**Example Problem on Aquifer Pumping and Steady State Drawdown Calculation****Problem Statement:**Water is pumped from a well tapping an unconfined aquifer at a rate of 2400 m³/day. A no-drawdown boundary exists at a distance of 5 km from the well center. Assuming the well to be fully penetrating, compute the steady-state drawdown at the well face. **Given Data:**- Initial saturated thickness \(b_0\) = 50 m- Hydraulic conductivity \(K\) = 20 m/day- Effective well radius \(r_w\) = 1 m**Step-by-Step Solution:**1. **Understand the given data and parameters:** - **Pumping rate (Q):** This is the rate at which water is extracted from the well, given as 2400 m³/day. - **No-drawdown boundary distance (R):** This is the distance at which there is no drop in the water table, located 5 km (5000 meters) from the well center. - **Initial saturated thickness (b_0):** The initial depth or thickness of the aquifer from which water is being pumped, given as 50 m. - **Hydraulic conductivity (K):** This is the measure of the aquifer's ability to transmit water, given as 20 m/day. - **Effective well radius (r_w):** The radius of the well where the drawdown is to be calculated, given as 1 meter.2. **Apply the formula for steady-state drawdown in an unconfined aquifer:** The general formula is: \[ s = \frac{Q}{2 \pi K} \ln\left(\frac{R}{r_w}\right) \]3. **Substitute the given values into the formula:** - Pumping rate (Q) = 2400 m³/day - Hydraulic conductivity (K) = 20 m/day - No-drawdown boundary distance (R) = 5000 m - Effective well radius (r_w) = 1 m \[ s = \frac{2400}{2 \pi \times 20} \ln\left(\frac{5000}{1}\right) \]4. **Calculate the logarithm term:** \[ \ln\

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Water is pumped from a well tapping an unconfined aquifer at a rate of 2400 m³/day. A no-drawdown boundary exists at a distance of 5 km from the well centre. Assuming the well to be fully penetrating, compute the steady state drawdown at the well face. Given: Initial saturated thickness = 50 m, hydraulic conductivity = 20 m/day, effective well radius = 1 m.

Water is pumped from a well tapping an unconfined aquifer at a rate of 2400 m³/day. A no-drawdown boundary exists at a distance of 5 km from the well centre. Assuming the well to be fully penetrating, compute the steady state drawdown at the well face. Given: Initial saturated thickness = 50 m, hydraulic conductivity = 20 m/day, effective well radius = 1 m.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Problem 3.5 A well of diameter 0.3 m is constructed to the full depth of an unconfined aquifer of thickness of 150 m. The water table is 10 m below the ground surface. A pumping test of 12 m'/hour has resulted a drawdown of 10 m. Assuming ro = 400 m; calculate the coefficient of permeability of the aquifer. If the flow rate increases to 18 m'/hour, and in the absence of any other data, what will be the best estimate for the drawdown in the well?
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TELEGRAM Q1: A 30cm diameter well penetrates vertically through an aquifer to an impermeable stratum located 18.0 m below the static water table. After a long period of pumping at a rate of 1.2 m3/min, the drawdown in test holes 11m and 35 m from the pumped well is found to be 3.05 and 1.62m respective. What is the hydraulic conductivity of the aquifer? Express in meters per day. What is its transmissivity? Express in cubic meters per day per meter. What is the drawdown in the pumped well?