Given Parameters

The problem provides the following parameters for a confined aquifer system:

• Tubewell diameter: $D = 0.2 \text{ m}$
• Aquifer thickness: $b = 20 \text{ m}$
• Hydraulic conductivity: $K = 20 \text{ m/day}$
• Radius of influence: $R = 1000 \text{ m}$
• Distance between wells: $d = 300 \text{ m}$
• Maximum drawdown: $s_0 = 10 \text{ m}$
• Number of wells: 3
• Well radius: $r_w = D/2 = 0.1 \text{ m}$

Single Well Discharge Calculation

The discharge ($Q_0$) of a single tubewell operating independently in a confined aquifer is given by the Thiem equation (steady-state):

$ s_0 = \frac{Q_0}{2 \pi K b} \ln\left(\frac{R}{r_w}\right) $

Here, $s_0$ is the drawdown, $Q_0$ is the discharge, $K$ is hydraulic conductivity, $b$ is aquifer thickness, $R$ is the radius of influence, and $r_w$ is the well radius.

Simultaneous Well Operation Analysis

When multiple wells operate simultaneously, their cones of depression interfere. The total drawdown at the location of any well is the sum of the drawdown caused by its own pumping and the drawdowns caused by the other pumping wells (superposition principle).

Let $Q_{actual}$ be the discharge of each well when all three operate simultaneously. The drawdown at the location of the first well ($s_1$) is:

$ s_1 = \frac{Q_{actual}}{2 \pi K b} \ln\left(\frac{R}{r_w}\right) + \frac{Q_{actual}}{2 \pi K b} \ln\left(\frac{R}{d}\right) + \frac{Q_{actual}}{2 \pi K b} \ln\left(\frac{R}{d}\right) $

Simplifying this equation:

$ s_1 = \frac{Q_{actual}}{2 \pi K b} \left[ \ln\left(\frac{R}{r_w}\right) + 2 \ln\left(\frac{R}{d}\right) \right] $

Calculating Actual Discharge and Loss

Assuming the maximum drawdown at each well location remains the same ($s_1 = s_0 = 10$ m), we can find the ratio of the actual discharge ($Q_{actual}$) to the original discharge ($Q_0$).

From the single well equation: $10 = \frac{Q_0}{2 \pi K b} \ln\left(\frac{R}{r_w}\right)$

From the simultaneous operation equation: $10 = \frac{Q_{actual}}{2 \pi K b} \left[ \ln\left(\frac{R}{r_w}\right) + 2 \ln\left(\frac{R}{d}\right) \right]$

Equating the two expressions (since both equal 10):

$ \frac{Q_0}{2 \pi K b} \ln\left(\frac{R}{r_w}\right) = \frac{Q_{actual}}{2 \pi K b} \left[ \ln\left(\frac{R}{r_w}\right) + 2 \ln\left(\frac{R}{d}\right) \right] $

Solving for the ratio $\frac{Q_{actual}}{Q_0}$:

$ \frac{Q_{actual}}{Q_0} = \frac{\ln\left(\frac{R}{r_w}\right)}{\ln\left(\frac{R}{r_w}\right) + 2 \ln\left(\frac{R}{d}\right)} $

Now, substitute the values:

• $R/r_w = 1000 / 0.1 = 10000$
• $R/d = 1000 / 300 = 10/3 \approx 3.333$
• $\ln(R/r_w) = \ln(10000) \approx 9.2103$
• $\ln(R/d) = \ln(10/3) \approx 1.2040$

$ \frac{Q_{actual}}{Q_0} = \frac{9.2103}{9.2103 + 2 \times 1.2040} = \frac{9.2103}{9.2103 + 2.4080} = \frac{9.2103}{11.6183} \approx 0.79275 $

Percent Loss Calculation

The percent loss in discharge is calculated as:

$ \text{Percent Loss} = \left(1 - \frac{Q_{actual}}{Q_0}\right) \times 100 $

$ \text{Percent Loss} = (1 - 0.79275) \times 100 = 0.20725 \times 100 \approx 20.73\% $