Question 7 of 7: Energy budget equation and infiltration simulation

Problem 7: Energy budget equation 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.

Part (i) — Three terms of the energy budget equation and their hydrologic meaning (10 marks)

The equation printed on the paper is a statement of conservation of energy at a surface. Everything on the right-hand side is an energy flux per unit area of surface per unit time, with dimensions M T⁻³ and SI units of W/m²; the first bracket collects the fluxes delivered to the surface and the second those removed from it, and the difference $E_s$ is what remains stored in the surface layer. Because latent heat is the currency in which evaporation and melt are paid, an energy budget is simultaneously a water budget, and that correspondence is what the question asks about.

The three terms selected here are $R_t$, $E_e$ and $H_n$; each is defined, then connected to the hydrologic cycle.

Term 1 — $R_t$, the incoming short-wave (solar) radiation absorbed at the surface. This is the flux of solar radiation reaching the surface, comprising the direct beam and the diffuse sky component, after depletion by scattering and absorption in the atmosphere and by cloud. It varies with latitude, season, time of day, slope and aspect of the surface, atmospheric turbidity and cloud cover, and in Canadian conditions its daily mean ranges from a few tens of W/m² in midwinter at high latitude to over 300 W/m² on a clear summer day. In the printed equation it appears as a gain, and the companion term $R_r$ in the loss bracket is the portion reflected away, determined by the surface albedo — roughly 0.05–0.10 for open water, 0.15–0.25 for vegetation and soil, and 0.4–0.9 for snow depending on age and contamination. Correspondence with the hydrologic cycle: absorbed short-wave radiation is the principal energy source driving the entire cycle. It supplies the latent heat that evaporates water from oceans, lakes, soil and interception storage and that transpires it through vegetation, thereby powering the atmospheric limb of the cycle; and in spring it supplies the heat that ripens and melts the snowpack, initiating the freshet that dominates the annual hydrograph of most Canadian rivers. The strong dependence of $R_r$ on albedo is why melt accelerates non-linearly once a snowpack begins to darken: less reflection means more absorption, which means faster melt and further darkening.

Term 2 — $E_e$, the long-wave radiation emitted by the surface. Every surface above absolute zero radiates in the thermal infrared according to the Stefan–Boltzmann law, $E_e = \varepsilon\sigma T_s^4$, where $\varepsilon$ is the surface emissivity (0.95–0.99 for water, snow, soil and vegetation), $\sigma = 5.67\times10^{-8}$ W m⁻² K⁻⁴ and $T_s$ is the absolute surface temperature. For a melting snow surface at 273 K, taking an emissivity of 0.98, this amounts to about 310 W/m², a very large loss, which is partly offset by the atmospheric counter-radiation $E_a$ in the gain bracket. The net long-wave exchange $E_a - E_e$ is therefore usually negative under clear skies and near zero or positive under a warm overcast, which is the physical reason that clear cold nights produce refreezing of a snowpack and that overcast, humid conditions produce rapid melt. Correspondence with the hydrologic cycle: emitted long-wave radiation is the mechanism by which the surface cools at night, and surface cooling to the dew point is what produces dew, frost and radiation fog — small but genuine terms in the precipitation and interception side of the cycle. More importantly for design, it governs the diurnal cycle of snowmelt, producing the characteristic daily rise and fall in the freshet hydrograph, with melt ceasing overnight when long-wave loss exceeds the daytime radiative gain.

Term 3 — $H_n$, the sensible heat flux between the surface and the atmosphere. This is the turbulent transfer of heat driven by the temperature difference between the surface and the overlying air, $H_n \propto \rho c_p C_H U (T_a - T_s)$, where $U$ is wind speed and $C_H$ a bulk transfer coefficient. It is a loss from the surface when the surface is warmer than the air (a summer afternoon over dry ground) and a gain when the air is warmer than the surface, which is exactly the case over a snowpack held at 0 °C beneath warm advected air. Its magnitude scales with wind speed, so it is the dominant melt term in a warm windy rain-on-snow event and a minor one in calm clear conditions. Correspondence with the hydrologic cycle: sensible heat is the partner of latent heat in the surface energy partition, and the way the available energy divides between the two — the Bowen ratio $\beta = H/L_E$ — directly determines how much of the incoming radiation goes into evaporating water rather than warming air. A low Bowen ratio over a lake or a well-watered forest means a large evapotranspiration term in the basin water balance; a high one over dry ground means a small one. In snow hydrology, sensible heat delivered by warm winds is the term that turns an ordinary melt into a destructive rain-on-snow flood, the single most important flood-generating mechanism in coastal British Columbia.

For completeness, the remaining terms are: $E_a$, the incoming long-wave counter-radiation from atmospheric water vapour, carbon dioxide and cloud; $R_r$, the reflected short-wave radiation, equal to the albedo times $R_t$; and $R_1$, the residual flux group comprising latent heat carried away by evaporation or sublimation, conduction to the ground beneath, and heat advected by rainfall. The residual $E_s$ is the rate of change of heat stored in the surface layer, and it is this stored energy that first warms a snowpack to the isothermal condition and then, once the pack is ripe, supplies the latent heat of fusion for melt.

Part (ii) — Estimating peak snowmelt runoff from the energy budget (5 marks)

The link between the energy budget and runoff is the latent heat of fusion: energy that remains at the surface of a ripe snowpack, after all the gains and losses of part (i) have been counted, converts snow to water at 334 kJ per kilogram. The melt depth over an interval is therefore

with $Q_m$ the net energy flux available for melt in W/m², $\rho_w = 1000$ kg/m³ and $\lambda_f = 334\,000$ J/kg. Converting that melt depth into a peak discharge then requires the assumed information the question refers to: the basin area, the fraction of the basin that is snow-covered and ripe, a runoff factor accounting for retention within the pack and infiltration into frozen or unfrozen ground, and the daily distribution of melt.

Given (assumed, representative values for a spring day on a small Canadian basin). Net energy available for melt $Q_m = 100$ W/m², sustained over the $\Delta t = 10$ h of significant daytime energy input; latent heat of fusion $\lambda_f = 334$ kJ/kg; density of water $\rho_w = 1000$ kg/m³; basin area $A = 25$ km², fully snow-covered with the pack already ripe and isothermal at 0 °C; runoff factor 0.9, the ground being frozen so that infiltration is small; melt assumed uniform over the 10-hour period, with negligible routing delay for this order-of-magnitude estimate.

Find. The daily melt depth and the resulting peak runoff at the basin outlet.

1. Convert the net energy flux into energy received per unit area. Over 10 hours, $$E = Q_m\,\Delta t = 100\ \text{W/m}^2 \times (10 \times 3600)\ \text{s} = 3.60\times10^{6}\ \text{J/m}^2$$
2. Convert energy to melt depth through the latent heat of fusion. $$M = \frac{E}{\rho_w \lambda_f} = \frac{3.60\times10^{6}}{1000 \times 334\,000} = 0.01078\ \text{m} = \boxed{10.8\ \text{mm}}$$ Cross-check with the degree-day method, the empirical shortcut used when radiation data are unavailable: with a melt factor $C_m = 4$ mm per °C per day and an air temperature 3 °C above the 0 °C base, $M = C_m(T_a - T_b) = 12$ mm/day, the same order as the energy-budget estimate, which confirms that the assumed flux is realistic.
3. Express the melt as a rate. Spread uniformly over the 10-hour melt period, $$m = \frac{10.78\ \text{mm}}{10\ \text{h}} = 1.078\ \text{mm/h}$$
4. Convert the melt rate over the basin into a discharge. Applying the runoff factor and the basin area, $$Q_p = \frac{0.9 \times (1.078\times10^{-3}\ \text{m/h}) \times (25\times10^{6}\ \text{m}^2)}{3600\ \text{s/h}} = \boxed{6.74\ \text{m}^3/\text{s}}$$ This is the peak of the melt-driven runoff, occurring in mid-afternoon when the energy input is greatest; the observed hydrograph would be this value attenuated and lagged by routing through the snowpack, the soil and the channel network.
5. State what would change the answer materially. Rain falling on the ripe pack adds both its own volume and its advected heat; a warm, windy, humid air mass raises the sensible and latent gains sharply and can double $Q_m$; partial snow cover reduces the contributing area; and unfrozen ground reduces the runoff factor well below 0.9. A design analysis would repeat the calculation for the critical combination and route the result rather than reporting the unrouted peak.

Part (iii) — The Horton infiltration model (5 marks)

Horton observed that the rate at which a soil can absorb water declines from a high initial value as rain continues, approaching a constant asymptote once the surface layer is saturated and gravity drainage governs. He represented this by an exponential decay of the infiltration capacity,

where $f(t)$ is the infiltration capacity at time $t$ (mm/h), $f_0$ the initial capacity of the dry soil at the start of rainfall (mm/h), $f_c$ the final or equilibrium capacity, approximately the saturated hydraulic conductivity (mm/h), and $k$ a decay constant with units of h⁻¹ that depends on soil type and initial moisture. Integrating gives the cumulative infiltration,

and the actual infiltration in any interval is the lesser of the capacity and the rainfall intensity, $f_{actual} = \min\left[i(t),\,f(t)\right]$, with the excess $i - f$ becoming the effective rainfall that the unit hydrograph of Problem 2 converts to direct runoff.

Given. A silty-loam soil with $f_0 = 75$ mm/h, $f_c = 12$ mm/h and $k = 2.5$ h⁻¹ (assumed, representative values).

Find. The infiltration capacity and the cumulative infiltration one hour into a ponded storm.

1. Evaluate the capacity from the decay equation. $$f(1) = 12 + (75-12)\,e^{-2.5(1)} = 12 + 63(0.08208) = \boxed{17.17\ \text{mm/h}}$$ The capacity has fallen to less than a quarter of its initial value within one hour, which is why the runoff from a long storm greatly exceeds what its average intensity minus a constant loss would suggest.
2. Evaluate the cumulative infiltration by integration. $$F(1) = 12(1) + \frac{75-12}{2.5}\left(1 - e^{-2.5}\right) = 12 + 25.2(0.91792) = 35.13\ \text{mm}$$ If the storm delivered 60 mm in that hour, the effective rainfall available for direct runoff would be $60 - 35.13 = 24.9$ mm.

Assumptions. (1) Rainfall intensity always exceeds the infiltration capacity, so the surface is ponded from the start and the soil infiltrates at capacity throughout — the equation describes capacity, not actual infiltration, and the two coincide only under ponding. (2) The soil is homogeneous and the process one-dimensional and vertical, with no lateral flow and no restricting layer within the depth wetted. (3) The three parameters are constant throughout the event, which presumes a stable surface and a single antecedent moisture condition. (4) Water is freely available at the surface and the surface itself does not change — no crusting, no macropore collapse, no frozen layer forming or thawing. (5) The soil is deep enough that the profile does not saturate from below during the event.

Limitations in engineering application. The model is empirical: its three parameters have no direct physical measurement and must be calibrated from infiltrometer tests or from rainfall-runoff records, which limits transferability between sites and makes them effectively fitting constants. Because the decay is written as a function of time rather than of accumulated infiltrated depth, the equation continues to decay during a rainless interval within a storm and cannot represent recovery of capacity during that hiatus without an explicit recovery routine — a serious defect for multi-peaked storms and for continuous simulation, and the reason SWMM and similar packages add an empirical regeneration term. It also gives a physically wrong answer for a storm whose intensity is less than the capacity, because the soil then infiltrates at the rainfall rate and the capacity decays more slowly than the equation predicts; the fix is to track cumulative infiltration and shift the time origin accordingly. It takes no account of soil-moisture profile, water-table depth, or the vertical redistribution of moisture, so it cannot predict the saturation-excess (Dunne) runoff that dominates in humid, low-relief and riparian settings — it represents only infiltration-excess (Hortonian) runoff. Finally, it cannot represent frozen ground, macropore or preferential flow, or spatial variability of soil across a catchment, all of which are first-order effects in Canadian basins during the spring freshet. Where a physically based alternative is warranted, the Green–Ampt model, $f = K\left(1 + \psi\,\Delta\theta/F\right)$, is preferred because its parameters — saturated conductivity, wetting-front suction head and moisture deficit — are measurable soil properties and it is expressed in terms of cumulative infiltration rather than time.