A confined aquifer with a transmissivity of 300 m2/day and a storativity of 0.0005 and a well radius of 0.3 m. Find the drawdown in the well at 100 days if the following pumping schedule is followed after a long period of time of no pumping. Period
1 2 3 4
Time (days) 0-20 20-50 50-90 90-100
Q (m3/day) 500 300 800 0

8.4627 m

Explanation:

Transmissivity( T ) = 300 m^2/day

Storativity( S )  = 0.0005

well radius ( r ) = 0.3m

Determine the drawdown in well at 100 days

Drawdown at 100 days = ∑ Drawdown at various period

We will use the equation : S = Q / U*π*T [ -0.5772 - In U ]   ( 1 )

where : Q = discharge , T = transmissivity

             S = drawdown ,

U = r^2*s / 4*T*t  --- ( 2 )

r = well radius , S = Storativity, t = time period

i) During 0-20

U1 = r^2*s / u*π*t  = 1.875 * 10^-9

Input values into equation 1

S1 = 2.5885

ii) During 20-50

U2 = r^2*s / 4*π*t = 0.3^2 * 30 / u * 300 * 30 = 1.25 * 10^-9

input values into equation 1

S2 = 1.5854 m

iii) During 50 -90

U3 = r^2*s / 4*π*t = 9.375 * 10^-10

input values into equation 1

S3 = 4.2888 m

iv) During 90-100

U4 = 0

s4 = 0

Drawdown at 100 days = ∑ Drawdowns at various period

                                       = s1 + s2 + s3 + s4 = 2.5885 + 1.5854 + 4.2888 + 0

                                       = 8.4627 m

energy can cross the boundaries of a closed system in the form of heat and mass

explanation: