Design an open well in fine sand to give a discharge of 0.03 cumec when worked under a depression head of 2.5 m.

Data Given,

H = 2.5 m

Q = 0.03 cumec (m³/s)

Q = 0.03 ∗ 3600 m³/hour

Q = 108 m³/hour

For open wells,

Q = (^k/_A)∙A∙H

108 = 0.50 * A * 2.5 [∵ For fine sand, (^k/_A) = 0.50 m³/hr]

∴ A = 86.4 m²

We know that,

A = ^∏/₄ * d²

86.4 = ^∏/₄ * d²

∴ d = 10.49 m