Question

A 60-cm diameter well is located in an aquifer with a transmissivity of 125 m2/day and a storativity of 5 x10-2. A fault, which acts as a barrier to flow, is located 300 m from the well pumping at 2700 m3/d. At the end of 1 year of pumping, what is the drawdown at the midpoint between the fault and the well

Answer 1

The drawdown at the midpoint is
`K = 29.11 \ m`

Step-by-step explanation:

From the question we are told that

The transmissivity is
`T = 125 \ m^3/day`

The diameter is
`d = 60 \ cm = 0.60 \ m`

The storativity is
`S = 5 *10^(-2)`

The distance of the vault from the pumping well is
`d = 300 \ m`

The pumping rate is
`R = 2700 \ m^3 / day`

Generally the radius is mathematically represented as

`r = ( d)/(2)`

=>
`r = ( 0.60 )/(2)`

=>
`r = 0.30 \ m`

Generally the time is 1 year = 365 days as stated in the question

Generally the drawdown at the midpoint between the fault and the well is mathematically represented as

`K = (R )/( 4 \pi * T ) * ln ((2.2459)*(T * t )/(r * S) )`

=>
`K = (2700 )/( 4* 3.142 * 125 ) * ln ((2.2459 ) * (125 * 365)/(03^2 * 5 *10^(-2)) )`

=>
`K = 29.11 \ m`