Solution Exercise 16
Variables:
[latex]T\ = \ 2.28\ x\ 10^{- 4}\ \frac{m^{2}}{s}[/latex]
[latex]S\ = \ 0.00003[/latex]
[latex]Q_{PW - 2}\ = \ 500\frac{m^{3}}{d} = 0.0058\ \frac{m^{2}}{s}[/latex]
[latex]r = 260\ m[/latex]
[latex]t = 631,152,000\ seconds\ \ \ (20\ years\ with\ 1\ year = 365.25\ days)[/latex]
Calculations:
Part 1 a:
[latex]u = \frac{r^{2}S}{4Tt} = \frac{{(260\ m)}^{2}\ \ 0.00003}{4\left( 2.28\ x\ 10^{- 4}\frac{m^{2}}{s} \right)\ 631,152,000\ seconds} = \ 3.5x10^{- 6}[/latex]
[latex]W(u)\ = \ 12[/latex]
Drawdown at 20 years:[latex]\ s = \ \frac{Q}{4\pi T}W(u) = \ \frac{0.0058\ \frac{m^{3}}{s}(12)}{4\ \pi\ 2.28x10^{- 4}\ \frac{m^{2}}{s}}\ = \ 24.3\ m[/latex]
Part 1 b:
Changed Variables:
[latex]Q_{PW - 5}\ = \ 200\frac{m^{3}}{d} = 0.0023\ \frac{m^{2}}{s}[/latex]
[latex]r\ = \ 150\ m[/latex]
[latex]t\ \ = \ 315,576,000\ seconds\ \ \ (10\ years\ with\ 1\ year = 365.25\ days)[/latex]
[latex]u = \frac{r^{2}S}{4Tt} = \frac{{(150\ m)}^{2}\ \ 0.00003}{4\left( 2.28x10^{- 4}\ \frac{m^{2}}{s} \right)\ 315,576,000\ seconds} = \ 2.3x10^{- 6}[/latex]
[latex]W(u)\ = \ 12.4[/latex]
Drawdown at [latex]10[/latex] years:[latex]\ s = \ \frac{Q}{4\pi T}W(u) = \ \frac{0.0058\ \frac{m^{3}}{s}(12.4)}{4\ \pi\ 2.28\ x\ 10^{- 4}\ \frac{m^{2}}{s}}\ = \ 10\ m[/latex]
Drawdown at 20 years due to both wells pumping: 24.3[latex]\ m + 10\ m =[/latex] 34.2 m
Head is 386.3 n ASL – 34.2 m = 352.1 m ASL which is well above the top of the aquifer at 55 m ASL.