16-Civ-B4 Engineering Hydrology · May 2016

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

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

Notes on this paper

Paper format. 98-Civ-B4 Engineering Hydrology, May 2016 — a three-hour closed-book examination; a candidate-prepared two-sided aid sheet and one approved Casio or Sharp calculator are permitted. The cover page states that “any five(5) questions constitute a complete paper” and that “each question is equally weighted at twenty (20) points”, so the seven printed Problems each carry 20 marks towards a 100-mark paper. The page-6 Marking Scheme confirms the sub-part split for all seven. Note 1 invites the candidate to state any assumptions made where a question is open to interpretation; this sitting needs that licence twice, and both places are flagged in a callout below. All seven Problems are worked here, because this set is a study resource rather than a timed sitting.

Reference texts. V. T. Chow, D. R. Maidment and L. W. Mays, Applied Hydrology (hydrologic cycle, unit hydrograph theory, Horton infiltration, level-pool and Muskingum routing, frequency analysis); W. Viessman and G. L. Lewis, Introduction to Hydrology, 5th ed. (areal precipitation, hydrograph analysis, conceptual watershed models); P. B. Bedient, W. C. Huber and B. E. Vieux, Hydrology and Floodplain Analysis, 5th ed. (rating curves, reservoir and river routing, urban design storms); R. S. Gupta, Hydrology and Hydraulic Systems, 4th ed. (groundwater recharge and discharge areas, streamflow measurement); L. W. Mays, Water Resources Engineering, 3rd ed. (Rational Method, IDF design practice). For the Canadian frame: Environment and Climate Change Canada IDF curve files, the Water Survey of Canada Hydrometric Manual (mid-section gauging to ISO 748), and the Transportation Association of Canada Drainage Manual for design-storm and runoff-coefficient practice.

Check — two source-data issues and one declared convention.

(1) Problem 1(iii) gives the IDF relation as i = 6.0 − 0.3 t_d without stating the units of i. The paper's own Problem 6(iii) figure plots rainfall on an axis labelled “Rainfall and Infiltration, mm/h” with a peak near 13, so mm/h is adopted and the alternative reading is carried through in the answer as a sensitivity.

(2) Problem 7(i) as printed cannot be satisfied: the stated area and river discharge fix the runoff depth at 5045.76 mm/a, which is 63 times the 80 mm/a of rain the question supplies, so the residual evapotranspiration comes out large and negative. The answer boxes the runoff depth, demonstrates that the balance cannot close, and then adopts a declared corrected precipitation. This is the response Note 1 asks for.

(3) Problems 3(iii), 5(iii), 6(i) and 7(iii) ask for method, not arithmetic; each is worked on a small dataset that is the solver's own representative example, clearly labelled as illustrative. Every number in those examples, and every number taken from the real source data.

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 (7 marks)

Given. A basin's area, its annual precipitation as printed, and the mean annual discharge of the river draining it:

Given data — Problem 7(i)
Quantity	Symbol	Value
Basin surface area	A	20 000 km² = 2.0 × 10¹⁰ m²
Annual precipitation, as printed	P	80 mm/a
Mean annual river discharge	Q	3200 m³/s
Basic equation of hydrology	—	P − R − G − E − T = ΔS

Find. The annual evapotranspiration depth in millimetres, as the residual of the water balance, with every assumption justified.

Approach. Convert the river discharge to a runoff depth over the basin area, reduce the balance to a residual for ET by justifying the neglect of the groundwater and storage terms, and test whether the result is physically admissible before quoting it.

Reduce the balance to a residual for ET. Over a full year on a basin of this size, two terms may be justifiably dropped. The change in storage $\Delta S$ averages to zero over a water year, provided the year is not part of a drought or recovery trend, because soil moisture, snowpack and groundwater all return to a similar state twelve months later. The net groundwater flux $G$ across the divide is also taken as zero: for a topographically closed basin the groundwater divide normally coincides with the surface divide, and the groundwater that does leave the basin does so through the gauged river as baseflow, where it is already counted in $R$. Combining evaporation and transpiration into ET, $$\text{ET} = E + T = P - R$$
Convert the river discharge to an equivalent depth over the basin. The annual volume carried by the river is $$V_R = Q\,t = 3200\ \text{m}^3\text{/s} \times 31\,536\,000\ \text{s} = 1.0092 \times 10^{11}\ \text{m}^3$$ using a 365-day year. Spreading that volume over the basin area gives the runoff depth $$R = \frac{V_R}{A} = \frac{1.0092 \times 10^{11}}{2.0 \times 10^{10}} = 5.0458\ \text{m} = \boxed{5045.8\ \text{mm/a}}$$ Equivalently, the specific discharge is $3200/20\,000 = 0.16$ m³/s per km², or 160 L/s per km² — a very wet basin, of the kind found on the outer Pacific coast, but a physically possible one.
Test the balance before quoting a residual. Substituting the printed precipitation, $$\text{ET} = P - R = 80 - 5045.8 = -4965.8\ \text{mm/a}$$ A negative evapotranspiration is impossible: it would require the basin to condense water out of the atmosphere at sixty times the rate at which rain falls on it. The data as printed are therefore internally inconsistent, and the inconsistency must be resolved before an estimate can be given. Note which term is at fault: $R$ follows from the area and the discharge alone, both of which are ordinary measured quantities, whereas 80 mm/a of precipitation is a desert value that no river of 3200 m³/s could drain. The printed precipitation is the term in error, and it appears to be short by roughly two orders of magnitude.
Adopt a declared corrected precipitation and complete the estimate. Conservation requires $P \ge R$, and a basin yielding 5046 mm/a of runoff is a wet Pacific-coast basin, for which an annual precipitation near 5500 mm and an annual evapotranspiration of 400 to 500 mm are the values reported in the literature. Adopting $$P = 5500\ \text{mm/a}$$ gives $$\text{ET} = 5500 - 5045.8 = \boxed{454.2\ \text{mm/a}}$$ with a runoff coefficient $R/P = 5045.8/5500 = 0.917$ — high, as it must be for a steep, thin-soiled, cool coastal basin where little of the rainfall is available for evaporation.
Sanity-check the adopted answer against climate. An ET of 454 mm/a is about 1.24 mm/d averaged over the year, which is consistent with a cool, cloudy maritime climate whose net radiation supports only modest evaporation, and it is well below the 700 to 900 mm/a typical of southern Ontario or the Prairies. The estimate is therefore self-consistent as well as conservation-consistent, which is the strongest statement available from a residual method.

Check — the printed data cannot close, and the assumption adopted. As printed (A = 20 000 km², P = 80 mm/a, Q = 3200 m³/s) the balance requires ET = −4966 mm/a, which is impossible. The answer above adopts P = 5500 mm/a, on the grounds that the runoff depth is fixed by the two measured quantities while the rainfall figure is not, and that the resulting ET and runoff coefficient are those of a real wet coastal basin. A second self-consistent reading is that the discharge carries the error: with Q = 320 m³/s and P = 800 mm/a the runoff depth is 504.6 mm/a and ET = 295.4 mm/a, with a runoff coefficient of 0.63 — a plausible interior basin. Either correction is defensible; what is not defensible is shipping a negative evapotranspiration. Note 1 on the cover page expressly asks the candidate to state such an assumption with the answer paper, and in an examination either reading would earn the marks provided the impossibility is identified and the correction declared.

Final results — Problem 7(i)
Quantity	Value
Annual runoff volume	1.0092 × 10¹¹ m³
Runoff depth over the basin, R	5045.8 mm/a
Specific discharge	160 L/s per km²
ET from the data as printed	−4965.8 mm/a — impossible
Adopted corrected precipitation	5500 mm/a (declared assumption)
Evapotranspiration, ET = P − R	454.2 mm/a
Runoff coefficient, R/P	0.917
Alternative reading (Q = 320 m³/s, P = 800 mm/a)	R = 504.6 mm/a, ET = 295.4 mm/a

(ii) The greenhouse effect from the energy budget equation (7 marks)

The energy budget at the land surface is a conservation statement exactly parallel to the water balance. Writing the net radiation absorbed at the surface and the ways it is spent, $$R_n = R_s(1 - \alpha) + R_L\!\downarrow - R_L\!\uparrow = H + LE + G_s$$ where $R_s$ is incoming short-wave (solar) radiation, $\alpha$ the surface albedo, $R_L\!\downarrow$ the long-wave radiation emitted downward by the atmosphere, $R_L\!\uparrow$ the long-wave radiation emitted upward by the surface, $H$ the sensible heat flux, $LE$ the latent heat flux that drives evaporation, and $G_s$ the heat conducted into the ground.

Figure 7.1 — The atmosphere passes short-wave radiation almost freely but absorbs the long-wave radiation the surface emits, and re-radiates part of it downward. That downward term is the greenhouse effect.

The greenhouse effect is the direct consequence of the two optical properties the question supplies. Because the atmosphere is transparent to solar radiation, short-wave energy reaches the surface almost undiminished and is absorbed there, apart from the fraction $\alpha$ reflected. The surface warms, and as a body near 288 K it radiates according to the Stefan-Boltzmann law $$R_L\!\uparrow = \varepsilon \sigma T_s^{4}$$ with the emission concentrated in the infrared, near a peak wavelength of about 10 µm. Because the atmosphere is opaque to infrared radiation — the absorption bands of water vapour, carbon dioxide, methane and nitrous oxide cover most of that spectrum — almost none of this emission escapes directly to space. It is absorbed within the atmosphere, which warms and re-radiates in all directions, and roughly half of that re-radiation is directed downward. That downward stream is $R_L\!\downarrow$, the back radiation, and it is a second energy input to the surface additional to the sun.

The consequence is a surface energy balance that closes at a higher temperature than solar radiation alone could sustain. With no infrared-absorbing atmosphere, the surface would need to satisfy $R_s(1 - \alpha) = \varepsilon \sigma T_s^4$, which gives an effective radiating temperature of about 255 K, or −18 °C — an ice-covered planet. Adding $R_L\!\downarrow$ to the left-hand side requires $T_s$ to rise until the increased upward emission balances the increased input, and the observed global mean is about 288 K, or 15 °C. The 33 K difference is the natural greenhouse effect, and it is the reason liquid water, and therefore the hydrologic cycle, exists at all.

Two hydrologic consequences follow, and they are why the topic appears on a hydrology paper. First, increasing the concentration of infrared-absorbing gases thickens the blanket: the atmosphere's effective emitting level rises to colder air, less energy escapes at a given surface temperature, $R_L\!\downarrow$ increases, and the balance re-closes warmer. Second, the extra energy does not appear only as temperature. Because the surface budget must still balance, a larger $R_n$ is partitioned into larger $H$ and $LE$, and a larger latent flux is an intensified hydrologic cycle — more evaporation, a warmer atmosphere holding about 7 per cent more water vapour per kelvin by the Clausius-Clapeyron relation, and therefore heavier extreme rainfall. This is the physical basis for the climate-change adjustment factors now applied to IDF curves in Canadian municipal design, and for the shift from snowmelt-dominated to rain-dominated flood regimes in coastal British Columbia.

(iii) Two key features of Horton's infiltration model (6 marks)

Given. Horton's model for infiltration capacity, as printed in the figure accompanying the question, together with an illustrative parameter set and storm that are the solver's own representative values:

Table 7.1 — Illustrative Horton parameters and storm (solver's own example data)
Quantity	Symbol	Value
Initial (dry-soil) infiltration capacity	f₀	45 mm/h
Final (saturated) infiltration capacity	f_c	6 mm/h
Decay constant	k_t	0.9 h⁻¹
Storm hyetograph, four 1-hour blocks	P	8, 22, 15, 5 mm (total 50 mm)
Observed direct runoff	R	18 mm

Find. Two key features of the model as a predictor of infiltration capacity, demonstrated on the illustrative data, and the φ index for the same storm.

Figure 7.2 — Reproduction of the figure accompanying Problem 7(iii). Horton's curve is the soil's capacity, falling from f₀ towards f_c; the φ index is the constant loss rate that reproduces the same runoff volume. The hatched blocks are the rainfall excess.

Approach. State Horton's equation, then demonstrate its two defining features — the exponential decay between two physical bounds, and its status as a capacity rather than an actual rate — and contrast it with the φ index shown in the same figure.

State the model. Horton's equation, as printed in the figure, is $$f(t) = f_c + (f_0 - f_c)\,e^{-k_t t}$$ where $f(t)$ is the infiltration capacity at time $t$ from the start of rainfall, $f_0$ its initial value on the antecedent soil, $f_c$ its asymptotic value, and $k_t$ a decay constant with units of inverse time.
Key feature 1: the capacity decays exponentially between two physically meaningful bounds. The model is an exponential relaxation from the dry-soil capacity to the saturated capacity, with the decay driven by the filling of surface pores, the swelling of colloids, the breakdown of soil aggregates by raindrop impact, and above all the flattening of the hydraulic gradient across the wetting front as it advances. The two bounds are the useful part: $f_0$ depends on the antecedent condition and so changes storm to storm, while $f_c$ approaches the saturated hydraulic conductivity of the soil and is a stable property of it. On the illustrative parameters, $$f(1\ \text{h}) = 6 + (45 - 6)e^{-0.9} = \boxed{21.9\ \text{mm/h}}, \qquad f(3\ \text{h}) = 6 + 39\,e^{-2.7} = 8.6\ \text{mm/h}$$ so the capacity falls to within 1 mm/h of $f_c$ after about 4.1 hours. Integrating gives the cumulative infiltration, which is what a runoff volume calculation actually needs: $$F(t) = f_c t + \frac{f_0 - f_c}{k_t}\bigl(1 - e^{-k_t t}\bigr), \qquad F(2\ \text{h}) = 12 + 43.33(1 - e^{-1.8}) = 48.2\ \text{mm}$$
Key feature 2: it predicts a capacity, not the actual infiltration rate. This is the feature most often misused. The actual rate is $$f_{\text{actual}}(t) = \min\bigl[\,i(t),\ f(t)\,\bigr]$$ so while rainfall intensity is below the curve, all the rain infiltrates and the curve is not being followed; the soil is wetting more slowly than its capacity allows, and the capacity decays more slowly than the equation predicts because the decay is really driven by cumulative infiltration rather than by clock time. The equation is exact only for ponded conditions from t = 0. In practice this means Horton's curve must be shifted or re-indexed against cumulative infiltration when applied to a real intermittent storm — the reason the physically based Green-Ampt model, which is written in terms of $F$ rather than $t$, is preferred in modern continuous simulation. The remaining practical limitation is that $f_0$, $f_c$ and $k_t$ are three parameters that must be fitted to a measured infiltrometer or rainfall-runoff record; they cannot be looked up with confidence.
Contrast with the φ index in the same figure. Where a full loss curve is unnecessary, the φ index replaces it with the single constant loss rate that reproduces the observed runoff volume: $$\sum \max\bigl[\,P_j - \phi\,\Delta t,\ 0\,\bigr] = R$$ Solving for the illustrative storm requires a short trial. Assuming all four blocks contribute gives $\phi = (50 - 18)/4 = 8.0$ mm/h, which is inconsistent because the 5 mm block lies below it; assuming three contribute gives $\phi = (8 + 22 + 15 - 18)/3 = 9.0$ mm/h, still inconsistent because the 8 mm block lies below it. Assuming only the two largest blocks contribute, $$\phi = \frac{22 + 15 - 18}{2} = \boxed{9.5\ \text{mm/h}}$$ which is consistent: both contributing blocks exceed 9.5 mm/h and both excluded blocks fall below it. The excess is then $(22 - 9.5) + (15 - 9.5) = 18$ mm as observed, and the total abstraction is 50 − 18 = 32 mm. The comparison is instructive: φ is a bookkeeping device that lumps retention and infiltration into one constant and guarantees the right volume, whereas Horton's model attempts to describe the physical process and therefore also gets the timing of the excess approximately right — which is what a design hydrograph, as opposed to a design volume, requires.

Final results — Problem 7(iii) illustrative case
Quantity	Value
Infiltration capacity at t = 1 h	21.9 mm/h
Infiltration capacity at t = 3 h	8.6 mm/h
Cumulative infiltration to t = 2 h, F	48.2 mm
Time for f to fall within 1 mm/h of f_c	4.1 h
φ index for the illustrative storm	9.5 mm/h (two contributing blocks)
Rainfall excess reproduced	18 mm (matches the observed runoff)
Total abstraction	32 mm of the 50 mm storm

Topic. The water balance as a residual method, the surface energy budget and the greenhouse effect, and Horton infiltration versus the φ index.

Key relations. Water balance $P - R - G - E - T = \Delta S$, reducing annually to $\text{ET} = P - R$; runoff depth $R = Qt/A$; energy budget $R_n = R_s(1 - \alpha) + R_L\!\downarrow - R_L\!\uparrow = H + LE + G_s$; Stefan-Boltzmann $R_L\!\uparrow = \varepsilon \sigma T_s^4$; Horton $f = f_c + (f_0 - f_c)e^{-k_t t}$ with $F = f_c t + \frac{f_0-f_c}{k_t}(1 - e^{-k_t t})$; φ index $\sum \max[P_j - \phi \Delta t, 0] = R$.

Why this works. All three parts are conservation arguments, and the value of a conservation argument is that it constrains what an answer is allowed to be. The water balance lets the least measurable flux, ET, be recovered from the two best measured ones — but only if the result is admissible, which is why the negative residual in part (i) is information rather than an inconvenience. The energy budget does the same for heat: adding a downward long-wave term forces the surface temperature up, because the budget must close. Horton's model is the one part that is empirical rather than conservative, and that is precisely why it must be checked against the actual rainfall rate: an empirical capacity curve carries no guarantee of consistency with the storm on the day.

Common pitfall. In the balance, forgetting to convert the discharge to a depth over the basin area (comparing m³/s with mm/a directly), or accepting a negative ET because the arithmetic was done without a physical check. In the energy budget, describing the greenhouse effect as trapped heat or as reflected sunlight instead of as absorbed and re-radiated infrared. In the Horton part, applying the curve as the actual infiltration rate when rainfall intensity is below it, which over-predicts total infiltration and under-predicts the excess.

Source. Chow, Maidment and Mays, Applied Hydrology, Ch. 1 and Ch. 2 (hydrologic and energy balances), §4.3 (Horton) and §5.4 (φ index); Viessman and Lewis, Introduction to Hydrology, Ch. 4 (infiltration) and Ch. 5 (evaporation); Gupta, Hydrology and Hydraulic Systems, Ch. 2; Environment and Climate Change Canada, Canadian climate normals for regional ET and precipitation values.

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