itp330 pembekuan pangan (freezing of foods) · •menurunkan laju reaksi kimia/biokimia ......
TRANSCRIPT
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Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP330/Fateta/IPB/Freezing of Foods11/25/2016
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
ITP330 PEMBEKUAN PANGAN(Freezing of Foods)
Purwiyatno Hariyadi, PhD• Department of Food Science & Technology, Bogor Agricultural
University, BOGOR, Indonesia
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
Outline
Freezing systems
Frozen food properties (density, thermal conductivity, enthalpy, apparent specific heat, and apparent thermal diffusivity)
Freezing time (freezing curve, Plank’s equation, prediction of freezing time, freezing rate)
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Aspek engineering
Design (keperluan refrigerasi, T) Laju pembekuan (the rate at which freezing
progress)
Mutu produkProduktivitas
PEMBEKUAN PANGAN
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
• Penyimpanan produk pada T < suhu beku
• Umumnya pada T < 28 °F (-2 °C), atau khususnya pada < 0 °F (-18 °C)
• Sebagian besar air (~95%) beku
• daya awet produk beku ` bbrp bulan --- tahun• Laju pembekuan dipengaruhi oleh bbrp faktor :
perlu dikendalikan
• Pertumbuhan mikroorganisme dihambat, bbrp bahkan inaktif
PurwiyatnoHariyadi/IPN/ITP/Fateta/IPB
PEMBEKUAN PANGAN
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Things to notice:
• Pressure and temperature both affect the phase of matter.
• All three phases of matter exist at the triple point
Melting/Freezing
Boiling/Condensating
pembekuan
PEMBEKUAN PANGAN
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
PENGARUH POSITIF
• Menurunkan/menghambat pertumbuhan m.o.• Menurunkan laju reaksi kimia/biokimia• Meningkatkan daya simpan produk (3-40 lipat untuk setiap penurunan suhu 10°C)
PEMBEKUAN PANGAN
Pengaruhnya pada Produk Pangan :
+
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PENGARUH NEGATIF
• Kerusakan kimia• Kerusakan fisik (textural)
PEMBEKUAN PANGAN
Pengaruhnya pada Produk Pangan :
+
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
PENGARUH NEGATIF
• Freezer burn ?• Package properly • Control temperature fluctuations in storage.
• Oxidation? • Off-flavors • Vitamin loss • Browning
• Recrystallization?
PEMBEKUAN PANGAN
Pengaruhnya pada Produk Pangan :
+
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ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
- Penurunan titik beku = f (konsentrasi, BM)
m.KT
m.BMTRT
f
A2
0Agf
kg
kJ335air
kg
kJ,pembekuanlatenpanas
K.mol
J314.8gastatankonsR
K273,Kair)A(murnipelarutbekutitikT
.pelarutmg1000
solutmolmolalitasm
g
0A
dimana:
BMA = Berat Molekul pelarutK = konstanta molal titik beku
Lar. X dlm air Tf = (1.86 m)oC
A
A0Ag
1
XlnT
1
T
1
R
XA = fraksi mol air1 = panas laten pembekuan
SIFAT PRODUK PANGAN BEKU
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
Ice cream mix dengan komposisi sbb:10% butterfat12% solid-not-fat (54.5%: laktosa)15% sukrosa0.22% stabilizer
37.22% Air = 62.78%
Ditanya Tf = ?
m.BMTRT
A2
0Ag
f
m = ?
solvenkg
solutmolm
Solut? sukrosa BM = 342laktosa BM = 342solut lain diabaikan !!
Asumsi bahwa hanya gula (laktosa + fruktosa) yang memp. Efek menurunkan titik beku) !!
Contoh :
SIFAT PRODUK PANGAN BEKU
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Fraksi gula = 0.15 + 0.12 (0.545) = 0.2154 g/gFraksi air = 0.6278 Konsentrasi gula dlm air =
airg1000
gulag1,343
airg
gulag3431.0
6278.0
2154.0
m003.1airg1000
gulamol342
1.343
m
kg
J335.1000
kg
mol003.1
mol
g18K273
g18
mol1
K.mol
J314.8
T
2
f
Tf = 1,86 K
Contoh (lanjutan):
SIFAT PRODUK PANGAN BEKU
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
Air murni = 335
Larutan solid x dlm air = (335 mw)
kg
kJ
kgkJ
mw = Fraksi massa airContoh:
Kadar air
Selada 94.8 316.3 Strawberi 90.8 289.6 Kacang panjang 88.9 297.0 Kentang 77.8 258.0 Daging kambing 58.0 194.0 Kacang merah, biji kering 12.5 41.9 Kurma kering 24.0 79.0
kg
kJ
Perhitungan berdasarkan pd rumus
= 335 mwkg
kJ
Air:mol
J6030
mol1
1018
kg
J10335
kg
kJ335
33
PANAS LATEN PEMBEKUAN
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T-t Diagram :
A schematic freezing curve for water, displayingsensible heat loss (Regions I and III) and latent heat loss (Region II).
KURVA PEMBEKUAN … untuk air murni
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
Removal of heat (Q) from Region I (sensible heat), II (latent heat), and III (sensible heat) :
(1) Q1 = mCp1T1
m = weight of food Cp1 =specific heat of food above freezing T = temperature difference
(2) Q2 = mw ........> mw = weight of water ........> = latent heat
(3) Q3 = mCp2T3........> m = weight of food ........> Cp2 = specific heat of frozen food ........> T3 = temperature difference
KURVA PEMBEKUAN … untuk air murni & energi
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Titik Beku air
Super cooling
Titik eutektikAir
Larutan
Suhu
Waktu
Titik beku = f(waktu)
Driving force for nucleation/crystallization(i.e. T = T – Tf)
Removal of latent heat
Removal of sensible heat
KURVA PEMBEKUAN … untuk produk pangan
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
Freezing process
AB: Food is cooled below its freezing point (Tf), below 0oC
B: water remains liquid although temp below Tf supercooling
BC: Temp rises rapidly to Tf as ice crystals begin to form and latent heat of crystallization is released.
CD: heat is continued to be removed from foods freezing point is depressed
DE: solute becomes supersaturated and crystallize eutectic point.
EF: Freezing continuous to freezer temperature
Titik Beku air
Super cooling
Titik eutektikAir
Larutan
Suhu
Waktu
Titik beku = f(waktu)
A
B
C
DE
F
Tf
Tm
Freezing Process
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T
Tf
Ice
Sugar-Ice solid
LiquidSolubility line
Sugarcrystal
glass
eutectic
PHASE DIAGRAMPhase Diagram
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
3-21
KURVA PEMBEKUAN … untuk produk pangan
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Ti
Tf
T
t
Setelah terjadi pembekuan, konsentrasi solute pada sisa larutan menjadi lebih tinggi .....> penurunan titik beku lebih besar .....> Tf
()
You can’t freeze all of the water(Still have unfrozen (unfreezable) water : 5-10%)
KURVA PEMBEKUAN … untuk produk pangan
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
You can’t freeze all of the water(Still have unfrozen (unfreezable) water : 5-10%)
KURVA PEMBEKUAN … untuk produk pangan
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KURVA PEMBEKUAN … untuk produk IKAN
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
KURVA PEMBEKUAN … untuk produk IKAN
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Buah anggur (grape) ........> kadar air 84.7%........> Tf = -1.8oC (271.2oK)
A
A0Ag
1
XlnT
1
T
1
R
1 = 6003
Rg = 8.314 K.mol
JmolJ
XA=?
AXlnK2.271
1K273
1
K.molJ
314,8
molJ
6003
Ln XA = - 0.01755XA = 0.9826 (effective mol fraction of water )
ml
grapem
SUHU AWAL PEMBEKUAN PROD PANGAN
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
XA = fraksi mol air = 0.9826
XA = 0.9826 =
EBM
3.15
18
7.8418
7.84
BME = 183.61
Juice anggur dapat dianggap bertingkah laku mirip/sama dgn- lar. x dlm air- BMx = 183.61
- XA = 09826- Xx = ........ dst
mol
g
SUHU AWAL PEMBEKUAN PROD PANGAN
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SUHU AWAL PEMBEKUAN PROD PANGAN
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Hubungan antara % air beku vs. suhu
PROD PANGAN BEKU
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• EQUIPMENT RELATED • rate of heat transfer• size of refrigeration unit
• FOOD/PRODUCT QUALITY• slow freezing
• result in formation of few, large ice crystals•damaging to cell structure/quality
• rapid freezing• results in many small ice crystals•gives best product quality
• water ice: ~ 9% increase in volume
LAJU PEMBEKUAN
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LAJU PEMBEKUAN
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Slow Freezing
LAJU PEMBEKUAN
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Rapid Freezing
LAJU PEMBEKUAN
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LAJU PEMBEKUAN
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LAJU PEMBEKUAN
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LAJU PEMBEKUAN
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
• Slow Freezers 0.2 cm/h
- Still air and cold stores
• Quick Freezers 0.5-3 cm/h
- Air blast and plate freezers
• Rapid Freezers 5-10 cm/h
- Fluidized bed freezers
• Ultra rapid Freezers 10-100 cm/h
- Cryogenic freezers
-- Klasifikasi berdasarkan pada laju pergerakan “ice front”
LAJU PEMBEKUAN
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Freezing time
Freezing time: time required to lower the temperature of a food from an initial value to a predetermined final temperature at the thermal center.
Freezing time depends on: Size and shape of the product Thermal conductivity of the food material Area of the food available for heat transfer Surface heat transfer coefficient of the medium Temperature difference between the food and freezing
medium Type of packaging film in the case of packaged foods
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
- Pendugaan keperluan pembekuan ukuran sistem “mechanical
compression” evaluasi beban refrigerasi/pembekuan
- Disain peralatan + proses, untuk : memperoleh pembekuan yg diinginkan
- koef pindah panas- laju pembekuan
WAKTU/LAMA PEMBEKUAN
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• Time-temperature method• Time required to freeze between two temperatures
(usually T = -5oC or –10oC)• Velocity of ice front
- rate of freezing- must be able to see ice front
• Appearance of specimen- internal conditions
• Thermal methods- calorimetric techniques- not real-world condition
+Time-temp. methods most common+many people use time to freeze to
–10oC as standard.
WAKTU/LAMA PEMBEKUAN
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
• panas laten adalah energi utama yang hrs diperhitungkan pada proses pembekuan • ~ 75% total energi pd proses pembekuan
333.3 kJ/kg air144 BTU/lb air
• Terjadi perubahan sifat fisik bahan selama proses pembekuan ~ f (T,m)
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Plank’s Method (for infinite slab)
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.
WAKTU/LAMA PEMBEKUAN
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Tf
T1
Plank’s Method (for infinite slab)
WAKTU/LAMA PEMBEKUAN
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xfrozen frozen
Tf Tf
Ts
T1
Ts
T1
unfrozen
a
Plank’s Method (for infinite slab)
WAKTU/LAMA PEMBEKUAN
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
Convection:
q
hr
BTU= Qt = h (Ts – T1) ...... Pers. 1
h = convective heat transfer coeff. at the product surface.Conduction:
sff TTx
A.kq .......... Pers. 2
Tf = initial freezing pointx = x (t)
Combine 1&2:
h
1
k
xTT
q
f
1f
........... Pers. 3
Plank’s Method (for infinite slab)
WAKTU/LAMA PEMBEKUAN
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Ingat Pers 3 :
h
1
k
xTT
q
f
1f
Plank’s Method (for infinite slab)
WAKTU/LAMA PEMBEKUAN
Jumlah energi yang dibebaskan selama proses pembekuan
qdt = mi f = f dV f
qdt = f f A dx
so, q = f f A dx/dt .............. Pers. 4
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
dtTTdxh
1
k
x1f
fff
fT
01f
2
a
0 fff dtTTdx
h
1
k
x
h2
a
k8
a
TTt
2
if
fff
Pembekuan selesai lempeng jika x = a/2
Ti = Suhu PembekuanSuhu ruang pembeku
h1
kx
ATT
dt
dxA
f
1fff
Kombinasi Pers. 3 dan 4 ………………….>
Plank’s Method (for infinite slab)
WAKTU/LAMA PEMBEKUAN
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h
Pa
k
Ra
TTt
f
2
if
fff
Where:Infinite slab Sphere Infinite sylinder Cube
P 1/2 1/6 1/4 1/8R 1/8 1/24 1/6 1/24a Thickness Diameter Diameter Edge
f= latent heat of fusion [=] kJkg
kg
kJ water = 333.22 = 144
lbBTU
General Plank’s Equation :
WAKTU/LAMA PEMBEKUAN
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a
b c
P dan R untuk bentuk bata
a : dimensi terpendekc : dimensi terpanjang
2 = c/a1 = b/a
Lihat chart/diagram :dengan diketahui nilai 2 dan 1 maka dapat dibaca nilai P dan R
General Plank’s Equation :
WAKTU/LAMA PEMBEKUAN
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Chart providing P and R constants for Plank’s equation when applied to a brick or block geometry.
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.
General Plank’s Equation :
WAKTU/LAMA PEMBEKUAN
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Limitation of Plank’s method:
• no superheating or supercooling• thermal properties are constant• can’t incorporate a variable heat transfer
coeff.• can’t handle varying freezing point
General Plank’s Equation :
WAKTU/LAMA PEMBEKUAN
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1. AIR FREEZING - Products frozen by either "still" or "blast" forced air.
• cheapest (investment) • "still" slowest, more changes in product • "blast" faster, more commonly used
2. INDIRECT CONTACT - Food placed in direct contact with cooled metal surface.
• relatively faster • more expensive
3. DIRECT CONTACT - Food placed in direct contact with refrigerant (liquid nitrogen, "green" freon, carbon dioxide snow)
• faster • expensive • freeze individual food particles
METODE PEMBEKUAN
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• Blast freezing – a very cold air blasted on the food cools food very quickly.
• Close indirect contact – food is placed in a multi-plate freezer and is rapidly frozen.
• Immersion – food is placed into a very cold liquid (usually salt water – brine) or liquid nitrogen, this is known as cryonic freezing.
METODE PEMBEKUAN …commercial freezing
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• Mechanical Freezers- Evaporate and compress the refrigerant in a continuous cycle
• Cryogenic Systems- Use solid and liquid CO2, N2 directly in contact with the food
METODE PEMBEKUAN …freezing equipments
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METODE PEMBEKUAN …freezing equipments
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A typical fluidized bed freezer
METODE PEMBEKUAN …freezing equipments
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
METODE PEMBEKUAN …freezing equipments
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Batch Freezer
Source: Unit operations for food the food industries by: W.A. Gould
Blast Type
METODE PEMBEKUAN …freezing equipments
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
Source: Unit operations for food the food industries by: W.A. Gould
Hydraulic Pump
Top Pressure plate
Connecting Linkage
Corner Headers
Refrigerant hoses
Trays
Contact plates
Polyurethane and polystyrene insulated doors
METODE PEMBEKUAN …freezing equipments
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METODE PEMBEKUAN …freezing equipments
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METODE PEMBEKUAN …freezing equipments
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METODE PEMBEKUAN …freezing equipments
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
METODE PEMBEKUAN …freezing equipments
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ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
1. Water evaporation from the surface of the product can result in considerable weight loss in blast air freezers.
2. Individually quick frozen (IQF) products can be produced in belt freezers and fluidized bed freezers.
3. Cryogenic freezing results in frozen products with large ice crystals.
4. The rate of freezing in plate freezers is high.5. Water activity of a frozen product is only a function of
temperature.
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Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP330/Fateta/IPB/Freezing of Foods11/25/2016
ITP330 -- Freezing of Food - 2016phariyadi.staff.ipb.ac.id
6. Bound water does not freeze.7. Bound water affects water activity of a frozen product.8. As water in a solution changes to ice, solute
concentration increases and freezing point decreases.9. Heat capacity and thermal conductivity below the
freezing point change significantly with temperature.10.Thawing is slower than freezing because the thermal
conductivity of ice is lower than that of water.
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