Answer:
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
