Transcript Slide 1

FST 151
FOOD FREEZING
FOOD SCIENCE AND TECHNOLOGY 151
Food Freezing - Basic concepts (cont’d)
Lecture Notes
Prof. Vinod K. Jindal
(Formerly Professor, Asian Institute of Technology)
Visiting Professor
Chemical Engineering Department
Mahidol University
Salaya, Nakornpathom
Thailand
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Food Freezing Basic Concepts
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Food Freezing Basic Concepts
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Food Freezing Basic Concepts
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Food Freezing Basic Concepts
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Food Freezing Basic Concepts
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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
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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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
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Food Freezing Basic Concepts
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Food Freezing Basic Concepts
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Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Freezing Time Calculations
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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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.
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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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
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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PLANK’S EQUATION
• Plank’s 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.
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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• Plank’s equation considers only phase
change period during freezing process.
However, Plank’s 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:
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Fig. 1 Freezing of a slab
A(TF  Ta )
dx
q
 AL f
1
x
dt

h kf
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
Eqs. 7.1 – 7.3
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For conditions when t=0, x=0 and t=tf, x=a/2
(at the center of slab), this leads to
2
 a
a
tf 


(TF  Ta )  2h 8k f
L f



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.0C)
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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The general form of Plank’s equation is
 P 'a R 'a 2 
tf 



(TF  Ta )  h
k f 
L f
where P’ and R’ are constants accounting
for the product shape with P’=1/2, R’=1/8
for infinite plate; P’=1/4, R’=1/16 for infinite
cylinder; and P’=1/6 and R’=1/24 for sphere
or cube.
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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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, β1 and β2 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 Plank’s equation
Food Freezing Basic Concepts
when applied to(cont'd)
a brick
block
- Prof.or
Vinod
Jindalgeometry.
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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 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 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
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Example: Freezing time
• Solution: calculate the freezing time
H L
P' a R ' a 2
tF 
(

)
TF  T hc
k
1000kg / m 3  250kJ / kg
0.07 m
(0.07 m) 2
tF 
[

]
o
o
2
[1.25 C  (15 C )] 6  (50 W / m .K ) 24  (1.2 W / m. K )
3
3
kJ
 4 m .K
 4 m .K
 18182 3 o  [2.33 10
 1.7  10
]
W
W
m . C
 7.33 kJ / W
Since 1 KJ  1000J and 1W  1 J / s
7.33 1000 J
tF 
 7.33  103 s  2.04 hr
1 J /s
oC.
tF will be o.72 hr if the if Food
the air
temperature
is
assumed
40
Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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• Plank's equation results in the underestimation 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 ρf and kf to select.
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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• Despite the limitations, Plank’s equation is
the most popular method for predicting
freezing time.
• Most other available methods are based
on the modification of Plank’s equation.
• Because of data uncertainty alone,
freezing time estimates should be treated
as being accurate to within ±20% at best.
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Pham (1986) presented an improvement of Plank’s
equation for prediction of freezing times. The
approach is based on the following equations:
• The mean freezing temperature is defined as
Tfm  1.8  0.263Tc  0.105Ta
(7.8)
where Tc is final center temperature and Ta is
freezing medium temperature. The freezing
time is given by
d c  H1 H 2 
N Bi 
tF 

1 



E f h  T1
T2 
2 
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
(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)
H1  u cu (Ti  Tfm )
(7.10)
H2   f [ L f  c f (Tfm  Tc )]
(7.11)
 Ti  T fm 
  Ta
T1  
2


T2  T fm  Ta
(7.12)
(7.13)
ΔH1 = Enthalpy change during pre-cooling, J/m3
ΔH2 = Enthalpy change during phase change and post3 Basic Concepts
FoodJ/m
Freezing
cooling period,
(cont'd) - Prof. Vinod Jindal
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Freezing Time of Finite Shaped Objects
In Pham’s 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 Pham’s
approach and Plank’s equation.
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Freezing Time
• Alternate approach to determine the shape factor Ef in the calculation
of freezing time:
2
2
(1 
)
(1 
)
N Bi
N Bi
E  1

21
2 2
2
2
( 1 
) ( 2 
)
N Bi
N Bi
– 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)
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Freezing Time
• For different shapes e.g. ellipsoid, rectangular brick, finite cylinder etc.,
the shape factor can be calculated:
2
2
(1 
)
(1 
)
N Bi
N Bi
E  1

21
2 2
2
2
( 1 
) ( 2 
)
N Bi
N Bi
N Bi
hc R

k
• 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
V
A


• 1 and 2 can be determined:  1 
2
2
R
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
4
3
1 ( R 3 )
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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
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Solution: Freezing Time
(1) Determine shape factor:
A
0.25 0.6
0.25 0.6


 3.056
2
2
0.25 2
R
 (0.125)
(
)
2
V
0.25  0.6  1
2 

 5.999
4 3
4
1 ( R ) 3.056    (0.125 ) 3
3
3
(2) The Biot number is :
hc R
30  0.125
N Bi 

 2.5
k
1.5
1 
(3) Shape factor E:
2
2
2
2
)
(1 
)
(1  )
(1  )
N Bi
N Bi
2.5
2.5
E  1

 1

 1.197
2

2

2

3
.
056
2

5
.
999
(3.0562 
) (5.9992 
)
( 12  1 ) (  22  2 )
2.5
2.5
N Bi
N Bi
(1 
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Solution: Freezing Time
(4) T3 :
T3  1.8  0.263(10)  0.105(30)  3.98 o C
(5) H1: Cu(Ti-T3)
H 1  Cu (Ti  T3 )  (3500J / kg.K )  (1050 kg / m 3 )  [5  (3.98) o C ]
 33001500 J / m 3
H 2  L  C f (T3  T f )  (333.22 kJ / kg  1000J / kJ )  (0.745) (1050kg / m 3 )
 1800J / kg.K   (1050kg / m 3 )  (3.98  (10))
 272,039,145 J / m 3
(6) T1 and T2 :
(Ti  T3 )
(5  3.98)
T1 
 Ta 
 (30)  30.51 o C
2
2
T2  T3  Ta   3.98  (30)  26.02 o C
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Solution: Freezing Time
N Bi
R H 1 H 2
0.125 33001500 272039145 2.5
t

[

](
1

)

[

](1  )
(7) tslab : slab
h T1 T2
2
30
30.51
26.02
2
 108,156 s
(8) t = tslab/E;
t 
t slab
E
108,156

 90355 s  25.1 hr
1.197
Required time for lean beef (1 m  0.6m 0.25 m) will be
25.1 hours to freeze.
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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Prediction of Thawing Times
dc H10
tt 
( P1  P2 N Bi )
E f h (Ta  TF )
hdc
N Bi 
ku
(Ta  TF )
N Ste  u cu
H10
(TF  Ti )
N Pk   f c f
H10
P1  0.7754 2.2828NSte  NPk
2
P2  0.5(0.4271 2.122NSte  1.4847NSte
ΔH10 is the volumetric enthalpy change (J/m3) of the
product from 0 to -100C.
Food Freezing Basic Concepts
(cont'd) - Prof. Vinod Jindal
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