FST 151

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FST 151 FOOD FREEZING FOOD SCIENCE AND TECHNOLOGY
151 Food Freezing - Basic concepts
(contd) Lecture Notes Prof. Vinod K.
Jindal (Formerly Professor, Asian Institute of
Technology) Visiting Professor Chemical
Engineering Department Mahidol University Salaya,
Nakornpathom Thailand
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• The change from amorphous to glassy state? 
• An amorphous phase is a condensed phase (i.e.,
  liquid or solid) without a regular crystalline
  order.  Many foods crystallize only very slowly
  if at all so once they become supersaturated they
  will become increasingly viscous without
  crystallizing.  
• We can define two types of amorphous phase
  rubbery and glassy. The rubbery state is more
  mobile (and therefore softer and less viscous)
  than the glassy form. 
• The transition from a rubbery to a glassy state
  is governed by molecular mobility. A reduction in
  temperature or the amount of water will transform
  rubbery state to glass and the food will become
  more brittle and less chemically reactive.

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• What is the Glassy State?
• During processing and storage, the physical
  properties of food products change dramatically
  depending on water availability and temperature.
  
• Amorphous and partially amorphous structures
  in foods are formed in various processes such
  as baking, drying and freezing.
• The removal of water, by evaporation or
  separation as ice, produces supersaturated
  states of dissolved substances.
• Depending on temperature and water content,
  these unfrozen fractions can be in either a
  glassy or rubbery state.

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Typical Tg values for various food
products Food Product Tg (C)
orange juice - 30 to - 38 raspberry
juice - 37 to - 40 strawberry - 43
apple - 37 tomato - 21 carrot
- 32 egg white - 38 egg yolk -
32 beef muscle - 80/ - 35 ice creams
- 34
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• The values shown in Table indicate that the
  glassy state can be achieved only for a limited
  number of frozen food systems at temperatures
  commonly employed in the distribution chain.
• This leads to an important question as to what
  is the appropriate temperature for predicting
  the stability of frozen foods at temperatures
  above Tg?

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• Why is the Glassy State Important in
  Understanding Storage Stability?
• It is assumed that stability of foods, i.e.
  their shelf life and quality, is maintained in
  the glassy state. Below glass transition
  temperature the freeze-concentrated fraction is
  a glass and thus long-term stability may be
  expected.
• It is also assumed that changes occur above the
  glass transition and their kinetics depends upon
  the difference between the storage and the glass
  transition temperatures.
• Stability of frozen foods strongly depends upon
  the storage temperature. The deterioration rate
  may be multiplied by a factor of between 2 and
  30 for an increase of the storage temperature by
  10C.

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• Can Glass Transitions tell us about Product
  Stability above Tg?
• Attempts have been made to explain the drastic
  effect of temperature above Tg on rates of
  degradation. However, the experimental data of
  ice crystal growth in ice creams as a function
  of Tstorage -Tg cannot explain the influence of
  stabilizers which do not influence Tg.
• In practice this makes it difficult to explain
  the changes in the quality of frozen products by
  a single parameter e.g. Tg. However, an
  understanding of the mobility of reactive
  species in frozen foods and thus their stability
  as a function of temperature and formulation is
  important.

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Frozen-Food Properties

• Depend on thermal properties of the food product.
• Phase 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.

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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

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Fig. 1 Thawing loss of shrimp frozen by
air-blast freezing at 6 m/s air velocity and
-28oC
Sirintra Boonsumrej , Saiwarun Chaiwanichsiri ,
Sumate Tantratian , Toru Suzuki , Rikuo Takai
Effects of freezing and thawing on the quality
changes of tiger shrimp (ltceitalicgtPenaeus
monodonlt/ceitalicgt) frozen by air-blast and
cryogenic freezing
Journal of Food Engineering Volume 80, Issue 1
2007 292 - 299
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Fig. 3 TBA value in shrimps frozen under the
air-blast freezing at 6 m/s air velocity and -28
C
Sirintra Boonsumrej , Saiwarun Chaiwanichsiri ,
Sumate Tantratian , Toru Suzuki , Rikuo Takai
Effects of freezing and thawing on the quality
changes of tiger shrimp (ltceitalicgtPenaeus
monodonlt/ceitalicgt) frozen by air-blast and
cryogenic freezing
Journal of Food Engineering Volume 80, Issue 1
2007 292 - 299
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Freezing Time Calculations
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Freezing Time Calculation

• In freezing time calculations, the imprecise
  control of freezing conditions and uncertainty in
  thermal properties data of foods are mainly
  responsible for not so accurate predictions.
• The overall accuracy of prediction is governed
  more by the uncertainty in thermal properties
  data rather than the calculation procedure.

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• There are three alternatives for obtaining
  the thermal properties data of foods
• 1) Use data from literature
• 2) Direct measurement
• 3) Using prediction equations based on
  the composition information

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• PLANKS EQUATION
• Planks equation is an approximate analytical
  
• solution for a simplified phase-change
  model.
• Plank assumed that the freezing process
• (a) commences with all of the food unfrozen
• but at its freezing temperature.
• (b) occurs sufficiently slowly for heat
  transfer
• in the frozen layer to take place
  under steady-state conditions.

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• Planks equation considers only phase change
  period during freezing process. However, Planks
  approximate solution is sufficient for many
  practical purposes.
• This method when applied to calculate the time
  taken to freeze to the centre of a slab (Fig. 1)
  whose length and breadth are large compared with
  the thickness, results in the following equation

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Fig. 1 Freezing of a slab
Eqs. 7.1 7.3
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For conditions when t0, x0 and ttf, xa/2
(at the center of slab), this leads to
Eq. 7.5
Also Lf mm L (for a food material) where mm
moisture content of food (fraction)
L latent heat of fusion of water,
333.2 kJ/kg at 00C
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The general form of Planks equation is
where P and R are constants accounting for the
product shape with P1/2, R1/8 for infinite
plate P1/4, R1/16 for infinite cylinder and
P1/6 and R1/24 for sphere or cube.
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Brick-shaped solids have values of P and R
lying between those for slabs and those for
cubes, which can be obtained from the graph in
Fig. 2. In this figure, ß1 and ß2 are the ratios
of the two longest sides to the shortest. It does
not matter in what order they are taken.
Fig. 2 Chart providing P and R constants for
Planks equation when applied to
a brick or block geometry.
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Example Freezing time (Example 7.1)

• A spherical food product is being frozen in an
  air-blast wind tunnel. The initial product
  temperature is 10oC 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 oC
• Air temperature T? -15 oC (Not 40oC)
• Initial freezing temperature TF -1.25 oC
• Product diameter a 7 cm (0.07 m)
• Product density ? 1000 kg/m3
• Thermal conductivity of frozen product k 1.2
  W/m.k
• Latent heat HL 250 kJ/kg
• Shape constants for spheres P 1/6, R
  1/24
• Convective heat-transfer coefficient hc 50
  W/m2.k

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Example Freezing time

• Solution calculate the freezing time

tF will be o.72 hr if the air temperature is
assumed - 40oC.
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• Plank's equation results in the under-estimation
  of freezing times because of the assumptions made
  in its derivation.
• The initial freezing temperature (TF) for most
  foods is not reliably known. Although the initial
  freezing temperature is tabulated for many foods,
  the initial and final product temperatures are
  not accounted for in the computation of freezing
  times.
• Also we often do not know for sure what values of
  ?f and kf to select.

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• Despite the limitations, Planks equation is the
  most popular method for predicting freezing time.
  
• Most other available methods are based on the
  modification of Planks equation.
• Because of data uncertainty alone, freezing time
  estimates should be treated as being accurate to
  within 20 at best.

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• Pham (1986) presented an improvement of Planks
  equation for prediction of freezing times. The
  approach is based on the following equations
• The mean freezing temperature is defined as

(7.8)
where Tc is final center temperature and Ta is
freezing medium temperature. The freezing time is
given by
(7.9)
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where dc characteristic dimension r or
shortest distance Ef a shape factor (1
for slab, 2 for cylinder and 3
for sphere)
(7.10)
(7.11)
(7.12)
(7.13)
?H1 Enthalpy change during pre-cooling,
J/m3 ?H2 Enthalpy change during phase change
and post- cooling period, J/m3
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Freezing Time of Finite Shaped Objects
In Phams method, the value of Ef is adjusted
(Eq. 7.16) Ef G1 G2E1 G3E2 where
the values of G1, G2 and G3 are given in Table
7.1 and E1 and E2 are calculated from Eqs. 7.17
7.19 and Eqs. 7.18 7.20, respectively. We can
now follow Example 7.2 (Singh and Heldman) and
compare the freezing time calculations based on
Phams approach and Planks equation.
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Freezing Time

• Alternate approach to determine the shape factor
  Ef in the calculation of freezing time
• For infinite slab, the shape factor E 1 (since
  ?1infinite, ?2infinite)
• For an infinite cylinder, the shape factor E2
  (since ?11, ?2infinite)
• For a sphere, the shape factor, E 3 (?11,
  ?21)
• Alculation of

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Freezing Time

• For different shapes e.g. ellipsoid, rectangular
  brick, finite cylinder etc., 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 V
• ?1 and ?2 can be determined

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Example Freezing Time

• Lean 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) 0.745?333.22 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

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Solution Freezing Time

• (1) Determine shape factor
• (2) The Biot number is
• (3) Shape factor E

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Solution Freezing Time

• (4) T3
• (5) ?H1 Cu(Ti-T3)
• (6) ?T1 and ?T2

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Solution Freezing Time

• (7) tslab
• (8) t tslab/E

Required time for lean beef (1 m ? 0.6m ?0.25 m)
will be 25.1 hours to freeze.
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Prediction of Thawing Times
?H10 is the volumetric enthalpy change (J/m3) of
the product from 0 to -100C.