Particle Size Distribution Example

1

1. Seminar: Particle size distribution

Exercise sheet:

particle size

mass

mass fraction cumulative interval

fraction

fraction

width

frequency

mean

distribution

interval diameter

i

di-1 - di

mi

μ3,i

Q3,i

Δ di

q3,i

dm,i

d m,i ⋅ μ3,i

μ3,i

100

d m,i ⋅ 100

1 μ3 , i ⋅ d m3 ,i 100

n

∑d i =1

μ 3, i 3 m ,i

Q0(d)

q0(d)

⋅100

[mm]

[g]

[%]

[%]

[mm]

[%mm-1]

[mm]

[mm]

[mm-1]

[mm-3]

[mm-3]

[-]

[mm-1]

1

0-0,04

2,27

1,20

1,20

0,04

29,98

0,020

0,00024

0,6

1500

1500

0,797

20

2

0,04-0,063

5,31

2,80

4,00

0,023

121,91

0,0515

0,00144

0,544

205,0

1705,0

0,906

4,74

3

0,063-0,1

10,79

5,70

9,70

0,037

154,03

0,0815

0,00465

0,699

105,3

1810,3

0,962

1,51

4

0,1-0,25

64,02

33,80

43,50

0,15

225,34

0,175

0,0592

1,932

63,1

1873,4

0,996

0,22

5

0,25-0,4

40,34

21,30

64,80

0,15

141,99

0,325

0,0692

0,655

6,20

1879,7

0,999

0,02

6

0,4-0,63

36,56

19,30

84,10

0,23

83,93

0,515

0,0995

0,375

1,41

1881,0

1,000

0

7

0,63-1,0

13,44

7,10

91,20

0,37

19,18

0,815

0,0579

0,087

0,13

1881,1

1,000

0

8

1,0-2,5

15,34

8,10

99,30

1,5

5,40

1,750

0,142

0,046

0,02

1881,1

1,000

0

9

2,5-6,0

1,33

0,70

100,00

3,5

0,20

4,250

0,0298

0,002

0

1881,1

1,000

0

189,40

100,00

4,94

1881,1

1881,1

Σ

© Dr. Werner Hintz

2

1. Seminar: Particle size distribution

1) Calculation of the cumulative particle size distribution Q3(d) do

Q3 (d ) = ∫ q3 (d )d (d ) du

to sum up numerically in discrete intervals n

Q3 ( d ) = ∑ i =1

μ3,i

⋅ Δd i Δd i { { d (d ) q3 ( d )

μ3,i =

if

mi

- mass fraction,

mges

Δd i = d i − d i −1

- interval width,

n

Q3 ( d ) = ∑ μ3,i

summation from i=1...n...N

i =1

N – overall number of the intervals

→

results : see exercise sheet

Distribution functions are :

• monotone not decreasing, i.e. for d1 ≤ d2 is Q(d1) ≤ Q(d2), • steady, • scaling : for d ≤ du :

Q3(d) = 0

lower particle size limit

for d ≥ do :

Q3(d) = 1

upper particle size limit

Calculation of the particle size frequency distribution q3(d)

q3 ( d ) =

dQ3 ( d ) d(d )

numerically in discrete interval

q 3 (d i −1 ... d i ) =

Q3 (d i ) − Q3 (d i −1 ) d i − d i −1

=

μ3,i Δd i

2) normal and log – normal diagram of Q3(d) and q3(d) : see pictures

© Dr. Werner Hintz

3

1. Seminar: Particle size distribution

3) Calculation of the median particle size d50

read from the graphical diagram of Q3(d) :

d50 = 0,296 mm

Calculation of the modal particle size dh

read from the graphical diagram of q3(d) :

dh = 0,175 mm

4) Calculation of the mean particle size dm,3 do

d m,r = M r = ∫ dq r (d )d (d ) ( 1)

du

for a distribution related to the quantity mass r = 3 do

d m,3 = M 3 = ∫ dq 3 ( d ) d ( d ) (1)

du

in numerically form N

d m,3 = ∑ d m,i ⋅ μ3,i i =1

if the mean interval diameter is d m,i =

d i −1 + d i 2

see exercise sheet : dm,3 = 0,463 mm

5) from the graphics of Q3(d) in a logarithmical probability diagram

μln,3 = ln d 50,3 = ln 0,296 = − 1,217 σ ln,3 =

1 d 84( 3) 1 0,629 ln = ln = 0,796 2 d 16( 3) 2 0,128

graphical picture of Q3(d) in a RRSB – diagram

- linear correlation, curve is snapping off in the upper particle size range, not considered

d ′ = d 63 = 0,387 mm n = 1,84

by parallel displacement

n = tan α = tan 53,5° = 1,35 © Dr. Werner Hintz

determination of the slope angle --- deviation ???

4

1. Seminar: Particle size distribution

⎛ AS ,V , K ⋅ d ′ ⎞ using scale ⎜ ⎟ for calculating surface area ⎝ 1000 ⎠ i.e. specific surface area related to volume AS ,V , K AS ,V , K

(A =

S ,V , K

)

⋅ d ′ 1000 ⋅ 1000 d′

0,0107 ⋅ 10 3 0,0107 ⋅ 10 3 = = 0,387mm 0,387 ⋅ 10 − 3 m

m2 cm 2 = 27649 3 = 276,49 3 m cm

for Quarzit is ρs = 2650 AS ,m, K =

AS ,V , K

ρs

kg m3

= 10,4

m2 kg

Calculation of the Sauter - diameter dST and the specific surface area related the mass AS,m volume equivalent spheres d ST =

6 ⋅V AS , K

⇒ →

monodisperse particle collective

⇓

with equal specific surface area like real polydisperse particles d ST =

1 N

μ3,i

∑d i =1

=

1 = 0,202mm 4,94mm −1

m ,i

→ see exercise sheet

characteristic particle sizes :

⇒

∗ median particle size

d50,3 = 0,296 mm

∗ modal particle size

dh,3 = 0,175 mm

∗ mean particle size

dm,3 = 0,464 mm

∗ Sauter - diameter

dST = 0,202 mm

values of particle sizes are different !

© Dr. Werner Hintz

dST

5

1. Seminar: Particle size distribution

specific surface area

1 6 = d ST ψ A ⋅ d ST

AS ,V = f ⋅

N

μ3,i

i =1

d m,i

AS ,V = 6 ⋅ ∑

with ψ A ≈ 1 for spheres

= 6 ⋅ 4,94mm −1 = 29640

m2 m3

respectively: AS ,m =

AS ,V

m2 29640m 2 m 3 ρs = m 3 ⋅ 2650kg = 11,2 kg

in a good accordance with AS ,m = 10,9

m2 , see RRSB - diagram kg

7) Calculation of Q0(d) and q0(d)

∫ Q (d ) = ∫

d

du do

0

du

d −3 q3 (d )d (d ) d −3

∑ = q (d )d (d ) ∑ 3

n

i =1 N i =1

d m−3,i ⋅ μ 3 ,i d m−3,i ⋅ μ 3 ,i

i = 1...n...N

n – running number of intervals N – overall number of intervals

q0 ( d ) =

dQ0 ( d ) Q0 (d i ) − Q0 (d i −1 ) = Δd i d (d )

see working sheet

logarithmical probability diagram

μln,0 = ln d 50,0 = ln 0,022mm = − 3,82 σ ln,0 =

1 d 84 ,0 1 50μm = ln = 0,942 ln 2 d 16,0 2 7,6μm

not exact σ ln,0 = σ ln,3 = 0,882 caused by numerical deviations

− number distributions are shifted to the left, i.e. a lot of fine particles,

© Dr. Werner Hintz