Dr. Supatpong MattarajInstructor Department of Chemical EngineeringFaculty of EngineeringUbon Ratchathani University

1205 252 Unit Operation of Food Engineering

Presented to

Students from Food Engineering (Agricultural )

Semester 2/2545

Objective: Preservation of food by food freezing

Freezing temperature leveling below 0 oC causes a reduction in growth rates for microorganisms (for microbial activity).Influence of product quality:– Fluctuation in storage temperature– Freezing process (depend on product characteristics)– Frozen-storage conditions– Depend on products (short or long freezing time)

Introduction: Food Freezing

Freezing system Frozen-food properties Freezing time

Contents : Food Freezing

The products must exposed to a low-temperature medium for sufficient time to remove sensible heat and latent heat of fusion from the product. This reduces product temperature as well as a conversion of the water from a liquid to a solid state (ice).Approximately 20% of water remains in the liquid state at the storage temperature of the frozen food.Two types of freezing process :– indirect system– direct contact system– but depends on the product characteristics.

Freezing system

Indirect Contact SystemThe product and refrigerant are separated by a barrier (nonpermeable) throughout the freezing process

1. Plate freezers: – The product is frozen while held between two refrigerated plates.

(top and bottom under pressure)– The heat transfer through the barrier (plate) can be enhanced by

using pressure to reduce resistance to heat transfer across barrier.

– This can operate in both batch and continuous mode:– Batch with the product placed on the plates of a specific

residence time.– Continuous with moving the two plates holding the product

through the freezing process. The freezing is the total time required for the product to move from entrance to exit.

Indirect Contact System

2. Air-Blast Freezer:– Best alternative– Product placed in the room– Freezing time will be long due to lower air speeds over the product.– Inability to achieve contact between product and cold air– Most are continuous – move product through stream of high-velocity air

3. Freezers for liquid foods: – Removal of thermal energy from a liquid food before product package– Require heat exchanger to adjust pressure on low-pressure size of

refrigeration system– Can be batch or continuous – Residence time is sufficient to reduce the product temperature by

desired amount and other changes in the product before package.

Direct Contact SystemNo barrier to heat transfer between the refrigerant and product.May be low-temperature air at high speedsMaybe liquid refrigerants with phase change while in contact with the product surfaceIf require to rapid freezing, individual quick freezing (IQF) will apply

1. Air Blast– Low-temperature air at high speeds is a form of IQF– Combine low-temperature air, high convective heat-transfer coefficient,

small product shape to provide short freezing

2. Immersion– Immerse food product in liquid refrigerant– Product surface is reduced to a very low temperature– Freezing time is shorter than the air blast– Refrigerants are nitrogen, carbon dioxide, Freon– Cost of refrigerant is expensive because it change from liquid to vapor

during products freezing while vapor leaves from compartment

Frozen-Food PropertiesDepend on thermal properties of the food productPhase change: Liquid (water) change to solid, the density, thermal conductivity, heat content (enthalpy), specific heat of the product change as temperature decreases below the initial freezing point for water in the food.

1. Density– The density of solid water is less than that of liquid water– The density of a frozen food is less than the unfrozen product

Intensive properties– The magnitude of change in density is proportional to the moisture

content of the product

2. Thermal conductivity– The thermal conductivity of ice is about four times larger than that

of liquid water.– Same influence in the thermal conductivity of a frozen food

Frozen-Food Properties

3. Enthalpy (heat content)– Important parameter for refrigeration requirement– The heat content normally zero at -40 oC and increases with

increasing temperature– Significant changes in enthalpy occur in 10 oC below the initial

freezing temperature.4. Apparent specific heat– Depend on function of temperature and phase changes for water in

the product– The specific heat of a frozen food at a temperature greater than 20

below the initial point (-2.61 oC) 5. Apparent thermal diffusivity– The apparent thermal diffusivity increases as the temperature

decreases below the initial freezing point – Frozen product shows larger magnitude than unfrozen product

ExampleFreezing point (by reducing temperature) depends on molecular weight and concentration of solution:

)11(ln'

AAOA TTR

X

XA = mole fraction of A in solution’ = latent heat of fusion (J/mol)R = gas constant (8.314 J/mol.K)TAO = Initial freezing temperature of ATA = freezing temperature of A in solution

ExampleGrape fruit has moisture content of 84.7%. Calculate percent of freezing water in grape fruit; (Assume: 100 g of grape fruit)Given: initial temperature is 0 oC (= 273 K)

freezing temperature is -5.5 oC (= 273-5.5 = 267.5 K) latent heat of fusion (’ = 6003 J/mole) MW of grape fruit = 183.61 g/mole

)11(ln'

AAOA TTR

X

)5.267

12731(

./314.8/6003ln

KmoleJmoleJX A

05438.0ln AX

947.0AX

Example

Base on mole fraction:

61.183)7.84100(

18

18947.0

Mu

Mu

nnnX

BA

AA

gMu 8.26

MW of water = 18

Percent of water but not freezing in solution:

%681007.84

)8.267.84(

waterfreezingofPercent

percent of freezing water:

Freezing TimeTo ensure efficient of freezing time, methods for prediction of freezing times are very important.1. Plank’s Equation - Use for prediction of freezing time by Plank (1913)

Where, tf is the freezing time, is the density,

HL is the latent heat of fusion,

a is the size of product,

hc is the convective heat-transfer coefficient,

k is the thermal conductivity or frozen product,

P’ and R’ are used to account for product shape,

with P’=1/2, R = 1/8 for finite plate; P’=1/4, R’=1/16 for infinite cylinder; and P’=1/6, R = 1/24 for sphere.

The dimension a is product thickness for an infinite slab, diameter for an infinite cylinder, and diameter for a sphere.

)'(2'

kaR

haP

TTHt

cF

LF

A spherical food product is being frozen in an air-blast wind tunnel. The initial product temperature is 10 oC and the cold air -15 oC. The product has a 7-cm diameter with density of 1,000 kg/m3. The initial freezing temperature is -1.25 oC, and the latent heat of fusion is 250 kJ/kg. Compute the freezing time.

Given: Initial product temperature Ti = 10 oCAir temperature T = -15 oCInitial freezing temperature TF = -1.25 oCProduct diameter a = 7 cm (0.07 m)Product density = 1000 kg/m3

Thermal conductivity of frozen product k = 1.2 W/m.kLatent heat HL = 250 kJ/kgShape constants for spheres: P’ = 1/6, R’ = 1/24Convective heat-transfer coefficient hc = 50 W/m2.k

Example: Freezing timeExample: Freezing time

Solution: calculate the freezing time

)'(2'

kaR

haP

TTHt

cF

LF

Example: Freezing timeExample: Freezing time

hrssJ

Jt

sJWandJKJSinceWkJ

WKm

WKm

CmkJ

KmWm

KmWm

CCkgkJmkgt

F

o

ooF

04.21033.7/1

100033.7/1110001

/33.7

].107.1.1033.2[.

18182

])./2.1(24

)07.0()./50(6

07.0[)]15(25.1[

/250/1000

3

34

34

3

2

2

3

Freezing TimeOne dimensional Infinite Slab: To determine time to freeze infinite slab:

– R = characteristic dimension = thickness/2

– H1 = Cu (Ti-T3); Cu = specific heat capacity (unfrozen) (J/ m3.k)

– H2 = L + Cf (T3-Tf) ; Cu = specific heat capacity (frozen) (J/ m3.k)

– T1 = (Ti + T3)/2 – Ta; Ti = initial temperature,Tf = final temperature

– T2 = T3-Ta ; Ta = air temperature (T)

– T3 = 1.8 + 0.263Tf + 0.105Ta The above equation is valid within these range:

0.02 < NBiot<11, 0.11<NStefan<0.36, 0.03 <NPk(plank;s number)< 0.61

Where Stefan number, NSte = Cf(Tf-T)/H

Plank number, NPk = Cu(Ti-TF)/ H

)2

1]([2

2

1

1 Bislab

NTH

TH

hRt

Freezing Time

For thawing purposes, the following equation is recommended:

Applicable for thawing, Tf (final temp)= 0 oC. This is valid for these range.

0.3 < NBi<41, 0.08<NSte<0.77, 0.06 <NPk< 0.27

061.02712.00248.12

]125.025.0[7164.5 PkSteSteSteBiu

uslab NN

NNNkRC

t

Freezing TimeEllipsoid Shapes : To determine the freezing time (depend on shape factor E) :

– For infinite slab, the shape factor E = 1 (since 1=infinite, 2=infinite)

– For an infinite cylinder, the shape factor E=2 (since 1=1, 2=infinite)

– For a sphere, the shape factor, E = 3 (1=1, 2=1)

)2

(

)21(

)2

(

)21(1

222

121

Bi

Bi

Bi

Bi

N

N

N

NE

Freezing TimeShapes other than Ellipsoid (i.e. rectangular brick shape, finite cylinder) :The shape factor can be calculated:

Same characteristic dimension R: shortage distance from thermal center to the surface of the object.Smallest cross-sectional area A ; the smallest cross-section that incorporates R.Same volume V1 and 2 can be determined:

)2

(

)21(

)2

(

)21(1

222

121

Bi

Bi

Bi

Bi

N

N

N

NE

21 RA

)34( 3

1

2

R

V

kRh

N cBi

Example: Freezing TimeLean beef with 74.5% moisture content and 1 m length, 0.6 m width, and 0.25 m thickness is being frozen in an air-blast freezer with hc = 30 W/m2.K and air temperature of -30 oC. If the initial product temperature is 5 oC. Estimate the time required to reduce the product temperature to -10 oC. An initial freezing temperature of -1.75 oC. Has been measured for the product. The thermal conductivity of frozen beef is 1.5 W/m.K, and the specific heat of unfrozen beef is 3.5 kJ/kg.K. A product density of 1050 kg/m3 can be assumed, and a specific heat of 1.8 kJ/kg.K for frozen beef can be estimated from properties of ice.

– Product length d2 = 1 m– Product width d1 = 0.6 m– Product thickness a = 0.25 m– Convective heat-transfer coefficient hc = 30 W/m2.k– Air temperature T = -30 oC– Initial product temperature Ti = 5 oC– Initial freezing temperature TF = -1.75 oC– Product density = 1050 kg/m3– Enthalpy change (H) = 0.745333.22 kJ/kg = 248.25 kJ/kg (estimate)– Thermal conductivity k of frozen product = 1.5 W/m.K– Specific heat of product (Cpu) = 3.5 kJ/kg.K– Specific heat of frozen product (Cpf) = 1.8 kJ/kg.K

Solution: Freezing Time(1) determine shape factor:

(2) The Biot number is :

(3) Shape factor E:

056.3)125.0(6.025.0

)225.0(

6.025.02

221

RA

999.5)125.0(

34056.3

16.025.0

)34( 33

1

2

R

V

5.25.1125.030

kRh

N cBi

197.1)

5.2999.52999.5(

)5.2

21(

)5.2056.32056.3(

)5.2

21(1

)2

(

)21(

)2

(

)21(1

22222

121

Bi

Bi

Bi

Bi

N

N

N

NE

Solution: Freezing Time(4) T3 :

(5) H1: Cu(Ti-T3)

(6) T1 and T2 :

CT o98.3)30(105.0)10(263.08.13

3

331

/33001500

])98.3(5[)/1050()./3500()(

mJCmkgKkgJTTCH o

iu

3

3

332

/145,039,272))10(98.3()/1050(./1800

)/1050()745.0()/1000/22.333()(

mJmkgKkgJ

mkgkJJkgkJTTCLH ff

CTTT

CTTT

T

oa

oa

i

02.26)30(98.3

51.30)30(2

)98.35(2

)(

32

31

Solution: Freezing Time

(7) tslab :

(8) t = tslab/E;

s

NTH

TH

hRt Bi

slab

156,108

)25.21](

02.26272039145

51.3033001500[

30125.0)

21]([

2

2

1

1

hrsEt

t slab 1.2590355197.1

156,108

Require time for lean beef (1 m 0.6m 0.25 m) will be 25.1 hours to freeze