Groundwater – Interference of two fully penetrating wells in a confined aquifer (Thiem superposition) Two identical, fully penetrating wells are L metres apart in a homogeneous confined aquifer. At the midpoint (0.5L from each well), the drawdown is 2.0 m when only Well 1 is pumped at Q1 m^3/s. When both wells are pumped at identical steady rates Q2 m^3/s, the drawdown at the midpoint becomes 8.0 m. The drawdown at each well during pumping remains 10.0 m, and the radius of influence exceeds 0.5L. Determine Q1/Q2.

Introduction / Context:
In steady flow to wells in a homogeneous confined aquifer, the Thiem equation shows that drawdown is linearly proportional to discharge for a given observation point and radius of influence. When multiple wells pump, the principle of superposition applies: the total drawdown at any point is the sum of contributions from each well. This problem uses superposition at the midpoint to determine the ratio of discharges when one versus two wells operate.

Given Data / Assumptions:

• Confined aquifer; transmissivity T is constant.
• Two fully penetrating wells separated by L.
• Midpoint drawdown s_P = 2.0 m for single well at Q1; s_P = 8.0 m for two wells at equal Q2 each.
• Drawdown at a pumped well is 10.0 m; radius of influence R > 0.5L (large enough to use Thiem logs consistently).

Concept / Approach:

Thiem (steady) drawdown for one well at distance r: s(r) = (Q / (2πT)) * ln(R/r). At the midpoint P, distances are r = L/2 from each well. By superposition, when both wells pump at Q2, the total drawdown at P is twice the single-well drawdown created by Q2.

Step-by-Step Solution:

Verification / Alternative check:

Including the “10 m at the well” condition still leads to Q2 = 2 Q1 when self and interference drawdowns are summed; detailed algebra using the well drawdown adds terms that cancel in the final ratio. Thus, the midpoint data suffice.

Why Other Options Are Wrong:

8/5 and 4/3 imply the two-well case changes the proportionality constant—contrary to Thiem superposition with common R. 5/8 inverts the correct ratio. ln(L/2) is not a dimensionless discharge ratio.

Common Pitfalls:

Assuming the radius of influence changes appreciably between cases; for ratio-based evaluation at the same point, the logarithmic factor cancels.

Final Answer:

1/2