Question 7 of 7: The hydrologic equation, energy budget and infiltration simulation

Question 7: The hydrologic equation, energy budget and infiltration simulation (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) Annual evapotranspiration from the basic equation of hydrology (6 marks)

Given. The annual water balance of one large drainage basin, with the three quantities the question supplies collected below.

Find. The annual evapotranspiration ET = E + T from the watershed, in mm, using the basic equation of hydrology, with all assumptions stated and justified.

Approach. Reduce the BEH to a residual for ET by justifying the neglect of the storage and net groundwater terms over an annual period, convert the river discharge into an equivalent depth over the drainage area so that all terms share units, then close the balance — and test whether the closure is physically admissible before reporting it.

1. Reduce the BEH to a residual expression for ET. The printed form is $P-R-G-E-T=\Delta S$, so $$\mathrm{ET}=E+T=P-R-G-\Delta S$$ Two assumptions remove the last two terms. Assumption 1: $\Delta S \approx 0$ over a full year. Soil moisture, snowpack, lake and channel storage all cycle seasonally and return close to their starting values twelve months later, so the net annual change is small compared with the annual fluxes. This is the standard justification for annual water-balance work and is the reason such balances are struck on a water year rather than a calendar year. Assumption 2: $G \approx 0$, meaning there is no significant net groundwater flow across the topographic divide. Deep regional groundwater that leaves the basin without appearing in the river is genuinely lost from this balance, but for a basin of this size on typical Canadian terrain it is a small fraction of the total, and crucially the groundwater that discharges to the river as baseflow is already counted in the measured discharge and must not be counted again. With both terms dropped, $$\mathrm{ET}=P-R$$ Assumption 3: the gauge is at the basin outlet and its record represents the whole drainage area, so the measured discharge is the total surface-plus-baseflow runoff $R$.
2. Convert the river discharge into an equivalent depth over the drainage area. Depth and volume must be reconciled before the terms can be subtracted. The annual runoff volume is $$V_R=Q\,t=2200\ \text{m}^3/\text{s}\times 31\,536\,000\ \text{s}=6.938\times10^{10}\ \text{m}^3$$ and spreading it over the drainage area gives the runoff depth $$R=\frac{V_R}{A}=\frac{6.938\times10^{10}}{1.0\times10^{10}}=6.938\ \text{m} =\boxed{R = 6937.9\ \text{mm/a}}$$ Equivalently the basin yields a specific discharge of $2200/10\,000=0.22$ m³/s per km², that is 220 L/s/km².
3. Attempt to close the balance, and test the result. Substituting the printed precipitation, $$\mathrm{ET}=P-R=60-6937.9=-6877.9\ \text{mm/a}$$ Evapotranspiration cannot be negative: it is a flux out of the basin, bounded below by zero and above by the available energy. The printed data therefore do not describe a possible watershed. The ratio $R/P=115.6$ makes the point in one number — the river is discharging more than a hundred times the water the question says fell on the basin.
4. Identify which term is in error, and state the conservation requirement. The runoff depth is not an independent datum: it follows from the drainage area and the gauged discharge alone, both of which are routinely measured to within a few per cent. The precipitation, by contrast, is a single printed figure. Conservation of mass requires $$P \ge R + \mathrm{ET} \ge R$$ so any admissible precipitation for this basin must be at least 6937.9 mm/a. Under Note 1 of the examination instructions, the candidate is entitled to declare a corrected value and solve with it, which is done below.
5. Adopt a declared correction and report the evapotranspiration. Two repairs are admissible; both are carried through so the assumption is auditable. Reading A — the printed rainfall is mis-scaled and the discharge is as printed. A specific discharge of 220 L/s/km² belongs to a very wet Pacific-coast basin, and pairing it with $P=7400$ mm/a gives $$\mathrm{ET}=7400-6937.9=\boxed{462.1\ \text{mm/a}}$$ with a runoff coefficient of $R/P=0.938$. The evapotranspiration is 1.27 mm/d, a defensible figure for a cool, cloudy maritime climate where available energy rather than water limits the flux. Reading B — a single-digit slip in the discharge as well. Taking $Q=220$ m³/s gives $R=693.8$ mm/a at 22 L/s/km², which is squarely in the typical range, and pairing it with $P=1000$ mm/a gives $\mathrm{ET}=306.2$ mm/a with a runoff coefficient of 0.694 — typical values for a boreal-Shield basin. Reading A changes one printed number and is adopted as the answer; Reading B is the more typical Canadian basin and is reported alongside it.

(ii) The energy budget, the global radiative balance and the greenhouse effect (6 marks)

The water cycle is driven by energy, and the same budget that governs evaporation governs the planet's temperature. Written for a surface, the energy budget equation is

$$R_n = H + \lambda E + G_s + \Delta S_e$$

where $R_n$ is net radiation, $H$ the sensible heat flux into the air, $\lambda E$ the latent heat flux carried away by evaporating water ($\lambda\approx 2.45$ MJ/kg at 20 °C), $G_s$ the conduction into the ground and $\Delta S_e$ the change in heat storage. Net radiation is itself the difference between absorbed shortwave and net longwave, $R_n=(1-\alpha)R_s+R_{ld}-R_{lu}$.

Why a global radiative balance exists. The Earth has only one significant energy input, solar shortwave radiation, and only one significant output, longwave radiation to space. If the outgoing flux were smaller than the incoming, the planet would warm; a warmer body radiates more, by the Stefan-Boltzmann law $E=\sigma T^{4}$, so the outgoing flux would rise until it matched the input again. The balance is therefore not a coincidence but a stable equilibrium enforced by the strong temperature dependence of emission. Averaged over the globe and over a year, the top-of-atmosphere budget is $$\frac{S_0}{4}(1-\alpha)=\sigma T_e^{4}$$ where $S_0=1361$ W/m² is the solar constant, the factor of four converts the intercepting disc into the emitting sphere, and $\alpha\approx 0.30$ is the planetary albedo. The absorbed flux is $340.25(0.70)=238.2$ W/m², so the effective radiating temperature is $$T_e=\left(\frac{238.2}{5.67\times10^{-8}}\right)^{1/4}=\boxed{254.6\ \text{K}=-18.6\ {}^{\circ}\text{C}}$$

Why the surface is warmer than that — the greenhouse effect. The observed global mean surface temperature is about 288 K (15 °C), which is 33.4 K warmer than the effective radiating temperature just computed. The difference is the greenhouse effect. The atmosphere is largely transparent to incoming shortwave radiation, which passes through to be absorbed at the surface, but it is strongly absorbing in the thermal infrared because water vapour, carbon dioxide, methane and the other polyatomic trace gases have vibrational and rotational transitions at those wavelengths. The surface therefore radiates upward into a partially opaque atmosphere; the atmosphere absorbs that radiation and re-emits it both up and down, and the downward component ($R_{ld}$ in the budget above) is an additional energy input to the surface. Equilibrium at the top of the atmosphere is still satisfied at 254.6 K, but that emission now comes on average from an altitude of roughly 5 km rather than from the ground; because temperature falls with height along the lapse rate, the surface beneath must be correspondingly warmer. Increasing the concentration of an absorbing gas raises the effective emission altitude, and the surface warms further to restore the balance.

Where latent heat and differential solar heating come in. The radiative balance holds globally, but not locally, and that is what drives the circulation of both atmosphere and water. Because the Earth is a sphere, low latitudes receive solar radiation at near-normal incidence and absorb far more than they emit, while high latitudes receive it at a glancing angle, are more reflective, and emit more than they absorb. This differential solar heating creates a persistent equator-to-pole energy surplus and deficit, and the atmosphere and oceans transport energy poleward to close the gap. A large share of that transport is carried as latent heat: the $\lambda E$ term removes energy from the tropical ocean surface as water evaporates, the vapour is advected poleward and upward, and the energy is released as sensible heat wherever the vapour condenses. Condensation is thus the mechanism that couples the water cycle to the energy cycle, and it is why the same budget that predicts evaporation also predicts precipitation. Two further consequences follow directly. First, latent heat is the reason evaporation is self-limiting: the $\lambda E$ term competes with $H$ for the same net radiation, so a wet surface stays cool and a dry one heats up. Second, water vapour is itself the strongest greenhouse gas, and because a warmer atmosphere holds more of it (Clausius-Clapeyron, roughly 7 per cent per kelvin), the water cycle supplies the dominant positive feedback on any radiative forcing — which is why hydrology and climate cannot be treated as separate subjects.

(iii) Infiltration simulation, and the Saturation and Transition Zones (8 marks)

A potential infiltration simulation approach. Three levels of model are available, and the choice is a trade-off between physical fidelity and the data the site can support. The most rigorous is a numerical solution of the Richards equation, which is Darcy's law combined with continuity for unsaturated flow,

$$\frac{\partial \theta}{\partial t} =\frac{\partial}{\partial z}\left[K(\theta)\left(\frac{\partial \psi}{\partial z}+1\right)\right]$$

with $\theta$ the volumetric water content, $\psi$ the matric suction and $K(\theta)$ the unsaturated hydraulic conductivity described by a van Genuchten or Brooks-Corey relation. Solved numerically with a ponded upper boundary, it reproduces the whole profile drawn in the figure. It also demands soil hydraulic characterisation that is rarely available for a drainage design.

The practical middle course, and the one recommended here, is the Green-Ampt model, which idealises the profile as a sharp wetting front separating saturated soil above from soil at initial moisture below — a deliberate simplification of exactly the figure shown. Applying Darcy's law across the wetted depth $L$ under a ponded head $h_0$ and a suction head $\psi_f$ at the front gives the infiltration rate

$$f=K\,\frac{h_0+L+\psi_f}{L}, \qquad\text{with cumulative infiltration } F=K t+\psi_f\,\Delta\theta\,\ln\!\left(1+\frac{F}{\psi_f\,\Delta\theta}\right)$$

solved iteratively at each time step, where $\Delta\theta=\theta_s-\theta_i$ is the moisture deficit. Its three parameters — saturated conductivity, wetting-front suction and initial moisture deficit — are all tabulated by soil texture, which is why it is the model embedded in HEC-HMS, SWMM and most Canadian municipal practice. The simplest level is the empirical Horton equation,

$$f(t)=f_c+(f_0-f_c)\,e^{-kt}, \qquad F(t)=f_c t+\frac{f_0-f_c}{k}\bigl(1-e^{-kt}\bigr)$$

which fits the observed decay of infiltration capacity without representing the profile at all. For a silt loam with $f_0=75$ mm/h, $f_c=12$ mm/h and $k=2.5$ h⁻¹, one hour of ponding gives

$$f(1\ \text{h})=12+63e^{-2.5}=\boxed{17.2\ \text{mm/h}}, \qquad F(1\ \text{h})=12+\frac{63}{2.5}\bigl(1-e^{-2.5}\bigr)=35.1\ \text{mm}$$

The essential point common to all three is that under ponded conditions the soil, not the rainfall, controls the rate: infiltration proceeds at the capacity, which starts high, decays as the profile wets and the suction gradient weakens, and approaches the saturated hydraulic conductivity asymptotically.

Two characteristics of the Saturation Zone. (1) It is a thin layer — typically only a centimetre or two — immediately beneath the ponded surface in which the water content is at or extremely close to the saturated value $\theta_s$, so essentially all pore space is water-filled and the hydraulic conductivity is at its saturated maximum $K_s$. (2) Its thickness barely grows as infiltration continues; it remains shallow while the wetted profile below it deepens, and because it is saturated the pressure head within it is positive and equal to the ponded depth, so it transmits the ponded head downward almost without loss. It is also the zone in which surface sealing, crusting and entrapped air have their effect, all of which reduce the effective $K_s$ and therefore the whole infiltration rate.

Two characteristics of the Transition Zone. (1) It is the region of steep gradient immediately below the saturation zone, across which the water content falls rapidly from $\theta_s$ to the nearly uniform value that characterises the transmission zone below — it is, in the figure, the short reach where the profile curve bends away from the saturation line. (2) Like the saturation zone it stays thin and roughly constant in thickness as infiltration proceeds; it simply translates downward with the wetted front while the transmission zone lengthens. Because $\theta$ falls through it, the unsaturated conductivity $K(\theta)$ falls sharply through it as well, so it is the layer across which the largest part of the resistance to flow near the surface is concentrated. Together the two zones occupy only a small fraction of the wetted depth, which is precisely the observation that justifies the Green-Ampt idealisation of a single sharp front.