Given Data,
Time of recuperator (Tr) = 90 min. = 90/60 hr = 1.5 hr
Initial depression head (H1) = 2.5 m
Final description head (H2) = 1.25 m
Area (A) = ∏/4 ∗ d2 = ∏/4 ∗ (4)2 = 12.567 m2
1. Specific capacity per unit well area
Ks = (1/Tr) ∙ loge(H1/H2)
Ks = (1/1.5) ∙ loge(2.5/1.25)
Ks = 0.462 h-1
2. Yield of well for safe drawdown of 2.5 m
Q = Ks ∙ A ∙ H
Q = 0.462 ∗ 12.567 ∗ 2.5
Q = 14.52 m3/hr
