---
title: "Permeability and transmissibility from pumping test"
subject: "Engineering Hydrology"
chapter: 1
key: "semester_5/engineering_hydrology:well-pumping-test-unconfined-30cm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/well-pumping-test-unconfined-30cm
site: BCT Engineering
---

# Permeability and transmissibility from pumping test

Engineering Hydrology · Chapter 1: Introduction

- Asked in: 2067 Shrawan (old course), 10 marks
- Video: [Hydraulic Conductivity and Transmissivity](https://www.youtube.com/watch?v=WA7G5Yoe2fM) — Dane Andersen
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/well-pumping-test-unconfined-30cm
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-1

## Question

A 30 cm well fully penetrates an unconfined aquifer of saturated depth 25 m. When a discharge of 2100 lpm was being pumped for a long time, observation wells at radial distances of 30 and 90 m indicated drawdowns of 5 and 4 m respectively. Estimate the co-efficient of permeability and transmissibility of the aquifer. What is the drawdown at the pumping well?

## Short answer

Dupuit's equation for unconfined aquifer: K = Q ln(r2/r1) / [π (h2²−h1²)] = 2.99×10⁻⁴ m/s, T = K H = 7.46×10⁻³ m²/s, and well drawdown from h1²−hw² = (Q ln(r1/rw))/(πK) yields s_w = 10.78 m.

## Answer

**Given:** unconfined aquifer, saturated depth $H = 25$ m; well diameter 30 cm ($r_w = 0.15$ m); $Q = 2100$ lpm $= 0.035$ m³/s; observation wells $r_1 = 30$ m, $s_1 = 5$ m and $r_2 = 90$ m, $s_2 = 4$ m.

Water-table depths: $h_1 = 25 - 5 = 20$ m, $h_2 = 25 - 4 = 21$ m.

#### (a) Coefficient of permeability
For a steady unconfined flow (Dupuit):

$$
Q = \frac{\pi K (h_2^2 - h_1^2)}{\ln(r_2/r_1)}
$$

$$
K = \frac{Q\ln(r_2/r_1)}{\pi (h_2^2-h_1^2)} = \frac{0.035 \times \ln(3)}{\pi (441 - 400)} = \frac{0.035\times 1.0986}{128.8}
$$

$$
K = 2.985\times10^{-4}\ \text{m/s} \;(= 25.8\ \text{m/day})
$$

#### (b) Transmissibility
$$
T = K H = 2.985\times10^{-4} \times 25 = 7.46\times10^{-3}\ \text{m}^2/\text{s} \;(\approx 645\ \text{m}^2/\text{day})
$$

#### (c) Drawdown at the pumping well
Apply the same equation between $r_w$ and $r_1$:

$$
h_1^2 - h_w^2 = \frac{Q \ln(r_1/r_w)}{\pi K} = \frac{0.035 \times \ln(200)}{\pi \times 2.985\times10^{-4}} = 197.7\ \text{m}^2
$$

$$
h_w^2 = 400 - 197.7 = 202.3 \Rightarrow h_w = 14.22\ \text{m}
$$

$$
s_w = 25 - 14.22 = 10.78\ \text{m}
$$

**Answer: K = 2.985 × 10⁻⁴ m/s (25.8 m/day); T = 7.46 × 10⁻³ m²/s (about 645 m²/day); drawdown at the well = 10.78 m.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula gives K for unconfined flow with two observation wells?
   A. K = Q ln(r2/r1) / [2π T (s1 - s2)]
   B. K = Q / (π r^2)
   C. K = Q ln(r2/r1) / [π(h2^2 - h1^2)]
   D. K = Q (h2 - h1)/ln(r2/r1)
   Answer: C. Dupuit's unconfined equation uses squares of water-table heights. (Key point: Equation)

2. What are h1 and h2?
   A. 5 m and 4 m
   B. 25 m and 25 m
   C. 30 m and 90 m
   D. 20 m and 21 m (depth minus drawdown)
   Answer: D. h = 25 - s. (Key point: Heights)

3. T is found as:
   A. K x H = 7.46 x 10^-3 m^2/s
   B. K / H
   C. K x Q
   D. K x r
   Answer: A. 2.985e-4 x 25 = 7.46e-3. (Key point: Transmissibility)

4. How is drawdown at the pumping well found?
   A. It equals 5 m
   B. Apply the same equation between r_w and r1 to get h_w, then s_w = 25 - h_w = 10.78 m
   C. It equals 4 m
   D. It equals 25 m
   Answer: B. h_w = 14.22 m. (Key point: Well drawdown)

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