Question

A well pumps at 400 L/s from a confined aquifer whose thickness is 24 m. If the drawdown 50 m from the well is 1 m and the drawdown 100 m from the well is 0.5 m, then calculate the hydraulic conductivity and transmissivity of the aquifer. Do you expect the drawdowns at 50 m and 100 m from the well to approach a steady state? Explain your answer. If the radius of the pumping well is 0.5 m and the drawdown at the pumping well is measured as 4 m, then calculated the radial distance to where the drawdown is equal to zero. Why is the state-state drawdown equation not valid beyond this distance?

Answer 1

Answer:

the calculations and their justification are found in the explanation section.

Explanation:

The average transmissivity is

$T=\frac{Q_{w} }{2\pi (s_{1}-s_{2} ) } ln(\frac{r_{2} }{r_{1} } )$

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

$T=\frac{400000}{2\pi (1-0.5)} ln(\frac{100}{50} )=88254.24m^{2} /s$

the hydraulic conductivity is

$K=\frac{T}{b}$

b = thickness = 24 m

Replacing

$K=\frac{88254.24}{24} =3677.26m/s$

no steady state for this drawdown.

The radial distance is

$s=s_{w} -\frac{Q_{w} }{2\pi T} ln(\frac{r}{r_{w} } )$

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

$0=4-\frac{400000}{2\pi *88254.24 } ln(\frac{r}{0.5} )\\4=0.721(lnr+ln0.5)\\r=exp(\frac{3.5}{0.721} )=128.3m$

It can be said that the steady state reduction is not valid at a distance beyond that calculated because the reduction would become negative.