Solution Exercise 10

Variables:

[latex]r[/latex]= 150 m

[latex]Q\ = \frac{300\frac{\ \ m^{3}}{d}}{86400\ \frac{s}{d}} = 0.00347\ \frac{m^{3}}{s}[/latex]

Match points:

[latex]t[/latex] = 1 minute (60 seconds)

[latex]s[/latex] = 1 meter

u = 0.31

W(u) = 7

Solve for T:

[latex]s = \frac{Q}{4\pi T}W(u)[/latex] rearrange for [latex]T = \frac{Q}{4\pi s}W(u)[/latex]

[latex]u\ = \frac{r^{2}S}{4Tt}[/latex] rearrange for [latex]S = \frac{u\ 4Tt}{\ r^{2}}[/latex]

[latex]T\ = \ \frac{0.00347\frac{{\ \ m}^{3}}{s}}{4\ \ \pi\ \ 1\ m}\ (7)\ = 0.0019\ \frac{{\ \ m}^{2}}{s}[/latex]

[latex]S = \ \frac{0.31\ (4)\ 0.0019\ \frac{{\ \ m}^{2}\ }{s}\ 1\ \min\frac{60s\ }{\min}}{\ (150m)^{2}} = 6.4\ x\ 10^{- 6}[/latex]

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Basic Hydrogeology: An Introduction to the Fundamentals of Groundwater Science Copyright © 2025 by Matthew M. Uliana. All Rights Reserved.

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