Next, calculate the settling velocities of six different grains of quartz in wat
ID: 802347 • Letter: N
Question
Next, calculate the settling velocities of six different grains of quartz in water. Assume that the density of quartz is 2.7 g/cm3 (grams per cubic centimeter) and that the density of liquid water is 1.0 g/cm3. Gravitational acceleration (g) is 980 cm/s2 (centimeters per second squared). Viscosity has dimensions of mass, length and time and can be expressed in terms of g/(cm sec). Assume for this exercise that the viscosity of water is 1 x 10-2 g/(cm sec) (Hint: you can enter this in Excel as either 0.01 or 1E-02) and that the grains are spherical. (Tabular flakes of mica or clays will settle at slower velocities than spheres.) Note: Set up your calculations in Excel by creating separate columns for each component of the equations and submit your calculations with your answers.
Radius (cm)
2.
Typical turbulent streams have upward velocities of about 200 cm/s. Based upon your calculations above, are any of the particles going to settle in a turbulent stream? If yes, which ones? If no, why not?
3.
It is important to note that Stokes' Law is technically valid only for fine sand to clay sized particles (less than 0.2 mm in diameter). In fact, experiments have shown that particles larger than about 0.2 mm settle somewhat slower than predicted by Stokes' Law. (However, the effect is so small that Stokes' Law is still a good estimate.) Can you think of a reason why larger grains might settle more slowly than predicted by Stokes' Law?
Radius (cm)
Grain 1 0.0002 (clay) Grain 2 0.002 (silt) Grain 3 0.02 (fine sand) Grain 4 0.2 (coarse sand) Grain 5 2.0 (granule) Grain 6 12 (pebble)Explanation / Answer
Answer-1
Terminal Velocity= g x d2x (p-m)/ 18x mf
V = Terminal Velocity
D = Diameter of a Particle
g = Acceleration Of Gravity
mf = Viscosity of Medium
p = Density of Particle
m = Density of Medium
Given data
Density of Quartz
2.7
gm/cc
Gravitation acceleration
980
cm/sec2
Viscosity of water
0.01
gm/cm-sec
Density of medium
1
gm/cc
Reynolds Number= VD/f
r = density of medium
V = Velocity,
d = diameter and
m = viscosity
Grain type
Name of Grain
Radius (cm)
Diameter
(cm)
Settling
Velocity
(cm/sec)
Reynolds
number
Type of Flow
Grain 1
clay
0.0002
0.0004
0.00148
0.00006
Laminar
Grain 2
silt
0.002
0.004
0.14809
0.05924
Laminar
Grain 3
fine sand
0.02
0.04
14.81
59.24
Laminar
Grain 4
coarse sand
0.2
0.4
1,480.89
59,235.56
Turbulent
Grain 5
granule
2
4
1,48,088.89
5,92,35,555.56
Turbulent
Grain 6
pebble
12
24
53,31,200.00
12,79,48,80,000.00
Turbulent
Answer 2-
Typical turbulent streams have upward velocities of about 200 cm/s.
Reynolds Number= VD/f
According above calculations Grain4, Grain5 and Grain 6 are going to settle in a turbulent stream.
Would help predict the change in flow type
If the value is less than about 2000 then flow is laminar, if greater than 4000 then turbulent and in between these then in the transition zone.
This value is known as the Reynolds number, Re:
Laminar flow: Re < 2000
Transitional flow: 2000 < Re < 4000
Turbulent flow: Re > 4000
Answer 3-
The particle will be initially accelerate due to gravity, but soon the gravitational and drag forces reach equilibrium, resulting in a constant “Terminal Fall Velocity
Density of Quartz
2.7
gm/cc
Gravitation acceleration
980
cm/sec2
Viscosity of water
0.01
gm/cm-sec
Density of medium
1
gm/cc
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