The velocity at which a spherical particle will settle out of suspension (fall t
ID: 1778126 • Letter: T
Question
The velocity at which a spherical particle will settle out of suspension (fall to the bottom) is defined by Stokes' Law: 9 where g = acceleration of gravity (m. R- radius of particle (m) P = mass density ofthe sphere (kg/m3) mass density of the fluid (kg/m3) = viscosity of the fluid (kg·m-1-S1) 1. Using Stoke's Law (Eq. 5-1), calculate the velocity (m/s) at which the smallest sand (0.05 mm) and silt (0.002 mm) particles will settle in pure water at 20°C. Assume that the fluid density is 998.2071 kgm-3, fluid viscosity is 0.001005 kg-m-1s-1, gravity = 9.81 ms-2 and mass density is 2650 kg.m-3. Shows these calculations. 2. Using the velocity that you calculated in Step 1, calculate the amount of time it would take for the smallest sand particle to settle to the bottom from the top of a 100 cm deep column of water. Show your calculations. 3. Using the velocity that you calculated in Step 3, calculate the amount of time it would take for the smallest silt particle to settle to the bottom from the top of a 100 cm deep column of water. Show your calculations 4. If soil particles are uniformly suspended in a 100 cm deep column of water, then the settling distance ranges from 0 (particles already at the bottom) to 100 cm (particles suspended at the top of the fluid column). Assuming the average settling distance is 50 cm, recalculate the settling time for the smallest sand and silt particles. Show your calculations. 5.Assume that you suspended a 50 gram sample of soil. After all of the sand settled out, 39 grams of material remained suspended. Calculate the % of sand in the soil sample (Mass Soil-Mass Sand)/Mass Soil x 100%. Show your calculationExplanation / Answer
given,
diameter of samd particle = 0.05 mm
diameter of silt particle = 0.002 mm
density of fluid = 998.2071 kg/m^3
density of density of particles = 2650 kg/m^3
putting values in stoke's equation
Vsand = 2 * (2650 - 998.2071) * 9.8 * (0.05 * 10^-3 / 2)^2 / (9 * 0.001005)
Vsand = 0.002237 m/s
Vsilt = 2 * (2650 - 998.2071) * 9.8 * (0.002 * 10^-3 / 2)^2 / (9 * 0.001005)
Vsilt = 3.57934 * 10^-6 m/s
speed = distance / time
0.002237 = 1 / time
time = 447.027 sec
amount of time smallest sand particle would take = 447.027 sec
3.57934 * 10^-6 = 1 / time
time = 279381.114954 sec
amount of time smallest silt particle would take = 279381.114954 sec
settling time for sand particle = 0.5 / 0.002237
settling time for sand particle = 223.5136 sec
settling time for silt particle = 0.5 / (3.57934 * 10^-6)
settling time for silt particle = 139690.557 sec
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