Analytical Chem Soil HW Solutions 8

Analytical Chem Soil HW Solutions 8 - The total of the...

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8 Reading off from the curve, D 10 0.0035 mm (3.5 μ m) D 60 0.52 mm Hence the uniformity coefficient U = D 60 /D 10 150 (148.6) The soil is approximately 40% silt, 50% sand and 10% fine gravel: this makes it a sandy SILT according to the system given in Table 1.5. The soil is poorly (almost gap-) graded. Q1.7 A sieve analysis on a sample of initial total mass 411g gave the following results: Sieve size, mm 6.3 1.2 0.3 0.063 Mass retained, g 0 60 126 92 A sedimentation test on the 121 g of soil collected in the pan at the base of the sieve stack gave: Size, μ m <2 2-10 10-60 % of pan sample 33 24 43 Plot the particle size distribution curve and classify the soil using the system given in Table 1.5. On the PSD diagram, sketch a suitable curve for a granular filter to be used between this soil and a drainage pipe with 3 mm perforations. Q1.7 Solution
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Unformatted text preview: The total of the masses retained is 278g, which together with the 121g collected in the pan gives 399g. Thus there is a shortfall of 12g, which is attributable to sieve losses. Take the total mass of the sample as 399g. The % by mass of the sample passing each sieve is given by the total sample mass (399g) minus the cumulative mass of soil retained on larger size sieves. Hence Sieve size, mm 6.3 1.2 0.3 0.063 Mass retained, g 0 60 126 92 Cumulative mass retained, g 0 60 186 278 Mass passing, g 399 339 213 121 % passing 100 85.0 53.4 30.3 The sedimentation test data are again already part-processed, with the mass of soil in each size range expressed as a percentage of the 121g collected in the pan. The fraction of the pan sample smaller than a given size is equal to the sum of the percentages in this and the smaller...
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