16-Civ-B4 Engineering Hydrology · December 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examination, December 2015, 98-Civ-B4 Engineering Hydrology, three hours’ duration, closed book with one two-sided candidate-prepared aid sheet (8½″ × 11″) and one approved Casio or Sharp calculator whose model designation must be written 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 will be marked; Note 5 weights each problem at twenty (20) points, so the examinable total is 5 × 20 = 100 points. All seven problems are solved here, because this set is a study resource rather than a timed sitting. Sub-part mark values below are the printed ones from the page-6 marking scheme.
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, stormwater management, reservoir operation); W. Viessman and G. L. Lewis, Introduction to Hydrology, 5th ed. (measurement, areal precipitation, energy budget, snowmelt); C. W. Fetter, Applied Hydrogeology, 4th ed. (Darcy’s law, hydraulic conductivity, aquifer flow); V. T. Chow, Open-Channel Hydraulics (1959) (flood-wave propagation and unsteady flow). Canadian practice references: Environment and Climate Change Canada / Water Survey of Canada HYDAT archive and the ECCC Engineering Climate Datasets (short-duration rainfall IDF curves and the IDF_CC climate-adjustment tool); ISO 1100-2 / WMO Manual on Stream Gauging (stage-discharge rating practice); and the Canadian Dam Association Dam Safety Guidelines (inflow design flood and flood-wave routing).
Check — illustrative data are the solver’s own. Every problem on this paper is a discussion question; none supplies numerical data. Where a short calculation is shown below it exists to demonstrate the method, and its input values are stated explicitly in a Given line as assumed, representative Canadian values. They are not exam data, and a candidate who assumed different but reasonable values and carried them through consistently would earn the same marks.
Check — page-6 marking-scheme typo on Problem 6. The printed scheme reads “6. (i) 7, (ii) 7 marks, 6 marks total”, which omits sub-part (iii) and states a total of 6. The page-5 margin marks give (7)(7)(6) and page-1 Note 5 states that every question is weighted at twenty (20) points, so Problem 6 is marked at 20 here, exactly as the other six problems are.
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.
Darcy’s law is the constitutive equation of saturated porous-media flow. It was obtained experimentally by Henry Darcy in 1856 from sand-column tests and states that the volumetric flow rate through a porous medium is directly proportional to the cross-sectional area normal to flow and to the hydraulic gradient driving the flow. In the form printed on this paper,
$$Q \;=\; A\,K\left(\frac{\Delta h}{\Delta x}\right)$$each term carries the following meaning and dimensions. A single consistent system is used throughout — here SI, with length in metres, time in seconds and the fundamental dimensions written as L for length and T for time.
| Term | Name and physical meaning | Dimensions | Typical SI unit |
|---|---|---|---|
| $Q$ | Volumetric discharge through the porous medium — the volume of water crossing the section per unit time | L3T−1 | m3/s (also m3/d in well practice) |
| $A$ | Gross cross-sectional area of the aquifer measured normal to the direction of flow — for a confined aquifer of saturated thickness $b$ and flow width $w$, $A = b\,w$. It is the total area of solids plus voids, not the pore area. | L2 | m2 |
| $K$ | Hydraulic conductivity of the porous medium — the discharge per unit area under a unit hydraulic gradient. It depends on both the medium (grain size, sorting, connectivity of pores) and the fluid (density and viscosity), through $K = k\rho g/\mu$ where $k$ is the intrinsic permeability. | LT−1 | m/s (also m/d) |
| $\Delta h$ | Difference in hydraulic head between the two ends of the flow path. Head is the mechanical energy per unit weight, $h = z + p/\gamma$; the velocity head is negligible in porous media, so head is elevation plus pressure head only. | L | m |
| $\Delta x$ | Distance measured along the flow path between the two points at which the heads are observed. | L | m |
| $\Delta h/\Delta x = i$ | Hydraulic gradient — the rate of head loss per unit distance along the flow path. It is dimensionless, which is what makes the equation dimensionally homogeneous. | dimensionless | m/m |
Dimensional homogeneity is easily checked: $A\,K\,i$ has dimensions L2 × LT−1 × 1 = L3T−1, which is exactly the dimension of $Q$. A negative sign is often written in front of the right-hand side, $Q = -KA\,\mathrm{d}h/\mathrm{d}x$, to record that flow proceeds from high head to low head, that is, down the gradient; in the arithmetic form used here the magnitude of the head drop is substituted and the sign is carried by inspection.
Two derived quantities follow immediately and are frequently confused with each other. Dividing the discharge by the gross area gives the Darcy velocity or specific discharge, $q = Q/A = K i$, which has the dimensions of a velocity but is a flux per unit gross area rather than the speed of any water particle. The speed at which a tracer or contaminant actually travels is the seepage or linear velocity, $v = q/n_e$, where $n_e$ is the effective porosity; because $n_e$ is typically 0.15–0.35 in a sand aquifer, the seepage velocity is three to seven times the Darcy velocity. For a confined aquifer it is also convenient to combine $K$ and $b$ into the transmissivity $T = K b$ (dimensions L2T−1), so that $Q = T\,w\,i$.
Darcy’s law is valid only where flow is laminar, that is, where the pore Reynolds number is below roughly unity to ten. It therefore applies to essentially all flow in sands, silts and fractured rock of ordinary aperture, but fails immediately adjacent to a heavily pumped well, in coarse gravels or karst conduits, and in very low-permeability clays where a threshold gradient may exist.
To compute the flow of water through an aquifer with this equation the procedure is as follows.
A short worked illustration fixes the procedure. Suppose a confined sand aquifer is 12 m thick, that the section of interest is 500 m wide, that a pumping test gives $K = 25$ m/d, and that two observation wells 1500 m apart along the flow path stand at heads differing by 3.0 m.
Given. $b = 12$ m, $w = 500$ m, $K = 25$ m/d, $\Delta h = 3.0$ m, $\Delta x = 1500$ m, effective porosity $n_e = 0.25$ (all assumed, representative values).
Find. The discharge through the section and the speed at which a dissolved tracer would migrate.
| Quantity | Symbol | Value |
|---|---|---|
| Cross-sectional area | $A$ | 6000 m2 |
| Hydraulic gradient | $i$ | 0.0020 (dimensionless) |
| Aquifer discharge | $Q$ | 300 m3/d = 3.47 × 10−3 m3/s |
| Darcy (specific) velocity | $q$ | 0.050 m/d |
| Seepage (linear) velocity | $v$ | 0.20 m/d |
The diagram printed with the question is the classical picture of the hydrologic cycle: moisture condenses and falls as precipitation, part of it runs off the land surface into a lake and then into a stream, part percolates below the water table and moves as groundwater flow toward the ocean, and water returns to the atmosphere by evaporation from open water and soil and by transpiration from vegetation. To convert that picture into an equation, the correct step is to draw a control volume around the parcel of land under study — here a basin, bounded laterally by the topographic divide, above by the ground surface and below by a depth at which vertical exchange can be accounted for — and to apply conservation of mass to it.
Conservation of mass over a stated time interval $\Delta t$ requires that
$$\sum \text{inflow} \;-\; \sum \text{outflow} \;=\; \text{change in storage}$$For the basin, precipitation is the only inflow, and surface runoff, groundwater outflow and evapotranspiration are the outflows, so that using only the parameters the question supplies plus evapotranspiration,
$$\boxed{\,P - R - G - E \;=\; \Delta S\,}$$or, in the equivalent forward form used for prediction, $\Delta S = P - (R + G + E)$ and hence $S_2 = S_1 + P - R - G - E$. Written as an instantaneous rate rather than over an interval the same statement reads $\mathrm{d}S/\mathrm{d}t = P - R - G - E$.
The consistent dimensional system is the point of the question. Two internally consistent choices exist, and mixing them is the classic error.
| Parameter | Meaning | Volume system | Depth system (preferred) |
|---|---|---|---|
| $P$ | Precipitation reaching the ground surface over the basin | L3 (m3) | L (mm over the basin area) |
| $R$ | Surface runoff leaving the basin at its outlet | L3 (m3) | L (mm) |
| $G$ | Net groundwater outflow across the basin boundary (deep percolation that leaves the control volume rather than returning as baseflow) | L3 (m3) | L (mm) |
| $E$ | Evapotranspiration — evaporation from soil, water and interception plus plant transpiration | L3 (m3) | L (mm) |
| $\Delta S$ | Change in water stored in the control volume: soil moisture, depression and interception storage, snowpack, and shallow saturated storage | L3 (m3) | L (mm) |
| $A_b$ | Basin area, used to convert between the two systems | L2 (km2) | L2 (km2) |
| $\Delta t$ | Accounting interval — the equation is meaningless until it is stated | T | T |
In the volume system every term is a volume in cubic metres accumulated over $\Delta t$; in the depth system every term is that volume divided by the basin area and so is a depth in millimetres, and the conversion is simply depth = volume / $A_b$. Canadian practice reports precipitation and runoff as depths in millimetres, so the depth system is normally preferred; a discharge record in m3/s is converted to runoff depth by $R = Q\,\Delta t / A_b$. If rates rather than totals are wanted, each term becomes L T−1 (mm/d, or m3/s in the volume system) and the storage term becomes $\mathrm{d}S/\mathrm{d}t$.
Two refinements are worth stating because the question invites additional parameters. First, the surface runoff $R$ measured at a gauge is not purely overland flow: it contains a baseflow component $Q_b$ fed from groundwater, so where a distinction matters the equation is written $P - (R_s + Q_b) - G_{net} - E = \Delta S$, with $R_s$ the direct runoff. Second, the size of $\Delta S$ relative to the other terms depends entirely on the length of the accounting interval, and this is what makes the equation useful. Over a single storm, $E$ is negligible and $\Delta S$ is large — it is the infiltrated water. Over a full water year in a basin with no long-term trend in storage, $\Delta S \approx 0$ and the equation collapses to the annual water balance $P \approx R + G + E$, which is how evapotranspiration is estimated as a residual when the other three are measured. Over a decade, a non-zero $\Delta S$ is the signature of groundwater mining or of a shrinking snowpack or glacier.
A numerical illustration: for an assumed southern-Ontario basin with $P = 900$ mm, $R = 350$ mm, $G = 120$ mm and $E = 400$ mm in one water year, $\Delta S = 900 - 350 - 120 - 400 = +30$ mm, that is, storage rose by 30 mm of equivalent depth over the year. The runoff coefficient for the year is $R/P = 350/900 = 0.39$.
