Question

If the diameter of the well is doubled, the water yield will increase by about:

• A. 5%
• B. 8%
• C. 10%
• D. 15%

Hint:

The yield of a well is logarithmically dependent on its radius (\(Q \propto 1/\ln(R/r)\)). This means that doubling the radius gives only a small, incremental increase in yield (typically 10-15%), making it generally uneconomical to drill a much larger well just for a minor increase in discharge.

Answer 1

The Correct Option is C

Step 1: Use the Thiem equation for discharge from a well in an unconfined aquifer. The steady-state discharge (yield) \(Q\) is given by: \[ Q = \frac{\pi k (H^2 - h^2)}{\ln(R/r)} \] where \(k\) is permeability, \(H\) is the initial saturated thickness, \(h\) is the head in the well, \(R\) is the radius of influence, and \(r\) is the radius of the well.
Step 2: Analyze the effect of doubling the well radius.
Let the initial yield be \(Q_1\) with radius \(r_1\). Let the new yield be \(Q_2\) with radius \(r_2 = 2r_1\).
The ratio of the yields will be: \[ \frac{Q_2}{Q_1} = \frac{\ln(R/r_1)}{\ln(R/r_2)} = \frac{\ln(R/r_1)}{\ln(R/(2r_1))} = \frac{\ln(R) - \ln(r_1)}{\ln(R) - \ln(2r_1)} = \frac{\ln(R) - \ln(r_1)}{\ln(R) - \ln(2) - \ln(r_1)} \] Step 3: Evaluate the increase using typical values.
The radius of influence \(R\) is typically much larger than the well radius \(r\). Let's assume some realistic values, e.g., \(R = 200\) m and \(r_1 = 0.2\) m. \[ \frac{Q_2}{Q_1} = \frac{\ln(200/0.2)}{\ln(200/0.4)} = \frac{\ln(1000)}{\ln(500)} = \frac{6.907}{6.214} \approx 1.11 \] This represents an increase of about 11%. This shows that doubling the diameter does not double the yield but results in a relatively small increase. Among the given options, 10% is the most plausible and commonly cited approximate increase.