Answer:
The drawdown in a confined aquifer under transient conditions can be estimated using the Theis solution for the non-equilibrium radial flow of water. This solution is given by:
s = Q / (4πT) * W(u),
where s is the drawdown, Q is the pumping rate, T is the transmissivity, and W(u) is the well function (also called the Theis function) which depends on the variable u, where:
u = r²S / (4Tt),
where r is the distance from the pumping well and t is the time since pumping began.
Given T = 2500 m²/day, S = 1.0 x 10-3, and Q = 500 m³/day, we can calculate the drawdown at 10 m (r1 = 10 m) and 50 m (r2 = 50 m) for t = 150 days.
For (i) r1 = 10 m:
u1 = r1²S / (4Tt) = (10 m)² * 1.0 x 10-3 / (4 * 2500 m²/day * 150 days) = 0.000667
s1 = Q / (4πT) * W(u1) = 500 m³/day / (4π * 2500 m²/day) * W(0.000667).
For (ii) r2 = 50 m:
u2 = r2²S / (4Tt) = (50 m)² * 1.0 x 10-3 / (4 * 2500 m²/day * 150 days) = 0.01667
s2 = Q / (4πT) * W(u2) = 500 m³/day / (4π * 2500 m²/day) * W(0.01667).
Step-by-step explanation:
Unfortunately, the well function W(u) cannot be evaluated directly without more specialized knowledge or tools. The well function is related to the exponential integral function, which requires numerical computation. You would typically use a table of values, a calculator with this function, or a computer program to evaluate it. After obtaining W(u), multiply it by the remaining fraction to find the drawdowns.