16-Civ-B4 Engineering Hydrology · May 2015

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examination, May 2015, 98-Civ-B4 Engineering Hydrology, three hours’ duration, closed book with one two-sided candidate-prepared aid sheet (8½″ × 11″) and an approved Casio or Sharp calculator whose model must be declared in the work book. Seven problems are printed; page-1 Note 4 states that any five (5) questions constitute a complete paper and that only the first five answers appearing in the work book are marked. Each problem is weighted at twenty (20) points, so the examinable total is 5 × 20 = 100 points. The page-6 marking scheme gives the sub-part split for every problem. All seven problems are solved here, because this set is a study resource rather than a timed sitting; the sub-part mark values shown below are the printed ones.

Reference texts. V. T. Chow, D. R. Maidment and L. W. Mays, Applied Hydrology (hydrologic cycle, unit hydrographs, routing, frequency analysis, infiltration); L. W. Mays, Water Resources Engineering, 3rd ed. (design application, rainfall–runoff, reservoir operation); W. Viessman and G. L. Lewis, Introduction to Hydrology, 5th ed. (measurement, areal precipitation, energy budget); V. T. Chow, Open-Channel Hydraulics (1959) (flood-wave propagation, dam-break and gradually varied unsteady flow); C. W. Fetter, Applied Hydrogeology, 4th ed. (Darcy’s law, hydraulic conductivity, advective transport). Canadian practice references: Environment and Climate Change Canada / Water Survey of Canada HYDAT archive and the ECCC Engineering Climate Datasets (IDF curves and the IDF_CC climate-adjustment tool); ISO 1100 / WMO Manual on Stream Gauging (stage–discharge rating practice); and the Canadian Dam Association Dam Safety Guidelines (inflow design flood, dam-break inundation mapping).

Check — Problem 7(i) source data are internally inconsistent. As printed, the 10 000 km² basin receives 50 mm of rain in a year while its river carries 200 m³/s, which is a runoff depth of 630.72 mm — about 12.6 times the stated rainfall. No basin can discharge more water than it receives, so the printed precipitation is in error (almost certainly a lost order of magnitude), not the discharge. Problem 7(i) below boxes the runoff depth first, demonstrates that the balance cannot close, and then adopts a declared corrected annual precipitation of 1000 mm/a to complete the estimate. Every number is flagged where the correction is used.

Question 7: 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 (10 marks)

Given.

Quantity	Symbol	Value
Watershed surface area	$A$	10 000 km² = 1.0 × 10¹⁰ m²
Annual precipitation (as printed)	$P$	50 mm/a
Annual mean river discharge at the outlet	$Q$	200 m³/s
Averaging period	$\Delta t$	1 year = 365 × 86 400 = 31 536 000 s
Governing relation	BEH	$P - R - G - E - T = \Delta S$

Find. The annual evapotranspiration depth $E + T$ over the watershed, in millimetres, together with a justified statement of the assumptions that make the estimate possible.

Figure 7.1 — The watershed control volume with the annual fluxes. Closing the balance requires $P \ge R$; the printed precipitation violates that requirement.

Approach. Convert the gauged discharge to a runoff depth over the basin, impose the two standard annual-balance assumptions ($\Delta S \approx 0$ and $G \approx 0$) to reduce the BEH to $E + T = P - R$, then test whether the printed data can satisfy conservation of mass before evaluating the residual.

Convert the annual mean discharge to an annual runoff volume. The river carries a mean of $Q = 200\ \text{m}^3/\text{s}$ for a full year, so the volume leaving the basin is $$V_R = Q\,\Delta t = 200\ \text{m}^3/\text{s} \times 31\,536\,000\ \text{s} = 6.3072 \times 10^{9}\ \text{m}^3$$
Express that volume as a depth over the watershed. Every term in the BEH must be a depth over the same area, so divide by $A$: $$R = \frac{V_R}{A} = \frac{6.3072 \times 10^{9}\ \text{m}^3}{1.0 \times 10^{10}\ \text{m}^2} = 0.63072\ \text{m} = \boxed{R = 630.72\ \text{mm/a}}$$ Equivalently the basin yields a specific discharge of $200/10\,000 = 0.02\ \text{m}^3/\text{s}$ per km², i.e. 20 L/s per km² — a value typical of a humid, snowmelt-fed Canadian Shield or Cordilleran basin, which is a useful sanity check that the discharge figure itself is reasonable.
Reduce the BEH with the two annual-balance assumptions. Over a complete water year, snowpack has melted, soil moisture has returned to a similar state and groundwater levels have recovered, so the change in storage is small compared with the fluxes: $\Delta S \approx 0$. The topographic divide is taken as a no-flow boundary for the subsurface as well, so there is no net groundwater import or export: $G \approx 0$. Both assumptions are standard for an annual balance on a basin of this size and both are stated as assumptions, as the question requires. The BEH then reduces to $$P - R - E - T = 0 \qquad\Longrightarrow\qquad E + T = P - R$$
Test conservation of mass before evaluating the residual. Substituting the printed precipitation: $$E + T = 50 - 630.72 = -580.72\ \text{mm/a}$$ Evapotranspiration cannot be negative — that would mean the atmosphere delivering water to the basin by a route other than precipitation. The printed data are therefore internally inconsistent: the runoff depth exceeds the stated rainfall by a factor of $$\frac{R}{P} = \frac{630.72}{50} = 12.6$$ whereas physically the runoff coefficient $R/P$ must lie between 0 and 1. Since $R$ follows from the area and the discharge alone — two independent, mutually consistent figures — the term in error is the printed precipitation, almost certainly by a lost order of magnitude. No answer can be reported from the data as printed.
Adopt and declare a corrected precipitation, then evaluate. Conservation requires $P \ge R = 630.72$ mm/a. Taking $P = 1000\ \text{mm/a}$ — a representative annual precipitation for a humid Canadian basin whose specific discharge is 20 L/s/km² — $$E + T = P - R = 1000 - 630.72 = \boxed{369.3\ \text{mm/a}}$$
Check the result for physical plausibility. The implied runoff coefficient is $$C = \frac{R}{P} = \frac{630.72}{1000} = 0.63$$ which is high but entirely credible for a cold, humid basin where a large fraction of the annual precipitation falls as snow and runs off rapidly at melt while the ground is still frozen. The ET of roughly 370 mm/a is consistent with published actual-evapotranspiration values for boreal and Shield catchments (typically 300–450 mm/a), where ET is energy-limited rather than moisture-limited. The estimate is therefore self-consistent under the declared correction.

Check — declared correction. The 369.3 mm/a result rests on a corrected annual precipitation of 1000 mm/a, not on the 50 mm printed in the question. The printed value is demonstrably impossible (it implies a runoff coefficient of 12.6). In an examination, the correct response is to show the mass-balance contradiction explicitly, state the conservation requirement $P \ge R$, and proceed with a clearly declared assumption — which is exactly what page-1 Note 1 of this paper invites the candidate to do. Reporting a negative evapotranspiration would be a serious error; reporting 369.3 mm/a without flagging the correction would be a different one.

Quantity	Symbol	Result
Annual runoff volume	$V_R$	6.307 × 10⁹ m³/a
Annual runoff depth over the basin	$R$	630.72 mm/a
Specific discharge	$q$	20 L/s per km²
ET from the printed data (P = 50 mm/a)	$E+T$	−580.72 mm/a — impossible; balance does not close
Required condition for closure	—	$P \ge R = 630.72$ mm/a
ET with the declared corrected P = 1000 mm/a	$E+T$	369.3 mm/a
Implied runoff coefficient	$C = R/P$	0.63
Assumptions invoked	—	$\Delta S \approx 0$ over the water year; $G \approx 0$ across the divide

(ii) The energy budget equation for predicting runoff in cold climates (5 marks)

In a cold-climate basin the water is present but immobilised: the winter’s precipitation is stored as snow and ice, and runoff is controlled not by rainfall but by the energy available to melt that store. A water balance alone cannot predict the timing of the freshet; an energy balance can. The snowpack energy budget is

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

where $Q_m$ is the energy available for melt, $Q_{sn}$ the net short-wave (solar) radiation, $Q_{ln}$ the net long-wave radiation, $Q_h$ the sensible-heat flux from the air, $Q_e$ the latent-heat flux (condensation adds energy, sublimation removes it), $Q_g$ the ground heat flux, $Q_p$ the heat advected by rain falling on snow, and $\Delta U$ the change in the pack’s internal (cold-content) energy. Melt depth follows by dividing the melt energy by the latent heat of fusion and the pack’s thermal quality $B$:

$$M = \frac{Q_m}{\rho_w \lambda_f B}, \qquad \lambda_f \rho_w = 334\ \text{kJ per mm per m}^2 = 0.334\ \text{MJ/m}^2\text{ per mm}$$

Worked example. A 60 km² alpine sub-basin is fully snow-covered in late April. Meteorological measurements over one day give net short-wave radiation of 8.2, net long-wave of −2.6, sensible heat of +1.4, latent heat of −2.6, and ground heat of +0.1 MJ/m²/d, so the net melt energy is $Q_m = 4.5\ \text{MJ/m}^2/\text{d}$. With a ripe pack of thermal quality $B = 0.97$,

$$M = \frac{4.5}{0.334 \times 0.97} = \boxed{13.9\ \text{mm/d}}$$

Converting to a basin yield with the “other watershed information” — the snow-covered area, the snow-water equivalent from snow-course or snow-pillow surveys, and the routing lag:

$$Q_{\text{melt}} = \frac{M\,A}{\Delta t} = \frac{0.0139\ \text{m} \times 60 \times 10^{6}\ \text{m}^2}{86\,400\ \text{s}} = \boxed{9.6\ \text{m}^3/\text{s}}$$

of meltwater generation, which is then reduced by any refreeze and retention in the pack, routed through the soil and channel network with the basin’s unit hydrograph, and added to baseflow to give the predicted discharge hydrograph. Repeating the calculation day by day through the melt season, and depleting the measured snow-water equivalent as melt proceeds, produces the full freshet hydrograph — the calculation that Canadian river-forecast centres run every spring.

Why the energy approach matters in cold climates. It captures the phenomena a temperature-index (degree-day) method misses: a rain-on-snow event, where $Q_p$ and greatly increased $Q_h$ and $Q_e$ under warm, humid, windy conditions produce melt rates several times the radiation-driven value and generate the most severe floods in coastal British Columbia; the cold-content term, which explains why a pack must first be ripened to isothermal 0 °C before any melt leaves it, delaying runoff by days; the effect of albedo, which falls from about 0.9 for fresh snow to 0.4 for old, dirty snow and thereby accelerates melt late in the season; and aspect and forest-cover differences, which the radiation terms represent directly. The trade-off is data: the energy budget needs radiation, humidity and wind measurements that a temperature-index model does not, which is why the degree-day method $M = C_m(T_a - T_b)$ remains in operational use where only temperature is available.

(iii) Key features of the Green–Ampt infiltration model in a developing watershed (5 marks)

Green–Ampt is a physically based infiltration model, in contrast to the empirical Horton equation. It idealises the wetting process as a sharp front separating soil that is saturated behind it from soil at its initial moisture content ahead of it, and then applies Darcy’s law across the wetted zone. That idealisation gives

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

where $f$ is the infiltration rate, $F$ the cumulative infiltration, $K_s$ the saturated hydraulic conductivity, $\psi$ the suction head at the wetting front and $\Delta\theta = \theta_s - \theta_i$ the moisture deficit.

Figure 7.2 — (a) The Green–Ampt idealisation: a sharp front with saturation behind it. (b) Infiltration capacity decays toward $K_s$; runoff begins at ponding, when capacity first falls below the rainfall intensity.

Key features. The infiltration capacity decays with cumulative infiltration, not with clock time. As $F$ grows, the wetting front deepens, the suction gradient across the wetted zone weakens, and $f$ falls asymptotically toward $K_s$ — gravity-driven flow. This is the physical mechanism behind the empirical Horton curve, and it means the model responds correctly to intermittent rainfall, which a purely time-based curve does not. Parameters are physical, not fitted. $K_s$, $\psi$ and $\theta_s$ are soil properties available from texture-class tables (Rawls et al.), so the model can be applied to a soil map without a calibration record — decisive for a developing watershed with no gauged history of the post-development condition. Antecedent moisture enters explicitly through $\Delta\theta$, so a wet basin correctly infiltrates less. Ponding is predicted, not assumed: runoff begins when the capacity first falls to the rainfall intensity, at cumulative infiltration and time

$$F_p = \frac{\psi\,\Delta\theta}{(i/K_s) - 1}, \qquad t_p = \frac{F_p}{i}$$

Illustration. With $K_s = 10$ mm/h, $\psi = 110$ mm, $\Delta\theta = 0.25$ (so $\psi\Delta\theta = 27.5$ mm) and a steady rainfall of $i = 25$ mm/h, the capacity once 30 mm has infiltrated is $f = 10(1 + 27.5/30) = 19.2$ mm/h, and ponding begins at

$$F_p = \frac{27.5}{(25/10) - 1} = 18.3\ \text{mm}, \qquad t_p = \frac{18.3}{25} = 0.73\ \text{h} = \boxed{44\ \text{minutes}}$$

after which the excess $i - f$ becomes runoff.

Application to a developing watershed. The model quantifies exactly what development changes. Impervious cover removes infiltration entirely over its footprint, so the runoff coefficient is a straightforward area-weighted combination of pervious Green–Ampt cells and impervious cells. Construction compacts the remaining pervious soil, reducing $K_s$ by an order of magnitude and shortening $t_p$ dramatically — a mechanism that a simple curve-number approach hides inside a lumped parameter but that Green–Ampt represents directly, and that explains why post-development flooding is often worse than the impervious fraction alone suggests. Because the parameters are physical, low-impact-development measures can be evaluated on the same basis: an infiltration trench or bioswale is modelled by restoring $K_s$ and $\Delta\theta$ over its contributing area, and the reduction in runoff volume comes out of the same equations. The model’s limitations should be stated alongside: it assumes a homogeneous soil profile with uniform initial moisture and a genuinely sharp front, so it does not represent layered soils, macropore or preferential flow, surface crusting, or frozen ground — the last a significant restriction in Canada, where infiltration into frozen soil during a spring melt is small and must be handled by a separate frozen-ground routine.

Topic. Conservation of mass and energy applied to a basin — annual water balance, snowmelt energy budget, and physically based infiltration.

Key relations. BEH $P - R - G - E - T = \Delta S$; runoff depth $R = Q\Delta t/A$; snowmelt $M = Q_m/(\rho_w\lambda_f B)$ with $\rho_w\lambda_f = 0.334$ MJ/m² per mm; Green–Ampt $f = K_s(1 + \psi\Delta\theta/F)$ and $F_p = \psi\Delta\theta/((i/K_s)-1)$.

Why this works. All three parts are conservation statements written for different quantities and different control volumes. The annual balance conserves mass over a basin and a year, which is what lets the hardest-to-measure flux (ET) be recovered as a residual from the two easiest (rainfall and discharge). The snowmelt budget conserves energy over the snowpack, which is the right currency in a cold climate because water availability is not the constraint — the phase change is, and 334 kJ is required per kilogram regardless of how the energy arrives. Green–Ampt conserves mass in the soil column while applying Darcy’s law across the wetted zone, so the declining infiltration capacity emerges from physics rather than curve-fitting. Because each is a conservation law, each also supplies its own error check: a negative residual, as in part (i), is proof the inputs are wrong rather than a result to be reported.

Common pitfall. Reporting a residual ET without testing it for physical plausibility. A residual absorbs every error in every other term, so it must always be checked against the sign and magnitude expected for the climate — a negative ET, or a runoff coefficient above one, means the data are inconsistent and the analysis must stop and say so. In the snowmelt calculation, the corresponding pitfall is applying a melt rate to a pack that has not yet ripened: until the cold content is satisfied, meltwater refreezes within the pack and no runoff appears at the base.

Source. Chow, Maidment & Mays, Applied Hydrology, Ch. 2 (hydrologic balance), Ch. 3 (evaporation and the energy budget), Ch. 4 (Green–Ampt infiltration, ponding time); Viessman & Lewis, Introduction to Hydrology, 5th ed., Ch. 5–6 (snow hydrology and the energy budget); Rawls, Brakensiek & Miller (1983) for Green–Ampt parameters by soil texture.

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