Step-by-step solved example
Want 4 m of drawdown at the well in a confined T=0.01 m²/s, R/rw=200. Q?
1. ln
ln200=5.30.
2. Q
Δh=Q ln /(2πT) → Q=Δh 2π T / ln =4×2π×0.01/5.30=0.047 m³/s.
Answer: Q ≈ 0.047 m³/s per well (steady Thiem)
Want 4 m of drawdown at the well in a confined T=0.01 m²/s, R/rw=200. Q?
1. ln
ln200=5.30.
2. Q
Δh=Q ln /(2πT) → Q=Δh 2π T / ln =4×2π×0.01/5.30=0.047 m³/s.
Answer: Q ≈ 0.047 m³/s per well (steady Thiem)
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