Unit Food Freezing.ppt
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Transcript of Unit Food Freezing.ppt
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