2. Micromeritics is the science of small particles. It is
the study of a number of characteristics, including
paticle size and size distribution, shape, angle of
repose, porosity, true volume, apparent demsity nd
bulkiness.
3. Particles are any unit of
matter having defined
physical dimensions.
Physical state of particles
can be altered by physical
manipulation
Particle characteristics
can alter therapeutic
effectiveness.
4. MICROMERITICS
FACTORS AFFECTING FLOW PROPERTIES
1. PARTICLE SIZE AND SHAPE
250-2000m = free flowing
75 – 250 m = flow freely or cause problem depending on shape
Very fine particles (less than 10 m) = do not flow freely as large
particles
Particle shape and flow properties
Spherical shape flow better than needle particles
Elongated or flat particles tend to pack resulting to high
porosity powders
2. POROSITY AND DENSITY
High density, low porosity = FREE FLOWING
3. SURFACE ROUGHNESS
Leads to poor flow characteristics
5. MICROMERITICS
Techniques of determining particle size
MICROSCOPIC METHOD(OPTICAL MICROSCOPY)
Uses an ordinary microscope for particle
measurement in the range of 0.2 m to 100 m.
Presence of agglomeration and particles of more than
one component may be detected
The diameter is obtained only from two dimensions:
length and breadth, the thickness/depth in not
measured.
The microscopic method can include counting not
fewer than 200 particles in a single plane using
calibrated ocular on a microscope.
6. Particle Size
SIZE OF MIDDLE # OF “ND” 1. Given the following
PARTICLE VALUE PARTICLE
IN uM “D”(um) PER data, what is the
GR ”N” average diameter of
40-60 50 15 750 the particles?
60-80 70 25 1750 dav= ∑ nd / ∑n
80-100 90 95 8550 dav = 36,850 / 355
100-120 110 140 15400 dav = 103.8uM
120-140 130 80 10400
∑n=355 ∑nd=
36,800
7. PARTICLE SIZE
SIEVING - uses standard sieves; generally used for grading
coarser particles. May be employed for screening materials
as fine as 44 m (No. 325 sieve)
POWDERS OF VEGETABLE AND ANIMAL DRUGS ARE
OFFICIALLY DEFINED AS:
VERY COARSE (#8) – all particles pass through no.8 sieve and
not more than 20 % through sieve no. 60.
COARSE (#20) – all particles pass through no.20 sieve and
not more than 40 % through sieve no. 60.
MODERATELY COARSE (#40) - all particles pass through
no.40 sieve and not more than 40 % through sieve no. 80.
FINE (#60) - all particles pass through no.60 sieve and not
more than 40 % through sieve no. 100.
VERY FINE (#80) – all particles pass through a no. 80 sieve.
There is no limit as to greater fineness.
8. PARTICLE SIZE
POWDERS OF CHEMICAL DRUGS ARE OFFICIALLY
DEFINED AS:
COARSE (#20) – all particles pass through no.20 sieve
and not more than 40 % through sieve no. 60.
MODERATELY COARSE (#40) - all particles pass through
no.40 sieve and not more than 60 % through sieve no.
60.
FINE (#80) - all particles pass through no.80 sieve and
there is no limit as to greater fineness.
9. PARTICLE SIZE
0.5 um – 10um – suspensions and fine
emulsions
10 um – 50um - upper limit of subsieve
range; coarse emulsion particles;
flocculated suspension particles
50 um – 100um – lower limit of sieve range;
fine powder range
150 um – 1000um – coarse powder range
1000 um – 3360um –average granule size
10. A stack of sieve is arranged in order, the powder placed in the
top sieve, the stack shaken, the quantity of powder resting on
each sieve weighed, and this calculation performed:
dav= ∑(% retained)x(ave size) / 100
dav = 29.232 / 100 = 0.2923mm
11. PARTICLE SIZE
Other ways of particle size determonation elutriation,
centrifugation, permeation, adsorption, the Coulter Counter, and
light obstruction and the use of Andreasen pipet.
dst = 18 h /(i - e)gt
Angle of repose
A relative simple technique for estimating the flow properties of a
powder. It can easily be determined by allowing a powder to flow
though a funnel and fall feerly onto a surface. The height and
diameter of the resulting cone are measured and the angle of repose
calculated as:
tan Ѳ = h/r
h is the height of the powder cone and r is the radius of the powder cone
Low angle of repose – flow freely; high angle of repose – flow poorly
12. PARTICLE SIZE
Charateristics used to describe powder –
porosity, true volume, bulk volume,
apparent density, true density, and
bulkiness.
Void= Vbulk - V ; Porosity = Vbulk - V x 100
Vbulk Vbulk
Apparent Density pa = Weight of the sample ;
Vbulk
true density p = Weight of the sample
V
Bulkiness, B = 1/pa
13. PARTICLE SIZE
EXAMPLE
A selected powder has a true density (p)
of 3.5g/cc. Experimentally, 2.5 g of the
powder measures 40 mL in a cylindrical
graduate. Calculate the true volume,
void, porosity, apparent density, and
bulkiness.
14. PARTICLE SIZE
Closest packing:rhombus-triangle – angles
of 60 deg and 120 deg: void=o.26 ;
porosity=20%
Open type of packing:cubical – cubes
packed at 90 deg angle: void=0.47;
porosity=47%
Importance of packing and flow:
a. Affects the size of the container
b. The flow of granulation
c. Efficiency of filling apparatues for
making tablets and capsules.
d. Ease of working with powders.
15. Particle Size Reduction
Comminution, reduction of the particle size of a
solid substance to a finer state, is used :
1. Facilitate crude drug extraction.
2. Increase the dissolution rates of a drug
3. Aid in the formulation of acceptable
pharmaceutical forms.
4. Enhance absorption of drugs.
The reduction in the particle size of a solid is accompanied
by a great increase in the specific surface area of the
substance.
16. Particle Size Reduction
EXAMPLE
Increase in number of particles
If a powder consists of 1mm on edge and it is reduced to
paricles 10 um on edge, what is the number of particles
reduced?
1. 1mm equals 1000um.
2. 1000um/10um = 100 pieces produced on each edge; that
is, if the cube is sliced into 100 pieces on the x-axis,each
10um long, 100 pieces result.
3. If this repeated on the x- and y-axis, the result is
100x100x100 = 1 million particles produced, each 10um
on edge, for each original particle 1mm on edge. This can
also be written an 106 .
17. Particle Size Reduction
Increase in surface area
What increase in the surface area of the powder is produced by decreasing the
particle size from 1mm to 10um?
1. The 1mm cube has 6 surfaces, each 1mm on edge. Each face has a surface area
of 1 mm2. Because there are 6 surfaces, this 6 mm2 surface area per particle.
2. Each 10um cube has 6 surfaces, each 10um on edge. Each face has a surface
area of 10x10 = 100 um2 . Because there are 6 faces, this is 6x100 um2 or 600
um2 surface area per particle. Since 106 particles resulted from comminuting
the 1m cube, each 10um on edge, the surfaces are now is 600 um2 x 106, or 6 x
108um2 .
3. To get everything in the same units for ease of comparison, convert the 6 x
108um2 in to square mm as follows.
4. Since there are 1000um/mm, there must be 1000 2 ,or 1 million um2 / mm2 .
This is more appropriately express as 106 um2 / mm2,
6 x 108um2 = 6 x 102mm2 The surface area have been increased from
106 um2 / mm2 6 mm2 to 600 mm by the reduction in
Particle size of cubes 1mm to cubes 100um
on edge, an hundred fold inc in surface area.
