Question

# (a) A constant head permeability test was performed for the soil specimen. The sample length was...

(a)
A constant head permeability test was performed for the soil specimen. The sample length was 90 mm and sample diameter was 75 mm. The total head loss across the sample was measured to be 635 mm. The flow rate was measured and it took 278 seconds to fill up a 1-litre beaker. Estimate the coefficient of permeability (k), in m/s, based on results of the constant head test.
(b)
If a falling head test was conducted on the same soil specimen in part (a) (with same dimensions), with a standpipe diameter of 4 mm and initial total head difference of 1.05 m, how long would it take for the water level to drop by 1 m through the standpipe? Is it practical to use this test to determine k value of the soil?
(c)
The soil specimen in parts (a) and (b) were obtained from a project site, where the soil layer was 10 m thick, overlying an impermeable rock layer. A pumping test was conducted at the site.
Before the pumping started, the water table was found to be 2 m below ground surface. At the pumping rate (flow rate, q) of 0.015 m3/s, the water table dropped by 1.5 m at a borehole 10 m away from the pumping well. What would be the drop in water table at another observation borehole, located 5 m away from the pumping well?

Solution (a) :

We have ,for constant head permeability test:

q = k * i * a = k * ( h/L) * A

k = q*L/(h.A)

Data given : Discgarge (q) = 10-3/(278) m3/s = 3.60 * 10-6 m3/s

Sample length = 90 mm = 0.09 m

Head loss (h) = 635 mm = .635 m

Area = (3.14/4) * d2 = 4.42 * 10-3 m2

substitute these value in above equation ,

k = 3.6* 10-6 * .09/ ( .635 * 4.42 * 10-3)

k = 1.154 * 10-4 m/s.

Permeability is given by, k = (aL/At) * ln( h1/h2)

Given data : Area of stand pipe (a) = (3.14/4) * 0.0042 = 1.26 * 10-5 m2.

initial head difference ( h1) = 1.05 m & final head difference (h2) = 1.05-1=0.05 m

Putting all the given values in equation we have,

1.154 *10-4 = ((1.26 * 10-5 * .09)/ (4.42 * 10-3 *t)) * ln( 1.05/.05)

t = 6.768 seconds.

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