---
title: "Average yield of well from recuperation test"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:well-recuperation-test"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/well-recuperation-test
site: BCT Engineering
---

# Average yield of well from recuperation test

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2067 Mangsir (old course), 8 marks
- Video: [Recuperation Test l Yield of Well l Part 2 l Numerical l SM Class](https://www.youtube.com/watch?v=-q4BxSI1NKM) — SM Class
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/well-recuperation-test
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

In a recuperation test, the static water level in an open well was depressed by pumping 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 yield of the pump.

## Short answer

Specific capacity K = (1/1) ln(3.0/1.5) = 0.693 h⁻¹. With well area A = 7.07 m² and safe depression 2.4 m, average pump yield is Q = 0.693 × 7.07 × 2.4 = 11.8 m³/h (3.27 L/s).

## Answer

A **recuperation test** gives the **specific capacity** of the well, $K$: the rate at which water enters the well per unit area per unit depression head.

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

#### Specific capacity
$$
K = \frac{1}{t}\ln\frac{h_1}{h_2} = \frac{2.303}{1}\log_{10}\frac{3.0}{1.5} = 0.693\ \text{h}^{-1}
$$

#### Area of the well
$$
A = \frac{\pi}{4}D^2 = \frac{\pi}{4}(3.0)^2 = 7.07\ \text{m}^2
$$

#### Yield
$$
Q = K\,A\,H = 0.693\times7.07\times2.4 = 11.76\ \text{m}^3/\text{h}
$$

$$
Q = \frac{11.76}{3.6} = 3.27\ \text{L/s} \approx 282\ \text{m}^3/\text{day}
$$

**Answer: Average yield of the pump $\approx 11.8$ m$^3$/h (3.27 L/s).**

## Check yourself

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

1. What does a recuperation test give?
   A. The storage coefficient of the aquifer
   B. The specific capacity K of the well, the inflow rate per unit area per unit depression head
   C. The permeability of the pump
   D. The well diameter
   Answer: B. K measures how fast water enters the well per unit area per unit depression. (Key point: Specific capacity)

2. What are h1 and h2 in K = (1/t) ln(h1/h2)?
   A. h1 = 3.0 m, h2 = 3.0 + 1.5 = 4.5 m
   B. h1 = 2.4 m, h2 = 1.5 m
   C. h1 = 3.0 m, h2 = 3.0 - 1.5 = 1.5 m
   D. h1 = 1.5 m, h2 = 3.0 m
   Answer: C. Initial depression 3 m reduces by the 1.5 m recovery to 1.5 m, so K = ln 2 = 0.693 per hour. (Key point: Depression after recuperation)

3. Which expression gives the yield at the safe depression head?
   A. Q = K A h1 = 0.693 x 7.07 x 3.0
   B. Q = A H / K
   C. Q = K D H
   D. Q = K A H = 0.693 x 7.07 x 2.4 = 11.76 m^3/h
   Answer: D. Yield uses the safe working depression head H = 2.4 m and area pi/4 D^2 = 7.07 m^2. (Key point: Yield)

## Related in Engineering Hydrology

- [Specific yield and specific capacity, recuperation test](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Awell-recuperation-specific-yield) — Engineering Hydrology, chapter 1