---
title: "Specific yield and specific capacity, recuperation test"
subject: "Engineering Hydrology"
chapter: 1
key: "semester_5/engineering_hydrology:well-recuperation-specific-yield"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/well-recuperation-specific-yield
site: BCT Engineering
---

# Specific yield and specific capacity, recuperation test

Engineering Hydrology · Chapter 1: Introduction

- Asked in: 2066 Magh (old course), 6 marks
- Video: [Specific Yield and Specific Retention | Specific yield of well | Specific yield and storativity](https://www.youtube.com/watch?v=cc4Cs6jb2AU) — VIP Civil Guru
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/well-recuperation-specific-yield
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-1

## Question

In a recuperation test, the static water level in an open well was depressed using pumps by 3 m and it recuperated 1.5 m in 1 hour. If the diameter of the well is 3.0 m and the safe working depression head is 2.4 m, find out the average specific yield of the soil and specific capacity of the soil.

## Short answer

Recuperation test yields specific capacity constant K_i = ln(h1/h2)/t = 0.693 h⁻¹ (0.693 m³/h per m² per m depression). Well specific capacity = K_i × area = 4.90 m³/h per m depression; safe yield at 2.4 m depression = 11.76 m³/h.

## Answer

A **recuperation test** measures how fast water rises in a depressed open well. The rate of recovery is proportional to the depression head $h$ at that time:

$$
\frac{dh}{dt} = -\frac{Q}{A} = -K_i h \;\Rightarrow\; K_i = \frac{1}{t}\ln\frac{h_1}{h_2}
$$

where $K_i$ is the **specific capacity (per unit area, per unit depression)** of the soil, also called the *constant of the well/soil*, and $A$ is the well area.

**Given:** initial depression $h_1 = 3$ m; after $t = 1$ h the depression $h_2 = 3 - 1.5 = 1.5$ m; well diameter 3.0 m; safe depression head $H = 2.4$ m.

$$
K_i = \frac{1}{1}\ln\frac{3}{1.5} = 0.693\ \text{h}^{-1}
$$

Well area: $A = \frac{\pi}{4}(3.0)^2 = 7.069$ m².

#### Specific capacity
Specific capacity of the well (discharge per unit depression):

$$
\frac{Q}{H} = K_i A = 0.693 \times 7.069 = 4.90\ \text{m}^3/\text{h per m of depression}
$$

The specific capacity of the soil per unit area is $K_i = 0.693$ m³/h per m² per m depression ($0.693\ \text{h}^{-1}$).

#### Specific (safe) yield at the working depression
$$
Q = K_i A H = 0.693 \times 7.069 \times 2.4 = 11.76\ \text{m}^3/\text{h}
$$

**Answer: Specific capacity = 0.693 m³/h/m² of well area per metre depression (4.90 m³/h per metre depression for this well); safe yield (average yield) = 11.76 m³/h (about 282 m³/day).**

## Check yourself

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

1. In a recuperation test the rate of recovery is proportional to:
   A. The well diameter only
   B. Time squared
   C. The depression head h at that time
   D. The initial level
   Answer: C. dh/dt = -K_i h. (Key point: Principle)

2. K_i is found from:
   A. (h1 - h2)/t
   B. t ln(h1 h2)
   C. (h1/h2)^2
   D. (1/t) ln(h1/h2)
   Answer: D. Integration of dh/dt = -K_i h. (Key point: Formula)

3. After 1 h of 1.5 m recovery from 3 m depression, h2 is:
   A. 1.5 m
   B. 4.5 m
   C. 3 m
   D. 0.5 m
   Answer: A. 3 - 1.5 = 1.5 m. (Key point: Depression)

4. Safe yield is:
   A. K_i / A
   B. K_i A H = 11.76 m^3/h
   C. K_i x h1
   D. 4.90 m^3/h
   Answer: B. 0.693 x 7.069 x 2.4 = 11.76. (Key point: Yield)

## Related in Engineering Hydrology

- [Average yield of well from recuperation test](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Awell-recuperation-test) — Engineering Hydrology, chapter 4