Hydrologic Equation, Infiltration Simulation and Energy Budget: 16-Civ-B4 Engineering Hydrology, December 2014, Q6

Problem 6: Hydrologic Equation, Infiltration Simulation and Energy Budget (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.

Part (i) — Annual evapotranspiration as the residual of the water balance (7 marks)

Given.

Quantity	Symbol	Value as printed
Watershed surface area	A	10 000 km² = 1.0 × 10¹⁰ m²
Annual precipitation	P	150 mm/a
Annual mean river discharge	Q	300 m³/s
Seconds in a year	t	365 × 86 400 = 3.1536 × 10⁷ s

Find. The annual evapotranspiration depth ET = E + T in mm, obtained as the residual of the basic equation of hydrology.

Approach. Convert the river discharge to an equivalent depth over the basin, then close the annual water balance with the storage change and deep groundwater flux set to zero, leaving ET as the only unknown — and test whether the result is physically admissible.

State the balance and justify the terms that are dropped. Over a full year the basic equation of hydrology is $$P-R-G-E-T=\Delta S$$ Two assumptions close it. First, $\Delta S \approx 0$: over a complete annual cycle the basin returns to a comparable soil-moisture and groundwater state, so storage change is small relative to the annual fluxes. Second, $G \approx 0$: deep groundwater leakage out of the basin is neglected, which is standard for a topographically closed basin whose groundwater divide coincides with the surface divide. With ET = E + T, this leaves $$\text{ET}=P-R$$
Convert the river discharge to a runoff depth over the basin. The annual volume leaving through the river is $$V_R=Q\,t=300\times 3.1536\times10^{7}=9.4608\times 10^{9}\ \text{m}^{3}$$ and spreading that volume over the drainage area gives the equivalent depth $$R=\frac{V_R}{A}=\frac{9.4608\times 10^{9}}{1.0\times 10^{10}}=0.94608\ \text{m}$$ $$\boxed{R = 946.1\ \text{mm/a}}$$ This value depends only on the discharge and the area, both of which are unambiguous, so it is the most reliable number in the question.
Attempt the balance with the printed precipitation, and test the result. Substituting the printed P = 150 mm/a, $$\text{ET}=P-R=150-946.1=-796.1\ \text{mm/a}$$ which is physically impossible: evapotranspiration cannot be negative, since that would require the atmosphere to deliver water to the land surface by a route other than precipitation. The data as printed are internally inconsistent — the basin discharges 6.3 times more water than it receives.
Identify which datum is in error. The runoff depth follows from discharge and area alone and cannot be adjusted. Conservation of mass requires $P \ge R$, so the precipitation figure must be too small. A runoff of 946 mm/a corresponds to a specific discharge of 30 L s⁻¹ km⁻², which is characteristic of a wet mountain or coastal basin — in Canada, the Coast Mountains or the Columbia headwaters. Such basins receive of order 1200–2500 mm/a. The printed value of 150 mm/a is short by very nearly a factor of ten, consistent with a lost digit in the question.
Adopt a declared corrected precipitation and close the balance. Taking $P = 1500$ mm/a — the printed figure with the missing zero restored, and squarely within the observed range for such a basin: $$\text{ET}=P-R=1500-946.1$$ $$\boxed{\text{ET} = 554\ \text{mm/a}}$$ The corresponding runoff coefficient is $$C=\frac{R}{P}=\frac{946.1}{1500}=0.63$$ Both results are physically sound: an annual ET of 554 mm is typical of a forested temperate Canadian basin, and a runoff coefficient of 0.63 is what a wet, steep, shallow-soiled basin should produce.

Check — the printed data cannot be solved as given. With P = 150 mm/a and Q = 300 m³/s over 10 000 km², the water balance returns ET = −796 mm/a, which violates conservation of mass. Since R = 946.1 mm/a follows unambiguously from discharge and area, the precipitation figure is the term in error and is short by about one order of magnitude. The solution above therefore: (1) reports R = 946.1 mm/a, which is unaffected by the inconsistency; (2) demonstrates that the balance cannot close as printed; and (3) adopts a declared corrected value P = 1500 mm/a to complete the calculation. In an examination, stating the inconsistency, showing the impossibility, and solving with a justified assumption earns full marks — reporting a negative evapotranspiration does not.

Quantity	Symbol	Value
Annual river volume	V_R	9.461 × 10⁹ m³
Runoff as an equivalent depth	R	946.1 mm/a
ET using the printed P = 150 mm/a	ET	−796 mm/a (impossible)
Adopted corrected precipitation	P	1500 mm/a (declared)
Annual evapotranspiration	ET	554 mm/a
Runoff coefficient	C	0.63

Part (ii) — An infiltration model and three assumptions behind it (8 marks)

Model selected: the Green–Ampt model, chosen in preference to Horton’s equation because its parameters are physically measurable soil properties rather than pure fitting constants, which is what an engineer needs when transferring the model to an ungauged site. Green–Ampt idealises infiltration as a sharp wetting front advancing into the soil, and gives the infiltration rate as

$$f = K\left(1+\frac{\psi\,\Delta\theta}{F}\right)$$

where K is the saturated hydraulic conductivity, $\psi$ the suction head at the wetting front, $\Delta\theta$ the moisture deficit (porosity minus initial moisture content), and F the cumulative infiltration. The rate starts high and decays towards K as F grows, reproducing the observed decay shown above. Horton’s empirical alternative, $f = f_c + (f_0-f_c)e^{-kt}$, produces the same shape but with parameters that must be fitted to infiltrometer data.

Assumption 1: the soil is homogeneous, isotropic and of semi-infinite depth, with a uniform initial moisture content. Real profiles are layered, and this matters more than any other departure. A low-conductivity layer — a plough pan, a clay lens, a compacted subgrade under a construction site, or a frozen layer — will throttle infiltration long before the model predicts, and the true rate will fall to the conductivity of the restricting layer rather than to that of the surface soil. Frozen ground is the Canadian case that matters most: an impermeable frost layer during spring melt can reduce infiltration essentially to zero on soils that would otherwise absorb the entire event, which is why prairie snowmelt floods are so much larger than their water equivalent alone suggests.

Assumption 2: a sharp, piston-like wetting front separates saturated soil above from soil at the initial moisture content below. The real transition is a gradual moisture profile, not a step. The idealisation is what makes the model analytically tractable, and it is adequate for coarse soils where the front is genuinely sharp; it is poorest in fine-textured soils with strong capillarity, where the transition zone is thick. The engineer should also recognise that this assumption excludes macropore and preferential flow — cracks, root channels, worm holes — which in structured clay soils can convey a large fraction of the water at rates the matrix model cannot reproduce.

Assumption 3: ponded surface supply, with negligible ponding depth, and no surface sealing. The equation as written gives infiltration capacity, the rate at which soil could absorb water if enough were available. It is valid only after ponding begins; before that, the actual infiltration rate equals the rainfall rate, and applying the capacity curve from the start of the storm over-estimates the losses badly. Neither is the surface static: raindrop impact breaks down aggregates and forms a surface crust, and fine sediment clogs pores, so measured field rates commonly fall below the model during a long storm. The engineer must also confirm that the parameters used reflect current land condition — compaction during construction can reduce infiltration capacity by an order of magnitude.

Part (iii) — Energy and mass balance for a watershed adjacent to an ocean (5 marks)

The energy and water budgets are two accounts of the same system, coupled through latent heat: every kilogram of water evaporated removes about 2.45 MJ from the surface and carries it into the atmosphere, and releases it again on condensation. Evaporation is therefore simultaneously an entry in the water balance and an entry in the energy balance, and it is impossible to change one without changing the other. The two statements are

$$R_n = H + \lambda E + G_s \qquad\text{and}\qquad P - R - \text{ET} = \Delta S$$

where $R_n$ is net radiation, H sensible heat, $\lambda E$ latent heat and $G_s$ ground heat flux. The available energy $R_n$ sets a hard ceiling on evaporation: no basin can evaporate more water than its radiation budget can supply the latent heat for, which is why potential evapotranspiration is fundamentally an energy-limited quantity and why the Penman–Monteith equation is built from the energy balance rather than from the water balance.

Next to an ocean, this coupling acquires a strong spatial structure, because land and water partition the same incoming radiation quite differently. Water has a high heat capacity, is transparent to some depth, and mixes, so it stores the absorbed energy and warms slowly; a large share of its net radiation goes into $\lambda E$, making the ocean an effectively unlimited moisture source. Land has a low heat capacity and, once its surface dries, limited water, so more of its net radiation goes into sensible heat H and the surface warms rapidly by day.

The consequences for the adjacent watershed follow directly. The differential heating creates a horizontal pressure gradient, driving the sea breeze: moist marine air is advected inland. As that air is lifted — by the coastal topography that almost always accompanies such a watershed, or by convergence over the heated land — it cools at the adiabatic lapse rate, reaches saturation, condenses, and releases its latent heat, which invigorates the ascent and produces precipitation. This is the orographic mechanism that gives coastal British Columbia among the highest precipitation totals in Canada, with the corollary rain shadow immediately to the lee. The energy imbalance between land and sea is thus the direct cause of the precipitation term in the watershed’s water balance.

Two further couplings complete the picture. The ocean moderates temperature extremes on the coastal watershed, so a larger fraction of winter precipitation falls as rain rather than snow, which shifts the annual runoff regime from a spring-freshet type to a rainfall-dominated winter-peaking one — a change of flood season, not merely of volume. And the moist marine air suppresses the vapour-pressure deficit over the land, reducing evapotranspiration below what the radiation alone would allow, so the coastal basin retains a higher runoff coefficient than an inland basin of the same precipitation. The basin analysed in Part (i), with its 946 mm of runoff and a runoff coefficient of 0.63, is exactly such a basin.

Topic. The basic equation of hydrology as a closure tool, physically based infiltration modelling, and the energy–water coupling that governs coastal precipitation.

Key relations. Water balance $P-R-G-\text{ET}=\Delta S$; runoff as a depth $R=Qt/A$; Green–Ampt $f=K(1+\psi\Delta\theta/F)$; Horton $f=f_c+(f_0-f_c)e^{-kt}$; surface energy balance $R_n=H+\lambda E+G_s$; latent heat of vaporisation $\lambda \approx 2.45$ MJ/kg at 20 °C.

Why this works. Solving for ET as a residual is attractive because ET is the hardest term to measure directly at basin scale, while P and Q are routinely gauged. The price is that the residual inherits every error in the other terms — which is exactly what this question demonstrates, since a single wrong precipitation figure drives ET negative. That sensitivity is a feature worth understanding rather than a nuisance: a negative residual ET is a reliable diagnostic that one of the measured terms is wrong, and it is a check worth running on any real water balance. The energy balance provides the independent constraint, since $\lambda E$ can never exceed the available energy $R_n - G_s$.

Common pitfall. Reporting a negative evapotranspiration, or silently taking the absolute value, rather than recognising an impossible result and diagnosing which input is wrong. In infiltration modelling, the parallel error is applying the infiltration capacity curve from the beginning of a storm rather than from the moment of ponding, which over-estimates losses and under-estimates runoff on every low-intensity event.

Source. Chow, Maidment & Mays, Applied Hydrology, Ch. 1 (water balance), Ch. 4 (Green–Ampt) and Ch. 3–4 (energy balance and evaporation); Viessman & Lewis, Introduction to Hydrology, Ch. 3 and Ch. 6; Mays, Water Resources Engineering, Ch. 3.

Question 6 of 7: Hydrologic Equation, Infiltration Simulation and Energy Budget

Problem 6: Hydrologic Equation, Infiltration Simulation and Energy Budget (20 marks)

Part (i) — Annual evapotranspiration as the residual of the water balance (7 marks)

Part (ii) — An infiltration model and three assumptions behind it (8 marks)

Part (iii) — Energy and mass balance for a watershed adjacent to an ocean (5 marks)