Specific yield and specific capacity, recuperation test

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).