What does a recuperation test give?

The specific capacity K of the well, the inflow rate per unit area per unit depression head

The storage coefficient of the aquifer

What are h1 and h2 in K = (1/t) ln(h1/h2)?

h1 = 3.0 m, h2 = 3.0 - 1.5 = 1.5 m

h1 = 3.0 m, h2 = 3.0 + 1.5 = 4.5 m

Which expression gives the yield at the safe depression head?

Q = K A H = 0.693 x 7.07 x 2.4 = 11.76 m^3/h

Q = K A h1 = 0.693 x 7.07 x 3.0

2067 Mangsir (old course) · 8 marks

Average yield of well from recuperation test

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).

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.

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).

✓ Answer checked by Claude Sonnet 5.5

Specific yield and specific capacity, recuperation test · Engineering Hydrology, chapter 1

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