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