Question 3 of 5: Retarded Transport Velocity and a PCE Column Breakthrough with Biodegradation

Question 3: Retarded Transport Velocity and a PCE Column Breakthrough with Biodegradation (equal value)

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) Linear (seepage) velocity $v=16\ \text{cm/day}$, $K_d=6.6\ \text{mL/g}$, $n=0.37$, solids density $\rho_s=2.64\ \text{g/cm}^3$. (b) Column diameter 6 cm, length $L=110\ \text{cm}$, $n=0.36$, bulk density $\rho_b=1.85\ \text{g/cm}^3$, $f_{oc}=0.015$, influent $C_0=50\ \text{mg/L}$ PCE, $\log K_{ow}=2.88$, $\log K_{oc}=-0.21+\log K_{ow}$, flow rate $Q=0.35\ \text{L/hr}$, dispersivity $\alpha_L=0.06\ \text{m}$, effective diffusion $D^{*}=10^{-10}\ \text{m}^2/\text{s}$, $t=3\ \text{days}$. (c) Same column, first-order biodegradation rate $\lambda=0.09\ \text{day}^{-1}$.

Find. (a) Retarded transport (contaminant) velocity. (b) PCE effluent concentration after 3 days, ignoring decay. (c) The same, with first-order biodegradation.

Approach. (a) Build the retardation factor from $K_d$ and the bulk/porosity ratio, then divide groundwater velocity by it. (b) Estimate $K_{oc}$ (and hence $K_d$, $R$) from the given log-linear correlation, compute pore velocity and dispersion coefficient from the column's flow geometry, then evaluate the Ogata-Banks 1-D advection-dispersion solution at the column outlet. (c) Re-evaluate the same solution with its first-order-decay extension, using the identical retarded transport parameters.

1. Part (a) — retardation factor and transport velocity. Bulk density from porosity and grain density: $\rho_b=\rho_s(1-n)=2.64(1-0.37)=1.663\ \text{g/cm}^3$. $$R=1+\frac{\rho_b}{n}K_d=1+\frac{1.663}{0.37}(6.6)=\boxed{30.7}.$$ $$v_c=\frac{v}{R}=\frac{16}{30.7}=\boxed{0.522\ \text{cm/day}}.$$ The contaminant moves about 30× slower than the water itself, since it repeatedly partitions onto the solid phase as it is carried along.
2. Part (b) — sorption parameters for PCE in the column. $\log K_{oc}=-0.21+2.88=2.67\Rightarrow K_{oc}=10^{2.67}=467.7\ \text{mL/g}$. $$K_d=K_{oc}f_{oc}=467.7(0.015)=7.02\ \text{mL/g},\qquad R=1+\frac{\rho_b}{n}K_d=1+\frac{1.85}{0.36}(7.02)=\boxed{37.1}.$$
3. Flow, pore velocity, and dispersion coefficient. Column cross-section $A=\pi(3\ \text{cm})^2=28.27\ \text{cm}^2$. Darcy flux $q=Q/A=(0.35\ \text{L/hr}\times1000\times24)/28.27=297.1\ \text{cm/day}$, pore (seepage) velocity $v=q/n=297.1/0.36=825.2\ \text{cm/day}$. Dispersion coefficient $D_L=\alpha_L v+D^{*}=(6\ \text{cm})(825.2)+(10^{-6}\ \text{cm}^2/\text{s})(86400\ \text{s/day})=4951.6+0.09\approx4952\ \text{cm}^2/\text{day}$ (molecular diffusion is negligible next to mechanical dispersion at this flow rate). Retarded parameters: $$v_R=\frac{v}{R}=\frac{825.2}{37.1}=\boxed{22.3\ \text{cm/day}},\qquad D_R=\frac{D_L}{R}=\frac{4952}{37.1}=\boxed{133.6\ \text{cm}^2/\text{day}}.$$ At this retarded velocity the column's own breakthrough time ($C/C_0=0.5$) is $L/v_R=110/22.3=4.94\ \text{days}$ — so the 3-day observation point asked for sits before the midpoint of the breakthrough curve.
4. Effluent concentration after 3 days (Ogata-Banks). $$\frac{C}{C_0}=\frac12\text{erfc}\!\left(\frac{L-v_Rt}{2\sqrt{D_Rt}}\right)+\frac12\exp\!\left(\frac{v_RL}{D_R}\right)\text{erfc}\!\left(\frac{L+v_Rt}{2\sqrt{D_Rt}}\right).$$ With $L=110\ \text{cm}$, $t=3\ \text{d}$, $v_R=22.3$, $D_R=133.6$: the first argument is $(110-66.8)/(2\sqrt{400.8})=1.079$ (erfc $\approx0.128$ from Table G.2), so the first half-term contributes $\tfrac12(0.1272)=0.0636$. The second argument is $(110+66.8)/(2\sqrt{400.8})=4.415$ and $v_RL/D_R=18.33$, so $\exp(v_RL/D_R)\,\text{erfc}(4.415)=0.0390$ (evaluated in scaled form, $e^{a-b^2}\text{erfcx}(b)$, to avoid overflow) and its half-term adds $\tfrac12(0.0390)=0.0195$. Summing the two half-terms, $$\frac{C}{C_0}=0.0636+0.0195=\boxed{0.0831},\qquad C=0.0831\times50\ \text{mg/L}=\boxed{4.15\ \text{mg/L}}.$$
5. Part (c) — adding first-order biodegradation. With decay rate $\lambda=0.09\ \text{day}^{-1}$ acting on the same retarded transport, the governing PDE gains a $-\lambda C$ sink term, and the analytical (van Genuchten/Bear) solution becomes $$\frac{C}{C_0}=\frac12\exp\!\left[\frac{v_R L}{2D_R}(1-\gamma)\right]\text{erfc}\!\left(\frac{L-v_Rt\gamma}{2\sqrt{D_Rt}}\right)+\frac12\exp\!\left[\frac{v_R L}{2D_R}(1+\gamma)\right]\text{erfc}\!\left(\frac{L+v_Rt\gamma}{2\sqrt{D_Rt}}\right),\quad \gamma=\sqrt{1+\frac{4\lambda D_R}{v_R^2}}.$$ Here $\gamma=\sqrt{1+4(0.09)(133.6)/22.3^2}=1.047$, and evaluating the same way (again in overflow-safe scaled form) gives $$\frac{C}{C_0}=\boxed{0.0657},\qquad C=0.0657\times50\ \text{mg/L}=\boxed{3.28\ \text{mg/L}}.$$ Biodegradation lowers the 3-day effluent concentration from 4.15 to 3.28 mg/L (about 21% lower) — the front itself is still advancing at the same retarded velocity $v_R$ (sorption and decay are independent processes here), but every parcel of water loses mass to biodegradation as it travels, so less PCE survives to reach the outlet.