Food Freezing (2009)

7/30/2019 Food Freezing (2009)

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FOOD FREEZING

Mechanisms, Food Quality, and

Engineering Aspects


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Understanding Freezing

Retard deterioration of foods - preservation

Chemical, physical, microbial, etc.

Improve organoleptic properties Desirable characteristics


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Controlling Freezing

Maximal quality of product

Initial freezing process

Storage and distribution conditions Efficient and economic processing


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Food Freezing

For a food to freeze, must lower the temperaturebelow its freezing point

Foods are mixtures of various ingredients, some

of which affect phase behavior of water Sugars, salts, proteins, fats, flavors, etc.

Freezing point depression

Dependent on composition

Particularly smaller molecular weightingredients like sugars and salts


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Average freezing point of some food categories

Food Moisturecontent (%)

Tf, oC

Vegetables 78 - 92 -0.8 to -2.8

Fruits 87 - 95 -0.9 to -2.7

Apple juice 87.2 -1.44Apple sauce 82.8 -1.67

Apple Juice concentrate 49.8 -11.3

Meat 55 - 70 -1.7 to -2.2

Milk 87 -0.5

Egg 74 -0.5


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Freezing Process

As freezing proceeds, heat is released andconcentration of unfrozen liquid phase increases

Phase change (333.2 kJ/kg of ice) causes

temperature of local environment to increaseTemperature increase depends on amount

of ice freezing and the rate of heatremoval

Freeze concentration of remaining fluid phasecauses decrease in freezing point of remainingliquid


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Freezing Point Depression

-6

-5

-4

-3

-2

-1

0

0 10 20 30 40

FreezingP

oint,C

Concentration, %

Glucose

Sucrose

Corn Syrup


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Freezing Process

As more heat is removed, the unfrozen phase

continues to become more concentrated

Continued freezing causes decrease in

molecular mobility (increase in viscosity of

unfrozen phase)

Molecules move more slowly

Approaches glassy state where molecular

mobility is very low


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Freezing Process

The endpoint of freezing is either:

When freezing point temperature reaches freezertemperature

Product temperature goes below the glasstransition temperature and the unfrozen phasebecomes glassy

A state diagram helps understand which willoccur

Follow trajectory of freezing process


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State Diagram

State diagrams tell us where to expect the systemto head for phase equilibrium

At a given storage temperature, the system will

move to approach the equilibrium curveMaximum amount of ice formed

Any point other than on the freezing pointdepression curve, including the glassy state, isnonequilibrium - metastable

Stability depends on process/storage conditions


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Freezing Rate

But freezing rate determines how much of the

allowable water freezes in a food

Slow freezing - equilibrium ice formation

Follows freezing point depression

Fast freezing - any amount of ice, depending

on freezing rate

Any trajectory


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Water Frozen

Amount of water frozen into ice thus depends on

freezing rate

Slow freezing - maximum ice

Fast freezing - any ice content


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Unfreezable Waterassuming phase equilibrium

All products have some water that remainsunfrozen even at very low temperatures (


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Freezing Rate Due to conduction heat transfer, the freezing

rate is also a function of the position in the

food Center sees much slower freezing rate than surface

Mechanisms of freezing may be different Ice distribution may also be different at surface from interior

Temperature differential allows moisture migration


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Freezing Rate

Freezing rate defined as: Ratio between the minimum distance from the surface to

the thermal center, and the time elapsed between thesurface reaching 0C and the thermal center 10C colderthan the temperature of initial ice formation. (InternationalInstitute of Refrigeration, as quoted by Zaritzky, 2000)

Typical food freezing rates 0.2 - 0.5 cm/h slow static

0.5 - 3 cm/h quick air blast and plate

5 - 10 cm/h rapid IQF fluidized bed 10-100 cm/h ultra-rapid cryogenic


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Question

Freezing rate has many impacts on a freezing

operation - how many can you list?

Product quality

Throughput rate

Refrigeration costs

Equipment costs Others?


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Freezing Mechanisms

The process of freezing requires these steps:

Subcooling - bring temperature down belowfreezing temperature

Nucleation - formation of the smallest crystalsfrom the liquid state

Growth - increase in size of those nuclei until thesystem approaches phase equilibrium

Ripening - change in dispersion of crystal sizeswith time due to thermodynamic effects


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Subcooling

Nuclei do not form under most circumstances

until temperature is lowered substantially

below the melting point (Tm)

Related to an energy barrier to be overcome to form

a stable nucleus

The temperature at which nuclei form depends on

process conditions Cooling rate, agitation, etc.


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Subcooling

High T (20-30C)

Rapid freezing

High nucleation rate

Many nuclei formed

Low T (1-5C)

Slower freezing Lower nucleation rate

Fewer nuclei formed

T = the difference between freezing point and freezing temperature


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Nucleation Onset of nuclei formation in a frozen food is

when the water molecules attain the correctenergy and position to form into a crystallattice


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Nucleation Mechanisms

Homogeneous nucleation - water molecules cluster together

Heterogeneous nucleation - dust particles promote nucleation


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Growth

After nuclei form, they grow until all

subcooling has been relieved Equilibrium temperature and product temperature are the same

Mechanisms

Heat removal rate Counter-diffusion of solutes


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Effects of Freezing

on Food Quality

Numerous changes take place during all stages

of freezing that can affect food quality

Prefreezing conditions

Freezing rate

Storage conditions


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Prefreezing

If initial temperature is well above the freezingpoint when a product is frozen

Water migration occurs due to thermal gradientsduring cooling and freezing

The warm water inside migrates toward surface

Redistribution of solutes

May be a problem in regions of different watercontent, e.g., crumb and crust

Can cause separation and unsightly appearance


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Freezer Bloom Freezing of frozen cakes with sugar frosting

Freezing from warm state causes watermigration, which carries dissolved sugar

When the water evaporates (or ice

sublimes) leaves unsightly spots


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Freezing Freezing affects properties of the food

Effects on cell structure

Osmotic pressure differences between intracellular andextracellular fluid cause moisture migration

o May lead to cell lysis (rupture)

Moisture migration

Osmotic differences; thermal gradients; etc.

Volume expansion of ice may rupture cells

Freeze concentration of solutes in unfrozen phase

Salts, sugars, etc. may lead to crystallization

Protein denaturation Freeze concentration of solutes like salts


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Slow vs. Rapid Freezing

Rapid freezing leads to formation of manymore and smaller crystals

Fewer internal changes in structures (cells,etc.)

Smoother product


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Freezer Storage

Over time, changes can occur in the frozen

product that cause product quality todeteriorate

Equilibration of ice phase volume

Changes in ice crystal dispersion due to ripening

Starch retrogradation

Protein denaturation

Water migration and loss


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Phase Equilibration

If phase equilibrium was not attained during

freezing, the system will drive towards that

equilibrium over time if T > Tg

Increase in ice content

Changes in ice crystal

size distribution


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Recrystallization

Definition:

"Any change in the number, size, shape, orientation

or perfection of crystals following completion of

initial solidification." (Fennema, Powrie and Marth,

1973)

Enhanced ramatically

by fluctuating

temperatures during

storage


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Fluctuation in Ice Content


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Recrystallization

Effects of recrystallization

Increase in mean size causes disruption of

microstructure and loss of texture

Smooth frozen product becomes coarse


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Moisture Migration and Freezer Burn

During freezing and frozen storage, regions ofdifferent water activity tend to equilibrate

Crust and crumb will change during storage

Bread and icing or filling

Pizza crust and sauce

Freezer burn - color and quality change Loss of moisture to air


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Summary

An understanding of the physico-chemical

factors that affect quality during freezing

allows production of the highest quality

product with the most efficient process


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Engineering Aspects

Freezing process has a dramatic affects on the

thermal properties of food products

As the water within the product changes from

liquid to solid, the density, thermal

conductivity, heat content (enthalpy), and

apparent specific heat change gradually as the

temperature decreases below the initialfreezing point for water in the food.


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Density

The density of solid water (ice) is less than thedensity of liquid water. Thus the density of a frozen

food will be less than the unfrozen product.

950

960

970

980

990

1000

1010

1020

1030

1040

1050

1060

-45 -35 -25 -15 -5 5 15 25

ProductDensity,

kg/m3

Temperature, C

Strawberries

Water content 89.3%

Init. Freezing temp = -0.89C


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Thermal Conductivity Thermal conductivity of ice is approximately four time larger

than that of liquid water.

This relationship has a similar influence on the thermal

conductivity of frozen food.

The majority of thermal conductivity increase occurs within

10oC below the initial freezing temperature of the product.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

-40 -30 -20 -10 0 10

ThermalCondu

ctivity,W/m-oC

Temperature, C

Froozen lean beef


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Enthalpy The heat content (enthalpy) of a froozen food is an important property in

computations of refrigeration requirement for freezing of product. The heat content normally zero at -40C and increases with increasing

temperature.

Significant changes in enthalpy occur at 10oC below the initial freezing

temperature, when most of the phase change in product water occurs.

0

50

100

150

200

250

300

350

400

-45 -35 -25 -15 -5 5 15

Entha

lpy(kJ/kg)

Temperature, C

Sweetcheries

Moisture content = 77%

Init. Freezing temp. = -2.61C


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Apparent Specific heat

Specific heat of a frozen food at a temperature greater than

20oC below the initial freezing point is not significantlydifferent from the specific heat of the unfrozen product.

0

5

10

15

20

25

30

35

-45 -35 -25 -15 -5 5 15

ApparentSpecificHeat,kJ/k

g.oC

Temperature, C

Sweet cherries

Moisture content = 77%Init.freezing temp. = -2.61oC


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Freezing Process Prefreezing

Phase change

Postfreezing


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Prefreezing : The temperature of the waterdecreases to the freezing point as sensible heat isremoved. Small amount of supercooling takeplace; but once nucleation occurs and ice crystals

begin to form, the freezing point increases to 0o

C. Phase change : The temperature remains at thefreezing point until a complete phase changeoccurs. Latent heat of fusion is removed from

liquid water to convert it into solid ice. Postfreezing : When all the liquid water haschanged into solid ice, the temperature of the icedecreases rapidly as sensible heat is removed.

Freezing Process of Pure Water


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Prefreezing : Temperature decreases during prefreezing

as sensible heat is removed. Nucleation and ice crystalsbegin to form at lower temperature that that of purewater.

Phase change : After a brief supercooling, latent heat isgradually removed with decreasing temperature. This

is due to concentration effect during freezing of food. As water in the food converts into ice, the remaining

water becomes more concentrated with solutes anddepresses the freezing point.

After this time, sensible heat is further removed untilpreselected endpoint is reached (Typically -18oC forfruits and vegetables and -25oC for foods with higherfat contents such as ice cream, fatty fish, ect.)

Freezing Process of Potato


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Freezing Time

The most important calculation in the designof freezing process is the determination offreezing time.

Freezing time is the most critical factor

associated with the selection of a freezingsystem to ensure optimum product quality.

Freezing time requirements help establishsystem capacity.

Methods of prediction : Planks Equation andPhams Method.


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Planks Equation

Proposede by Plank (1913) and adapted to food by

Ede (1949).

This equation describes only the phase change

period of the freezing process.


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Planks Equation

Assume the homogeneous slab is

at initial freezing temperature (TF).

The slab is exposed to a freezing

merdium at temperature Ta.

After some time, the will be 3

layers; 2 frozen layers at the

surface and a middle unfrozen

layer.

The moving front inside the slab

separates the frozen from the

unfrozen region.

As water is converted into ice at

the moving front, latent heat of

fusion (L) is generated.


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The latent heat of fusion generated at themoving front must be transfered through thefrozen layer to the surface (by conduction).

The heat is then transfered to the freezingmedium via convection. The convective heattransfer coefficient at the surface of the slab ish.

The temperature of the unfrozen regionremains at TF until the freezing front moves allthe way to the center plane of the food.

Formulation of Planks Equation


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Two layers of heat transfer : conduction through the

frozen layer and a convective boundary layer.


Ak

XqTT

X

TTAkq sF

Fscond

.)(

)(.

Ah

qTTTTAhq asasconv

.)()(.

hk

X

A

qTTTT assF

1)()(

k

X

h

TTAq aF

1

)(


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The heat released due to advancement of the

freezing front :

where :A = cross section area, m

L = latent heat of fusion, kJ/kg

f = density of water, kg/m3dx/dt = velocity of moving front, m/s


dt

dxLAq f


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Since all the heat released at the freezing front must be

transfered out to the surrounding medium, then :

The freezing process is completed when the moving front

advances to the center of the slab (a/2). Then


k

X

h

TTA

dt

dxLA aFf

1

)(

2/

001

a

faF

ft

dxkx

hTT

Ldtf

faF

f

f

k

a

h

a

TT

Lt

82

2 Freezing time for

infinite slabs


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For food material with moisture content m,Planks equation becomes :


faF

f

fk

a

h

a

TT

mLt

82

2

Where :

L = latent heat of fusion for water (333.2 kJ/kg)

m = moisture content of food, fraction

f = density of frozen product, kg/m3

TF = freezing temperature, C

Ta = medium temperature, C

a = thickness of the slab, m

h = convective coefficient, W/m2.K

kf = thermal conductivity of frozen product, W/m.K


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General Form of Planks Equation

where :

f = density of the frozen maerial

kf = conductivity of frozen material

P and R = shape factors


faF

f

fk

aR

h

aP

TT

mLt

2'.'...


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Values for P and R

Infinite slab : P = R = 1/8

Infinite cylinder: P = R = 1/16

Sphere : P = 1/6 R = 1/24

For finite slab and cylinder, use diagram to

determine the values for P and R.


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Material geometry constants P and R for a brick-shaped object for the Planks Freezing

time equation with dimensions a, b, c; where a is the shortest side.

E l #1


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Example #1

A spherical food product is being frozen in an air-blast freezer. The initial rpoduct temperature is 10oC

and the cold air -40oC. The product has a 7 cm

diameter with density of 1000 kg/m3, the initial

freezing temperature is -1.25oC, the thermalconductivity of the frozen product is 1.2 W/m.K, and

the latent heat of fusion is250 kJ/kg. Convective heat

transfer coefficient is 50 W/m2.K. Compute the

freezing time.

E l #1


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Given:Ti = 10

oC

Ta = -40oC

TF = -1.25oC

a = 0.07 m

f= 1000 kg/m3

kf= 1.2 W/m.K

HL = 250 kJ/kg

P = 1/6

R = 1/24

Example #1

E l #1


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Using Planks Equation :

Example #1

hoursWJt

KmW

m

KmW

mx

C

kgkJmkgt

F

oF

72.02600/2600

)./2.1(24

)07.0(

)./50(6

07.0

)]40(25.1[

)/250()/1000(

2

3

E l #2


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Blueberries are to be frozen in 2 liters cardboard boxes having thefollowing dimensions: 6 cm x 15 cm x 22.5 cm

Blueberries have the following characteristics:moisture content m = 85%Density = 1015 kg/m3Thermal conductivity

frozen: kf= 1.95 W/m.Kunfrozen: k

u= 3.85 W/m.K

Freezing temperature: Tf= - 0.9CThe boxes are made of cardboard 0.8 mm thick with a thermalconductivity kcard = 0.05 W/m.KThe blueberries are frozen in an air blast freezer where the

convective heat transfer coefficient (hc) between the cold air andthe boxes surface is equal to 25 W/m2.K.

The temperature of the cold air in the freezer can be adjustedbetween -20C and -40C.

Example #2

E l #2


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Box dimensions: 0.06 m x 0.15 m x 0.225 m

a = 0.06 m b = 0.15 m c = 0.225 m

b/a = 2.5 c/a = 3.75

Example #2


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E l #2


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From above chart we get: R = 0.085 P = 0.304

Evaluation of the overall heat transfer coefficient:

= 1/25 + 0.8 x 10-3/0.05 = 0.056

U = 17.86 W/m2.Kkf= 1.95 W/m.K

Latent heat of fusion for blueberries = 85% x latent heat of fusion of waterL . m = 335 x 0.85 = 284.75 kJ/kg = 284750 J/kg

We want the blueberries to freeze in less than 2.5 hours: q f= 2.5 x 3600 = 9000 s

Example #2

E l #2


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Example #2

CT

xxx

k

aR

U

aP

t

mLTT

kaR

UaP

t

mLTT

o

a

ff

f

Fa

ff

faF

7.38

95.1

06.0085.0

86.17

06.0304.0

9000

10152847509.0

''..

''..

2

2

2

Ph M th d


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Phams Method

Proposed by Q. T. Pham in 1986

Can be used for finite-size objects of irregular

shapes by approximating them to be similar to

an ellipsoid.

Easy to use yet provides answer with

reasonable accuracy.

Ph M th d


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Basic Assumptions:

Environmental conditions are constant

Initial temperature, Ti is constant

Final temperature, Tc

is fixed

Convective heat transfer at the surface of

the object obeys Nowtons law of cooling

Phams Method

C li d F i Di


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Cooling and Freezing Diagram

Ph M th d


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Based on experimental data for variety of foods

where Tc is final center temperature and Ta is mediumtemperature.

Freezing time for simple-shaped objects :

where dc is the shortest distance to the center, or radius. h isconvective heat transfer coefficien, Efis shape factor (Ef= 1 forinfinite slab, 2 for infinite cylinder, and 3 for sphere

Phams Method

acfm TTT 105.0263.08.1

2

12

2

1

1 Bi

f

c N

T

H

T

H

hE

dt


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k

lhNBi

.

Food Freezing (2009)

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