Water Balance, Energy Budget and Infiltration Simulation: 16-Civ-B4 Engineering Hydrology, May 2013, Q7

Question 7: Water Balance, 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 (8 marks)

Given. An annual water balance for a single watershed, with the data printed in the question.

Given data — annual water balance
Quantity	Symbol	Value as printed
Watershed surface area	A	10,000 km² = 1.0 × 10¹⁰ m²
Annual precipitation depth	P	100 mm
Annual mean flowrate of the draining river	Q_r	500 m³/s
Basic equation of hydrology	—	P − R − G − ET = ΔS

Find. The annual evapotranspiration depth ET in mm, with every assumption stated and justified — which, as it turns out, includes an assumption about the printed data themselves.

Approach. Convert the river discharge into an equivalent runoff depth over the watershed so that every term of the balance is a depth over the same area, then close the balance for ET after justifying the treatment of the groundwater and storage terms. Finally, test the result for physical sense.

State the two closure assumptions the balance requires. Over a full year on a natural watershed, the change in storage ΔS — soil moisture, groundwater, snowpack and channel storage — is small compared with the annual fluxes, and it is conventional to set it to zero when the balance is written over a complete hydrologic year that starts and ends in the same season. Second, if the watershed divide is also the groundwater divide, then all subsurface flow leaves through the river section where the discharge is gauged and is already counted in R; the separate deep-groundwater term G is then zero. Both assumptions are standard and both are stated here as required by the question: $$\Delta S = 0, \qquad G = 0 \qquad\Longrightarrow\qquad ET = P - R$$
Convert the river discharge to an equivalent depth over the watershed. Taking a year as 365 days, the annual runoff volume and the depth it represents are $$V_R = Q_r\,\Delta t = 500\ \frac{\text{m}^3}{\text{s}} \times 31\,536\,000\ \text{s} = 1.5768 \times 10^{10}\ \text{m}^3$$ $$\boxed{R = \frac{V_R}{A} = \frac{1.5768 \times 10^{10}\ \text{m}^3}{1.0 \times 10^{10}\ \text{m}^2} = 1.5768\ \text{m} = 1576.8\ \text{mm}}$$ This number is firm: it follows from two unambiguously printed values and nothing else.
Substitute into the balance exactly as printed, and test the result. With P = 100 mm as printed, $$ET = P - R - G - \Delta S = 100 - 1576.8 - 0 - 0 = -1476.8\ \text{mm}$$ A negative evapotranspiration is physically impossible — it would require the atmosphere to deliver water to the watershed by a route other than precipitation. The failure is not marginal: the runoff depth exceeds the stated rainfall by a factor of $$\frac{R}{P} = \frac{1576.8}{100} = 15.8$$ so the printed data cannot be reconciled by any adjustment to the assumptions. No choice of ΔS or G rescues them, because both of those terms can only make ET smaller, not larger.
Identify which datum must be in error, and correct it. The conservation requirement is P ≥ R for any watershed over a long averaging period. Since R is fixed by the area and the discharge, and the area is the one value corroborated by the phrase “10,000 km² surface area” twice implied in the question, the printed precipitation depth is the term that must be wrong — almost certainly a lost digit. A basin of 10,000 km² that actually yields 1576.8 mm of runoff is a wet maritime basin; the coastal watersheds of British Columbia are of exactly this character, with annual precipitation of roughly 2000–3000 mm and runoff ratios above 0.7. Adopting P = 2000 mm/a as representative of such a basin: $$\boxed{ET = P - R = 2000 - 1576.8 = 423.2\ \text{mm/a}}$$ which corresponds to a runoff coefficient R/P = 0.788 and an annual evapotranspiration volume of 4.232 × 10⁹ m³. An ET of about 420 mm/a is a credible value for a cool, wet, forested coastal basin, where evapotranspiration is limited by available energy rather than by available water.
State the method independently of the numbers. The general result — and the one the question is actually testing — is that annual evapotranspiration is obtained as the residual of the water balance, $$ET = P - \frac{Q_r\,\Delta t}{A} - G - \Delta S$$ Because ET is computed as a residual, it inherits the errors of every other term. That is precisely why the physical-plausibility test in Step 3 is not optional: a residual method has no internal check of its own, so the analyst must supply one.

Check — data inconsistency in the printed question, and the correction adopted. As printed, the three values (10,000 km², 100 mm/a of rain, 500 m³/s of river flow) cannot coexist: they require the basin to discharge 15.8 times more water than it receives. The runoff depth of 1576.8 mm is retained because it follows from the area and the discharge, and the precipitation figure is taken to be a transcription error for a value of at least 1577 mm; P = 2000 mm/a has been adopted as representative of a coastal-BC basin with this runoff yield, giving ET = 423.2 mm/a. Page-1 Note 1 of the paper expressly invites this: “If doubt exists as to the interpretation of any question, the candidate is urged to submit with the answer paper, a clear statement of any assumptions made.” In an examination, full marks are earned by computing the runoff depth correctly, showing that the balance cannot close, and stating the correction — not by reporting a negative evapotranspiration as an answer.

Final results — Question 7(i)
Quantity	Basis	Result
Annual runoff volume	Q_r × 31,536,000 s	1.5768 × 10¹⁰ m³
Annual runoff depth	V_R / A	1576.8 mm
ET from the data exactly as printed	P = 100 mm	−1476.8 mm — physically impossible
Runoff-to-rainfall ratio as printed	R / P	15.8 (must be ≤ 1)
ET, corrected reading	P = 2000 mm/a assumed	423.2 mm/a (4.232 × 10⁹ m³/a; runoff coefficient 0.788)

(ii) Using the energy budget to predict snowmelt infiltration (6 marks)

The energy budget of a snowpack states that the energy available to melt ice is whatever remains of the net energy exchange after the pack’s internal energy has changed. Written per unit area over a time interval,

$$Q_m = Q_{sn} + Q_{ln} + Q_h + Q_e + Q_g + Q_p - \Delta Q_s$$

where Q_sn is net shortwave radiation (incoming solar less the fraction reflected by the albedo), Q_ln is net longwave radiation, Q_h and Q_e are the turbulent sensible- and latent-heat exchanges with the air, Q_g is ground heat conducted from below, Q_p is heat advected by rain falling on the pack, and ΔQ_s is the change in the internal energy of the pack. The melt depth follows by dividing the melt energy by the latent heat required per unit depth of water,

$$M = \frac{Q_m}{\rho_w\,\lambda_f\,B}$$

with ρ_w = 1000 kg/m³, λ_f = 334 kJ/kg the latent heat of fusion, and B the thermal quality of the snow (the mass fraction that is ice, typically 0.95–0.97 for a ripe pack).

Worked example. Consider a April thaw on an open agricultural basin in southern Manitoba. A snow survey gives a snow water equivalent of 120 mm. Over a warm 24-hour period, an automatic weather station and a radiometer at the site give net shortwave radiation of 5.0 MJ/m² (the pack has aged and its albedo has fallen to about 0.6), net longwave of +0.8 MJ/m² under a warm overcast sky, sensible heat of 2.6 MJ/m² and latent heat of −0.4 MJ/m² from a 5 m/s southerly wind at 8 °C, ground heat of 0.2 MJ/m², and negligible rain-on-snow. The pack is already isothermal at 0 °C, so ΔQ_s = 0. The melt energy is then about 8.2 MJ/m², and dividing by ρ_wλ_fB ≈ 3.24 × 10⁵ J per mm of water per m² gives a melt of roughly 25 mm for the day.

That melt depth is the water arriving at the soil surface, and this is where the “other information” that the question refers to enters. The melt is not the infiltration. To predict infiltration, three further pieces of information are needed. First, the melt must be routed through the snowpack itself, which delays and smooths the delivery, and the pack’s liquid-water holding capacity (about 3–5 % of its mass) must be satisfied before any water leaves the base — a cold pack must also be ripened to 0 °C throughout before it produces any outflow at all. Second, the infiltration capacity of the soil must be evaluated for its actual condition, using an infiltration model such as Green–Ampt from part (iii); for a spring thaw this means the frozen soil condition, in which the ice-filled pores can reduce the saturated conductivity by one to two orders of magnitude, and the antecedent moisture is high from autumn recharge. Third, the arriving water is compared with that capacity at each time step: the smaller of the two infiltrates, and the excess becomes surface runoff.

In the example above, if the frozen silt loam accepts only 6 mm/d, then of the 25 mm of melt about 6 mm infiltrates and roughly 19 mm becomes overland flow. That partition — a large melt input meeting a small frozen-soil infiltration capacity — is the reason prairie spring snowmelt produces runoff ratios far higher than any summer rainstorm of comparable depth, and it is the direct engineering consequence of coupling the energy budget to an infiltration model. Where instrumentation is unavailable, the same chain is run with a calibrated degree-day melt estimate, M = C_m(T_a − T_b) with C_m of order 2–5 mm per degree-day, in place of the full energy budget; the energy-budget form is preferred whenever radiation, wind or rain-on-snow dominate, because a temperature index cannot represent any of them.

(iii) How the Green–Ampt model simulates infiltration (6 marks)

Figure 7.1 — The Green–Ampt idealisation. The wetted zone is taken as fully saturated down to a sharp front at depth L; as the front descends, the driving head becomes a smaller fraction of the travel length, so the infiltration capacity decays toward the saturated conductivity.

Green–Ampt is a physically based, approximate solution of unsaturated flow that replaces the smooth moisture profile of the exact (Richards) solution with a sharp front. The soil above the front is assumed fully saturated at moisture content θ_s; the soil below it is at its initial content θ_i; and the front advances as a piston. Applying Darcy’s law across the wetted zone, the driving head is the ponded depth plus the depth of the front plus the capillary suction head ψ acting just below it, and the travel length is the front depth L. Taking the ponded depth as negligible, this gives the infiltration capacity

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

where K_s is the saturated hydraulic conductivity of the wetted zone, Δθ = θ_s − θ_i is the moisture deficit, and F(t) = LΔθ is the cumulative infiltrated depth. Combining with continuity, dF/dt = f, and integrating gives the implicit Green–Ampt equation for cumulative infiltration,

$$F(t) - \psi\,\Delta\theta\,\ln\left(1 + \frac{F(t)}{\psi\,\Delta\theta}\right) = K_s\,t$$

which is solved for F at each time step by Newton iteration or successive substitution, after which f follows directly from the first equation.

The way the model is used in a rainfall–runoff simulation matters as much as its algebra. Three parameters are required — K_s, ψ and Δθ — and all three are estimable from soil texture: standard tables give ψ and θ_s by USDA texture class, θ_i comes from the antecedent moisture condition, and K_s from texture with a reduction factor (commonly one-half) to account for entrapped air in the wetted zone. This is the model’s practical advantage over an empirical relation such as Horton’s: its parameters have physical meaning and can be assigned to an ungauged catchment from a soil map rather than fitted to an observed hydrograph.

At each time step the simulation compares the rainfall or snowmelt supply rate with the current capacity f(t). While the supply is smaller, all of it infiltrates and no runoff is generated; the model must then track cumulative infiltration under a supply-limited rather than a capacity-limited condition. When the supply first exceeds the capacity, ponding begins — the model computes the ponding time explicitly from the cumulative depth at which f falls to the rainfall rate — and from that instant onward the infiltration equals the capacity and the surplus becomes rainfall excess. Integrating the excess over the storm gives exactly the P_e of Question 1(ii), which then drives the unit hydrograph. Between storms the model redistributes and drains the profile so that θ_i is updated for the next event.

The idealisations should be stated whenever the model is used. A sharp front is an approximation to a real, gradually varying moisture profile, and it works best in coarse, uniform soils where the front genuinely is abrupt; in layered profiles a lower-conductivity layer, or a frozen layer as in part (ii), controls the rate once the front reaches it, and the model must be applied layer by layer. Macropores, cracks and root channels can conduct water far faster than the matrix conductivity implies. And the model represents Hortonian (infiltration-excess) runoff only: on a humid, vegetated basin where runoff is generated by saturation from below, a saturation-excess formulation is the physically correct choice instead. Where those conditions are not met, the SCS curve-number method is the common alternative — simpler and thoroughly tabulated for Canadian and US soil groups, but purely empirical, with no representation of intensity or of the time distribution of the loss.

Topic. The annual water balance, the snowpack energy budget, and physically based infiltration modelling.

Key relations. Basic equation of hydrology P − R − G − ET = ΔS, with runoff depth R = Q_rΔt/A; snowmelt energy budget and M = Q_m/(ρ_wλ_fB) with λ_f = 334 kJ/kg; degree-day index M = C_m(T_a − T_b); Green–Ampt capacity f = K_s(1 + ψΔθ/F) and its implicit cumulative form.

Why this works. The water balance is nothing more than conservation of mass applied to a control volume bounded by the watershed divide and a datum below the active groundwater; because ET is the hardest term to measure directly and the easiest to obtain as a residual, this is how regional ET is routinely estimated. The energy budget is conservation of energy applied to the snowpack, and it converts a radiation and turbulent-flux measurement into a water flux through the latent heat of fusion. Green–Ampt is Darcy’s law — the same law used in Question 1(iii) — applied vertically across a saturated wetted zone, which is why the decay of infiltration capacity with time falls out of the geometry rather than being fitted.

Common pitfall. Mixing units in the water balance — leaving the discharge as a volumetric rate while the precipitation is a depth — and, more seriously, reporting a residual ET without testing it for physical sense; a residual method has no internal error check, so a negative or absurdly large ET is the signal that an input, not the arithmetic, is wrong. In snowmelt work, the recurring error is treating the computed melt as though it were all infiltration, ignoring the pack’s liquid-water holding capacity and the reduced conductivity of frozen soil.

Source. V. T. Chow, D. R. Maidment and L. W. Mays, Applied Hydrology, §1.5 (the hydrologic equation), §4.2–4.3 (energy balance and evaporation) and §4.4 (Green–Ampt infiltration); W. Viessman and G. L. Lewis, Introduction to Hydrology, 5th ed., Ch. 5 (snow hydrology and the energy budget); L. W. Mays, Water Resources Engineering, 3rd ed., Ch. 7 (infiltration and abstractions).

Question 7 of 7: Water Balance, Energy Budget and Infiltration Simulation

Question 7: Water Balance, Energy Budget and Infiltration Simulation (20 marks)

(i) Annual evapotranspiration from the basic equation of hydrology (8 marks)

(ii) Using the energy budget to predict snowmelt infiltration (6 marks)

(iii) How the Green–Ampt model simulates infiltration (6 marks)