1) (10 points) A spill of a hazardous chemical has occurred on a road parallel to a river. Two rows of wells were installed parallel to the road/river. The hydraulic head of all wells in row 1 (near the road) is 200m and in row 2 (near the river) 150m. The contaminated water is flowing from the spill site towards the river. The wells of row 2 are planned to be used to pump the contaminated water away so that it does not reach the river. The pump rate must be at least as high as the rate of groundwater flow towards the river.
a) (5 points)
Calculate this rate
by using Darcy's law. The hydraulic conductivity is 1x10-3m/s,
the distance between the two rows of wells 2000m, the thickness of the
aquifer 100m and the width of the aquifer 1000m.
Q= A k (h2 -h1) /L
where A = cross-sectional area ( 100 m x 1000m)
K = hydraulic conductivty (1 x 10 -3 m/s)
h2 = hydraulic head of well #2 (150 m)
h1 = hydraulic head of well #1 (200m)
L = distance between the wells (2000m)
so Q= (100m x 1000m) (1 x 10 -3 m/s) (150 m - 200m) / 2000m = 2.5 m3/s
b) (5 points) The road is 2000m distant from the wells in row 2. How much time will pass after the spill until the contaminated water will reach the wells in row 2? The porosity of the aquifer is 0.3. (time = distance/velocity!)
The time that will pass until the contaminated water reaches the wells in row 2:
q = Q/A = 2.5 m3/s /100,000 m 2 = 2.5 x
10
-5
m/s
V = q /n = 2.5 x 10 -5 m/s / 0.3 = 8.3 x 10 -5 m/s
t = d/v = 2000 m/ 8.3 x 10 -5 m/s = 2.41 x 10 7 seconds or 278.9
days