Sedimentation of Particles in Water or Air
Example of WWT exercises
Uniform straight-line motion
The case is for example a particle sedimentation in the gravitational field
when the particle has a constant falling speed.
Balance of forces act to a particle - Steady speed motion is important for our
solution (particles sedimentation).
FD
- A particle moves with a constant falling
FLF
, 
d
uSC
speed uSC. Forces acting to the particle
have to be in balance (there is not
P
acceleration or deceleration).
- The steady speed is reached for time
G
  ;
in praxis final speed u = 0,99*uSC is used.
We suppose a spherical particle with following acting forces:
G
Gravitational force
 *d3
FLF 
Lifting force
6
 *d3
6
Drag force
FD  C D *
Inertial force
FIN = 0
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* P * g
* *g
 * d 2 uS 2
4
*
2
*
P. Hoffman
Datum tisku: 06.02.上午五
The forces have to be in balance
G = FLF + FD
 *d3
6
* P * g =
u S2 
 *d3
6
*  * g + CD *
2
 * d 2 uS
4
*
2
*
4 d * (P  ) * g
*
3
CD * 
Value of the drag coefficient CD depends on Re number and thereby on unknown sedimentation speed uS too. Therefore it is impossible to set the coefficient. But we can use relations valid for laminar and turbulent regions of particles sedimentation.
- For laminar (Stokes) region it is valid:
CD 
24
Re S
Re S 
where
uS * d * 
 0,2  0,3 ..... for accuracy  +/- 0,5 %

 2 .................. for accuracy  +/- 5 %
- For turbulent (Newton) region it is valid:
3*105  ReS  400 – 500
CD = 0,44
- For transient region it is valid:
CD = 18,5 * ReS-0,6
or
24
CD 
Re S
 Re 2S / 3 

* 1 
6


After a substitution we can set the sedimentation speed of the particle in the
laminar region:
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P. Hoffman
Datum tisku: 06.02.上午五
4 d * (  P   ) * g * uS * d * 
u S2  *
3
24 *  * 
and after modification it is
d 2 * (P   ) * g
uS 
18 * 
Analogously after substitution we can set the sedimentation speed of the particle
in the turbulent region:
4 d * (P  ) * g
uS2  *
3
0,44 * 
u S  1,74 *
and after modification it is
d * ( P  ) * g

- Setting of a sedimentation region
A dimensionless term is used for this setting. The term contains only known values, it is parameters of the particle and its ambient (water, air etc.).
4 d 3 *(P  ) *  * g
CD * Re  *
3
2
2
S
For L region it is valid
CD*ReS2  12 – 48 (accuracy +/- 0,5 – 5 %)
For T region it is valid
1,1*105  CD*ReS2  4*1010
Note: Sludge and fine sand particles usually settle in the laminar region  by reason of
time saving you can calculate the speed according equations for L region and
than check up the Re number.
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P. Hoffman
Datum tisku: 06.02.上午五
Ex. 1: Sedimentation of sludge particles in water.
Given: Sludge d = 0,2 mm; p = 1020 kg/m3;  = 998 kg/m3; = 1*10-3 Pa*s
Calculate a sedimentation speed uS = ?, time  = ? when the particle
Task:
reaches the speed u = 0,99 uS and a distance h = ? that is covered during
the time .
Setting of a sedimentation region
4 (0,2 *10 3 ) 3 * (1020  998) * 998 * 9,81
C D * Re  *
 2,30  12  48
3 2
3
(1*10 )
2
S
 L region
Setting of the sedimentation speed of the particle
(0,2 *10 3 ) 2 * (1020  998) * 9,81
uS 
 4,80 *10 4 m / s  1,73m / h
3
18 *1 *10
Setting of the time when the particle reaches the sedimentation speed u =
0,99*uS
u
d 2 * P
 
* ln( 1   )
18 * 
uS
(0,2 * 10 3 ) 2 * 1020
 
* ln( 1  0,99)  1,04 * 10  2 s
3
18 * 1 * 10
Setting of the distance that the particle covers during the time 
h
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č
Re
*d * S

18
4
u
u 
*    ln( 1   )
uS 
 uS
P. Hoffman
Datum tisku: 06.02.上午五
h
1020
0,0958
* 0,2 * 10 3 *
* 0,99  ln( 1  0,99)  3,93 * 10 6 m  3,93 * 10 3 mm
998
18
uS * d *  4,80 *104 * 0,2 *103 * 998
Re S 

 0,0958  2

1*103
where
Similar calculations are for sedimentation of gypsum or sand.
Ex. 2: Sedimentation of gypsum particles in water.
G:
T:
d = 0,1 mm; P = 1800 kg/m3;  = 998 kg/m3; = 1*10-3 Pa*s
Calculate a sedimentation speed uS = ?, time  = ? when the particle
reaches the speed u = 0,99 uS and a distance h = ? that is covered during
the time .
Setting of a sedimentation region
4 (0,1 * 10 3 ) 3 * (1800  998) * 998 * 9,81
C D * Re 2S  *
 10,5  12  48  L region
3
(1 * 10 3 ) 2
Setting of the sedimentation speed of the particle
uS 
(0,1 * 10 3 ) 2 * (1800  998) * 9,81
 4,37 * 10 3 m / s  15,74m / h
3
18 * 1 * 10
Setting of the time when the particle reaches the sedimentation speed
u = 0,99*uS
 
(0,1 * 10 3 ) 2 * 1800
* ln( 1  0,99)  4,6 * 10 3 s
3
18 * 1 * 10
Setting of the distance that the particle covers during the time 
h
where
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1800
0,436
* 0,1 * 10 3 *
* 0,99  ln( 1  0,99)  1,58 * 10 5 m  0,0158mm
998
18
4,37 *10 3 * 0,1 *10 3 * 998
Re S 
 0,436  2
1 *10 3
5
P. Hoffman
Datum tisku: 06.02.上午五
Ex. 3: Sedimentation of sand particles in water.
G:
T:
d = 0,1 mm; P = 2300 kg/m3;  = 998 kg/m3; = 1*10-3 Pa*s
uS = ?,  = ? for u = 0,99 uS and h = ? for time  .
Setting of a sedimentation region
4 (0,1 *10 3 ) 3 * (2300  998) * 998 * 9,81
C D * Re  *
 17  12  48  L region
3
(1 *10 3 ) 2
2
S
Setting of the sedimentation speed of the particle
(0,1*10 3 ) 2 * (2300  998) * 9,81
uS 
 7,10 *10 3 m / s  25,5m / h
3
18 *1*10
Setting of the time when the particle reaches the sedimentation speed
u = 0,99*uS
(0,1 * 10 3 ) 2 * 2300
 
* ln( 1  0,99)  5,9 * 10 3 s
3
18 * 1 * 10
Setting of the distance that the particle covers during the time 
h
2300
0,709
* 0,1 * 10 3 *
* 0,99  ln( 1  0,99)  3,28 * 10 5 m  0,0328mm
998
18
where
7,10 *10 3 * 0,1 *10 3 * 998
Re S 
 0,709  2
1 *10 3
Ex. 4: Sedimentation of sand particles in water.
G:
T:
d = 0,14 mm; P = 2300 kg/m3;  = 998 kg/m3;  = 1*10-3 Pa*s
uS = ?,  = ? for u = 0,99 uS and h = ? for time  .
Setting of a sedimentation region
C D * Re 2S 
4 (0,14 * 10 3 ) 3 * (2300  998) * 998 * 9,81
*
 46,6  48  L region
3
(1 * 10 3 ) 2
(boundary between L and transient regions)
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P. Hoffman
Datum tisku: 06.02.上午五
Note:
For a result accuracy c. +/- 5 % ..... limit of application of relations for the laminar region.
Note:
For the same conditions but sand particle diameter d = 1,9 mm we reach a beginning of the turbulent
region (CD*ReP2 = 1,17*105).
Setting of the sedimentation speed of the particle
(0,14 *10 3 ) 2 * (2300  998) * 9,81
uS 
 13,9 *10 3 m / s  50,1m / h
3
18 *1 *10
Setting of the time when the particle reaches the sedimentation speed
u = 0,99*uS
(0,14 * 10 3 ) 2 * 2300
 
* ln( 1  0,99)  11,5 * 10 3 s
3
18 * 1 * 10
Setting of the distance that the particle covers during the time 
h
2300
1,94
* 0,14 * 10 3 *
* 0,99  ln( 1  0,99)  1,26 * 10  4 m  0,13mm
998
18
Re S 
where
13,9 * 10 3 * 0,14 * 10 3 * 998
 1,94  2
1 * 10 3
Ex. 5 Sedimentation of sand (dust) particles in air.
G:
T:
d = 0,075 mm; P = 2300 kg/m3;  = 1,190 kg/m3;  = 1,82*10-5 Pa*s
uS = ?,  = ? for u = 0,99 uS and h = ? for time  .
Setting of a sedimentation region
4 (0,075 *10 3 ) 3 * (2300  1,19) *1,19 * 9,81
C D * Re  *
 45,6  48  L region
3
(1,82 *10 5 ) 2
2
S
Setting of the sedimentation speed of the particle
(0,075 *10 3 ) 2 * (2300  1,19) * 9,81
uS 
 0,387m / s
18 *1,82 *10 5
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P. Hoffman
Datum tisku: 06.02.上午五
Setting of the time when the particle reaches the sedimentation speed
u = 0,99*uS
(0,075 * 10 3 ) 2 * 2300
 
* ln( 1  0,99)  0,182s
18 * 1,82 * 10 5
Setting of the distance that the particle covers during the time 
h
2300
1,90
* 0,075 * 10 3 *
* 0,99  ln( 1  0,99)  0,0553m  55,3mm
1,19
18
It is usually impossible to neglect the distance.
Re S 
where
0,387 * 0,075 * 10 3 * 1,19
 1,9  2
1,82 * 10 5
Ex. 6: Dust particle sedimentation in air.
G:
T:
d = 0,020 mm; P = 2300 kg/m3;  = 1,190 kg/m3;  = 1,82*10-5 Pa*s
uS = ?,  = ? for u = 0,99 uS and h = ? for time  .
Setting of a sedimentation region
4 (20 *10 6 ) 3 * (2300  1,19) *1,19 * 9,81
C D * Re  *
 0,84  12  48  L region
3
(1,82 *10 5 ) 2
2
S
Setting of the sedimentation speed of the particle
(20 *10 6 ) 2 * (2300  1,19) * 9,81
uS 
 0,0275m / s
5
18 *1,82 *10
Setting of the time when the particle reaches the sedimentation speed
u = 0,99*uS
(20 * 10 6 ) 2 * 2300
 
* ln( 1  0,99)  0,0129s
18 * 1,82 * 10 5
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P. Hoffman
Datum tisku: 06.02.上午五
Setting of the distance that the particle covers during the time 
h
0,0275 * 20 * 10 6 * 1,19
Re S 
 0,0359  2
1,82 * 10 5
where
Note:
2300
0,0359
* 20 * 10 6 *
* 0,99  ln( 1  0,99)  0,000277m  0,28mm
1,19
18
For the same conditions (sand particles in air) but diameter d = 1,0 mm it is C C*ReP2 = 1,08*105
it is that we are at the beginning of the turbulent region.
Ex.7: Free fall of a hailstone from a storm cloud
G:
T:
d = 20 mm; P = 918 kg/m3;  = 1,190 kg/m3;  = 1,82*10-5 Pa*s
Set a falling (sedimentation) speed uS = ?
Setting of a sedimentation region
4 (20 * 10 3 )3 * (918  1,19) * 1,19 * 9,81
CD * Re  *
 3,45 * 108
5 2
3
(1,82 * 10 )
2
S
 T region
It is valid for the region CD = 0,44
Setting of the Reynolds number
C D * Re 2S
3,45 *10 8
Re S 

 2,77 *10 4
CD
0,44
Setting of the constant sedimentation speed of the particle
u S  1,74 *
d * ( P  ) * g

20 *10 3 * (918  1,19) * 9,81
 1,74 *
 21,4m / s
1,19
Re*  2,77 *10 4 *1,82 *10 5
uS 

 21,2m / s  76km / h
d *
20 *10 3 *1,19
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Datum tisku: 06.02.上午五
Note 1:
The biggest in the Czech Republic observed hailstone had the diameter c. 120
mm. The corresponding falling speed is c. 52,4 m/s = 188 km/h.
Note 2:
As the falling time is relatively short the majority of the hail mass has to be
formed in clouds in rising air flows. These rising flows have to have approximately the same speed as the falling speed is.
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P. Hoffman
Datum tisku: 06.02.上午五