Solution Exercise 17
Drawdown at Observation Well in Figure 55 with no-flow boundary
Variables:
[latex]Q\ = \ 1000\frac{m^{3}}{d}\ or\ 0.0116\ \frac{m^{3}}{s}[/latex]
[latex]t = 315,576,000\ seconds\ (10\ years\ assuming\ 365.25\ day[/latex]s per year
[latex]T\ = \ Kb\ = \ 0.00002\frac{m}{s}\ 20\ m\ = \ 0.0004\ \frac{m^{2}}{s}[/latex]
[latex]S\ = \ 0.0001[/latex]
[latex]r_{pumping}\ = \ 200[/latex] m
[latex]r_{image}\ = 300[/latex]m
Drawdown at OW from the pumping well:
[latex]u\ = \frac{r^{2}S}{4Tt} = \frac{{(200\ m)}^{2}\ 0.0001}{\left\lbrack 4\ \left( 0.0004\frac{m^{2}}{s} \right)315,576,000\ seconds \right\rbrack} = \ {7.92x10}^{- 6}[/latex]
[latex]W(u)\ = \ 11.17[/latex]
[latex]s = \frac{Q}{4\pi T}W(u) = \frac{0.0116\frac{{\ \ m}^{3}}{s}}{4\ \pi\ 0.0004\frac{\ \ m^{2}}{s}}(11.17) = \ 25.78\ m[/latex]
Drawdown at OW from image well:
[latex]u\ = \frac{r^{2}S}{4Tt} = \frac{{\ \ (300\ m)}^{2}\ \ 0.0001}{4\ \ (0.0004\ \frac{m^{2}}{s})\ 315,576,000\ seconds} = {1.78x10}^{- 5}[/latex]
[latex]W(u)\ = \ 10.39[/latex]
[latex]s\ = \ \frac{0.0116\frac{{\ \ m}^{3}}{s}}{4\ \pi\ 0.0004\frac{\ \ m^{2}}{s}}(10.39) = \ 23.98\ m[/latex]
Combined Drawdown at OW:
[latex]s = \ 25.78\ m + 23.98\ m = \ 49.76\ m[/latex]
Drawdown at Observation Well in Figure 55
Calculations are identical except Q of the image well is injection [latex]Q = - 0.0116\frac{m^{3}}{s}[/latex]. For the total drawdown, the drawdown associated with the image well is subtracted from the real well drawdown rather than added. The solution is:
[latex]s = \ 25.78\ m + ( - 23.98\ m) = \ 1.8\ m[/latex]