MEMORIAL UNIVERSITY OF NEWFOUNDLAND
FACULTY OF ENGINEERING AND APPLIED SCIENCE
ENGI 9621 Soil Remediation Engineering
Assignment #1 (Due on May 29th, 2012)
1. A former metal plating shop is to be redeveloped but there is extensive contamination of
the soils by heavy metals. Which of the following technologies would you expect to be
most effective in remediating the soil contamination at this site: a) soil vapor extraction;
b) soil flushing; and c) bioremediation? Please explain. (15 marks)
Answer: The most effective remediation technology for this site should be (b) soil
flushing. It is because that
(1) Soil Vapor Extraction is always used for removal of volatile contaminants from the
subsurface environment, but heavy metals are non-volatile;
(2) Bioremediation is applied to treat the biodegradable contaminants, but heavy metals
are hard to be degradated; and
(3) Soil flushing is a process to use soil flushing reagent to wash the soil and possibly
remove not only organic matter but also heavy metals from the subsurface
environment.
2. A fully penetrating well discharges 100 gallon per minute (gpm) from an unconfined
aquifer. The original water table was recorded as 35 ft above mean sea level (MSL).
After a long time period the water table was recorded as 20 ft MSL in an observation well
located 75 ft away and 34 ft MSL at an observation well located 1500 ft away. Determine
the hydraulic conductivity of this aquifer in ft/s. (25 marks)
Solution: Q=100gpm, 1 gallon=0.134ft3, h1=20ft, r1=75ft, h2=34ft, r2=1500ft
In an unconfined aquifer, the steady radial flow
1 Ö
K = 2.81×10-4ft3 /s
3. A tank of TCE has ruptured, releasing 2.5 cubic meters of TCE. The tank is surrounded
by an impermeable dike of circular plan with a diameter of 2 meters. The TCE is held
within the containment dike, but completely infiltrates into the underlying sandy soil. The
geologic profile consists of the following: 3 meters of sand overlying a 10-cm layer of
clay which in turn overlies a gravel aquifer. The water table is one meter below ground
surface. Assume that ground surface is impermeable, the soil porosity is 0.25, and the
water filled porosity in the zone of aeration is 0.2 for the sand. Please answer the
following questions: (1) what is the total water filled volume in the subsurface sand layer;
(2) does TCE DNAPL reach the clay layer? (35 marks)
Solution:
(1) Calculate VW in the unsaturated zone
VT1 = π R2 h1 = 3.14 × (1m)2 × 1m = 3.14m3
VW1 = nw VT1 = 0.2 × 3.14 m3 = 0.628 m3
2 Calculate VW in the saturated zone above the clay layer
VT2 = π R2 h2 = 3.14 × (1m)2 × 2m = 6.28 m3
VW2 = n VT2 = 0.25 × 6.28 m3 = 1.57 m3
Calculate total Vw in the sand column
VW = VW1 + VW2 = 0.628 m3 + 1.57 m3 = 2.20 m3
(2) VW = 2.20 m3 < 2.5 m3 Î TCE DNAPL has reached the clay layer.
4. Dissolved chloride in a concentration of 1000 mg/L is being advected with flowing
groundwater at a velocity of 0.3 m/day in an aquifer with an effective porosity of 0.25.
Groundwater from the aquifer discharges into a river. What is the mass flux of chloride into
the river if the aquifer is 2 m thick and 100 m wide where it discharges into the river?
Solution:
The advective flux (mass per unit cross area per unit
time) FA = neVs C = 0.25 × 0.3m / day ×1000mg / L ×
1000 L 1g
= 75 g / m 2 / day
3
m 1000mg
Multiplied by cross sectional area, the mass flux (mass per unit time)
FA′ = FA × A = 75 g / m 2 / day × 2m × 100m = 15kg / day
5. At a landfill site, leachate accumulated over a very thick clay contains a chloride
concentration of 1000 mg/l. If the tortousity is 0.5, what would be the concentration of
chloride at a depth of 1m and 3m after 10, 100, and 1000 years of diffusion? Neglect the
effects of advection. (D0 for chlorine is 20.3 × 10-10 m2/s)
Solution: C0 = 1000mg / l , τ = 0.5 , x = 1m,3m , t = 10 yr ,100 yr ,1000 yr
The contaminant concentration at a distance x from the source at time t is calculated by
3 ⎛ x ⎞
C ( x, t ) = C0 × erfc ⎜
⎟
*
⎝2 D t ⎠
D* = D0τ = 20.3 ×10−10 m2 / s ×
( 365 × 24 × 60 × 60 ) s × 0.5 = 0.032m2 / yr
1yr
(1) At 1m, after 10 years
⎛
⎞
1m
⎜
⎟
C (1m,10 yr ) = 1000mg / l × erfc
⎜⎜
⎟
2
2 0.032m / yr ) (10 yr ) ⎟
⎝ (
⎠
= 1000mg / l × erfc ( 0.8839 ) = 1000mg / l × 0.2115 = 211mg / l
(2) At 1m, after 100 years
⎛
⎞
1m
⎜
⎟
C (1m,100 yr ) = 1000mg / l × erfc
⎜⎜
⎟
2
2 0.032m / yr ) (100 yr ) ⎟
⎝ (
⎠
= 1000mg / l × erfc ( 0.2795) = 1000mg / l × 0.6928 = 693mg / l
(3) At 1m, after 1000 years
⎛
⎞
1m
⎜
⎟
C (1m,1000 yr ) = 1000mg / l × erfc
⎜⎜
⎟⎟
2
2 0.032m / yr ) (1000 yr )
⎝ (
⎠
= 1000mg / l × erfc ( 0.0884 ) = 1000mg / l × 0.9006 = 901mg / l
(4) At 3m, after 10 years
⎛
⎞
3m
⎜
⎟
C ( 3m,10 yr ) = 1000mg / l × erfc
⎜⎜
⎟
2
2 0.032m / yr ) (10 yr ) ⎟
⎝ (
⎠
= 1000mg / l × erfc ( 2.6517 ) = 1000mg / l × 0.000183 = 0.183mg / l
(5) At 3m, after 100 years
⎛
⎞
3m
⎜
⎟
C ( 3m,100 yr ) = 1000mg / l × erfc
⎜⎜
⎟
2
2 ( 0.032m / yr ) (100 yr ) ⎟
⎝
⎠
= 1000mg / l × erfc ( 0.8385 ) = 1000mg / l × 0.2359 = 236mg / l
4 (6) At 3m, after 1000 years
⎛
C ( 3m,1000 yr ) = 1000mg / l × erfc ⎜
⎜⎜
2
⎝
⎞
⎟
2
( 0.032m / yr ) (1000 yr ) ⎟⎟⎠
3m
= 1000mg / l × erfc ( 0.2652 ) = 1000mg / l × 0.7078 = 708mg / l
5