18-Env-B3 Contaminant Transport · December 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2015 — 04-Env-B3 / Contaminant Transport. 3 hours duration; closed-book exam (any non-communicating calculator permitted). Five problems are printed, each worth 25 marks; per the exam’s own Note 3, only the first four as they appear in the answer book constitute a complete marked paper, and Note 5 states that the sub-parts (a)–(d) of each problem can be treated independently. All five problems are solved below for completeness.
Reference texts. Freeze & Cherry, Groundwater; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); Cooper & Alley, Air Pollution Control: A Design Approach (4th ed.); Wark, Warner & Davis, Air Pollution: Its Origin and Control (3rd ed.).
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.
Advection is the bulk transport of a dissolved or suspended chemical carried along with the moving fluid, at the fluid’s own velocity — mass moves because the parcel of water or air itself is moving, the way a leaf is carried downstream by a river current. It is directional, follows the mean flow field, and occurs regardless of whether a concentration gradient exists.
Diffusion (molecular diffusion, or its turbulent analogue, dispersion) is the net transport of mass down a concentration gradient, driven by random molecular motion (or, at the larger scale relevant to rivers and the atmosphere, random turbulent eddy motion). It occurs even in still fluid with zero bulk velocity, always moves from high to low concentration, and spreads/dilutes a plume rather than translating it bodily downstream.
The key differentiator: advection needs a bulk fluid velocity and carries mass at the flow’s own speed and direction irrespective of concentration; diffusion needs a concentration gradient and moves mass down that gradient irrespective of whether the fluid is moving at all.
Advective mass flux (one-dimensional, flux per unit cross-sectional area per unit time):
$$J_{adv} = u\,C$$
where u is the fluid (or particle) velocity and C is the local concentration.
Diffusive mass flux (Fick’s First Law):
$$J_{diff} = -D\,\dfrac{\partial C}{\partial x}$$
where D is the (molecular or turbulent) diffusion/dispersion coefficient; the negative sign encodes transport down-gradient. Combining both mechanisms into the one-dimensional advection–diffusion (mass-conservation) equation:
$$\dfrac{\partial C}{\partial t} = -u\,\dfrac{\partial C}{\partial x} + D\,\dfrac{\partial^{2} C}{\partial x^{2}}$$
Given. Stream depth H = 3 m; stream (horizontal) velocity u = 2 m/s; particle diameter d = 350 µm (radius r = 175 µm); particle density ρs = 2600 kg/m³; kinematic viscosity νf = 1.3×10−2 cm²/s = 1.3×10−6 m²/s at 10 °C; the exam’s own Stokes formula above.
Find. The horizontal distance the particle travels before it settles all the way to the river bottom.
The fluid (water) density ρf is not given explicitly — it is taken as the standard 1000 kg/m³, consistent with the implied dynamic viscosity μ = ρfνf ≈ 1.3×10−3 Pa·s, which matches water’s known viscosity near 10 °C. Separately, the resulting particle Reynolds number (computed in Step 1 below) is Re ≈ 22, above the Re < 1 range in which Stokes’ law is strictly valid. The exam supplies this formula explicitly and directs its use, so it is applied literally as instructed below; in practice, drag in this transitional regime is somewhat higher than Stokes predicts, so the true settling velocity would be a little lower and the true travel distance a little greater than the values boxed here.
Approach. Use the exam’s own Stokes formula to get the particle’s vertical settling velocity, then apply simple plug-flow kinematics: the horizontal distance travelled equals the stream velocity multiplied by the time needed to fall the full depth at that settling velocity.
| Quantity | Value |
|---|---|
| Settling velocity, wf | 0.0822 m/s |
| Time to settle full depth, t | 36.5 s |
| Horizontal travel distance, x | 73.0 m |
