Question

A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well is pumped at a constant rate of 500 gpm, what is the drawdown at a distance of 7.0 m from the well after 1 day of pumping?

Answer 1

S = 5.7209 M

Step-by-step explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

= 14.9 * 20.1 = 299.5 m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T* t )

= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

= 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S =
`(Q)/(4\pi T) * W (u)`

where : Q = 2725.5

T = 299.5

W(u) = 7.9

substitute the given values into equation above

S = 5.7209 M