Hi,
With this single solved example, you will refresh two concepts of soil mechanics, as follows.
- How to find the equivalent permeability of linearly stacked soil samples.
- How to find the hydraulic gradient in individual layers, and hence the head losses.
So, let’s move to the problem and its solution.
Problem
In a laboratory, a permeability test is as indicated in the figure below. The cross-sectional area of the sand is 200 cm2, and the permeability coefficients of sand I and sand II are 0.2 cm/s and 0.1 cm/s, respectively. The total hydraulic head loss is 30 cm. It is assumed that the seepage velocity in sand I is same as in sand II. Determine the height of the water in the standpipe as shown in the figure below.
Solution
For soil samples in series, Keq = H/(H1/K1 + H2/K2) = 80/(40/0.2 + 40/0.1) = 0.133 cm/s
Total head loss after water crossing the two sand layers, dH = 30 cm (read the water levels at top in two limbs)
Total Hydraulic Gradient, i = dH/L = 30/80 = 3/8
Discharge, q = keq*i = 0.133*3/8 = 1/20 = 0.05 cm/s
For a flow is series, discharge is equal at every cross-section at a single time, hence q = q1 = q2 = 1/20 cm/s
but q1 = k1*i1, so k1*i1 = 1/20 cm/s, => i1 = (1/20)/0.2 = 1/4
but i1 = dh1/l1 => dh1/40 = 1/4, or dh1 = 1*40/4 = 10 cm, hence Head loss in sand1 is 10 cm,
As the standpipe is at the section joining two sand layers, The height of water in the standpipe (piezometer)
= initial water elevation at top left limb – total head lost in two layers + head lost in upper layer
= top water level in right limb + head loss in upper sand layer = 20 + 40 + 10 = 70 cm
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If you have any doubt, please leave a comment in the comment box below.
Thank you!
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