8 Exercise Solutions

Exercise 1 Solution

Answer: The proportion of river water in piezometer P2 is calculated to be 25% using 2H values, and 20% using chloride concentrations. The corresponding proportions of regional groundwater in piezometer P2 are then 75% and 80% using 2H and chloride, respectively.

Details: Equation 7 can be re-arranged to \displaystyle f=\frac{c_m-c_2}{c_1-c_2}

where f is the proportion of end-member 1, c1 and c2 are the respectively end-member concentrations and cm is the concentration of the mixed sample. Using cm = 380 mg/L, c1 = 25 mg/L and c2 = 500 mg/L gives f = (-120)/(-475) = 0.25. Using cm = -26 ‰, c1 = -10 ‰ and c2 = -30 ‰ gives f = 4/20 = 0.2.

Return to exercise 1

Exercise 2 Solution

Answer: The velocity is 4.6 meters per year and the hydraulic conductivity is 16.3 meters per day.

Details: First use Equation 1 to calculate the groundwater age at each well. Using c0 = 90 pmC and λ = 1.21 × 104, gives t = 1077 years for the upstream well and 3921 years for the downstream well. The age gradient is then (3921 – 1077) / 13 000 years per meter = 0.219 y m1. From Equation 9, the horizontal velocity is the inverse of the age gradient, which equals 4.57 m y1.

Darcy’s Law gives the relationship between the seepage velocity and the hydraulic conductivity as:

v = Ki/ne

so

K = vne/i

i = (h1h2)/(distance between h1 and h2) = 2 m / 13000 m = 0.000154

K = 4.57 m/y × 0.2 / 0.000154 = 5943 m/y = 16.3 m/day

Return to exercise 2

Exercise 3 Solution

Answer: The recharge rate is 60 millimeters per year.

Details: Equation 9 can be re-arranged to give R=\frac{H\theta}{t}\ln \left(\frac{H}{H-z}\right)

Using H = 50 m, z = 10 m, θ = 0.2 and t = 39 years, gives R = 0.057 m/y or 57 mm/y or rounding 60 mm/yr.

Return to exercise 3

Exercise 4 Solution

Answer: The groundwater age is 7.3 years.

Working: In Equation 3, cp = 10 TU, cd = 5 TU and λ= 0.0558 y1. The groundwater age is therefore calculated to be 7.3 years.

Return to exercise 4

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