Size reduction
Theory
 Force for reduce the size of food
 Compression
 Impact
forces
forces
 Shearing(or attrition) forces
Stress-strain diagram for various foods
Relationship between stress and strain force in size reduction
 When stress (force) is applied to a food the
resulting internal strains are first absorbed, to
cause deformation of the tissues.
 If the strain does not exceed a certain critical level
named the elastic stress limit (E), the tissues
return to their original shape when the stress is
removed, and the stored energy is released as heat
(elastic region(O–E))
Relationship between stress and strain force in size reduction
 If the strain area exceeds the elastic stress limit,
the food is permanently deformed. If the stress is
continued, the strain reaches a yield point(Y).
 Above the yield point the food begins to flow (Y–
B) Finally, the breaking stress is exceeded at the
breaking point (B) and the
food fractures along a line of weakness.
 Part of the stored energy is then released as
sound and heat.
Relationship between stress and strain force in size reduction
 The size of the piece is reduced, there are fewer lines
of weakness available, and the breaking stress that
must be exceeded increases.
 When no lines of weakness remain, new fissures
must be created to reduce the particle size
Force for size reduction in food
 Friable or crystalline foods
 Compression force
 Fibrous foods


Impact force
Shearing force
Other factor
 Other factor which influence the energy input;
 Moisture content of the food
Optimum moisture-easily breakdown e.g. wet milled
 Exceed moisture-agglomeration of particles which block the mill


Heat sensitivity of the food

high speed mill temperature increase so necessary to cool the mill
Size-reduction Equipment
The size-reduction equipment
 Using to reduce the size of food materials


fibrous foods (as meats, fruits and
vegetables) to smaller pieces or pulps
dry particulate foods to powders.
Size reduction of fibrous foods
 slicing and flaking equipment
 dicing equipment
 shredding equipment
 pulping equipment
Slicing and flaking equipment
 used to slice the products including
cheeses, pizza toppings, cooked meats,
cucumber and tomato.
 Meats are also cut using circular rotary
knives with a blade at right angles to the
path of the meat.
 The blade advances with the product on the
conveyor to ensure a square cut edge
regardless of the conveyor speed or cut
length which can be adjusted.
Dicing equipment
 The products are first sliced and then cut
into strips by rotating blades.
 The strips are fed to a second set of rotating
knives which operate at right angles to the
first set and cut the strips into cubes
Shredding equipment
 is a modified hammer mill in which knives
are used instead of hammers to produce a
cutting action.
Pulping equipment
 This uses a combination of compression and
shearing forces for juice extraction from
fruits or vegetables, for cooking oil
production and for producing pureed and
pulped meats.
 For example; a rotary fruit crusher consists of
a cylindrical metal screen fitted internally
with high-speed rotating brushes or paddles
 For bowl chopper is used to chop meat and
harder fruits and vegetables into a pulp
(for example for sausagemeat or mincemeat
preserve).
 Food may be passed several times beneath the
knives until the required degree of size reduction
and mixing has been achieved.
Size reduction of dry foods
 Ball mills
 Disc mills
 Hammer mills
 Roller mills
Ball mills
 These have a slowly rotating, horizontal steel
cylinder
which is half filled with steel
balls 2.5–15 cm in diameter.

At low speeds, the small balls are used.

At higher speeds, the larger balls are used.
 They are used to produce fine powders, such as food
colourants.
Disc mills
• Single-disc mills in which food passes through an
adjustable gap between a stationary casing and a grooved
disc, which rotates at high speed.
• Double-disc mills which have two discs that rotate in
opposite directions to produce greater shearing forces.
 Pin-and-disc mills
which have intermeshing pins fixed either
to the single disc and casing or to double
discs. These improve the effectiveness of
milling by creating additional impact and
shearing forces.
Hammer mills
 These have a horizontal cylindrical
chamber, lined with a toughened steel
breaker plate.
 A high-speed rotor inside the chamber is
fitted with swinging hammers along its
length.
 Using for crystalline and fibrous materials
including spices and sugar.
Hammer mill
Roller mills
 Using to mill wheat.
 Two or more steel rollers revolve towards
each other and pull particles of food through
the ‘nip’ (the space between the rollers).
 The size of the nip is adjustable for different
foods and overload springs protect against
accidental damage from metal or stones.
Applications of size reduction equipment
Properties and applications of selected size reduction equipment
Energy for size reduction
The energy required to reduce the size of solid foods is
calculated using one of three equations, as follows:
 Kick’s law
 Rittinger’s law
 Bond’s law
Kick’s law
the energy required to reduce the size of particles is proportional
to the ratio of the initial size of a typical dimension to the final
size of that dimension
d 
E  K K ln 1 
 d2 
E(J.kg-1) = the energy required per mass of feed (W/(kg/s))
KK = Kick’s constant,
d1 (m) = the average initial size of pieces,
d2 (m) = the average size of ground particles.
d1/d2 = the size reduction ratio (RR) and is used to
evaluate the relative performance of different types of
equipment. Coarse grinding has RRs below 8:1, whereas
in fine grinding, ratios can exceed 100:1
Rittinger’s law
the energy required for size reduction is proportional to the
change in surface area of the pieces of food
 1 1
E  K R   
 d 2 d1 
E(J.kg-1) = the energy required per mass of feed (W/(kg/s))
KR = Rittinger’s constant,
d1 (m) = the average initial size of pieces,
d2 (m) = the average size of ground particles.
Bond’s law
E
100
100


W
d2
d1
E(J.kg-1) = the energy required per mass of feed (W/(kg/s))
W (J kg-1) = the Bond Work Index (40,000–80,000 J kg-1
for hard foods Such as sugar or grain)
d1 (m) = diameter of sieve aperture that allows 80% of
the mass of the feed to pass
d2 (m) = diameter of sieve aperture that allows 80% of
the mass of the ground material to pass.
 Kick’s law gives reasonably good results for coarse
grinding in which there is a relatively small increase in surface
area per unit mass.
 Rittinger’s law gives better results with fine grinding
where there is a much larger increase in surface area
 Bond’s law is intermediate between these two.
However,equations Rittinger’s law and Bond’s law were
developed from studies of hard materials (coal and limestone) and
deviation from predicted results is likely with many foods.
EXAMPLE 1
Food is milled from 6 mm to 0.0012 mm using a 10 hp motor.
Would this motor be adequate to reduce the size of the
particles to 0.0008 mm? Assume Rittinger’s equation and that
1 hp 745.7 W.
Given
1. d1 = 6 mm. = 6 x 10-3 m. ,
d2 = 0.0012 mm. = 0.0012x10-3 m.
E1 = 10 hp. x (745.7 W/hp.)
2. E2 = ? When d2 =0.0008 mm. = 0.0008 x10-3 m.
Assume rate of throughout no change
From Rittinger’s equation
 1 1
E  K R   
 d 2 d1 
1
1

 

(10hp.)(745.7W / hp)  K R 




3
3
0
.
0012
x
10
m
.
6
x
10
m
.

 

Therefore,
K R  0.0089W .m.
To produce particles of 0.0008 mm.
1
1

 

E 2  (0.0089W .m.)




3
3
 0.0008x10 m.   6 x10 m. 
E 2  11,123W  15hp.
Therefore the motor is unsuitable and an increase in power of 50% is required
EXAMPLE 2
Sugar is ground from crystals of which it is acceptable that 80%
pass a 500 m sieve (US Standard Sieve No.35), down to a size in
which it is acceptable that 80% passes a 88 m (No.170) sieve, and
a 5-horsepower motor is found just sufficient for the required
throughput. If the requirements are changed such that the grinding
is only down to 80% through a 125 m (No.120) sieve but the
throughput is to be increased by 50% would the existing motor
have sufficient power to operate the grinder? Assume Bond's
equation.
Given :
1st condition E1 = 5 hp. , rate of throughout = M kg./s.
d1 =500 m. = 500x10-6 m. , d2=88 m. = 88x10-6 m
2nd condition E2 = ?.
, rate of throughout = 1.5M kg./s.
d1 =500 m. = 500x10-6 m. , d2=125 m. = 125x10-6 m
Assume Bond's equation.
E
100
100


W
d2
d1


100
100
 5hp. 
1st condition  M   W  88x106 m.  500x106 m.   618.7899W




1


100
100
 E2 
2nd condition  1.5M   W  125x10 6 m.  500x10 6 m.   447.2136W




2
2
1
(E2/1.5M) = 447.2136 W
(5 hp./M) 618.7899 W
So, E2 = 5.42 hp.
So the motor would be expected to have insufficient power to
operate the grinder equal 5.42 hp.
Size Determination
Sieving
Sieving is a mechanical size
Separation process.
Widely used in the food
Industry for
* separating fine from
larger particle
* removing large solid
Particle from liquid stream
Sieve Analysis
Involves :
- Passing the material being sized through
openings of a particular standard size in a screen.
- The particle-size distribution is then reported
as the weight percentage retained on each of a
series of standard sieves of decreasing size and the
percentage passed of the finest size.
Sieving is a gravity-driven process. usually a stack of
sieves are used when fraction of various sizes are
to be produce from a mixture of particle size
The shaker may be in the form of an eccentric drive
which a screens a gyratory or oscillating motion or
vibrator which gives the screens small-amplitude,
high frequency, up and down motion
When the sieve are inclined, the
particles retained on the screen fall
off at the lower end and are
collected by a conveyor. Screening
and particle size separation can
thus be carried out automatically
Standard sieve size
Sieves may be designated by the opening size,
US sieve mesh or Tyler sieve mesh
The US-sieve mesh designation is the metrication
The Tyler mesh designation refer to the number of
opening per inch.
The two mesh designations have equivalent opening
size although the sieve number designations are not
exactly the same.
Standard US-sieve size
Method for sieve analysis
* The percentage frequency curve graph
*The cumulative percentage curve graph
or The probability curve graph
* Calculate method
The percentage frequency curve graph
Figure 2 : Schematic of relative percentage frequency
distribution curve.
The probability curve graph
 Plot opening sieve diameter against probability
percentage
 The diameter at 0.5 or 50% probability is particle size
The cumulative percentage curve graph
Figure 1 : Schematic of cumulative percentage
frequency distribution curve.
The probability curve graph
Figure 3 : Schematic of the probability curve
Calculate method
Method 1
Particle size
=
1
 Xi 
  Dpi 


Method 2
Particle size
=
  (Wt .x log dia 

log 

Wt . 

1
Example
The mass fraction of a sample of milled corn retained on each
of a series of sieves. Calculate a mean particle diameter which
should be specified for this mixture.
U.S.
Micron
Wt.
Sieve
Size
grams
6
3,360
8
X (%)
% accumulate
1.6
1.62
1.62
2,380
3.2
3.24
4.85
12
1,680
7.9
7.99
12.84
16
1,191
19.4
19.62
32.46
20
841
18
18.20
50.66
30
594
15
15.17
65.82
40
420
11.6
11.73
77.55
50
297
8
8.09
85.64
70
212
6.6
6.67
92.32
100
150
3.4
3.44
95.75
140
103
3.2
3.24
98.99
200
73
0.9
0.91
99.90
270
53
0.1
0.10
100.00
Pan
37
0
0.00
100.00
Sum.
98.9
98.9
100.00
The percentage frequency curve graph
Frequency curve
% Wt.(frequency)
25.00
20.00
15.00
10.00
5.00
0.00
0
500
1,000
1,500
2,000
seive
2,500
3,000
3,500
4,000
The accumulative percentage curve graph
accumulative arithmetic curve
100.00
90.00
80.00
Percent
70.00
60.00
50.00
40.00
30.00
20.00
10.00
0.00
3,360
2,380
1,680
1,191
841
594
420
297
seive
212
150
103
73
53
37
Method 1
U.S.
Micron
Wt.
Sieve
Size
grams
6
3,360
8
%
Xi
Dpi
Xi/Dpi
1.6
1.62
0.016
3,680
4.40E-06
2,380
3.2
3.24
0.032
2870
1.13E-05
12
1,680
7.9
7.99
0.080
2030
3.93E-05
16
1,191
19.4
19.62
0.196
1435.5
1.37E-04
20
841
18
18.20
0.182
1016
1.79E-04
30
594
15
15.17
0.152
717.5
2.11E-04
40
420
11.6
11.73
0.117
507
2.31E-04
50
297
8
8.09
0.081
358.5
2.26E-04
70
212
6.6
6.67
0.067
254.5
2.62E-04
100
150
3.4
3.44
0.034
181
1.90E-04
140
103
3.2
3.24
0.032
126.5
2.56E-04
200
73
0.9
0.91
0.009
88
1.03E-04
270
53
0.1
0.10
0.001
63
1.60E-05
Pan
37
0
0.00
0.000
45
0.00E+00
98.9
100.00
Sum.
1.87E-03
D = 1/(Xi/ Dpi) = 1/1.87E-03 = 535.75 m
Method
2
U.S.
Micron
Wt.
Sieve
Size
grams
6
3,360
8
log dia
Wt*log dia
1.6
3.526
5.642
2,380
3.2
3.377
10.805
12
1,680
7.9
3.225
25.480
16
1,191
19.4
3.076
59.673
20
841
18
2.925
52.646
30
594
15
2.774
41.607
40
420
11.6
2.623
30.430
50
297
8
2.473
19.782
70
212
6.6
2.326
15.354
100
150
3.4
2.176
7.399
140
103
3.2
2.013
6.441
200
73
0.9
1.863
1.677
270
53
0.1
1.724
0.172
Pan
37
0
1.568
0.000
Sum.
98.9
277.11
 277.11 




(
Wt
.
x
log
dia
1  
 98.9 

Dvg  log
 10
 633.72m.


Wt . 
