equal value; most questions require an answer involving calculations. All six questions are solved here as a complete study resource.
Given. A 15-minute unit hydrograph (direct runoff from 1 cm of excess precipitation uniformly over the watershed):
Time (min)
Runoff, U (m³/s)
0
0
15
2
30
5
45
2.5
60
1
75
0
A 45-minute storm with excess precipitation of 0.5, 1.0 and 2.0 cm in successive 15-minute increments.
Find. (a) The watershed area implied by the unit hydrograph's runoff volume. (b) The storm's direct-runoff rate at 30 minutes, the peak runoff rate, and the time of the peak.
Approach. The volume under a unit hydrograph equals its defining 1 cm of excess precipitation spread over the watershed area, which fixes the area; the storm hydrograph then follows by linear superposition (convolution) of the unit hydrograph, scaled and lagged by each 15-minute pulse of excess precipitation.
Volume under the unit hydrograph (trapezoidal rule, Δt = 15 min = 900 s).
$$V=900\left[\tfrac{0+2}{2}+\tfrac{2+5}{2}+\tfrac{5+2.5}{2}+\tfrac{2.5+1}{2}+\tfrac{1+0}{2}\right]=900(10.5)=\boxed{9450\ \text{m}^3}$$
Watershed area from the defining 1 cm of excess precipitation. The unit hydrograph is, by definition, the runoff produced by exactly 1 cm (0.01 m) of excess rainfall spread uniformly over the watershed, so its volume equals depth × area:
$$A=\frac{V}{0.01\ \text{m}}=\frac{9450}{0.01}=945000\ \text{m}^2=\boxed{94.5\ \text{ha}}$$
Convolve the storm's three 15-min pulses with the unit hydrograph. With pulses $P_1=0.5$, $P_2=1.0$, $P_3=2.0$ cm applied at $t=0$, $15$, $30$ min respectively, linearity gives the direct-runoff hydrograph as the lagged, scaled sum
$$Q(t)=P_1\,U(t)+P_2\,U(t-15)+P_3\,U(t-30)$$
where $U$ is the given unit-hydrograph ordinate (zero outside the table). At $t=30$ min:
$$Q(30)=0.5\,U(30)+1.0\,U(15)+2.0\,U(0)=0.5(5)+1.0(2)+2.0(0)=\boxed{4.5\ \text{m}^3/\text{s}}$$
Locate the peak by evaluating $Q(t)$ at every 15-min ordinate. Repeating the convolution of Step 3 across $t=0,15,\dots,105$ min gives the full storm hydrograph (plotted below alongside the unit hydrograph); the largest ordinate occurs at $t=60$ min:
$$Q(60)=0.5\,U(60)+1.0\,U(45)+2.0\,U(30)=0.5(1)+1.0(2.5)+2.0(5)=\boxed{13\ \text{m}^3/\text{s}}$$
Checking the neighbouring ordinates ($Q(45)=10.25$, $Q(75)=6.0\ \text{m}^3/\text{s}$) confirms $t=60$ min is the single maximum, so the time of peak runoff is 60 minutes after the start of runoff.
Grey dashed: the given 15-min unit hydrograph. Blue: the convolved direct-runoff hydrograph for the 0.5/1.0/2.0 cm storm, peaking at 13 m³/s at t = 60 min.
