Size reduction

Theory  Force for reduce the size of food  Compression forces  Impact 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 (elastic region(O–E)) ( E), the tissues return to their original shape when the stress is removed, and the stored energy is released as heat

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 E



K K

ln  

d d

1 2   E(J.kg

-1 ) = the energy required per mass of feed (W/(kg/s))

K

K = Kick’s constant,

d

1 (m) = the average initial size of pieces,

d

2 (m) = the average size of ground particles.

d

1 /

d

2 = the

size reduction ratio

(

RR

) and is used to evaluate the relative performance of different types of equipment. Coarse grinding has

RR

s 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 E



K R

  1

d

2  1

d

1   E(J.kg

-1 ) = the energy required per mass of feed (W/(kg/s))

K

R = Rittinger’s constant,

d

1 (m) = the average initial size of pieces,

d

2 (m) = the average size of ground particles.

Bond’s law

E W

 100 

d

2 100

d

1 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)

d

1 (m) = diameter of sieve aperture that allows 80% of the mass of the feed to pass

d

2 (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. d 1 = 6 mm. = 6 x 10 -3 m. , d 2 = 0.0012 mm. = 0.0012x10

-3 m.

E 1 = 10 hp. x (745.7 W/hp.) 2. E 2 = ? When d 2 =0.0008 mm. = 0.0008 x10 -3 m.

Assume rate of throughout no change

From Rittinger’s equation E



K R

  1

d

2  1

d

1   ( 10

hp

.)( 745 .

7

W

/

hp

) 

K R Therefore,

1 0 .

0012

x

10  3

m

.

K R

 0 .

0089

W

.

m

.

 1 6

x

10  3

m

.

To produce particles of 0.0008 mm.

E

2  ( 0 .

0089

W

.

m

.) 1 0 .

0008

x

10  3

m

.

 1 6

x

10  3

m

.

E

2  11 , 123

W

 15

hp

.

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 E 1 d 1 = 5 hp. , rate of throughout = M kg./s. =500



m. = 500x10 -6 m. , d 2 =88



m. = 88x10 -6 m 2nd condition E 2 d 1 = ?. , rate of throughout = 1.5M kg./s.

=500



m. = 500x10 -6 m. , d 2 =125



m. = 125x10 -6 m

Assume Bond's equation.

E W

 100 

d

2 100

d

1

1st condition 2nd condition

5

hp M E

2 1 .

5

M

.



W

   

W

   100 88

x

10  6

m

.

 100 125

x

10  6

m

.

 100 500

x

10  6

m

.

    618 .

7899

W

100 500

x

10  6

m

.

    447 .

2136

W

1 2 2 1

(E 2 /1.5M) = 447.2136 W (5 hp./M) 618.7899 W So, E 2 = 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 =

log  1  (

Wt

.

x

log 

Wt

.

dia

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. Sieve

6 8 12 16 20 30 40 50 70 100 140 200 270 Pan Sum.

Micron Size

3,360 2,380 1,680 1,191 841 594 420 297 212 150 103 73 53 37 98.9

Wt. grams

1.6

3.2

7.9

19.4

18 15 11.6

8 6.6

3.4

3.2

0.9

0.1

0 98.9

X (%)

1.62

3.24

7.99

19.62

18.20

15.17

11.73

8.09

6.67

3.44

3.24

0.91

0.10

0.00

100.00

% accumulate

1.62

4.85

12.84

32.46

50.66

65.82

77.55

85.64

92.32

95.75

98.99

99.90

100.00

100.00

The percentage frequency curve graph

Frequency curve 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

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. % Xi Dpi Xi/Dpi Sieve

6 8 12 16 20 30 40 50 70 100 140 200 270 Pan

Sum.

Size

3,360 2,380 1,680 1,191 841 594 420 297 212 150 103 73 53 37

grams

1.6

3.2

7.9

19.4

18 15 11.6

8 6.6

3.4

3.2

0.9

0.1

0

98.9

1.62

3.24

7.99

19.62

18.20

15.17

11.73

8.09

6.67

3.44

3.24

0.91

0.10

0.00

100.00

0.016

0.032

0.080

0.196

0.182

0.152

0.117

0.081

0.067

0.034

0.032

0.009

0.001

0.000

3,680 2870 2030 1435.5

1016 717.5

507 358.5

254.5

181 126.5

88 63 45 4.40E-06 1.13E-05 3.93E-05 1.37E-04 1.79E-04 2.11E-04 2.31E-04 2.26E-04 2.62E-04 1.90E-04 2.56E-04 1.03E-04 1.60E-05 0.00E+00

1.87E-03

D = 1/  (Xi/ Dpi) = 1/1.87E-03 = 535.75  m

Method 2 U.S. Sieve

6 8 12 16 20 30 40 50 70 100 140 200 270 Pan

Micron Size

3,360 2,380 1,680 1,191 841 594 420 297 212 150 103 73 53 37

Wt. grams

1.6

3.2

7.9

19.4

18 15 11.6

8 6.6

3.4

3.2

0.9

0.1

0 Sum.

D vg

 log  1 98.9

 (

Wt

.

x

 log

Wt

.

dia

log dia

3.526

3.377

3.225

3.076

2.925

2.774

2.623

2.473

2.326

2.176

2.013

1.863

1.724

1.568

Wt*log dia

5.642

10.805

25.480

59.673

52.646

41.607

30.430

19.782

15.354

7.399

6.441

1.677

0.172

0.000

277.11

 10 277 .

11 98 .

9  633 .

72 

m

.