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: Understanding the Question.
A well is pumped from an unconfined aquifer, and two observation wells at different radii show different drawdowns after a long period of steady pumping. We need the transmissivity of the aquifer, which describes how easily water flows through the full saturated thickness of the aquifer.

Step 2: Set up the heads at the two observation points.
The saturated thickness (height of the water table above the base of the aquifer) before pumping is \(H = 30\) m.
At \(r_1 = 10\) m, drawdown \(s_1 = 12\) m, so the head there is \(h_1 = H-s_1 = 30-12 = 18\) m.
At \(r_2 = 100\) m, drawdown \(s_2 = 11\) m, so the head there is \(h_2 = H-s_2 = 30-11 = 19\) m.

Step 3: Apply the Thiem equation for an unconfined (Dupuit) aquifer.
For steady radial flow in an unconfined aquifer, the discharge is related to the hydraulic conductivity \(K\) by
\[ Q = \frac{\pi K (h_2^2-h_1^2)}{\ln(r_2/r_1)} \]
Here, \(Q = 63\ \text{m}^3/\text{s}\), \(h_2^2-h_1^2 = 19^2-18^2 = 361-324 = 37\ \text{m}^2\), and \(\ln(r_2/r_1) = \ln(100/10) = \ln(10) = 2.3026\).

Step 4: Solve for the hydraulic conductivity \(K\).
\[ K = \frac{Q\ln(r_2/r_1)}{\pi(h_2^2-h_1^2)} = \frac{63\times2.3026}{\pi\times37} \]
\[ K = \frac{145.06}{116.24} \approx 1.248\ \text{m/s} \]

Step 5: Convert to transmissivity.
Transmissivity is the hydraulic conductivity multiplied by the full saturated thickness of the aquifer:
\[ T = K\times H = 1.248\times30 \approx 37.4\ \text{m}^2/\text{s} \]

Final Answer:
Rounded off to one decimal place, the transmissivity of the aquifer is approximately \(37.4\ \text{m}^2/\text{s}\).
\[ \boxed{T \approx 37.4\ \text{m}^2/\text{s}} \]