8 Exercise Solutions

Exercise 1 Solution

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

Details: Equation 7 can be re-arranged to

where f is the proportion of end-member 1, c₁ and c₂ are the respectively end-member concentrations and c_m is the concentration of the mixed sample. Using c_m = 380 mg/L, c₁ = 25 mg/L and c₂ = 500 mg/L gives f = (-120)/(-475) = 0.25. Using c_m = -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 × 10^–4, 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 m^–1. From Equation 9, the horizontal velocity is the inverse of the age gradient, which equals 4.57 m y^–1.

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

v = Ki/n_e

so

K = vn_e/i

i = (h₁–h₂)/(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

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, c_p = 10 TU, c_d = 5 TU and λ= 0.0558 y^–1. The groundwater age is therefore calculated to be 7.3 years.

Return to exercise 4