A municipal water supply well is to be installed in the aquifer shown in Figure 8.21. This well will have an 8-in diameter casing in a 24-in diameter boring with a submersible pump that draws the water level down to a depth of 62 ft below the ground surface. Compute the maximum flow rate that could

Question

A municipal water supply well is to be installed in the aquifer shown in Figure 8.21. This well will have an 8-in diameter casing in a 24-in diameter boring with a submersible pump that draws the water level down to a depth of 62 ft below the ground surface. Compute the maximum flow rate that could be produced by this well.

Accepted Answer

This is an unconfined aquifer, so we will use Equation 8.30 [latex]Q=\frac{\pi k\left(h^{2}_{0} - h^{2}_{w} ight) }{\ln \left\lgroup\frac{r_{0}}{r_{w}} ight group }[/latex].

[latex]h_{0} = 55 ft + 22 ft - 56 ft = 21 ft [/latex]

[latex]h_{w} = 55 ft + 22 ft - 62 ft= 15 ft [/latex]

[latex]k=\left(6 imes 10^{-3} ft/s ight) \left\lgroup\frac{30.5 cm}{1 ft} ight group =0.18 cm/s[/latex]

[latex]r_{0}=300\left(h_{0} - h_{w} ight) \sqrt{k} =300\left(21 ft - 15 ft ight) \sqrt{0.18 cm/s} =760 ft[/latex]

Using a unit conversion factor of [latex]449 gal/min per ft^{3}/s[/latex],

[latex]Q=\frac{\pi k\left(h^{2}_{0} - h^{2}_{w} ight) }{\ln \left\lgroup\frac{r_{0}}{r_{w}} ight group } =\frac{\pi \left(6 imes 10^{-3} ft/s ight) \left[\left(21 ft ight) ^{2} - \left(15 ft ight) ^{2} ight] }{\ln \left\lgroup\frac{760 ft}{1 ft} ight group } =0.61 ft^{3}/s[/latex] = 276 gal/min

Question 8.3: A municipal water supply well is to be installed in the aqui...

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