18-Env-A2 Hydrology and Municipal Hydraulics Engineering, December 2017, Q1

Problem 1 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(i) The natural hydrologic cycle (8 marks)

Fig. 1 — Four key components of the hydrologic cycle and the three interactions described below.

The hydrologic cycle is the continuous mass balance of water circulating between the atmosphere, land surface and subsurface. Four key components are shown above: precipitation (the atmospheric input, rain/snow reaching the ground), surface runoff (the fraction of precipitation that flows overland and through channels to streams, lakes and the ocean), evaporation and transpiration (the return of moisture to the atmosphere from open water, soil and vegetation), and infiltration and groundwater flow (water entering the soil, recharging the water table and discharging slowly as baseflow).

Three main interactions link these components. First, precipitation partitions at the ground surface between infiltration and runoff depending on soil moisture, land cover and rainfall intensity — once the infiltration capacity of the soil is exceeded, the excess becomes overland flow (rainfall excess), which is the process the Rational Method in part (ii) is built around. Second, evapotranspiration draws moisture from both the surface-runoff/storage pathway (open-water evaporation from lakes and channels) and the vegetated land surface (plant transpiration), returning it to the atmosphere to condense and fall again as precipitation, closing the loop. Third, infiltrated water that percolates below the root zone recharges the groundwater table and discharges gradually to streams as baseflow, sustaining flow between storm events and buffering the flashy response that surface runoff alone would produce.

(ii) 100-year design peak runoff by the Rational Method (7 marks)

Given.

Area	Area (ha)	C	t_c (min)
A1	20	0.5	70
A2	30	0.8	100

100-year IDF curve family (figure supplied with the exam, reproduced in shape below); A2's longer time of concentration means A2 lies downstream of / in series with A1, so the two candidate storms are "only A1 has reached the outlet" (duration = 70 min) and "the whole composite area is contributing" (duration = 100 min).

Find. The governing 100-year design peak runoff Q (m³/min) at the common outlet.

Fig. 2 — Reading the 100-year IDF curve at the two anchor durations bracketing t = 70 min and t = 100 min (chart values read to ±1–2 mm/hr, typical of graphical IDF interpolation).

Check: the intensities below are read graphically off the supplied IDF chart (log–log duration axis, no printed data table) at the 100-year curve: i ≈ 37 mm/hr at t = 60 min and i ≈ 29 mm/hr at t = 120 min (2 hr) — the two labelled gridlines bracketing the required 70 min and 100 min durations. A candidate reading the original chart with a straightedge would obtain values within a similar ±1–2 mm/hr band; the method and governing conclusion are insensitive to this level of reading tolerance.

Approach. Interpolate the 100-year curve log–log between the two chart anchors to get i(70) and i(100), then apply the metric Rational Formula $Q = \dfrac{CiA}{360}$ to both candidate storms and take the larger as the governing composite peak.

Interpolate the IDF curve (power-law fit through the two chart anchors). Between $t=60$ min ($i=37$) and $t=120$ min ($i=29$), fit $i = i_{60}(t/60)^{-m}$: $$m = \frac{\ln(37/29)}{\ln(120/60)} = 0.351$$ $$i(70) = 37\left(\frac{70}{60}\right)^{-0.351} = \boxed{35.0\ \text{mm/hr}} \qquad i(100) = 37\left(\frac{100}{60}\right)^{-0.351} = \boxed{30.9\ \text{mm/hr}}$$
Case 1 — storm duration 70 min, only A1 at the outlet (A1's own $t_c$). $$Q_1 = \frac{C_1 A_1\, i(70)}{360} = \frac{0.5\times20\times35.0}{360} = 0.97\ \text{m}^3/\text{s}$$
Case 2 — storm duration 100 min, both A1 and A2 contributing (A2's own $t_c$, the whole catchment engaged). $$Q_2 = \frac{(C_1A_1+C_2A_2)\,i(100)}{360} = \frac{(0.5\times20+0.8\times30)\times30.9}{360} = \frac{34\times30.9}{360} = \boxed{2.92\ \text{m}^3/\text{s}}$$
Governing case. $Q_2 > Q_1$, so the 100-year design peak occurs when the full composite area is contributing at $t=100$ min. Converting to the requested units: $$Q_{100} = 2.92\ \text{m}^3/\text{s}\times 60\ \text{s/min} = \boxed{175\ \text{m}^3/\text{min}}$$

Quantity	Value
$i(70\text{ min})$, 100-yr	≈ 35.0 mm/hr
$i(100\text{ min})$, 100-yr	≈ 30.9 mm/hr
$Q_1$ (A1 only @ 70 min)	0.97 m³/s
$Q_2$ (A1+A2 @ 100 min) — governing	2.92 m³/s
$Q_{100}$, design peak	≈ 175 m³/min

(iii) Meaning and control of the runoff coefficient C (5 marks)

$C$ is the fraction of rainfall intensity that appears as peak runoff intensity, i.e. $Q=CiA/360$ treats $C$ as a single lumped ratio of "runoff out" to "rainfall in" for the catchment; physically it bundles together imperviousness, depression storage, infiltration capacity, slope and antecedent soil moisture into one coefficient, so it is inherently an approximation rather than a measured physical constant, and tables of "typical" $C$ by land use (roofs, pavement, lawn, forest) are calibrated averages, not exact values for any specific site. Two methods to reduce its susceptibility to error in a large urban watershed: (1) use a composite, area-weighted C built from a detailed land-use/imperviousness inventory (GIS-delineated roof, pavement, lawn and open-space areas each with a literature or locally-calibrated sub-coefficient, weighted by area, $C_{composite}=\sum C_iA_i/\sum A_i$) rather than a single coefficient guessed for the whole watershed, since a large urban catchment is rarely homogeneous; and (2) calibrate/verify $C$ against observed rainfall–runoff data from a gauged sub-catchment or continuous flow monitoring, back-calculating an effective $C$ from measured storms and adjusting the design value to match local conditions, rather than relying solely on tabulated values that may not reflect the watershed's actual soil, slope and land-cover mix.

Topic: The hydrologic cycle and the Rational Method for a composite (in-series) catchment.

Key relations: $Q = \dfrac{CiA}{360}$ (metric Rational Method, $Q$ in m³/s, $i$ in mm/hr, $A$ in ha); composite $C_{composite}=\sum C_iA_i/\sum A_i$; IDF power-law interpolation $i=i_1(t/t_1)^{-m}$.

Why this works: when two sub-areas with different times of concentration drain to one outlet, the peak at that outlet is not simply each sub-area's individually-optimal peak added together (that double-counts intensity, since each would be evaluated at its own, different, storm duration) — it is the larger of (a) the fastest sub-area alone at its own $t_c$, or (b) the full composite area once the slowest sub-area has also started contributing, at the composite $t_c$. Checking both and taking the maximum is the only way to find the true governing peak.

Common pitfall: using only the largest sub-area's $t_c$ as "the" catchment $t_c$ without checking whether a shorter-duration, partial-area storm might govern instead; or summing each area's own-$t_c$ peak flow directly, which mixes two different storm durations in one sum.

Source: Linsley, Kohler & Paulhus, Hydrology for Engineers, Ch. on rainfall-runoff and the Rational Method.

Question 1 of 7

Problem 1 (20 marks)

(i) The natural hydrologic cycle (8 marks)

(ii) 100-year design peak runoff by the Rational Method (7 marks)

(iii) Meaning and control of the runoff coefficient C (5 marks)