Subtopic: Steady and unsteady flow in confined and unconfined aquifers
Question Statement:
A fully penetrating tubewell of 0.2 m diameter is operational in a confined aquifer of 20 m thickness. The hydraulic conductivity of the formation (K) is 20 m per day and the radius of influence is 1000 m. Water table is 50 m above the bottom of screen and the maximum drawdown is 10 m. Subsequently two more tubewells of the same size are installed in such a way that the three tubewells form an equilateral triangle of side 300 m. All three tubewells have equal discharge when operated independently. If all tubewells are operated simultaneously, the percent loss in the discharge of the first well is
Options:
(A) 20.73
(B) 35.75
(C) 48.85
(D) 61.90
Official Answer Key:
A
Step-by-Step Solution & Explanation
For a single confined well (Thiem equation): Q₀ = 2πKb·sw/ln(R/rw), with K=20 m/d, b=20 m, sw=10 m, R=1000 m, rw=0.1 m: Q₀ = 2π(20)(20)(10)/ln(10000) = 25133/9.210 = 2728.8 m³/d. When three identical wells operate together (by superposition), the same drawdown sw at well 1 must now be shared among the self-effect and the interference from the other two wells (each at distance 300 m, forming an equilateral triangle): sw = [Qc/(2πKb)]·[ln(R/rw) + 2·ln(R/300)], since the interference terms from the two neighbouring wells (each contributing ln(R/300)) add to the drawdown at well 1. Qc = sw·2πKb / [ln(R/rw) + 2ln(R/300)] = 1256.6 / [9.210 + 2(1.204)] = 1256.6/11.617 = 2163.2 m³/d. Percent loss = (Q₀−Qc)/Q₀ × 100 = (2728.8−2163.2)/2728.8 × 100 = 20.73%, matching option (A) exactly.
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