Column Breakthrough, Vapour-Water Partitioning, and Retarded/Decaying Transport: 18-Geol-B1 Contaminant Hydrogeology, December 2017, Q2

Question 2: Column Breakthrough, Vapour-Water Partitioning, and Retarded/Decaying Transport (20 marks: a-7, b-6, c-7)

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). Sand: $n=0.33$, $\rho_b=1.8\ \text{g/cm}^3$, $f_{oc}=0.007$. PCE: $\log K_{ow}=0.29$, linear sorption. Column: diameter 5 cm, length $x=80$ cm, flow rate $Q=5.2\ \text{mL/min}$, input $C_0=50\ \text{mg/L}$, dispersivity $\alpha_L=5$ cm, $D^{*}=10^{-6}\ \text{cm}^2/\text{s}$; evaluation time $t=4$ days.

Find (a). Effluent PCE concentration at $t=4$ days.

Approach. Convert $K_{ow}$ to $K_{oc}$ (Karickhoff correlation), get $K_d$ and the retardation factor $R$, compute the column's seepage velocity and dispersion coefficient from the pump rate, then apply the retarded Ogata-Banks solution.

Sorption and retardation. $\log K_{oc}=\log K_{ow}-0.21=0.29-0.21=0.08\Rightarrow K_{oc}=1.20\ \text{mL/g}$ (Karickhoff, 1981). $$K_d=K_{oc}f_{oc}=1.20(0.007)=0.00842\ \text{mL/g}.$$ $$R=1+\frac{\rho_b}{n}K_d=1+\frac{1.8}{0.33}(0.00842)=\boxed{1.046}.$$ PCE's low given $\log K_{ow}$ makes this sand only very weakly retarding.
Column seepage velocity and dispersion coefficient. Cross-section $A=\pi d^2/4=\pi(5)^2/4=19.6\ \text{cm}^2$. Darcy flux $q=Q/A=5.2/19.6=0.265\ \text{cm/min}=381\ \text{cm/day}$. $$v=\frac{q}{n}=\frac{381}{0.33}=\boxed{1156\ \text{cm/day}}.$$ $$D_L=\alpha_Lv+D^{*}=5(1156)+\left(10^{-6}\times86400\right)\approx5778\ \text{cm}^2/\text{day}.$$
Retarded transport parameters and pore volumes exchanged. $$v_r=\frac{v}{R}=\frac{1156}{1.046}=1105\ \text{cm/day},\qquad D_r=\frac{D_L}{R}=\frac{5778}{1.046}=5525\ \text{cm}^2/\text{day}.$$ In 4 days the retarded front travels $v_rt=1105(4)\approx4420\ \text{cm}$, i.e. $55.2$ column lengths. Equivalently, the delivered water volume $Qt=5.2(1440)(4)=29{,}950\ \text{mL}$ is $57.8$ pore volumes of the column's $\approx518\ \text{mL}$ pore space — either way, breakthrough is long past.
Effluent concentration (Ogata-Banks, retarded). Since the retarded front is ~54 column-lengths past the outlet, the column is essentially at full breakthrough well before day 4: $$C(80,4)=\boxed{50.0\ \text{mg/L}}\ (\approx C_0).$$

Check: at $Q=5.2$ mL/min the column's own pore volume (roughly $A\,L\,n\approx19.6\times80\times0.33\approx518$ mL) is flushed in only about 100 minutes even before retardation, so the true breakthrough time (a couple of hours) is far shorter than the 4-day question horizon. Reporting $C\approx C_0$ is the correct physical answer, not a rounding artifact of the erfc evaluation — the underlying error-function terms were confirmed to evaluate to within $10^{-6}$ of their asymptotic limits.

Given (b). Gas volume $V=1\ \text{m}^3$, TCE mass in gas $m=42\ \text{g}$, $MW=131\ \text{g/mol}$, $T=20^\circ\text{C}=293.15\ \text{K}$; dimensionless Henry's coefficient $H=0.42$; (context) $P_v^o=8590\ \text{Pa}$, solubility $=1100\ \text{mg/L}$.

Find (b). Partial pressure and equilibrium aqueous concentration of TCE.

Approach. The ideal-gas law gives the partial pressure from the molar TCE content of the gas phase; the given dimensionless Henry's coefficient ($C_{gas}/C_{water}$) then converts the gas-phase mass concentration to the equilibrium water concentration.

Moles of TCE and partial pressure. $$n_{TCE}=\frac{42}{131}=0.3206\ \text{mol}.$$ $$P_{TCE}=\frac{n_{TCE}RT}{V}=\frac{0.3206(8.314)(293.15)}{1}=\boxed{781\ \text{Pa}}.$$ Well below the pure-liquid vapour pressure ($8590$ Pa), confirming no pure-TCE vapour is present — consistent with a dissolved/sorbed-source vapour, not free product.
Gas-phase mass concentration and equilibrium aqueous concentration. $$C_{gas}=\frac{m}{V}=\frac{42\ \text{g}}{1\ \text{m}^3}=42\ \text{mg/L}.$$ $$C_{water}=\frac{C_{gas}}{H}=\frac{42}{0.42}=\boxed{100\ \text{mg/L}}.$$ Well below the 1100 mg/L solubility limit, confirming a physically consistent single-phase (dissolved) equilibrium with no residual TCE liquid.

Given (c). PCE from Q2(a) (same $K_{oc}$/$K_d$), first-order biodegradation half-life $t_{1/2}=600$ days; Type I boundary, transport along the plume centreline; evaluation point $x=30$ m, $t=1000$ days.

Find (c). PCE concentration 30 m from the landfill boundary after 1000 days.

Check: Q2(c) supplies no Darcy velocity, dispersivity, or boundary concentration of its own for this landfill/aquifer — it only ties the SOIL properties to Q2(a) ($n=0.33$, $\rho_b=1.8\ \text{g/cm}^3$, $f_{oc}=0.007$, hence the same PCE $K_d$ and $R=1.046$ as part (a)). Since Q1(a) is the only other "landfill leaking to an aquifer" scenario in this paper, its Darcy velocity (0.35 m/day), dispersivity (10 m) and boundary concentration (100 mg/L) are adopted here as the transport/boundary inputs, per the exam's own Note 1 inviting a stated assumption when data is ambiguous. A different assumed velocity/dispersivity/$C_0$ rescales the numbers below but not the method.

Approach. Extend the Ogata-Banks solution with retardation ($v\to v/R$, $D\to D/R$) and first-order decay via the decay factor $\gamma=\sqrt{1+4\lambda D_r/v_r^2}$ (Bear's retarded-decay solution).

Decay constant and retarded transport parameters. $$\lambda=\frac{\ln 2}{600}=1.155\times10^{-3}\ \text{day}^{-1}.$$ Using $v=1.00\ \text{m/day}$, $D_L=10.0\ \text{m}^2/\text{day}$ (Q1(a)) and $R=1.046$ (Q2(a)): $$v_r=\frac{v}{R}=0.956\ \text{m/day},\qquad D_r=\frac{D_L}{R}=9.56\ \text{m}^2/\text{day}.$$
Decay factor. $$\gamma=\sqrt{1+\frac{4\lambda D_r}{v_r^2}}=\sqrt{1+\frac{4(1.155\times10^{-3})(9.56)}{0.956^2}}=\boxed{1.024}.$$
Retarded-decay concentration at $x=30$ m, $t=1000$ days. $$\frac{C}{C_0}=\frac{1}{2}e^{\frac{v_rx}{2D_r}(1-\gamma)}\operatorname{erfc}\!\left(\frac{x-v_rt\gamma}{2\sqrt{D_rt}}\right)+\frac{1}{2}e^{\frac{v_rx}{2D_r}(1+\gamma)}\operatorname{erfc}\!\left(\frac{x+v_rt\gamma}{2\sqrt{D_rt}}\right).$$ $$C(30,1000)=100(0.965)=\boxed{96.5\ \text{mg/L}}.$$ At $x=30$ m the retarded plume front (which has advanced roughly $v_rt\gamma\approx980$ m in 1000 days) is far downgradient, so the near-boundary concentration has barely fallen below $C_0$ despite the decay term — 30 m is simply too close to the source, relative to the plume's travel distance, for either dispersion or 600-day decay to matter much.

Question 2 — Final Results
Item	Result
2(a) $K_d$, $R$	0.00842 mL/g, $R=1.046$
2(a) Effluent PCE concentration at 4 days	50.0 mg/L (full breakthrough)
2(b) TCE partial pressure	781 Pa
2(b) Equilibrium aqueous TCE concentration	100 mg/L
2(c) PCE concentration at $x=30$ m, $t=1000$ days	96.5 mg/L (assumed inputs, see the check note)

Concept & Theory

Topic. Sorption-retarded column transport, vapour-water equilibrium partitioning, and decay-extended 1-D solute transport.

Key relations. $\log K_{oc}=\log K_{ow}-0.21$ (Karickhoff, 1981); $K_d=K_{oc}f_{oc}$; $R=1+\dfrac{\rho_b}{n}K_d$; ideal gas $PV=nRT$; dimensionless Henry's law $C_{gas}=H\,C_{water}$; retarded-decay transport $\gamma=\sqrt{1+4\lambda D_r/v_r^2}$.

Why this works. Linear equilibrium sorption partitions the solute instantaneously between water and solids in fixed proportion $K_d$, which is mathematically equivalent to slowing the solute's own advection and dispersion by the constant factor $R$ — the retarded solute obeys the SAME transport equation as the unretarded one, just with $v/R$ and $D/R$ in place of $v$ and $D$. First-order decay adds a sink term that is handled by replacing the plume's effective advance rate with $v_r\gamma$ (faster than $v_r$ alone) while attenuating the leading exponential, which reduces to the plain Ogata-Banks solution as $\lambda\to0$ ($\gamma\to1$).

Common pitfall. Assuming a lab column with a high pump rate behaves like a slow natural aquifer — a small-diameter column driven at a few mL/min can flush dozens of pore volumes per day, so "after 4 days" may simply mean "long past breakthrough," not "partway through it." A second pitfall is dividing only $v$ by $R$ while forgetting $D$, which desynchronizes the front's spreading from its advance.

Source. Fetter, Contaminant Hydrogeology Ch.2 (K_ow-K_oc correlations, retardation) & Ch.6 (Henry's law, gas-water partitioning); Domenico & Schwartz, Physical and Chemical Hydrogeology Ch.4 (retarded, decay-extended transport).

Question 2 of 5: Column Breakthrough, Vapour-Water Partitioning, and Retarded/Decaying Transport

Question 2: Column Breakthrough, Vapour-Water Partitioning, and Retarded/Decaying Transport (20 marks: a-7, b-6, c-7)