Analytical Chem Soil HW Solutions 45

# Analytical Chem Soil HW Solutions 45 - h A ≈ ~ 10 m –...

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45 (a) The flowrate q is calculated from q ( m 3 /s per metre length) = k t .H.N F /N H where k t , the equivalent permeability of the transformed section, = (k v .k h ) (see main text Section 3.14); k t = 4 × 10 -6 m/s; H, the overall head drop = 10 m (from the groundwater level on the retained side to the floor of the excavation); N F , the number of flowtubes =5 (note the flownet is NOT symmetrical in this case); and N H , the number of equipotential drops, = 8 Hence metre s m m s m q / / 10 5 . 2 8 5 ) ( 10 ) / ( 10 4 3 5 6 × = × × × = or q = 0.025 litre/sec per metre length (b) the point A is roughly ¼ of the way between the third and fourth equipotential lines after h = 10 m. Each equipotential drop is 10/8 = 1.25 m. Interpolating, the head at A is approximately
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Unformatted text preview: h A ≈ ~ 10 m – (3.25 1.25 m) 5.94 m The point A is about 24.5 m below the datum for the measurement of head, giving u A = γ w .(24.5 + 5.94) 300 kPa (see main text Section 3.10 and Example 3.7 for details of the calculation of pore water pressures from flownets). The upward hydraulic gradient between the sheet piles is (scaling from the flownet) approximately 1.25 m ÷ 3 m or 0.42, which is comfortably below the critical value of about 1. Thus our dewatering scheme, which involves installing pumped wells with sufficient capacity to draw down the groundwater level within the excavation to formation level, should be adequate to ensure the stability of the base....
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## This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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