Sedimentation of Particles in Water or Air

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22 Φεβ 2014 (πριν από 3 χρόνια και 4 μήνες)

61 εμφανίσεις

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1

P.
Hoffman



Datum tisku:
22.02.
上午四

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 pa
rticle

-

Steady speed motion is important for our


solution (particles sedimentation).

-

A particle moves with a constant falling


speed u
SC
. Forces acting to the particle


have to be in balance (there is not


acce
l
eration or deceleration).

-

The steady speed is reached for time









†
i渠n牡xis⁦湡l⁳e敤†甠㴠〬u9*u
SC

is used.


We suppose a spherical particle with following acting forces:


Gravitational force



g
d
G
P
*
*
6
*
3





Lifting force




g
d
F
LF
*
*
6
*
3






D
rag force







*
2
*
4
*
*
2
2
S
D
D
u
d
C
F



Inertial force




F
IN

= 0


d


P


,


G

F
LF

F
D

u
SC

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2

P.
Hoffman



Datum tisku:
22.02.
上午四

The forces have to be in balance



G = F
LF

+ F
D


g
d
P
*
*
6
*
3



=
g
d
*
*
6
*
3



+


*
2
*
4
*
*
2
2
S
D
u
d
C






*
*
)
(
*
*
3
4
2
D
P
S
C
g
d
u




Value of the d
rag coefficient C
D

depends on Re number and thereby on u
n-
known sedimentation speed u
S

too
.
Therefore it is impossible to set the coeff
i-
cient. But we can use relations valid for laminar and turbulent regions of part
i-
cles sediment
a
tion.


-

For laminar (Stok
es) region it is valid:



S
D
C
Re
24



where


3
,
0
2
,
0
*
*
Re





d
u
S
S

..... for accuracy



-

0ⰵ‥




†††††


2‮⸮⸮⸮⸮⸮⸮⸮⸮.⁦潲⁡捣畲ac†



-

5‥


-

F爠瑵牢rl敮琠⡎ewt温⁲ngi渠n琠ts⁶ali携



C
D

= 0,44


3*10
5




S



400


500


-

For transient region it is valid:



C
D

= 18,5 * Re
S
-
0,6

or










6
Re
1
*
Re
24
3
/
2
S
S
D
C


After a substitution we can set the
sedimentation speed of the particle in the
la
m
inar region:


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3

P.
Hoffman



Datum tisku:
22.02.
上午四







*
*
24
*
*
*
*
)
(
*
*
3
4
2
d
u
g
d
u
S
P
S



and after modific
a
tion
it is






*
18
*
)
(
*
2
g
d
u
P
S





Analogously after substitution we can set the
sedimentation speed of the particle
in the turbulent region:







*
44
,
0
*
)
(
*
*
3
4
2
g
d
u
P
S





and after mod
i
fication it is








g
d
u
P
S
*
)
(
*
*
74
,
1







-

Setting of a sedi
mentation region


A dimensionless term is used for this setting. The term contains only known va
l-
ues, it is parameters of the particle and its ambient (water, air etc.).



2
3
2
*
*
)
(
*
*
3
4
Re
*




g
d
C
P
S
D




For L region it is valid


C
D
*Re
S
2



12


48 (accuracy +/
-

0,5


5 %)


For T region it is valid


1,1*10
5



C
D
*Re
S
2



4*10
10




Note:

Sludge and fine sand particles usually settle in the laminar region


b礠e慳潮

瑩te⁳慶n朠⁹潵⁣慮⁣慬捵a慴a⁴桥⁳灥⁡c潲楮朠gqu慴楯a猠潲⁌⁲e杩潮⁡

瑨慮⁣heck⁵p

瑨t⁒umber.


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P.
Hoffman



Datum tisku:
22.02.
上午四

Ex. 1: Sedimentation of sludge particles in water.


Given: Sludge d = 0,2 mm;

p

= 1020 kg/m
3
;


㴠=98g/m
3
;

㴠=*10
-
3

Pa*s

Task:


Calculate a sedimentation speed u
S

= ?, time


㴠=w桥渠h桥⁰慲hicl攠

††
牥r捨cs⁴桥sp敥搠u


㴠=
ⰹ9⁵
S

and a distance h = ? that is covered during


the time

.


Setting of a sedimentation region


48
12
30
,
2
)
10
*
1
(
81
,
9
*
998
*
)
998
1020
(
*
)
10
*
2
,
0
(
*
3
4
Re
*
2
3
3
3
2







S
D
C



䰠牥杩潮


Setting of the sedimentation speed of the particle



h
m
s
m
u
S
/
73
,
1
/
10
*
80
,
4
10
*
1
*
18
81
,
9
*
)
998
1020
(
*
)
10
*
2
,
0
(
4
3
2
3









Setting of the time when the part
icle reaches the sedimentation speed u



0ⰹ9*u
S


)
1
ln(
*
*
18
*
2
S
P
u
u
d










s
2
3
2
3
10
*
04
,
1
)
99
,
0
1
ln(
*
10
*
1
*
18
1020
*
)
10
*
2
,
0
(










Setting of the distance that the particle covers during the time














)
1
ln(
*
18
Re
*
*
S
S
S
č
u
u
u
u
d
h






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P.
Hoffman



Datum tisku:
22.02.
上午四



mm
m
h
3
6
3
10
*
93
,
3
10
*
93
,
3
)
99
,
0
1
ln(
99
,
0
*
18
0958
,
0
*
10
*
2
,
0
*
998
1020










where

2
0958
,
0
10
*
1
998
*
10
*
2
,
0
*
10
*
80
,
4
*
*
Re
3
3
4









d
u
S
S


Simi
lar calculations are for sedimentation of gypsum or sand.


Ex. 2: Sedimentation of gypsum particles in water.


G:

d = 0,1 mm;

P

= 1800 kg/m
3
;


㴠=98g/m
3
;

㴠=*10
-
3

Pa*s

T:

Calculate a sedimentation speed u
S

= ?, time


㴠=w桥渠瑨攠pa牴i捬攠

牥r捨敳
瑨攠獰敥搠


㴠=ⰹ9⁵
S

and a distance h = ? that is covered during


the time

.


卥瑴ng映愠s敤em敮瑡ti渠n敧in


†
48
12
5
,
10
)
10
*
1
(
81
,
9
*
998
*
)
998
1800
(
*
)
10
*
1
,
0
(
*
3
4
Re
*
2
3
3
3
2







S
D
C
†


䰠牥gin


卥瑴ng映瑨攠s敤em敮瑡ti渠獰n敤映瑨攠灡牴icle



h
m
s
m
u
S
/
74
,
15
/
10
*
37
,
4
10
*
1
*
18
81
,
9
*
)
998
1800
(
*
)
10
*
1
,
0
(
3
3
2
3









卥瑴ng映
瑨攠瑩t攠睨敮⁴桥⁰a牴icl攠牥a捨cs⁴桥s敤em敮瑡瑩渠獰敥搠

u


㴠=ⰹ9*u
S



s
3
3
2
3
10
*
6
,
4
)
99
,
0
1
ln(
*
10
*
1
*
18
1800
*
)
10
*
1
,
0
(










Setting of the distance that the particle covers during the time






mm
m
h
0158
,
0
10
*
58
,
1
)
99
,
0
1
ln(
99
,
0
*
18
436
,
0
*
10
*
1
,
0
*
998
1800
5
3










where

2
436
,
0
10
*
1
998
*
10
*
1
,
0
*
10
*
37
,
4
Re
3
3
3






S

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P.
Hoffman



Datum tisku:
22.02.
上午四


Ex. 3: Sedimentation of sand p
articles in water.


G:

d = 0,1 mm;

P

= 2300 kg/m
3
;


㴠=98g/m
3
;

㴠=*10
-
3

Pa*s

T:

u
S

= ?,


㴠=⁦潲⁵


㴠=ⰹ9⁵
S

and h = ? for time


.


卥瑴ng映愠s敤em敮瑡ti渠n敧in



48
12
17
)
10
*
1
(
81
,
9
*
998
*
)
998
2300
(
*
)
10
*
1
,
0
(
*
3
4
Re
*
2
3
3
3
2







S
D
C
†


䰠牥杩潮


卥瑴ng映瑨攠s敤em敮瑡ti渠獰n敤e

瑨攠灡牴icle



h
m
s
m
u
S
/
5
,
25
/
10
*
10
,
7
10
*
1
*
18
81
,
9
*
)
998
2300
(
*
)
10
*
1
,
0
(
3
3
2
3









卥瑴ng映瑨攠瑩m攠睨敮⁴桥⁰a牴icl攠牥a捨cs⁴桥s敤em敮瑡瑩渠獰敥搠

u


㴠=ⰹ9*u
S



s
3
3
2
3
10
*
9
,
5
)
99
,
0
1
ln(
*
10
*
1
*
18
2300
*
)
10
*
1
,
0
(










Setting of the distance that the particle covers during the time





mm
m
h
0328
,
0
10
*
28
,
3
)
99
,
0
1
ln(
99
,
0
*
18
709
,
0
*
10
*
1
,
0
*
998
2300
5
3










w桥牥h

2
709
,
0
10
*
1
998
*
10
*
1
,
0
*
10
*
10
,
7
Re
3
3
3






S


䕸⸠4㨠
卥摩m敮瑡ti渠潦⁳a湤npa牴i捬敳i渠wat敲
.




搠㴠〬4m㬠

P

= 2300 kg/m
3
;


㴠=98kg/m
3
;


㴠=*10
-
3

Pa*s

T:

u
S

= ?,


㴠=⁦潲⁵


㴠=ⰹ9⁵
S

and h = ? for time


.


卥瑴ng映愠s敤em敮瑡ti渠n敧in


†
48
6
,
46
)
10
*
1
(
81
,
9
*
998
*
)
998
2300
(
*
)
10
*
14
,
0
(
*
3
4
Re
*
2
3
3
3
2






S
D
C



䰠牥gin







⡢(畮uar⁢整w敥渠䰠慮搠瑲a湳i敮琠牥ti湳)

lovinggudgeon_fca5119f
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P.
Hoffman



Datum tisku:
22.02.
上午四

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 begin
ning of the turbulent

region (C
D
*Re
P
2

= 1,17*10
5
).


Setting of the sedimentation speed of the particle



h
m
s
m
u
S
/
1
,
50
/
10
*
9
,
13
10
*
1
*
18
81
,
9
*
)
998
2300
(
*
)
10
*
14
,
0
(
3
3
2
3









Setting of the time when the particle reaches the sedimentation speed

u


㴠=ⰹ9*u
S



s
3
3
2
3
10
*
5
,
11
)
99
,
0
1
ln(
*
10
*
1
*
18
2300
*
)
10
*
14
,
0
(










Setting of the

distance that the particle covers during the time






mm
m
h
13
,
0
10
*
26
,
1
)
99
,
0
1
ln(
99
,
0
*
18
94
,
1
*
10
*
14
,
0
*
998
2300
4
3










w桥牥h

2
94
,
1
10
*
1
998
*
10
*
14
,
0
*
10
*
9
,
13
Re
3
3
3






S


䕸⸠5
卥im敮瑡tin映a湤n⡤(s琩⁰a牴i捬es⁩渠air
.




搠㴠〬75m㬠;

P

= 2300 kg/m
3
;


㴠=ⰱ,0g/m
3
;


㴠=ⰸ㈪10
-
5

Pa*s

T:

u
S

= ?,


㴠=

景爠r


㴠=ⰹ9⁵
S

and h = ? for time


.


卥瑴ng映愠s敤em敮瑡ti渠n敧in



48
6
,
45
)
10
*
82
,
1
(
81
,
9
*
19
,
1
*
)
19
,
1
2300
(
*
)
10
*
075
,
0
(
*
3
4
Re
*
2
5
3
3
2






S
D
C
†


䰠牥杩潮


卥瑴ng映瑨攠s敤em敮瑡ti渠獰n敤映瑨攠灡牴icle



s
m
u
S
/
387
,
0
10
*
82
,
1
*
18
81
,
9
*
)
19
,
1
2300
(
*
)
10
*
075
,
0
(
5
2
3









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21256703ae27.doc

8

P.
Hoffman



Datum tisku:
22.02.
上午四

Setting of the time when the particle reaches the sedimenta
tion speed

u


㴠=ⰹ9*u
S



s
182
,
0
)
99
,
0
1
ln(
*
10
*
82
,
1
*
18
2300
*
)
10
*
075
,
0
(
5
2
3









Setting of the distance that the particle covers during the time






mm
m
h
3
,
55
0553
,
0
)
99
,
0
1
ln(
99
,
0
*
18
90
,
1
*
10
*
075
,
0
*
19
,
1
2300
3









I琠ts⁵獵all⁩灯ssi扬攠e敧l散琠瑨攠摩s瑡湣攮


w桥牥h

2
9
,
1
10
*
82
,
1
19
,
1
*
10
*
075
,
0
*
387
,
0
Re
5
3





S


䕸⸠6㨠:us琠灡牴rcl攠s敤ime
湴n瑩渠楮nair.




搠㴠〬20m㬠;

P

= 2300 kg/m
3
;


㴠=ⰱ,0g/m
3
;


㴠=ⰸ㈪10
-
5

Pa*s

T:

u
S

= ?,


㴠=⁦潲⁵


㴠=ⰹ9⁵
S

and h = ? for time


.


卥瑴ng映愠s敤em敮瑡ti渠n敧in


†
48
12
84
,
0
)
10
*
82
,
1
(
81
,
9
*
19
,
1
*
)
19
,
1
2300
(
*
)
10
*
20
(
*
3
4
Re
*
2
5
3
6
2







S
D
C



䰠牥gin


卥瑴ng映瑨攠s敤em敮瑡ti渠獰n
敤映瑨攠灡牴icle



s
m
u
S
/
0275
,
0
10
*
82
,
1
*
18
81
,
9
*
)
19
,
1
2300
(
*
)
10
*
20
(
5
2
6







卥瑴ng映瑨攠瑩m攠睨敮⁴桥⁰a牴icl攠牥a捨cs⁴桥s敤em敮瑡瑩渠獰敥搠

u


㴠=ⰹ9*u
S



s
0129
,
0
)
99
,
0
1
ln(
*
10
*
82
,
1
*
18
2300
*
)
10
*
20
(
5
2
6











lovinggudgeon_fca5119f
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ece4
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P.
Hoffman



Datum tisku:
22.02.
上午四

Setting of the distance that the particle covers during the time






mm
m
h
28
,
0
000277
,
0
)
99
,
0
1
ln(
99
,
0
*
18
0359
,
0
*
10
*
20
*
19
,
1
2300
6









w桥
牥r

2
0359
,
0
10
*
82
,
1
19
,
1
*
10
*
20
*
0275
,
0
Re
5
6





S



Note:

For the same conditions (sand particles in air) but diameter d = 1,0 mm it is C
C
*Re
P
2

= 1,08*10
5


it is that we are at the beginning of the turbulent region.



Ex.7: Free fall of a hailstone from a storm cloud


G:

d
= 20 mm;

P

= 918 kg/m
3
;


㴠=ⰱ,0g/m
3
;


㴠=ⰸ2*10
-
5

Pa*s

T:

Set a falling (sedimentation) speed u
S

= ?


Setting of a sedimentation region


8
2
5
3
3
2
10
*
45
,
3
)
10
*
82
,
1
(
81
,
9
*
19
,
1
*
)
19
,
1
918
(
*
)
10
*
20
(
*
3
4
Re
*





S
D
C



吠牥gin


I琠ts⁶ali搠爠瑨攠r敧i渠nC
D

= 0,44


Setting of the Reynolds number



4
8
2
10
*
77
,
2
44
,
0
10
*
45
,
3
Re
*
Re



D
S
D
S
C
C


Setting of the constant sedimentation speed of the particle


s
m
g
d
u
P
S
/
4
,
21
19
,
1
81
,
9
*
)
19
,
1
918
(
*
10
*
20
*
74
,
1
*
)
(
*
*
74
,
1
3











h
km
s
m
d
u
S
/
76
/
2
,
21
19
,
1
*
10
*
20
10
*
82
,
1
*
10
*
77
,
2
*
Re*
3
5
4











lovinggudgeon_fca5119f
-
ece4
-
46b8
-
a951
-
21256703ae27.doc

10

P.
Hoffman



Datum tisku:
22.02.
上午四


Note 1:

The biggest in the Czech Republic observed hailstone had the diameter c. 120
mm. The corresponding falling spee
d 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 approx
i-
mately the same speed as the falling speed is.