Question 1 of 5: Solute Dispersion – Continuous and Instantaneous Sources, Degradation Half-Life

Question 1: Solute Dispersion – Continuous and Instantaneous Sources, Degradation Half-Life (20 marks: a-7, b-5, c-8)

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.

Given (a). Continuous source $C_0=100\ \text{mg/L}$; Darcy velocity $q=0.35\ \text{m/day}$; porosity $n=0.35$; longitudinal dispersivity $\alpha_L=10\ \text{m}$; effective diffusion coefficient $D^{*}=10^{-10}\ \text{m}^2/\text{s}$; observation point $x=1000\ \text{m}$; times $t=900$ and $1100$ days.

Find (a). $C(x,t)$ at $t=900$ and $1100$ days, and the conditions justifying a 1-D treatment.

Approach. This is a continuous (Type I, constant-concentration) source along a 1-D flow path, so the seepage velocity and dispersion coefficient feed directly into the Ogata-Banks solution.

1. Seepage velocity and dispersion coefficient. $$v=\frac{q}{n}=\frac{0.35}{0.35}=1.00\ \text{m/day}.$$ $$D_L=\alpha_L v+D^{*}=10(1.00)+\left(10^{-10}\times86400\right)=10.0+8.64\times10^{-6}\approx\boxed{10.0\ \text{m}^2/\text{day}}.$$ Molecular diffusion is five orders of magnitude smaller than the mechanical-dispersion term and is negligible here.
2. Ogata-Banks continuous-source solution. $$\frac{C(x,t)}{C_0}=\frac{1}{2}\left[\operatorname{erfc}\!\left(\frac{x-vt}{2\sqrt{D_Lt}}\right)+\exp\!\left(\frac{vx}{D_L}\right)\operatorname{erfc}\!\left(\frac{x+vt}{2\sqrt{D_Lt}}\right)\right].$$ The exponential-times-erfc term is evaluated with the scaled complementary error function $\operatorname{erfcx}(z)=e^{z^2}\operatorname{erfc}(z)$ to avoid overflow, since $vx/D_L=100$.
3. Evaluate at $t=900$ days. $vt=900\ \text{m}$, so the plume front has not yet fully reached $x=1000\ \text{m}$ (pure-advection travel time is $x/v=1000$ days). $$C(1000,900)=\boxed{24.9\ \text{mg/L}}.$$
4. Evaluate at $t=1100$ days. $vt=1100\ \text{m}$, past the advective arrival time, so the concentration has risen well past the mid-breakthrough value. $$C(1000,1100)=\boxed{77.2\ \text{mg/L}}.$$ The three-fold jump between 900 and 1100 days is expected, not an error — $x=1000$ m sits almost exactly on the advective front ($t=x/v=1000$ days), the steepest part of the breakthrough curve, so a $\pm10\%$ change in travel time produces a large swing in concentration.

Discussion — when is a 1-D solution appropriate? A 1-D advection-dispersion solution is valid when the contaminant source is effectively infinite (or very large) in the two directions transverse to flow relative to the travel distance of interest, so that concentration gradients perpendicular to flow are negligible and no transverse spreading dilutes the plume along the centreline — e.g. a landfill whose footprint is wide compared with the 1000 m travel distance, feeding a planar, continuous source across the full aquifer cross-section (and, for a real 3-D aquifer, also fully penetrating the aquifer thickness so there is no vertical spreading). If the source is narrow (a point or line source small compared with the travel distance), transverse and/or vertical dispersion becomes significant and a 2-D or 3-D solution (as used in part c) is required instead.

Given (b). First-order degradation, rate coefficient $k=0.001\ \text{day}^{-1}$.

Find (b). Time for 90% degradation.

Approach. First-order decay, $C=C_0e^{-kt}$; solve for $t$ when $C/C_0=0.10$.

1. Solve the first-order decay equation. $$C=C_0e^{-kt}\ \Rightarrow\ t=\frac{\ln(C_0/C)}{k}=\frac{\ln(1/0.10)}{0.001}=\frac{\ln(10)}{0.001}=\boxed{2303\ \text{days}}\ (\approx6.3\ \text{years}).$$

Given (c). Instantaneous release, dissolved chloride concentration $C_0=3500\ \text{mg/L}$ over infiltration area $A=100\ \text{m}^2$; Darcy velocity $q=0.1\ \text{m/day}$; porosity $n=0.36$; $\alpha_L=15\ \text{m}$, $\alpha_T=2\ \text{m}$; $D^{*}=10^{-10}\ \text{m}^2/\text{s}$; evaluation time $t=100$ days.

Find (c). Maximum chloride concentration and its location at $t=100$ days, and the concentration 100 m from the spill centre (along the flow centreline) at $t=100$ days.

Approach. Model the derailment as an instantaneous 2-D (horizontal) Gaussian-puff source of total mass $M=C_0\,A\,b\,$, advected at the seepage velocity and spread by longitudinal/transverse dispersion (Domenico & Schwartz's 2-D instantaneous-source solution).

1. Seepage velocity and dispersion coefficients. $$v=\frac{q}{n}=\frac{0.1}{0.36}=0.278\ \text{m/day}.$$ $$D_x=\alpha_Lv+D^{*}\approx15(0.278)=4.17\ \text{m}^2/\text{day},\qquad D_y=\alpha_Tv+D^{*}\approx2(0.278)=0.556\ \text{m}^2/\text{day}.$$
2. Instantaneous mass (with the disclosed unit-thickness assumption). $$M=C_0\,A\,b=(3.5\ \text{kg/m}^3)(100\ \text{m}^2)(1\ \text{m})=\boxed{350\ \text{kg}}.$$
3. 2-D instantaneous-source concentration field. $$C(x,y,t)=\frac{M}{4\pi t\,n\,b\sqrt{D_xD_y}}\exp\!\left[-\frac{(x-vt)^2}{4D_xt}-\frac{y^2}{4D_yt}\right].$$ The peak always sits on the flow centreline ($y=0$) at the plume's advected centroid, $x=vt$.
4. Location and value of the maximum at $t=100$ days. $$x_{\max}=vt=0.278(100)=\boxed{27.8\ \text{m downgradient of the spill, on the centreline}}.$$ $$C_{\max}=\frac{350}{4\pi(100)(0.36)(1)\sqrt{4.17\times0.556}}=\boxed{509\ \text{mg/L}}.$$
5. Concentration 100 m from the spill centre (centreline, $y=0$) at $t=100$ days. The point $x=100\ \text{m}$ lies well beyond the plume centroid ($x_{\max}=27.8\ \text{m}$), on the leading (low-concentration) tail: $$C(100,0,100)=\boxed{22.2\ \text{mg/L}}.$$