# 8 Exercise Solutions

##### Exercise 1 Solution

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

**Details:** Equation 7 can be re-arranged to

where *f* is the proportion of end-member 1, *c*_{1} and *c*_{2} are the respectively end-member concentrations and *c*_{m} is the concentration of the mixed sample. Using *c*_{m} = 380 mg/L, *c*_{1} = 25 mg/L and *c*_{2} = 500 mg/L gives *f* = (-120)/(-475) = 0.25. Using *c*_{m} = -26 ‰, *c*_{1} = -10 ‰ and *c*_{2 }= -30 ‰ gives *f* = 4/20 = 0.2.

##### 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 *c*_{0} = 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* = v*n*_{e}*/**i*

*i* = (*h*_{1}–*h*_{2})/(distance between *h*_{1} and *h*_{2}) = 2 m / 13000 m = 0.000154

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

##### 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.

##### 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.