Respuesta :
Answer:
the calculations and their justification are found in the explanation section.
Explanation:
The average transmissivity is
[tex]T=\frac{Q_{w} }{2\pi (s_{1}-s_{2} ) } ln(\frac{r_{2} }{r_{1} } )[/tex]
Qw = pumping rate = 400 L/s = 400000 m³/s
s₁ = drawdown of the well = 1 m
s₂ = 0.5 m
r₁ = piezometric level = 50 m
r₂ = 100 m
Replacing values
[tex]T=\frac{400000}{2\pi (1-0.5)} ln(\frac{100}{50} )=88254.24m^{2} /s[/tex]
the hydraulic conductivity is
[tex]K=\frac{T}{b}[/tex]
b = thickness = 24 m
Replacing
[tex]K=\frac{88254.24}{24} =3677.26m/s[/tex]
no steady state for this drawdown.
The radial distance is
[tex]s=s_{w} -\frac{Q_{w} }{2\pi T} ln(\frac{r}{r_{w} } )[/tex]
here
sw = 4 m = drawdown at the pumping well
s =drawdown of the radial distance = 0
rw = radius of the pumping well = 0.5
Clearing r
[tex]0=4-\frac{400000}{2\pi *88254.24 } ln(\frac{r}{0.5} )\\4=0.721(lnr+ln0.5)\\r=exp(\frac{3.5}{0.721} )=128.3m[/tex]
It can be said that the steady state reduction is not valid at a distance beyond that calculated because the reduction would become negative.
