In an area, one tubewell of 15 cm diameter was operating such that groundwater table is 40 m above the bottom of screen. The groundwater formation has permeability (K) as 20 m·d⁻¹ and well has radius of influence of 980 m with a maximum drawdown of 12 m. After some time, a second tubewell of same size was installed at a distance of 500 m from the first tubewell. If both the tubewells are running simultaneously and have equal discharge, the loss of discharge in the first tubewell due to installation of the second tubewell, in %, is __________. (Rounded off to 2 decimal places)
Official Answer Key:
6.60 to 6.70
Step-by-Step Solution & Explanation
This is a well-interference problem in an unconfined aquifer (Dupuit–Thiem).
Step 1 — single well acting alone:
H = 40 m (saturated thickness), drawdown = 12 m ⇒ h = 40 − 12 = 28 m
r = 0.15/2 = 0.075 m ; R = 980 m
Q₁ = πK(H² − h²)/ln(R/r)
= π × 20 × (1600 − 784)/ln(980/0.075)
= π × 20 × 816/ln(13,066.7)
= 51,272/9.4779 = 5409.6 m³·d⁻¹
Step 2 — two identical wells, same drawdown, spacing d = 500 m. Each well’s drawdown is now the sum of its own drawdown plus the interference from its neighbour, which adds a second logarithmic term. For two equal wells:
Q₂ = πK(H² − h²)/[ln(R/r) + ln(R/d)]
= 51,272/[9.4779 + ln(980/500)]
= 51,272/[9.4779 + 0.6729]
= 51,272/10.1508 = 5050.9 m³·d⁻¹
Step 3 — percentage loss of discharge:
Loss = (5409.6 − 5050.9)/5409.6 × 100 = 358.7/5409.6 × 100 = 6.63 %
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