Question

A fully-penetrating well of 20 cm diameter is provided in an unconfined aquifer. The height of the ground water table is 30 m from the bottom of the aquifer. After a long period of pumping at a rate of 63 m\(^3\)/s, the drawdown in the observation wells at 10 m and 100 m from the pumped well is 12 m and 11 m, respectively. The transmissibility (in m\(^2\)/s) of the aquifer is (rounded off to one decimal place).

Hint:

Use the Thiem equation for an unconfined aquifer with the two observation-well drawdowns to find the hydraulic conductivity K, then multiply by the saturated thickness to get transmissivity.

Answer 1

Correct Answer: 37.4

Step 1: Key Formula or Approach.
For an unconfined aquifer, transmissivity is not constant, since it falls as the water table drops. One common method treats the problem as if it were confined, but first corrects the drawdown values using Jacob's correction:
\[ s' = s - \frac{s^2}{2H} \]
where $H$ is the initial saturated thickness. Once corrected, the ordinary confined-aquifer Thiem equation applies directly:
\[ Q = \frac{2\pi T (s_1'-s_2')}{\ln(r_2/r_1)} \]

Step 2: Correct the two drawdown values.
With $H = 30$ m:
\[ s_1' = 12 - \frac{12^2}{2\times30} = 12 - \frac{144}{60} = 12-2.4 = 9.6\text{ m} \]
\[ s_2' = 11 - \frac{11^2}{2\times30} = 11 - \frac{121}{60} = 11-2.017 = 8.983\text{ m} \]
\[ s_1'-s_2' = 9.6-8.983 = 0.617\text{ m} \]

Step 3: Detailed Explanation.
Plug into the confined-form Thiem equation, with $Q=63\text{ m}^3/\text{s}$ and $\ln(r_2/r_1)=\ln(10)=2.3026$:
\[ T = \frac{Q\ln(r_2/r_1)}{2\pi(s_1'-s_2')} = \frac{63\times2.3026}{2\pi\times0.617} \]
\[ T = \frac{145.06}{3.877} \approx 37.4\text{ m}^2/\text{s} \]
This matches the direct unconfined-flow calculation, since Jacob's correction is only an algebraic rearrangement of the same underlying physics, but it lets us reuse the simpler confined-aquifer formula once the drawdowns are corrected.

Step 4: Final Answer.
\[ \boxed{T \approx 37.4\text{ m}^2/\text{s}} \]