Question

A 10 inch diameter well is drilled to the bottom of an unconfined aquifer. The water table is originally 500 ft above the bottom of the aquifer. If 100 gal/min is pumped from the well, the equilibrium depth of the water in the well is 224 ft and the radius of influence is 1000 ft. (1) Determine the transmissivity T of the aquifer and (2). How much drawdown should be expected at a distance of 300 ft from the well?

Answer 1

1 Q=100 gol/min. Page. No. 2 k loinch Fitta ک Original Water table Sell H=500 ft Unconfined aquifer hw=224 ft -R=1000ft 12=300ft 5 12 Jinbervious Boundoing k Confined aquifer Given data, Diameter of well, 25w = 10 inch [ rw = 5inch = 0.4167ft Original depths from bottom of- aquifer, H= 500 ft 100 gal/min. = 100 x0.134 Pumping ratene 13.4 #S =0-2233 113 Equilibrium depth of water in the well, hw = 224 ft Radias of influence, 'R = 1000 ft min S

Page. No. 2 o Transmissivity of the aquifer, Pumping out Q = TTK (h²_h?) from well Cunconfined aquifer) ln ( fil Put, h = H, h = hw r2 = R, n=tw Q = TK (H263) ln (R rw 0.2233 = TTXK x (5002-2242) en (1000 10.4167 K= 2.7685 x 10 ft/sec Transmissibity., TE KH = 2:7685x100x500 fty's T = 138 43 x 15-3 ?/ = 11 9 6 fli/day

Page No 3 2. Draw down (S2), Distance from the well, 1 =300ft Put, h₂ h2, hi ha t2 =12, r=rw in equation o 12 Q = TTK (2 - 4 In a → 0.2233 = TTK (F22-2242) In (300 is 3:4167 Pat K = 2:7685 x100 ft/s (Permeability constant for same 012233 = TT X 2.7685x100x (re? 2242) same well ln Com 300 0.4167 → fiz –2242 = 168913.9474 h2 = 219089.9474 h2 = 468.0705 ft S2 =H-h2 5 500 -468.0705 S2 = 31.9295 ft Drawdown at a distance of 300ft from the wella