The paper instructs candidates to answer any THREE of the FIVE questions in Section A and any TWO of the THREE questions in Section B. All eight questions are answered in full below, since this solution set is used as a complete study resource.
Question B-1: Pump-and-Treat Clean-Up Time for a TCE Spill (20 marks)
Given. Spilled volume of TCE, aquifer porosity, average groundwater (Darcy) flux, TCE aqueous solubility, TCE specific gravity, and the fraction of the spilled mass residing in the 1 m3 control volume.
Given data
Quantity
Symbol
Value
Volume of TCE spilled
Vspill
30 L
Specific gravity of TCE
SG
1.47
Aquifer porosity
n
0.30
Average groundwater (Darcy) flux
v
0.03 m/d
TCE aqueous solubility
Cs
1,100 mg/L
Fraction of spilled mass in the 1 m3 control volume
f
20%
Find. The time required for pump-and-treat to remove the TCE mass present in the stated 1 m3 control volume of aquifer.
Mass balance on the 1 m³ control volume: the TCE mass initially present is removed by extracted groundwater flowing through the volume at the solubility-limited concentration.
Approach. Compute the TCE mass residing in the control volume, then divide by the maximum (solubility-limited) mass-removal rate that the extracted groundwater flow can achieve.
Total TCE mass spilled. Specific gravity converts the spilled volume directly to mass:
$$M_{spill} = V_{spill}\times SG = 30\ \text{L}\times 1.47\ \tfrac{\text{kg}}{\text{L}} = 44.1\ \text{kg}$$
Mass present in the 1 m3 control volume. Only 20% of the spilled mass is assumed to reside within the volume being treated:
$$M_{cv} = 0.20\times 44.1\ \text{kg} = 8.82\ \text{kg} = 8{,}820{,}000\ \text{mg}$$
Extraction flow through the control volume. The stated average velocity is the Darcy flux (specific discharge); through the unit 1 m2 face of the 1 m3 cube it gives a volumetric flow rate:
$$Q = v\times A = 0.03\ \tfrac{\text{m}}{\text{d}}\times 1\ \text{m}^2 = 0.03\ \tfrac{\text{m}^3}{\text{d}} = 30\ \tfrac{\text{L}}{\text{d}}$$
Solubility-limited mass removal rate. Pump-and-treat cannot extract TCE faster than it dissolves into the flowing water, and the maximum achievable concentration in that water is the aqueous solubility:
$$\dot{M}_{removed} = C_s\times Q = 1{,}100\ \tfrac{\text{mg}}{\text{L}}\times 30\ \tfrac{\text{L}}{\text{d}} = 33{,}000\ \tfrac{\text{mg}}{\text{d}}$$
Clean-up time. Dividing the mass present by the removal rate gives the time to flush the control volume clean:
$$t = \frac{M_{cv}}{\dot{M}_{removed}} = \frac{8{,}820{,}000\ \text{mg}}{33{,}000\ \text{mg/d}} = \boxed{267\ \text{days}\ (\approx 8.9\ \text{months})}$$
