Question

Compute the settling velocity for the following particles: very coarse sand ( diameter = 1.5 mm ) , (diameter=1.5 mm), medium sand (0.4 mm), very fine sand (0.075 mm), and clay (0.001 mm). Estimate the time for each particle to fall 6 in. in water.

Assume: SG: coarse sand = 2.65, medium sand = 2.67, fine sand = 2.70, clay = 2.8 & water temperature = 60°F

Note:The other solution on chegg is wrong.

Answer 1

Salution:

Given data:

Very coarse sand diameter d1 = 1.5 mm

SG of coarse sand = G1 =2.65

Medium sand diameter d2 = 0.4 mm

SG of medium sand = G2= 2.67

Very fine sand diameter d3 =0.075 mm

SG of fine sand = G3 = 2.70

clay particals diameter d4 = 0.001 mm

SG of clay = G4 = 2.8

g = 9.81 m/sec2

Say kinematic viscosity of water =   = 1.787 x 10-6 m2 s-1

(a)

=>

Very coarse sand diameter d1 = 1.5 mm

SG of coarse sand = G1 =2.65

Partical diameter =d1 = 1.5 mm > 1 mm

Turbulent settling velocity = Vs =1.8  { g d (G-1) } = 1.8  { g d1 (G1-1) } = 1.8 { 9.81 x (1.5/1000) (2.65-1) }

Vs = 0.2805 m/sec

Velocity for very coarse sand ( diameter = 1.5 mm ) is 0.2805 m/sec.....................(Answer)

Fall = h = 6" = 0.1524 meter

6" Fall time =t = h/Vs = 0.1524 m / 0.2805 m/sec = 0.5433 second........................(Answer)

(b)

=>

Medium sand diameter d2 = 0.4 mm

SG of medium sand = G2= 2.67

Water Temparature = T =60°F = { (60°F − 32) × 5/9 } = 15.556°C

Partical diameter =d2 = 0.47 mm , 0.1 mm <d2 < 1 mm

Settling velocity = Vs =418 (G-1 ) d {(3T+70)/100}  = 418 x (2.67-1 ) x (0.4/1000 m) x {((3x15.556)+70)/100}

Vs = 0.3258 m/sec

Velocity for medium sand (0.4 mm) is 0.3258 m/sec.....................(Answer)

Fall = h = 6" = 0.1524 meter

6" Fall time =t = h/Vs = 0.1524 m / 0.3258 m/sec = 0.468 second........................(Answer)

(c)

=>

Very fine sand diameter d3 =0.075 mm

SG of fine sand = G3 = 2.70

Water Temparature = T =60°F = { (60°F − 32) × 5/9 } = 15.556°C

Partical diameter =d3 = 0.075 mm ,   < 0.1 mm

Settling velocity = Vs =418 (G-1 ) d^2 {(3T+70)/100}  = 418 x (2.70-1 ) x (0.075/1000 m)^2 x {((3x15.556)+70)/100}

Vs = 4.663 x 10^-6 m/sec

Velocity for very fine sand (0.075 mm) is 4.663 x 10^-6 m/sec.....................(Answer)

Fall = h = 6" = 0.1524 meter

6" Fall time =t = h/Vs = 0.1524 m / 4.663 x 10^-6 m/sec = 32680 second. = 9 hours 4 minuts 40 Second....(Answer)

(d)

=>

Clay particals diameter d4 = 0.001 mm

SG of clay = G4 = 2.8

Water Temparature = T =60°F = { (60°F − 32) × 5/9 } = 15.556°C

Partical diameter =d4 = 0.001 mm ,   < 0.1 mm

Settling velocity = Vs =418 (G-1 ) d^2 {(3T+70)/100}  = 418 x (2.80-1 ) x (0.001/1000 m)^2 x {((3x15.556)+70)/100}

Vs = 8.778 x 10^-10 m/sec

Velocity for clay (0.001 mm) is 8.778 x 10^-10 m/sec.....................(Answer)

Fall = h = 6" = 0.1524 meter

6" Fall time =t = h/Vs = 0.1524 m / 8.778 x 10^-10 m/sec = 173613873 second

= 5.50526 years ........................(Answer)