Solution Exercise 14
Part 1: Drawdowns would be greater than expected due to the well loss associated with the lower efficiency.
Part 2: Well loss is directly proportional to a power (greater than 1) of the pumping rate with greater well loss at higher pumping rates. The well efficiency at a higher pumping rate (i.e., at 600 m3/d) would therefore be lower than that at the previous pumping rate (500 m3/d).
Part 3:
- The drawdown at 24 hours is 14.44 m, this is calculated from the Theis Equation with [latex]u = [/latex] 8.68×10‑8 and [latex]W(u) = [/latex] 15.68.
- The well efficiency is 85 percent (calculated as 14.44 m / 17 m). The well loss is 2.56 m (calculated as (17 m – 14.44 m).
Part 4:
The answer is (c) test results underestimate the formation transmissivity.
The transmissivity estimated from the test is based upon the measured drawdowns in the well. Lower efficiency means greater drawdown than expected from a well with 100 percent efficiency. Since the observed drawdowns are greater than expected, the calculated transmissivity would be less than expected, and therefore the test results would underestimate the actual transmissivity.