Solutions for Questions used in Lectures
Solid Processing 3-1 Particle characterization
Example:
Calculate the equivalent volume sphere diameter xv, equivalent surface sphere diameter xs,
and equivalent surface to volume sphere diameter xsv for a cuboid of side lengths 1,3,5 and
cylinder of diameter 3 and length 1.
For the cuboid:
The volume of the cuboid = 1 3 5 = 15
The surface area of the cuboid = 2 (1 3) + 2 (3 5) + 2 (1 5) = 46
 x 3v
(i) So Vsphere =
= 15
6
So xv = 3.06
1
(ii) Asphere =  x 2s = 46
x s = 3.82
So
A sphere
(iii)
Vsphere
So
x 2s
6
46
 3

3.07
15
x v x sv

x sv 1.95
For the cylinder R = 3/2, H=1 
The volume of the cylinder =  1 = 7.07

The surface area of the cylinder =
 x 3v
(i) Vsphere =
= 7.07
6
2




+ 2  1  23.55
So xv = 2.38
(ii) Asphere = x s = 23.55
2
x s  2.74
(iii)
A sphere
Vsphere
x 2s
6
23.55
 3

3.33
7
.
07
x v x sv

So
x sv 1.80
2
Example
Convert the cumulative surface distribution described by the following equation to a cumulative
volume distribution given that Fv(45) = 1:
Fs = (
x 2
)
45
From:
fs(x) = ksx2fN(x) and
fv(x) = kvx3fN(x)
kv
x fs ( x )
obtain: fv(x) =
kS
x
Fv ( x )  (
0
fs ( x )
kv
) x fs ( x )dx
ks
and
fs ( x ) 
dFs
dx
d x 2 2x
( ) 
dx 45
( 45)2
assuming kv and ks are independent of x and plugging fs(x) into the integral obtain:
2 x3 k v
Fv ( x )  (
)
3 ( 45 )2 k s
Fv(x) = 1.096 10-5 x3
which using Fv(45)=1 gives
kv
 0.0333
ks
3

Note:
 f ( x )dx  1
since the sum of fractions must equal 1 and this can be used to find ks,v,m
0
Caution: since assumption concerning constancy of shape, density are made in relating different
kinds of distributions, errors can propagate in converting one distribution to another, thus it is
better to choose if possible the measurement method giving the required distribution from the start.
Example
Based on the microscopic particle size analysis shown in the table below find the
number length mean diameter, the number surface mean diameter and the number volume
mean diameter.
particle size interval
1-1.4
1.4-2.0
2.0-2.8
Totals
middle size d
1.2
1.7
2.4
number length mean diameter
number surface mean diameter
number volume mean diameter
frequency of occurrence n
2
5
14
21
nd
nd2
nd3
2.4
8.5
34
44.9
2.9
14.5
81
98.4
3.5
24.6
194
222.1
Nx
xNL (  i i )  2.1
Ni
x NS
Ni xi2 0.5

(
)
 2.16
N
 i
x NV
Ni xi3 0.333

(
)
 2.20
N
 i
5
Solid Processing 3-2 Particle Size Production
Example
A material consisting originally of 25 mm particles is crushed to an average size of 7 mm and
requires 20 kJ/kg for this size reduction. Determine the energy required to crush the material
from 25 mm to 3.5 mm assuming (a) Rittinger’s postulate, (b) Kick’s postulate, (c) Bond’s
postulate.
6
(a)
E  CR [
1 1
 ]
x 2 x1
Rittingers postulate
1 1
20 CR [  ] gives CR = 194.4 (kJ mm)/kg
7 25
E 194.4[ 1  1 ]  47.8kJ / kg
3.5 25
hence:
(b)
x1
Kick’s postulate
)
x2
7
20  Ck ln( ) gives Ck = 15.7 kJ/kg
25
E  Ck ln(
hence: E  15.7 ln(
(c) E  CB [
10
10

] Bonds postulate
0.5
0.5
( x2 )
( x1)
20  CB [
hence:
3.5
)  30.9kJ / kg
25
10
10

] gives CB = 11.2 (kJ (mm)0.5)/kg
0.5
0.5
(7 )
(25 )
10
10
E  11.2[

]37.5 kJ / kg
0.5
0.5
(3.5)
(25)
7
Example
Values of breakage distribution function b(i,j) and specific rates of breakage Sj for a particular
material in a ball mill are shown in Table 1. To test the validity of these values, a sample of the
material with the size distribution indicated in Table 2 is to be ground in a ball mill. Use the
information in these tables to predict the size distribution of the product after one minute in the
mill. Note: Sj values in the table are based on one minute grinding time.
8
Table 1
Size interval (mm)
212-150 150-106 106-75
75-53
53-37
37-0
Interval No.
Sj
b(1,j)
b(2,j)
b(3,j)
b(4,j)
b(5,j)
b(6,j)
1
0.7
0
0.32
0.3
0.14
0.12
0.12
4
0.35
0
0
0
0
0.6
0.4
5
0.3
0
0
0
0
0
1.0
6
0
0
0
0
0
0
0
2
0.6
0
0
0.4
0.2
0.2
0.2
3
0.5
0
0
0
0.5
0.25
0.25
Table 2
Interval No. (j)
Fraction
1
0.2
2
0.4
3
0.3
4
0.06
5
0.04
6
0
change of fraction in interval 1
dy1
 0  S1 y1  0  0.7 0.2  0.14
dt
so new
y1 = 0.2 – 0.14 = 0.06
change of fraction in interval 2
dy2
 b(2,1)S1y1  S2 y 2  (0.32 0.7 0.2)(0.6 0.4)   0.1952
dt
so new y2 = 0.4 – 0.1952 = 0.2048
9
change of fraction in interval 3
dy3
 [b(3,1)S1y1  b(3,2)S2 y 2 ]  S3 y3  [(0.3 0.7 0.2)](0.4 0.6 0.4)]  (0.5 0.3)   0.012
dt
so new y3 = 0.3-0.012=0.288
change of fraction in interval 4
dy4
 [b( 4,1)S1y1  b( 4,2)S2 y 2  b( 4,3)S3 y3 ]  S4 y 4
dt
= [(0.14 0.7 0.2) + (0.2 0.6 0.4) + (0.5 0.5 0.3)]- 0.35 0.06 = 0.1216
so new y4 = 0.06 + 0.1216 = 0.1816
change of fraction in interval 5
dy5
 [b(5,1)S1y1  b(5,2)S2 y 2  b(5,3)S3 y3  b(5,4)S4 y 4 ]  S5 y5
dt
=[(0.12 0.7 0.2) + (0.2 0.6 0.4) + (0.25 0.5 0.3) + (0.6 0.35 0.06)]- 0.3 0.04 = 0.1029
so new y5 = 0.1029 + 0.04 = 0.1429
change of fraction in interval 6
dy6
 [b(6,1)S1y1  b(6,2)S2 y2  b(6,3)S3 y3  b(6,4)S4 y 4  b(6,5)S5 y5 ]  S6 y6
dt
10
= [(0.12 0.7 0.2) + (0.2 0.6 0.4) + (0.25 0.5 0.3) + (0.4 0.35 0.06) + (1 0.3 0.04)] – 0 = 0.1227
so new y6 = 0 + 0.1227 = 0.1227
check:
sum of predicted product interval mass fractions = y1 + y2 + y3 + y + y5 + y6 = 1.0
Interval No. (j)
Fraction
1
0.06
2
0.2048
3
0.288
4
0.1816
5
0.1429
6
0.1227
Thus with a set of S and b values for a given feed material, the product size distribution after a
given time in a mill may be determined.
11
Example
Solid Proc 3 - 3 Solid in fluid
Calculate the upper limit of the diameter of a spherical particle which has a density
rp = 2000 kg/m3 falling with a terminal velocity UT through air given that Stokes law holds and
the single particle Reynolds number is Rep  0.3. Repeat the calculation for the case where the
particle is falling through water.
data: rair = 1.2 kg/m3; rwater= 1000 kg/m3; mair = 1.84 10-5 Pa s; mwater= 0.001 Pa s
The upper limit of particle diameter in the Stokes regime is governed by the upper limit of single
particle Reynolds number:
Rep 
xUrf
= 0.3
m
In the Stokes regime the terminal velocity is given by: UT 
x 2 (rp  rf )g
18m
solving these two equations for xmax we have
18 m 2
xmax (0.3 g(r r )r )0.333
p
f f

m2
0.82( g(r r )r )0.333
p
f f
plugging in the density of the particle and the values of the density and viscosity
for air obtain: xmax through air: 42.7 mm
plugging in the density of the particle and the values of the density and viscosity
for water obtain: xmax through water: 82.1 mm
12
Example
A sphere of density 2500 kg/m3 falls freely under gravity in a fluid of density 700 kg/m3 and
viscosity 0.5 10-3 Pa s. Given that the terminal velocity of the sphere is 0.15 m/s, calculate its
diameter. What would be the edge length of a cube of the same material falling in the same
fluid at the same terminal velocity?
To calculate size x, for a given UT:
CD 4gmrp rf )

Rep
3U3Tr2f
CD 4 (9.81) ( 0.5 10 3 )  2500 700 )
3


7
.
12
10
Rep
3 (0.15 )3 ( 700 )2
log(CD) = log (7.12 10-3) + 1 log(Rep)
For plotting the relationship:
Rep
CD
100
1000
10000
0.712
7.12
71.2
13
Drag coefficient, CD
These values can now be plotted on the standard drag curves for particles of different sphericity.
For the first part of this problem we are examining a sphere so want the Y=1 curve.
Drag curves for particles of different sphericities
4
10
1000
8
6
4
2
100
Y = 0.125
Y = 0.22
10
Y = 0.6
Y  0.80
Y  1.0
1
0.1
0.001
0.01
0.1
1
10
100
1000
10 4
105
106
0.4
Single particle Reynolds number, Rep (Rep: equiv. volume d)
0.2 0.60.8
which gives then an Rep = 130
and through:Rep 
x vUrf
= 130, obtain xv = 619 mm
m
14
For a cube having the same terminal velocity under the same conditions, the same CD vs Rep
applies but need to use the standard drag curve for a cube.
Sphericity: Ycube 
surface area of sphere of equal volume to theparticle
surface area of theparticle
cube of edge length a has V = a3 and S.A. = 6a2
If xv is the diameter of a sphere having the same volume as the cube, then:
x 3v
 a3 and
6
therefore
(6)0.333 a
xv 
( )0.333
(6)0.333 a 2
[
]
0.333
( )
Ycube 
 0.806
2
6a
so looking at our drag curve plots again:
15
Drag curves for particles of different sphericities
Drag coefficient, CD
104
1000
8
6
4
2
100
Y = 0.125
Y = 0.22
10
Y = 0.6
Y  0.80
Y  1.0
1
0.1
0.001
0.01
0.1
1
10
100
1000
10 4
105
106
0.4
Single particle Reynolds number, Rep (Rep: equiv. volume d)
0.2 0.60.8
Find Rep = 310 and since Rep uses the equivalent volume sphere diameter:
310 (0.5 10 3 )
xv 
1.48 10 3
0.15 700
so
Vparticle
 x 3v

1.66 10 9 m3
6
and the edge length of the cube a = (1.66 10-9 m3)0.333 = 1.18 10-3 m
16
Example
A sphere of diameter 10 mm and density 7700 kg/m3 falls under gravity at terminal conditions
through a liquid of density 900 kg/m3 in a tube of diameter 12 mm. The measured terminal
velocity of the particle is 1.6 mm/s. Calculate the viscosity of the fluid. Stokes law applies.
Convert measured terminal velocity to the equivalent velocity which would be achieved by
the sphere in a fluid of infinite extent:
x
fw (1  )2.25
D
Rep  0.3; x  0.97
D
(x/D) = (10/12) = 0.833
UT
1
00 
56.34 so the terminal velocity for the particle in a fluid of infinite extent
x
UT
2
.
25
D (1 D )
UT  UT (56.34)  0.0901m / s
00
D
x2 (rp  rf )g (10 10 3 )2 (7700  900 ) 9.81
Stokes regime terminal velocity given by: UT00 

18m
18 m
so: m = 4.11 Pa s
Check for Stokes law validity: Rep 
x U rf
 0.197
m
which is less than 0.3
17
Example
A height-time curve for the sedimentation of a suspension, of initial suspension concentration 0.1,
in a vertical cylindrical vessel is shown in figure 1.
Determine:
Height of interface of suspension
with clear liquid (cm)
a) the velocity of the interface between clear liquid and suspension of concentration 0.1.
b) the velocity of the interface between clear liquid and a suspension of concentration 0.175
c) the velocity at which a layer of concentration 0.175 propagates upwards from the base of the
vessel.
d) the final sediment concentration
40
30
20
10
0
0
25
50
75
Time from start of test (s)
100
125
18
Height of interface of suspension
with clear liquid (cm)
a) Since the initial suspension is 0.1, the velocity required is that of the AB interface given
by the slope of the straight line portion of the height-time curve.
40
-1.333 cm/s
30
20
10
Slope =
0
0
25
50
75
Time from start of test (s)
100
125
20  40
 1.333 cm / s
15  0
b) We must first find the point corresponding to the point at which a suspension of
concentration 0.175 interfaces with the clear suspension. From:
C h
C B 0
h1
we obtain h1 
CB h0
C
22.85
19
A line drawn through the point t=0, h= h1=22.85, tangent to the height time curve locates the
point containing the time at which a suspension of concentration 0.175 interfaces with the clear
suspension:
40
30
h1 = 22.85
20
h=15
10
t=26
0
0
25
50
the slope of the curve at this point is
the downward velocity of this interface :
75
100
125
15  22.85
 0.302cm / s
26  0
c) From (b) after 26 seconds the layer of concentration 0.175 has just reached the clear liquid
interface and has travelled a distance of 15 cm from the base of the vessel in this time.
Therefore the upward propagation velocity of this layer is:
h 15

 0.577
t 26
20
d) The value of the final sediment height h is found by drawing a tangent to the part of the curve
corresponding to the final sediment and projecting it to the h-axis:
40
30
20
10
hs=10
0
0
25
Having found hs we now use C 
50
75
100
125
CB h0 0.1 40

 0.4
hs
10
21
Example
A suspension in water of uniformly sized spheres (diameter 150 mm, density 1140 kg/m3) has
a solids concentration of 25% by volume. The suspension settles to a bed of solids concentration
of 55% by volume. Calculate:
a) the rate at which the water/suspension interface settles.
b) the rate at which the sediment/suspension interface rises. (assume water properties: density,
1000 kg/m3; viscosity, 0.001 Pa s)
a) Solids concentration of initial suspension, CB = 0.25
The velocity of the interface between the initial suspension B and clear liquid A can be
obtained through: Uint, AB 
UpA CA  UpB CB
CA  CB
(see p51)
CA = 0 so:
Uint,AB = UpB
UpB is the hindered velocity of particles relative to the vessel wall in batch settling and
is given by: Up = UTen
(See p 48-49)
For the case where Stokes law applies n=4.65 and the single particle terminal velocity is given by:
22
x 2 (rp  rf )g 9.81(150 10 6 )2 (1140  1000 )
UT 

1.717 10 3 m / s
18m
18 0.001
to check whether Stokes law assumption was valid:
(150 10 3 ) 1.717 10 3 1000
Rep 
 0.258
0.001
Rep 
x vUrf
m
which is <0.3, the limiting value for Stokes law
The voidage of the initial suspension, eB = 1 – CB = 0.75 so:
UpB = 1.717 10-3 0.754.65 = 0.45 10-3 m/s; i.e,. the AB interface is moving downwards with a
velocity of 0.45 mm/s
b)
We can employ the same equation again, for the velocity between the initial suspension B
and the sediment S. Thus:
Uint,BS 
Uint,BS 
UpB CB  UpS CS
CB  CS
UpB 0.25  0
0.25  0.55
with CB = 0.25 and CS = 0.55
and the velocity of sediment UpS = 0
 0.833 U pB from (a) UpB = 0.45 mm/s so Uint,BS = -0.375 mm/s
so the BS interface is moving upwards with a velocity of 0.375 mm/s.
23
Example
Solid proc 3 - 4 filtartion
Water flows through 3.6 kg of glass particles of density 2590 kg/m3 forming a packed bed of
depth 0.475 m and diameter 0.0757 m. The variation in frictional pressure drop across the bed
with water flowrate in the range 200-1200 cm3/min is shown in columns 1 and 2 table 1. The
viscosity of water is m=0.001 Pa s
a) Demonstrate that the flow is laminar
b) Estimate the mean surface-volume diameter of the particles
c)Show that the Reynolds number indeed verifies laminar flow even at the highest velocity
examined here
Table 1
Water flowrate
(cm3/min)
200
400
500
700
1000
1200
a)
Pressure drop
(mm Hg)
5.5
12.0
14.5
20.5
29.5
36.5
We will use Erguns equation so we will first need the superficial velocities. We also
should have compatible units so we transform (mm Hg) to Pascal (Pa).
24
3
200 cm ( 1m )3( 1min )
min 100 cm 60 sec  7.41 10  4 m
U
s
( 0.0757 )2m2
2
5.5mmHg
P(Pa)  760 mmHg 101325 Pa  734Pa
Table 2
Water flowrate
(cm3/min)
200
400
500
700
1000
1200
Pressure drop
(mm Hg)
5.5
12.0
14.5
20.5
29.5
36.5
U
(m/s 104)
7.41
14.81
18.52
25.92
37.00
44.40
Pressure drop
(Pa)
734
1600
1935
2735
3936
4870
if the flow is laminar, the pressure gradient across the packed bed should increase linearly
with superficial velocity (for constant voidage and fluid viscosity).
p
m U (1  e)2
150 2
Under laminar conditions Erguns equation reduces to : 
H
x sv e3
25
m H (1  e )2
For H, m, e constant –p vs U should give a straight line with slope = 150 2
x sv e3
Pressure drop (Pa)
5000
4000
m H (1  e )2
6
150 2

1
.
12
10
Pa . s / m
3
x sv e
3000
2000
1000
0
5
10 15 20 25 30 35 40 45
Superficial fluid velocity (m/s 10000)
b) Mass of particles = volume of bed [
volume of particles
] density of particles
volume of bed
Mass of particles = AH[1-e]rp so e  
3.6
 1  0.3497
0.0757 2
0.475 
) 2590
2
so substituting the known and established values for e, H and m into
m H (1  e )2
6
150 2

1
.
12
10
Pa . s / m and solving for x sv  792 mm
3
x sv e
xUrf
c) Re*
which is indeed less than the limiting value of 10
 5.4 for U  44.4
26
m(1  e)
for a laminar flow even at the maximum velocity
Example
The reactor of a catalytic reformer contains spherical catalyst particles of diameter 1.46 mm.
The packed volume of the reactor is to be 3.4 m3 and the void fraction is 0.25. The reactor
feed is a gas of density 30 kg/m3 and viscosity 2 10-5 Pa s flowing at a rate of 11320 m3/h.
The gas properties may be assumed constant. The pressure loss through the reactor is
restricted to 68.95 kPa. Calculate the cross-sectional area for flow and the bed depth.
Need to describe the relationship between gas velocity and pressure drop across the
packed bed.
p
m U (1  e)2
rf U2 (1  e)

150 2
 1.75
3
3
H
e
x sv e
x sv
with m = 2 10-5 Pa s, rf=30kg/m3, xsv=1.46 10-3 m, -p=68.95 kPa and e= 0.25
30 U2 (1  0.25 )
p
2 10 5 U (1  0.25 )2
1.75

150
3

3
2
3
H
(1.46 10 ) (0.25 )
1.46 10 3 (0.25 )
68.95 103
which gives:
 50666 U 1.726 10 6 U2
H
Reactor volume: V = A H = 3.4 m3
Gas volumetric flowrate, Q = U A =
11320
 3.144 m3 / s
3600
Substituting gives: 0.681 H2 + 21.467 H3 = 1.0
H = 0.35 m so A = 9.71 m2
27
Example
A leaf filter has an area of 0.5 m2 and operates at a constant pressure drop of 500 kPa. The
following test results were obtained for a slurry in water which gave rise to a filter cake regarded
as incompressible.
Volume of filtrate collected (m3)
0.1
0.2
0.3
0.4
0.5
Time (s)
140
360
660
1040
1500
Calculate:
(a) the time needed to collect 0.8 m3 of filtrate at a constant pressure drop of 700 kPa.
(b) the time required to wash the resulting cake with 0.3 m3 of water at a pressure drop of 400 kPa
(a) For filtration at constant pressure drop we can use:
t
rc  m
rc  m

V

Veq
2
V 2 A 2 ( p)
A ( p)
which when plotting t/V vs V will give a straight line
rc  m
rc  m
Veq
with slope
and
y-intercept
2
2
2 A ( p)
A ( p)
28
so transforming the given data into the form required for plotting:
V(m3)
0.1
0.2
0.3
0.4
0.5
t/V(s/m3)
1400
1800
2200
2600
3000
3000
rc  m
slope =
= 4000 s/m6
2
2 A ( p)
2500
t/V
2000
1500
y-intercept =
1000
which with A = 0.5 m2 and –p = 500 103 Pa
500
0
0.0
rc  m
Veq= 1000 s/m3
2
A ( p)
0.1
0.2
0.3
0.4
Volume of filtrate passed, V
thus our equation becomes:
0.5
gives rc  m  10 9 Pa s / m2 and Veq = 0.125 m3
t  0.5 109 (4V 1)
V (p)
which applies to the filtration of the
same slurry in the same filter at any
pressure drop
thus the time required to pass 0.8 m3 of filtrate at a pressure drop of 700 kPa is:
29
t = 3000 s (50 min).
(b) During filtration for constant pressure drop the increase in cake thickness (H)
needs to be counterbalanced by a decrease in the volumetric flowrate as
seen by –P = rc m U H. During washing the thickness is constant so a change in
pressure drop is now proportional to a change in volumetric flowrate. Thus if we find the
volumetric flowrate at the end of the filtration we can then find the volumetric flowrate
at any other pressure drop during washing where H is constant and consequently the time
needed to pass a certain volume of wash water through.
The volumetric flowrate at the end of filtration can be found from
1 dV
( p)A
where V = 0.8 m3

A dt rc m ( V  Veq )
3
Veq = 0.125 m
which gives:
dV
1.89 10  4
dt
m3
s
A = 0.5 m2
rc  m  109 Pa s / m2
–p = 700 103 Pa
for the filtration rate at the end of the filtration period
dV
so
  p
dt
400 103
400 103 
3
3
-4
flowrate at (400 10 Pa) = flowrate at (700 10 Pa)
= 1.89 10
1.08 10-4 m3/s
700 103
700 103
As discussed above while washing
so the time to pass 0.3m3 at this volumetric rate = 2778 s = 46.3 min
30
Example
Tests on a cyclone give the results shown below
Size range x (mm)
0-5
5-10
10-15
15-20
20-25
25-30
5
Feed size
analysis, m(g)
10
15
25
30
15
Coarse product size
analysis, mc (g)
0.1
3.53
18.0
27.3
14.63
5.0
(a) From these results determine the total efficiency of the cyclone
(b) Plot the grade efficiency curve and hence show that the x50 cut size is 10 mm
(c) The dimensionless constants describing this cyclone are: Eu = 384 and Stk 50 = 1 10-3.
Determine the diameter and number of cyclones to be operated in parallel to achieve this cut size
when handling 10 m3/s of a gas of density 1.2 kg/m3 and viscosity 18.4 10-6 Pa s, laden with
dust of particle density 2500 kg/m3. The available pressure drop is 1200 Pa.
(d) What is the actual cut size of your design?
31
(a) Mass of feed, M = 10 + 15 + 25 + 30 + 15 + 5 = 100 g
Mass of coarse product, Mc = 0.1 + 3.53 + 18.0 + 27.3 + 14.63 + 5.0 = 68.56 g
Total efficiency: ET  Mc  0.6856 (or 68.56%)
M
dF
dF
Mc ( c )
( c)
dx  E
dx
(b) G( x ) 
T
dF
dF
M( )
( )
dx
dx
mc
G(x) =
m
G(x)
0-5
0.01
5-10
10-15
15-20
20-25
25-30
0.235
0.721
0.909
0.975
1.00
1.0
0.8
0.6
G(x)
Size range x (mm)
here, we can use the data provided to us directly
to obtain G(x)
0.4
0.2
0.0
0
5
10
15
20
25
Particle size x (mm)
30
32
we can also calculate the size distributions of the feed
and coarse product
dF
(mass fraction of feed of size x)
dx
dFc
(mass fraction of coarse product of size x),
dx
Size range, x (mm)
0-5
dFc/dx
0.00146
dF/dx
0.1
5-10
10-15
15-20
20-25
25-30
0.0515
0.263
0.398
0.2134
0.0729
0.30
0.15
0.05
0.15
0.25
and verify the calculated G(x) values; ex:
dFc
0.263
G( x )  ET dx 0.6856
0.721
dF
0.25
dx
2 p 0.5 2 1200 0.5
) (
)  2.282 m / s
(c)   (
Eurf
384 1.2
If we have n cyclones in parallel and assuming equal distribution of the gas between
cyclones, then the flowrate to each cyclone q = Q/n
D ( 4 Q )0.5 ( 4 10 )0.5  2.362
n 
n  2.282
(n)0.5
33
using now: Stk 50 
10 3
2
x50
rp 
18 m D
(10 10  6 )2 2500 2.282

2.362
18 18.4 10  6 ( 0.5 )
(n)
giving n = 1.88
so 2 cyclones are required each having diameter:
D (
2.362
2.362
)

(
)  1.67m
0.5
0.5
(n)
(2)
(d) The actual cut size is calculated by inputing the established D and number of cyclones into:
so actual cut size
x 50
10 3 18 18.4 10  6 1.67 0.5
(
) 9.85 10  6 m
2500 2.282
(Stk 50 
2
x50
rp 
18 m D
So in summary 2 cyclones required (characterized by Eu = 384 and Stk50 = 10-3)
of diameter 1.67 m and operating at a pressure drop of 1200 Pa with cut size of 9.85 mm.
Example
Calculate the terminal radial velocity of a particle (rp = 1050 kg/m3; 60 mm in diameter) in
air at 260 Celcius (m = 2.7 10-5 kg/(m s); rf = 0.658 kg/m3) orbiting in a cyclone at r = 0.225
at a tangential velocity of 2 m/s.
18 m
r
x 2 (rp  rf ) U2 (60 10 6 )2 (1050  0.658 ) (2)2
x 
Ur or Ur 

0.138m / s
2
5
(rp  rf ) U
18m
r
0
.
225
18 2.7 10
2
34
)
Example
Solid proc 3 -5 mixing
A random mixture consists of two components A and B in proportions 60% and 40% by mass
respectively. The particles are spherical and A and B have particle densities 500 and 700 kg/m3
respectively. The cumulative mass distribution of the two components are shown in Table I.
Table I
Size x (mm)
FA(x)
2057
1676
1405
1204
1003
853
699
599
500
422
1.00
0.80
0.50
0.32
0.19
0.12
0.07
0.04
0.02
0
1.00
0.88
0.68
0.44
0.21
0.08
0
FB(x)
If samples of 1 g are withdrawn from the mixture, what is the expected value for the standard
deviation of the composition of the samples?
The expression for the standard deviation of a randomly mixed 2-component system is given by:
R (
p(1  p) 0.5
)
n
where p and (1-p) are the proportions of the two components in the mixture with n particles in
each sample
We know p and 1-p, (0.6 and 0.4) but need to find n, the number of particles in each sample.
35
So need to first find the number of particles per unit mass of A and B, then add them up and
multiply by the mass of the sampling (0.001 kg).
So we need to find the mass of particles in each size range dm.
This will allow us to find the number of particles in each size range through:
[mass of particles in each size range] = [number of particles in each size range] [mass of one particle]
dm
=
dn
rp  x 3
6
rp = particle density
x = arithmetic mean of adjacent sieve sizes
so for example between sizes 1676 and 1405 mm we have:
mean size of x = 1676 + 1405 = 1540.5
2
F = mass fraction less than size x = mass in a certain size range that is less than x
total mass
for dm: have by unit mass 0.80 A particles have sizes < 1676 and 0.50 A particles have sizes < 1405
so 0.30 of A particles have sizes 1676>x>1405
36
500  (1540 .5 10 6 )3
kg A

 9.571 10 7
6
6
particle A
kg A
0.3
A particles
kg total
 0.313 10 6
so dn =
kg A
kg total
9.571 10 7
A particle
rp  x3
so generating the data for A and B for all the size distributions, obtain Table II for A and Table III
for B particles:
Table II A particles
Mean size of range x (mm)
dm
dn
1866.5
1540.5
1304.5
1103.5
928
776
649
549.5
461
0.20
0.30
0.18
0.13
0.07
0.05
0.03
0.02
0.02
0.117
0.313
0.310
0.370
0.335
0.409
0.419
0.460
0.780
Totals
1.00
3.513 106
106
106
106
106
106
106
106
106
106
37
Table III B particles
Mean size of range x (mm) dm
dn
1866.5
1540.5
1304.5
1103.5
928
776
649
549.5
461
0
0
0.12
0.20
0.24
0.23
0.13
0.08
0
0
0
0.148
0.406
0.819
1.343
1.298
1.316
0
Totals
1.00
5.33 106
106
106
106
106
106
106
so we have
nA = 3.513 106 particles/kg, nB = 5.33 106 particles/kg
and in the 1g samples that are being withdrawn:
n = 0.001 kg (3.513 106 0.6 + 5.33 106 0.4) = 4240 particles
thus, the standard deviation is: R (
p(1  p) 0.5
0.6(0.4) 0.5
(
)  0.0075
)
4240
n
38
Solid proc 3 -6 Movement and storage
Example
A cylindrical hopper of diameter 1 m is filled to a depth of 4m with solids resulting in a bulk
density of 6000 kg/m3. The wall friction factor between the solids and the wall is 0.5 and the
ratio of horizontal to applied stress is 0.5 and this value does not vary with depth.
The stress at the top free surface is 105 Pa. What are the horizontal and vertical stresses at the
base of the hopper?
D = 1m
H=4m
rB = 6000 kg/m3
 {(
DrB g
v 
[1  e
4m w k
mw = 0.5
k = 0.5
v0 = 105 Pa
4m w k
)}H
D
]
v0 e
 {(
4m w k
)}H
D
kg
m
4 0.5 0.5
4 0.5 0.5
1m 6000 3 9.81 2
 {(
)} 4m
 {(
)} 4 m
1m
5
1m
m
s
v 
[1  e
]  10 Pa e
4 0.5 0.5
58660
v 
kg
m s2 [1 e 4 ]  105Pa e 4  59416 Pa
1
h = 0.5 59416 Pa = 29708 Pa
39
Example
A dilute phase pneumatic pipe transports 900 kg/hr of sand of particle density 2500 kg/m3 and
mean spherical particle size 100 mm using air having a superficial gas velocity of
14.82 m/s. The pipe diameter is 78 mm. You may take the friction factor for the gas as 0.005.
(a) What is the pressure drop across the pipe if it has a length of 30 meters and is horizontal?
(b) What is the pressure across the pipe if it has length of 10 meters and is vertical? For this part
You may assume that all the initial acceleration of the solids and the gas has already taken place
in a previous pipe section.
2 F L FpwL rpL(1 e)gsinr Le gsin
(a) p1  p2 0.5 e rf U2  0.5(1 e)rp Up
fw
f
f
(1)
(2)
(3)
(4)
(5)
(6)
Gas acceleration and particle acceleration should be considered here so 1 and 2 remain.
Term 3 in the dilute regime we can use the fanning friction equation.
For term 4 we employ Hinkle’s correlation
Terms 5 and 6 are 0 since  = 0
2 L
2 fg rf U2 LH 2 fp rp (1 eH)UpH
H
fs 
pH 0.5 eH rf U2  0.5(1 eH)rp U2 
pH
fH
D
D
H = values specific to the horizontal pipe
40
to use this equation need: eH, UfH and UpH
Using Hinkles expression:
UpH Ufs (1  0.0638 x 0.3 rp0.5 ) = 14.82 (1- 0.0638 0.0631 50) = 11.84 m/s
From continuity: Particle mass flux =
so
and
eH 1 
G
 0.9982
rp UpH
G
Mp
A
 rp (1  eH )UpH
U
UfH  e fs  14.82 14.85m / s
H 0.9982
Still need the friction factor fp which we can again obtain using Hinkles expressions:
fp 
U  UpH 2
3rf D
CD [ fH
]
8 rp x
UpH
The only unknown left above in order to obtain fp is CD which we can obtain at the relative
velocity (UfH – UpH) by either using one of the approximate correlations or from the appropriate
CD vs Re chart
Rep<1:
1<Rep<500:
CD = 24/Rep
CD  18.5 Rep0.6
500<Rep<2 105
CD = 0.44
41
so:
Rep 
rf (UfH  UpH )x
m
Which using:rfair = 1.2 kg/m3 and mair = 18.4 10-6 Pa s
Rep = 19.63
so
fp 
CD  18.5 (19.63 )0.6  3.1
3 1.2 0.078
14.85  11.84 2
3
.
1
[
]  0.0281
6
11.84
8 2500 100 10
pH 0.5 0.9982 1.2 (14.85 )2  0.5(1  0.9982 ) 2500 (11.84 )2

2 0.005 1.2 (14.82)2 30
0.078

2 0.0281 2500 (1  0.9982 ) (11.84 )2 30
0.078
 PH = 15097 Pa
(b) p1  p2 0.5 e rf U2f  0.5(1  e )rp Up2 FfwL Fpw L rpL(1  e )gsin   rf Le gsin 
(1)
(2)
(3)
(4)
(5)
(6)
Initial acceleration of solids and gas has already taken place so 1 and 2 are 0
The fanning friction factor can be used to calculate the pressure loss due to gas-to-wall friction term 3
For term 4 the modified Konno Saito correlation is used
42
For vertical transport  is 90 degrees in terms 5 and 6
2 fg rf U2 L v
fs  0.057GLv ( g )0.5  rp (1 ev )gLv  r ev gLv
pv 
f
D
D
v= values specific to the vertical sections
To evaluate this equation need to find ev
Since we are in the dilute regime the slip velocity Urel = Ufv - Upv will be equal to the single particle
terminal velocity UT. Moreover noting that the gas superficial velocity is the same in both the
horizontal and vertical pipe sections we have:
Upv 
Ufs
 UT
ev
and continuity or mass balance requires for the mass flowrate:
G = rp(1-ev)Upv
combining these two equations gives:
e2v UT  [UT  Ufs 
G
]e v  Ufs  0
rp
so UT will allow us to find ev which will then allow us to find pv
Can use the relationship CD Rep2 
4 x3rf (rp  rf )g
3m
2
since we do not know the flow regime
43
3
4
x
rf (rp  rf )g 4 (100 10 6 )3 (1.2) (2500  1.2) 9.81
2
CD Rep 

 116
2
6 2
3m
3 (18.4 10 )
Rep
CD
0.2
1
10
2900
116
1.16
Drag curves for particles of different sphericities
Drag coefficient, CD
104
1000
8
6
.
4
2
.
100
Y = 0.125
.
10
Y = 0.22
Y = 0.6
Rep = 3.39
1
0.1
0.001
0.01
.
Y  0.80
Y  1.0
44
0.1
1
10
100
1000
104
105
106
0.4 0.8 Single particle Reynolds number, Re (Re : equiv. volume d)
p
p
0.2 0.6
(100 10  6 ) UT 1.2
Rep 
 3.39
6
18.4 10
so UT = 0.52 m/s
e2v UT  [UT  Ufs 
G
]e v  Ufs  0
rp
e2v 0.52  [0.52  14.82 
52.35
]e v  14.82  0
2500
eV = 0.9985
pv 
pv 
2 fg rf U2fsL v
D
g
 0.057GLv ( )0.5  rp (1  ev )gLv  rf ev gLv
D
2 0.005 1.2 (14.82)210
0.078
0.057 52.35 10(
9.81 0.5
)
0.078
 2500 (1  0.9985 )9.81 10  1.2 0.9985 9.81 10  1158 Pa
45