Question: Explain the recuperation test.
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Mumbai University > Civil Engineering > Sem 7 > Irrigation Engineering

Marks: 5M

Year: Dec 2015

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modified 2.0 years ago  • written 2.0 years ago by gravatar for Juilee Juilee2.1k
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Through the constant level pumping test gives an accurate value of safe yield of an open well, it is sometimes very difficult to regulate the pump in such a way that constant level is maintained in the well. In such a circumstance, a recuperation test is resorted to. In the recuperation test, water is depressed to any level below the normal and the pumping is stopped. The time taken for the water to recuperate to the normal level is noted.

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Let aa =static water level in the well, before the pumping started.

bb = Water level in the well when the pumping stopped

h1= depression head in the well when the pumping stopped

cc = water level in the well at a time T after the pumping stopped

h2 = depression head in the well at a time T after the pumping stopped

h = = depression head in the well at a time t after the pumping stopped

dh = decrease in depression head in a time dt

t,T = time in hours.

In time dt after this, the head recuperates by a value dh meters.

Volume of water entering the well, when the head recuperates by dh is

dV = A dh …(1)

Where A = cross-sectional area of well at its bottom

Again, if Q is the rate of discharge in the well at the time t, under the depression head h, the volume of water entering the well in them t hours is given by

dV = Q dt

But Q is proportional to h

Or Q = Kh ….(2)

dV = K h dt …..(3)

where K is a constant depending upon the soil at the base of the well through which water enters.

Equating (1) and (3), we get

K h dt = -A dh

The minus sign indicates that h decreases as time t increases. Integrating the above between the limts : t = 0 when h = h1 ; t = T when h = h2 .

We get $\frac{K}{A} \int\limits_0^Tdt = - \int\limits_{h1}^{h2} \frac{dh}{h}$

Or $\frac{K}{A} \int\limits_0^T dt = - \int\limits_{h2}^{h1} \frac{dh}{h}$

From which $\frac{K}{A} T = [log_e⁡h ]_{h2}^{h1}$

$\frac{K}{A} = \frac{1}{T} log_e⁡ \frac{h1}{h2} = \frac{2.303}{T} log_{10} \frac{h1}{h2}$

Thus knowing the value of h1 ,h2 and T from recuperation test, the quality K/A can be calculated. K/A is known as the specific yield or specific capacity of an open well.

Type of soil K / A
Clay 0.25
Fine sand 0.5
coarse sand 1.00

Knowing the value of K/A by observation, the discharge from a well under a contant depression head H can be calculated as under:

$Q = K H$

$Q = [\frac{K}{A}] A.H.$

$Q = \frac{2.303}{T} log_{10} \frac{h1}{h2} A H m^3/hour$

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written 2.0 years ago by gravatar for Juilee Juilee2.1k
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