**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

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ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Field applications of hydraulic conductivity

Hydraulic conductivity is one of the hydraulic properties of soil defined as a measurement of the ability of a porous medium to permit water movement or fluid flow through it. It is denoted by the letter K. It is influenced by the density and viscosity of the fluid. It is also called permeability. Hydraulic conductivity is defined as the ratio of velocity to the hydraulic gradient. The different methods used to determine the hydraulic conductivity are laboratory methods (constant head permeability tests and falling head tests), in-situ field methods (pumping in and pumping out tests), small scale field tests (infiltration tests and slug tests), and indirect methods such as estimation from grain size, consolidation test data, and horizontal capillary test. The pedotransfer function (PTF) method is a specific empirical estimating approach employed in the soil sciences is also used to determine the hydraulic conductivity of the soil.

Question

**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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