Solution Exercise 5

Vincent E.A. Post; Craig T. Simmons

Solution Exercise 5

It can be seen that the density does not vary between the screens, so this means the point water hydraulic heads can be used directly in the calculation of q_zwith Darcy’s law. Given the negative numbers, the smallest absolute value is the highest hydraulic head so the hydraulic head increases with depth, thus flow is upward from screen 3 to screen 2. The factor of 1000 in the equation converts meters to millimeters.

$\displaystyle q_{z}=-20\frac{(-3.08--3.04)}{(-96.44--118.93)}\times 1000=36\;\textrm{mm}\;\textrm{d}^{-1}$

While from screen 2 to 1 it is

$\displaystyle q_{z}=-20\frac{(-3.13--3.08)}{(-76.41--96.44)}\times 1000=50\;\textrm{mm}\;\textrm{d}^{-1}$

For the more general case where the density varies, one would first convert the hydraulic heads to freshwater heads using $h_{f}=z_{i}+(h_{i}-z_{i})\frac{\rho _{i}}{\rho _{f}}$ , which yields (three decimals are provided to prevent large rounding errors in the calculation of q_z):

Screen	h_f
1	−2.815
2	−2.679
3	−2.542

Using $q_{z}=-K_{f}\left ( \frac{\Delta h_{f}}{\Delta z}+\frac{\rho _{mean}-\rho _{f}}{\rho _{{f}}} \right )$ one finds for the flow between piezometer 3 and 2:

$\displaystyle q_{z}=-20\left ( \frac{(-2.679--2.542)}{(-96.44--118.93)}+\frac{1004.3-1000}{1000} \right )\times 1000=36\;\textrm{mm}\;\textrm{d}^{-1}$

And for the flow between piezometer 2 and 1

$\displaystyle q_{z}=-20\left ( \frac{(-2.815--2.679)}{(-76.41--96.44)}+\frac{1004.3-1000}{1000} \right )\times 1000=5\;\textrm{mm}\;\textrm{d}^{-1}$

Which is the same answer as before (note that the unrounded result has a 0.43 percent error because K_f was equated to the hydraulic conductivity).

Return to Exercise 5

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