Step-by-step solved example
k=1e-4 m/s, b=20 m, h=18 m at 50 m, h=12 m at 10 m. Q?
1. ln
ln(50/10)=ln5=1.609.
2. Q
2π(1e-4)(20)(6)/1.609=0.0469 m³/s.
Answer: Q ≈ 0.047 m³/s
k=1e-4 m/s, b=20 m, h=18 m at 50 m, h=12 m at 10 m. Q?
1. ln
ln(50/10)=ln5=1.609.
2. Q
2π(1e-4)(20)(6)/1.609=0.0469 m³/s.
Answer: Q ≈ 0.047 m³/s
1.Unconfined Thiem/Dupuit replaces b(h2−h1) with
2.T transmissivity is
3.If r2/r1=e, ln=
4.Darcy’s law is the local basis: q=
5.Radius of influence R is
6.Storativity S appears in
7.k=10^{-5} m/s is typical of
8.Two observation wells are needed to
9.Pumping increases drawdown at r=rw the most because
10.Confined b=10 m, k=2e-4, Δh=4 m, ln(r2/r1)=2: Q=