Question 1 of 7: Hydrologic Cycle, Effective Rainfall and Groundwater Flow

Question 1: Hydrologic Cycle, Effective Rainfall and Groundwater Flow (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) How the hydrologic cycle works, and three processes that govern a culvert design (7 marks)

The hydrologic cycle is the closed, solar-driven circulation of water among the atmosphere, the land surface and the subsurface. Net radiation supplies the latent heat that evaporates water from oceans, lakes and wet surfaces and that drives transpiration from vegetation; the combined flux is evapotranspiration. Moist air is advected, cooled and condensed, and the water returns to the surface as precipitation. At the surface the arriving depth is partitioned: part is intercepted by canopy and litter, part fills surface depressions, part infiltrates into the soil, and the remainder becomes overland flow that concentrates into rills, channels and finally the stream network. Infiltrated water either returns to the atmosphere from the root zone, moves laterally as interflow, or percolates to the water table and travels as groundwater toward a stream, lake or coastline, where it re-emerges as baseflow. Because the cycle is closed globally, every engineering intervention only redistributes water in space and time — it does not remove it, which is why storage and timing dominate drainage design.

For a hydraulic collection or conveyance system such as a highway culvert, three processes are decisive.

Precipitation intensity and its frequency. The culvert is sized for a design storm, not for an average one. What matters is the rainfall intensity sustained over a duration equal to the catchment’s time of concentration, at the return period set by the road classification — in Canadian practice typically a 10-year event for minor roadway drainage and a 50- or 100-year event for a highway crossing, read from an Environment and Climate Change Canada IDF curve for the nearest station and, increasingly, adjusted for a warming climate with the IDF_CC tool. Under-estimating intensity under-sizes the barrel; over-estimating it wastes structure and can accelerate the outlet.

Abstraction — interception, depression storage and infiltration. Only the effective rainfall reaches the inlet. The infiltration capacity of the catchment soils, their antecedent moisture, the depth of surface depressions and the canopy interception together determine the runoff coefficient. This is also the process that changes most when the catchment is developed: paving the contributing area removes almost all abstraction, so a culvert sized for the pre-development condition is quickly overwhelmed by upstream subdivision. A design must therefore state the land-use assumption it was based on.

Runoff concentration and routing — the timing of the flow. Two catchments with identical areas and identical rainfall can deliver very different peaks if their times of concentration differ. Overland-flow length, slope, channel roughness and any upstream storage set the shape of the inflow hydrograph, and the culvert responds to the peak of that hydrograph, not to the volume. Where the approach embankment ponds water, the culvert is a storage-release element and the outflow peak is attenuated below the inflow peak — a real design benefit, but only if the headwater depth stays within the allowable limit for the roadway and for upstream property.

Two further consequences follow directly from the cycle and belong in the same design decision: the baseflow contributed by groundwater fixes the low-flow condition the barrel must pass without silting, and the sediment and ice regime of the contributing area governs whether the inlet will remain hydraulically effective at the moment the design storm arrives.

(ii) A symbolic equation for effective rainfall on an undeveloped watershed (7 marks)

Effective rainfall (also called rainfall excess) is the depth of precipitation that appears as direct surface runoff. It is what is left after the abstractions have been satisfied. For an undeveloped watershed the three key processes that consume the arriving depth, in the order in which they act, are interception by vegetation and litter, storage in surface depressions, and infiltration into the soil. Writing all quantities as depths over the watershed and as cumulative totals for the storm,

where P is the gross rainfall depth (mm), I_c is the cumulative interception depth held on canopy, stems and litter, S_d is the depression-storage depth trapped in surface hollows, and F is the cumulative infiltration depth over the storm duration. All terms are non-negative and P_e is bounded below by zero: if the abstractions are not fully satisfied, no direct runoff is generated at all.

The infiltration term is the one that varies during the storm, so it is normally expressed through an infiltration-capacity function rather than as a constant. Using the instantaneous infiltration capacity f(t),

with i(t) the rainfall intensity and t_d the storm duration; the minimum operator enforces the physical rule that a soil cannot absorb faster than it is being supplied. In routine design the three abstractions are lumped into a single initial abstraction plus a constant loss rate, which gives the familiar phi-index form

where the phi-index is the uniform loss rate that makes the computed excess volume equal the measured direct-runoff volume. The two forms are the same statement at different levels of resolution: the first names the processes, the second calibrates them against an observed hydrograph. Note that evapotranspiration does not appear in either — over the few hours of a design storm it is negligible, although it dominates the same balance written over a season.

(iii) Daily groundwater flow through the confined aquifer (6 marks)

Given. A confined aquifer between two observation wells, with the data printed in the question and shown on the figure below.

Find. The daily volumetric groundwater discharge Q through the full cross-section of the aquifer, in m³/d, and the physical meaning of the hydraulic conductivity that produced it.

Approach. Apply Darcy’s law in the form printed on the paper: compute the hydraulic gradient from the two piezometric heads, compute the flow cross-section from the aquifer thickness and width, and multiply both by the hydraulic conductivity.

1. Establish the hydraulic gradient. The gradient is the head loss per unit distance travelled along the flow path, $$i = \frac{\Delta H}{\Delta L} = \frac{h_1 - h_2}{L} = \frac{100\ \text{m} - 90\ \text{m}}{1000\ \text{m}} = 0.0100$$ a dimensionless number: the piezometric surface falls one metre for every hundred metres of travel.
2. Establish the flow cross-section. Because the aquifer is confined, its saturated thickness equals the geological thickness everywhere, so the area normal to flow is constant, $$A = b\,w = 33\ \text{m} \times 7000\ \text{m} = 231\,000\ \text{m}^2$$ This is the step where the “depth of aquifer” given in the question is used; it is the thickness b in the printed formula, not a depth below ground.
3. Apply Darcy’s law. Substituting the gradient and the area into the printed expression, $$Q = K\,b\,\frac{\Delta H}{\Delta L}\,w = K\,i\,A$$ $$\boxed{Q = 1\ \frac{\text{m}}{\text{d}} \times 0.0100 \times 231\,000\ \text{m}^2 = 2310\ \text{m}^3/\text{d}}$$
4. Check the result two ways and convert. The Darcy flux (specific discharge) is $$q = K\,i = 1\ \text{m/d} \times 0.0100 = 0.0100\ \text{m/d}$$ and multiplying that flux by the same 231 000 m² returns 2310 m³/d, confirming the arithmetic. In SI flow units this is 2310 / 86 400 = 0.0267 m³/s, or 843 150 m³ per year — a modest but far from negligible yield, comparable to the demand of a small municipality.

The physical basis of hydraulic conductivity follows from the same calculation. K is the constant of proportionality between the Darcy flux and the hydraulic gradient; it has units of velocity because it is the flux that the medium would pass under a unit gradient. It is a property of the porous medium and of the fluid together,

where k is the intrinsic permeability of the matrix (units m², set by grain size, sorting, packing and the connectivity of the pore throats), and the fluid properties are density and dynamic viscosity. Physically, groundwater does not flow through the whole cross-section but through the tortuous, interconnected pore network only; K aggregates the resistance of that network into a single macroscopic coefficient. Two consequences matter in practice. First, K ranges over more than ten orders of magnitude — from about 10⁻¹¹ m/d for intact clay to 10³ m/d for clean gravel — so it is always the least certain number in a groundwater calculation and deserves a sensitivity check. Second, the actual pore-water velocity that governs contaminant travel time is faster than the Darcy flux by the reciprocal of the effective porosity, v = q / n_e; at n_e = 0.25 the water in this aquifer moves at 0.04 m/d, not 0.01 m/d.