Annlications Spring 2018 ou and your friend have begun working on the Klamath Ri

Question

Annlications Spring 2018 ou and your friend have begun working on the Klamath River. During high water events in the spring, the river becomes turbid and carries a substantial burden of fine grain sediment. The rivershed drains from b and sandy-rich soils. Clay soils shed sediment grains sediment with a size from 0.002 and 0.060 mm. oth clay-rich 0.002 mm and less in diameter. Sandy soils shed silt To work out an understanding of river basin ecology and resource cycling, you need to know which of the two soil regions loses the most material. Remembering your PHYX 118 lessons, you realize that these different sediments will ettle in water at different velocities. You devise an experiment to fill a 1 m tall glass cylinder and time how long it takes for the column of water to visibly change in opacity and indicate particle settlement. Clay and silt are both comprised of silicate minerals with a mass density of about 2,650 kg/m a) Calculate the settling velocity (terminal velocity) of particle at the break point size between silt and clay.

Explanation / Answer

According to the given problem,

a)The settling speed of particles is given by

Termnial velocity, v = 2(op - of)gR2/9, where is viscosity of fluid, of is density of fluid, op is density of particles, R is radius of oarticels, g is acceleration due to gravity

Now, for r = 0.002 mm = R, = 8.9*10-4, op = 2650 kg/m3, of = 1000 kg/m3

Substitue the values to get v,

v = 1.616629*10-5 m/s

b) Ans for sand, r = 0.06 mm

So, v = 0.0145496 m/s

As the speed for clay and sand has an order of magnitude difference of over 900

for average r = (0.06 + 0.002)/2 = 0.031 and v = 0.00388

For average sand partical, the speed of silt is order of magnitude, 240 times

So, to fall a depth of 5 cm in a beaker, time taken by sand = 5*10-3/0.00388 = 1.288 s

1.2 s we can see a visible change in opacity

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