Solution Exercise 4
The solution to Exercise 4 has two parts: a) and b).
- Using Equation (13) the minimum groundwater age is determined as:
The average precipitation‑weighted [latex]_{}^{3}H[/latex] for the last five years is 7.7 TU.
Agemin =[latex]\ -[/latex]ln(4.5/7.7)/(ln(2)/12.32)) = 9 years (rounded down to capture a minimum)
- Based on the plotted decay curve and [latex]_{}^{3}H[/latex] function, we can estimate a minimum groundwater age of nine years and a range of potential minimum age between four and 20 years.
The graphical solution steps, based on information in Section 3.4, are:
- Use the provided atmospheric [latex]_{}^{3}H[/latex] function.
- The data provided are annual averages and the problem statement indicates we can assume these are weighted by precipitation.
- Calculate the first point of the decay curve:
Year = 2012
Cm = 4.5 TU
- Calculate the second point of the decay curve:
C0 = Cm(exp(λΔt)
C0 = 4.5 TU (exp[latex]\left( \frac{ln(2)}{12.32\ yrs}(2012\ yrs - 1950\ yrs) \right)[/latex]
Year = 1950
C0 = 147 TU
- Therefore, the two points for plotting the decay curve are:
Year : [latex]_{}^{3}H[/latex] (TU)
2012 : 4.5 TU
1950 : 147 TU
More detail is provided in the following graphs and the spreadsheet provided on the web page for this book↗. Based on the plotted decay curve and the precipitation‑weighted [latex]_{}^{3}H[/latex] curve, the estimated minimum groundwater age is 2012−2003 = 9 years. Using the mean annual [latex]_{}^{3}H[/latex] curve, plus or minus one standard deviation, we can estimate uncertainty in minimum groundwater age. The decay curve crosses the annual mean minus one standard deviation line at about 2008. The decay curve crosses the annual mean plus one standard deviation line at about 1992. Thus, the full range of potential minimum ages is about 4 to 20 years.
A shorter period of the same graph, zoomed in on the period of time when there are intersections of the decay line and the recharge concentration curve.
