Solution Exercise 5
- The horizontal gradient, i, determined in the solution to Exercise 3 (0.00067) is used with the hydraulic conductivity of 0.05 m/s and effective porosity of 0.15 to calculate the average linear velocity between wells b and c.
[latex]v = \frac{q}{\varphi_{eff}} = \frac{K\ i}{\varphi_{eff}} = \frac{0.05\frac{m}{s}\ 0.00067}{0.15}\frac{86,400\ s}{1\ day} = 19.3\frac{m}{d}[/latex] from left to right in Figure 9
- Travel time is determined by dividing the distance to be traveled by the velocity just as is done to figure out how long it will take to drive somewhere at a set speed.
[latex]time = \frac{distance}{v} = \frac{1500\ m}{19.3\frac{m}{d}} = 78.125\ days[/latex]